Real Analysis: Theory of Measure and Integration (3rd Edition) [3rd Revised ed.] 9814578533, 9789814578530

Table of contents :
Contents
Preface to the First Edition
Preface to the Second Edition
Preface to the Third Edition
List of Notations
1 Measure Spaces
§0 Introduction
§1 Measure on a -algebra of Sets
[I] -algebra of Sets
[ll] Limits of Sequences of Sets
[III] Generation of -algebras
[IV] Borel -algebras
[V] Measure on a -algebra
[VI] Measures of a Sequence of Sets
[VII] Measurable Space and Measure Space
[VIII] Measurable Mapping
[IX] Induction of Measure by Measurable Mapping
§2 Outer Measures
[I] Construction of Measure by Means of Outer Measure
[II] Regular Outer Measures
[III] Metric Outer Measures
[IV] Construction of Outer Measures
§3 Lebesgue Measure on R
[I] Lebesgue Outer Measure on R
[II] Some Properties of the Lebesgue Measure Space
[III] Existence of Non-Lebesgue Measurable Sets
[IV] Regularity of Lebesgue Outer Measure
[V] Lebesgue Inner Measure on R
§4 Measurable Functions
[I] Measurability of Functions
[II] Operations with Measurable Functions
[III] Equality Almost Everywhere
[IV] Sequence of Measurable Functions
[V] Continuity and Borel and Lebesgue Measurability of Functions on R
[VI] Cantor Ternary Set and Cantor-Lebesgue Function
[VI.1] Construction of Cantor Ternary Set
[VI.2] Cantor-Lebesgue Function
§5 Completion of Measure Space
[I] Complete Extension and Completion of a Measure Space
[II] Completion of the Borel Measure Space to the Lebesgue Measure Space
§6 Convergence a.e. and Convergence in Measure
[I] Convergence a.e.
[II] Almost Uniform Convergence
[III] Convergence in Measure
[IV] Cauchy Sequences in Convergence in Measure
[V] Approximation by Step Functions and Continuous Functions
[V.1] Approximation of Lebesgue Measurable Functions by Step Functions
[V.2] Approximation of Lebesgue Measurable Functions by Continuous Functions
2 The Lebesgue Integral
§7 Integration of Bounded Functions on Sets of Finite Measure
[I] Integration of Simple Functions
[II] Integration of Bounded Functions on Sets of Finite Measure
[III] Riemann Integrability
§8 Integration of Nonnegative Functions
[I] Lebesgue Integral of Nonnegative Functions
[II] Monotone Convergence Theorem
[III] Approximation of the Integral by Truncation
§9 Integration of Measurable Functions
[I] Lebesgue Integral of Measurable Functions
[II] Convergence Theorems
[III] Convergence Theorems under Convergence in Measure
[IV] Approximation of the Integral by Truncation
[V] Translation and Linear Transformation of the Lebesgue Integral on R
[VI] Integration by Image Measure
§ 10 Signed Measures
[I] Signed Measure Spaces
[II] Decomposition of Signed Measures
[III] Integration on a Signed Measure Space
§ 11 Absolute Continuity of a Measure
[I] The Radon-Nikodym Derivative
[II] Absolute Continuity of a Signed Measure Relative to a Positive Measure
[III] Properties of the Radon-Nikodym Derivative
3 Differentiation and Integration
§ 12 Monotone Functions and Functions of Bounded Variation
[I] The Derivative
[II] Differentiability of Monotone Functions
[III] Functions of Bounded Variation
§ 13 Absolutely Continuous Functions
[I] Absolute Continuity
[II] Banach-Zarecki Criterion for Absolute Continuity
[III] Singular Functions
[IV] Indefinite Integrals
[V] Calculation of the Lebesgue Integral by Means of the Derivative
[V.1] The Fundamental Theorem of Calculus
[V.2] Integration by Parts
[V.3] Change of Variables
[VI] Length of Rectifiable Curves
§ 14 Convex Functions
[I] Continuity and Differentiability of a Convex Function
[II] Monotonicity and Absolute Continuity of a Convex Function
[III] Jensen's Inequality
4 The Classical Banach Spaces
§ 15 Normed Linear Spaces
[I] Banach Spaces
[II] Banach Spaces on R
[III] The Space of Continuous Functions C([a, b])
[IV] A Criterion for Completeness of a Normed Linear Space
[V] Hilbert Spaces
[VI] Bounded Linear Mappings of Normed Linear Spaces
[V.1] Continuous Linear Mappings
[V.2] Bounded Linear Mappings
[VI.3] Equivalence of Continuity and Roundedness of a Linear Mapping
[VI.4] Infimum of the Bounds of a Linear Mapping
[VI.5] Normed Linear Space of Bounded Linear Mappings
[VII.6] Dual of a Normed Linear Space
[VII] Baire Category Theorem
[VIII] Uniform Boundedness Theorems
[IX] Open Mapping Theorem
[X] Hahn-Banach Extension Theorems
[X.1] Hahn-Banach Extension Theorem for Real Linear Spaces
[X.2] Hahn-Banach Extension Theorem for Complex Linear Spaces
[X.3] Bounded Linear Mappings of a Dual Space into a Dual Space
[XI] Semicontinuous Functions
§ 16 The LP Spaces
[I] The P Spaces for p (0, )
[II] The Linear Spaces P for p [1, )
[III] The LP Spaces for p [1, )
[IV] The Space L
[V] The LP Spaces for p (0, 1)
[VI] Extensions of Holder's Inequality
§ 17 Relation among the LP Spaces
[I] The Modified LP Norms for LP Spaces with p [1, ]
[II] Approximation by Continuous Functions
[III] LP Spaces with p (0, 1]
[IV] The P Spaces
§ 18 Bounded Linear Functionals on the LP Spaces
[I] Bounded Linear Functionals Arising from Integration
[II] Approximation by Simple Functions
[III] A Converse of Holder's Inequality
[IV] Riesz Representation Theorem on the LP Spaces
§ 19 Integration on Locally Compact Hausdorff Space
[I] Continuous Functions on a Locally Compact Hausdorff Space
[II] Borel and Radon Measures
[III] Positive Linear Functionals on Cc(X)
[IV] Approximation by Continuous Functions
[V] Signed Radon Measures
[VI] The Dual Space of C(X)
5 Extension of Additive Set Functions to Measures
§20 Extension of Additive Set Functions on an Algebra
[I] Additive Set Function on an Algebra
[II] Extension of an Additive Set Function on an Algebra to a Measure
[III] Regularity of an Outer Measure Derived from a Countably Additive Set Function on an Algebra
[IV] Uniqueness of Extension of a Countably Additive Set Function on an Algebra to a Measure
[V] Approximation to a -algebra Generated by an Algebra
[VI] Outer Measure Based on a Measure
§21 Extension of Additive Set Functions on a Semialgebra
[I] Semialgebras of Sets
[II] Additive Set Function on a Semialgebra
[III] Outer Measures Based on Additive Set Functions on a Semialgebra
§22 Lebesgue-Stieltjes Measure Spaces
[I] Lebesgue-Stieltjes Outer Measures
[II] Regularity of the Lebesgue-Stieltjes Outer Measures
[III] Absolute Continuity and Singularity of a Lebesgue-Stieltjes Measure
[III.1] Absolute Continuity of Lebesgue-Stieltjes Measures
[III.2] Singularity of Lebesgue-Stieltjes Measures
[IV] Decomposition of an Increasing Function
§ 23 Product Measure Spaces
[I] Existence and Uniqueness of Product Measure Spaces
[II] Integration on Product Measure Space
[III] Completion of Product Measure Space
[IV] Convolution of Functions
[IV.1] Convolution of Integrable Functions
[IV.2] Convolution of LP-Functions
[IV.3] Approximate Identity in Convolution Product
[IV.4] Approximate Identity Relative to Pointwise Convergence
[V] Some Related Theorems
[V.1] Cavalieri's Formula and Extensions
[V.2] Minkowski's Inequality for Integrals
[V.3] Approximation by Product Simple Functions
6 Measure and Integration on the Euclidean Space
§24 Lebesgue Measure Space on the Euclidean Space
[I] Lebesgue Outer Measure on the Euclidean Space
[II] Regularity Properties of Lebesgue Measure Space on Rn
[III] Approximation by Continuous Functions
[IV] Lebesgue Measure Space on Rn as the Completion of a Product Measure Space
[V] Translation of the Lebesgue Integral on Rn
[VI] Linear Transformation of the Lebesgue Integral on Rn
§ 25 Differentiation on the Euclidean Space
[I] The Lebesgue Differentiation Theorem on Rn
[II] Differentiation of Set Functions with Respect to the Lebesgue Measure
[III] Differentiation of the Indefinite Integral
[IV] Density of Lebesgue Measurable Sets Relative to the Lebesgue Measure
[V] Signed Borel Measures on Rn
[VI] Differentiation of Borel Measures with Respect to the Lebesgue Measure
§ 26 Change of Variable of Integration on the Euclidean Space
[I] Change of Variable of Integration by Differentiable Transformations
[II] Spherical Coordinates in Rn
[III] Integration by Image Measure on Spherical Surfaces
7 Hausdorff Measures on the Euclidean Space
§27 Hausdorff Measures
[I] Hausdorff Measures on Rn
[II] Equivalent Definitions of Hausdorff Measure
[II.1] Covering by Closed Sets and by Open Sets
[II.2] Covering by Convex Sets
[III] Regularity of Hausdorff Measure
[IV] Hausdorff Dimension
§28 Transformations of Hausdorff Measures
[I] Hausdorff Measure of Transformed Sets
[II] 1-dimensional Hausdorff Measure
[III] Hausdorff Measure of Jordan Curves
§29 Hausdorff Measures of Integral and Fractional Dimensions
[I] Hausdorff Measure of Integral Dimension and Lebesgue Measure
[II] Calculation of the n-dimensional Hausdorff Measure of a Unit Cube in Rn
[III] Transformation of Hausdorff Measure of Integral Dimension
[IV] Hausdorff Measure of Fractional Dimension
A Digital Expansions of Real Numbers
[I] Existence of p-digital Expansion
[II] Uniqueness Question in p-digital Representation
[III] Cardinality of the Cantor Ternary Set
B Measurability of Limits and Derivatives
[I] Borel Measurability of Limits of a Function
[ll] Borel Measurability of the Derivative of a Function
C Lipschitz Condition and Bounded Derivative
D Uniform Integrability
[I] Uniform Integrability
[II] Equi-integrability
[III] Uniform Integrability on Finite Measure Spaces
E Product-measurability and Factor-measurability
[I] Product-measurability and Factor-measurability of a Set
[II] Product-measurability and Factor-measurability of a Function
F Functions of Bounded Oscillation
[I] Partition of Closed Boxes in Rn
[II] Bounded Oscillation in Rn
[III] Bounded Oscillation on Subsets
[IV] Bounded Oscillation on 1-dimensional Closed Boxes
[V] Bounded Oscillation and Measurability
[VI] Evaluation of the Total Variation of an Absolutely Continuous Function
Bibliography
Index

Citation preview

Sicsse Ty

REAL ANALYSIS

Theory of Measure and Integration J

Yeh

REAL ANALYSIS

Theory of Measure and Integration 3rd Edition

This page intentionally lett blank

3rd Edition

REAL ANALYSIS

Theory of Measure and Integration

J

Yeh

University of California, irvine

Ye NEW

JERSEY

» LONDON

» SINGAPORE

World Scientific + BEIJING

» SHANGHAI

» HONG

KONG

» TAIPEI

» CHENNAI

Published by World Scientific Publishing Co. Pte. Ltd.

5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Yeh, J. (James) Real analysis : theory of measure and integration/ by J. Yeh, University of California, Irvine, USA. — 3rd edition. pages cm Includes bibliographical references and index.

ISBN 978-981-4578-53-0 (hardcover : alk, paper) -- ISBN 978-981-4578-54-7 (pbk. : alk. paper)

1. Measure theory. 2. Lebesgue integral. 3. Integrals, Generalized. 4. Mathematical analysis. 5. Lp spaces. I. Title. QA312.¥44 2014 515'.42--1c23 2013049980

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2014 by World Scientific Publishing Co. Pte. Ltd. All rights reserved, This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the publisher.

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Printed in Singapore by World Scientific Printers.

To my wife Betty

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Contents Preface to the First Edition

xiii

Preface to the Second Edition

xvii

Preface to the Third Edition

xix

List of Notations

xxi

1 Measure Spaces §0 Introduction §1

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Measure on ao-algebraof Sets...

[I] o-algebra of Sets...

[II] Limits of Sequences of Sets...

[I] Generation of o-algebras. [IV] Borel o-algebras 2. 1

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§4

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[VII] Measurable Space and Measure Space .............-2000[VIII] Measurable Mapping... 2.6... ee ee ee ee [IX] Induction of Measure by Measurable Mapping .............-

17 19 22

Outer Measures... 6 [1] Construction of Measure by Means of Outer Measure

29 29

[II] Regular Outer Measures [I] Metric Outer Measures §3

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[V] Measure ona g-algebra. 2... [VI] Measures of aSequence of Sets...

§2

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2.0 2.

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[IV] Construction of Outer Measures...

2...

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33 35 38

Lebesgue MeasureonR.... 2... 0... ee [I] Lebesgue Outer Measureon RR... 2.2... 2. ee [II] Some Properties of the Lebesgue Measure Space ............0[II] Existence of Non-Lebesgue Measurable Sets ..............[IV] Regularity of Lebesgue Outer Measure ........... 0000006

42 42 47 51 53

[V] Lebesgue Inner MeasureonR .. 0... 0.0.0. ee cee ee eee Measurable Functions ............00 02.022 eee [1] Measurability of Functions 2... 1... ee ee eee [II] Operations with Measurable Functions .........02.0000 05%

60 72 72 76

[IIT] Equality Almost Everywhere 2.0... ee [IV] Sequence of Measurable Functions .. 1... 2... 000 eee vii

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81 82

viii

Contents

§5 §6

[V] Continuity and Borel and Lebesgue Measurability of FunctionsonR... [VI] Cantor Ternary Set and Cantor-Lebesgue Function ............ Completion of Measure Space 2.0... 6. ee eee [I] Complete Extension and Completion of a Measure Space ......... [11] Completion of the Borel Measure Space to the Lebesgue Measure Space Convergence a.e. and Convergence in Measure ................

86 88 99 99 102 104

[I] Convergence ae...

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104

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[II] Almost Uniform Convergence [Il] Convergence in Measure

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[IV] Cauchy Sequences in Convergence in Measure ............05 [V] Approximation by Step Functions and Continuous Functions ....... 2

116 119

The Lebesgue Integral

131

§7

Integration of Bounded Functions on Sets of Finite Measure..........

131

[I] Lebesgue Integral of Measurable Functions...

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177

.........

190

§8

§9

[I] Integration of Simple Functions ... 2... 20.2... 02.2002 eee [I] Integration of Bounded Functions on Sets of Finite Measure ....... [IIT] Riemann Integrability 2.0... ee een Integration of Nonnegative Functions... 2... ee eee ee [I] Lebesgue Integral of Nonnegative Functions .............2.004 [II] Monotone Convergence Theorem .............-.2-2045. [1] Approximation of the Integral by Truncation .............00Integration of Measurable Functions ................-2-2504[II] Convergence Theorems

2...

.. 1...

0.2...

[I] Convergence Theorems under Convergence in Measure

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[IV] Approximation of the Integral by Truncation ............... [V] Translation and Linear Transformation of the Lebesgue IntegralonR . [VI] Integration by Image Measure... 2... ee §10 Signed Measures ... 2.2... 2. es [T] Signed Measure Spaces .. 2... ce et [II] Decomposition of Signed Measures .. 2... 0... 0.0.0 ee eee [III] Integration on a Signed Measure Space... 2... 0. ee eee eee §11 Absolute Continuity ofa Measure 2... 0. ee [I] The Radon-Nikodym Derivative

2.2.1... 0... eee ee eee

[I] Absolute Continuity of a Signed Measure Relative to a Positive Measure [III] Properties of the Radon-Nikodym Derivative ............... 3

108

Differentiation and Integration

[OI] Functions of Bounded Variation...

186 191 . 196 201 212 212 218 227 235 235

236 247 257

§12 Monotone Functions and Functions of Bounded Variation [I] The Derivative...

[I] Differentiability of Monotone Functions

131 136 145 159 159 161 169 177

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257 257

263 274

§13 Absolutely Continuous Functions .............2020 000005 283 [I] Absolute Continuity... ee ee ee eee 283 [II] Banach-Zarecki Criterion for Absolute Continuity ............ 286

Contents

ix

[1] Singular Functions

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[IV] Indefinite Integrals 6. [V] Calculation of the Lebesgue Integral by Means of the Derivative [VI] Length of Rectifiable Curves

.............2

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§14 Convex Functions 2.0... te ee [1] Continuity and Differentiability of a Convex Function ........... [1] Monotonicity and Absolute Continuity of a Convex Function ....... [Il] Jensen’s Inequality... 2. eee 4 The Classical Banach Spaces ee ee ee ee ee ee eee ee ee ee ee eee eee ee ee ee ees

[II] The Space of Continuous Functions C([a,b])

.......2-..00-

[IV] A Criterion for Completeness of a Normed Linear Space 2...

311

323 323 332 335 339

§15 Normed Linear Spaces . 2.0... [I] Banach Spaces 0. [1] Banach Spaces on Re

[V] HilbertSpaces

289

ee 289 ..... 300

........

1...

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[VI] Bounded Linear Mappings of Normed Linear Spaces. .......... [VII] Baire Category Theorem... 1... ee ee [VI] Uniform Boundedness Theorems

339 339 342

345

347 349

350 360

............0022 000

363

[X] Hahn-Banach Extension Theorems... 2.2.0.0... 0000000008 [XI] Semicontinuous Functions... ..........0.22.0-.2000200§16 The L? Spaces 2... ee ee ee ee ee es [I] The £? Spaces for p€ (0,00) 2... ee eee eee [II] The Linear Spaces £? for p €[1,00) ©... . 2... cee ee eee [I] The L? Spaces for p €[1,c0) 2.2... . ceeee

373 386 392 392 395 400

[IX] Open Mapping Theorem...

[IV] The SpaceL®

2...

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[V] The L? Spaces forp € (0,1)...

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[VI] Extensions of Hilder’s Inequality .... §17 Relation among the L? Spaces... 1...

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[I] The Modified Z? Norms for L? Spaces with p € [l,oo]

[IV] The £2? Spaces 2. Bounded Linear Functionals on the L? Spaces

422 429

.........

429

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439 448

[II] Approximation by Continuous Functions ............200056 [Il] L? Spaces withpe€ (0,1].......----0---0 020000002 §18

366

[I] Bounded Linear Functionals Arising from Integration ........... [11] Approximation by Simple Functions .. 2... 0.20... 0.0 e eee [II] A Converse of Hélder’s Inequality... 2... ........0.2000[IV] Riesz Representation Theorem on the L? Spaces ............§19 Integration on Locally Compact Hausdorff Space ..............-[I] Continuous Functions on a Locally Compact Hausdorff Space ......

431 435

448 451 453 457 465 465

[II] Borel and Radon Measures .... 0.2... ce eee eee 470 [I] Positive Linear Functionals on C,(X) 2... ee 475

[IV] Approximation by Continuous Functions [V] Signed Radon Measures

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483

487

[VI] The Dual Space of C(X)

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5 Extension of Additive Set Functions to Measures §20 Extension of Additive Set Functions on an Algebra... [I] Additive Set Function on an Algebra

2.2...

2...

[1] Extension of an Additive Set Function on [1] Regularity of an Outer Measure Derived Function onan Algebra 2... [IV] Uniqueness of Extension of a Countably an Algebra toa Measure... 2. [V] Approximation to a o-algebra Generated

0.0005 ee

an Algebra to a Measure ... . from a Countably Additive Set ee Additive Set Function on ee by an Algebra .........

[VI] Outer Measure Based ona Measure... ...........222005 §21 Extension of Additive Set Functions ona Semialgebra 2... 2... ee [I] Semialgebras of Sets 2 1. ee

[0] Additive Set Function on a Semialgebra 2... ee ee [I] Outer Measures Based on Additive Set Functions on a Semialgebra . . . §22 Lebesgue-Stieltjes Measure Spaces 2... ee ee ee [I] Lebesgue-Stieltjes Outer Measures . 2... 1 ee [I] Regularity of the Lebesgue-Stieltjes Outer Measures ........... [II] Absolute Continuity and Singularity of a Lebesgue-Stieltjes Measure . . [IV] Decomposition of an Increasing Function... .........--045§23 Product Measure Spaces . 2... 0 ee es [I] Existence and Uniqueness of Product Measure Spaces ........... [II] Integration on Product Measure Space .. 6.0... eee ee ee [II] Completion of Product Measure Space... 0... eee ee eee [[V] Convolution of Functions 2... 20.0... ee [V] Some Related Theorems ...........--..---.-.----.

6

Measure and Integration on the Euclidean Space §24 Lebesgue Measure Space on the Euclidean Space

...........--05

[I] Lebesgue Outer Measure on the Euclidean Space...

..........

[I] Regularity Properties of Lebesgue Measure SpaceonR” ......... [1] Approximation by Continuous Functions .............2004 [IV] Lebesgue Measure Space on R” as the Completion of a Product Measure Space... 0. ee ee ee ee [V] Translation of the Lebesgue IntegralonR™ ...........-0005 [VI] Linear Transformation of the Lebesgue IntegralonR” .......... §25 Differentiation on the Euclidean Space... 1. ee ee ee ee [I] The Lebesgue Differentiation Theoremon RR” ...............

[01] Differentiation of Set Functions with Respect to the Lebesgue Measure [III] Differentiation of the Indefinite Integral... 2. .......02..4. [IV] Density of Lebesgue Measurable Sets Relative to the Lebesgue Measure

658 664 [VI] Differentiation of Borel Measures with Respect to the Lebesgue Measure 666 §26 Change of Variable of Integration on the Euclidean Space... ......... 673 [V] Signed Borel Measureson R?

2...

0...

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Contents [I] Change of Variable of Integration by Differentiable Transformations [I] Spherical CoordinatesinR® 2.2... .. cc ee [II] Integration by Image Measure on Spherical Surfaces ........... 7 Hausdorff Measures on the Euclidean Space §27

Hausdorff Measures .. 1... 0... ce [I] Hausdorff Measures on R?. 2... [11] Equivalent Definitions of Hausdorff Measure

[10] Regularity of Hausdorff Measure...

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[IV] Hausdorff Dimension ... 2... 20... 0.0.00. §28 Transformations of Hausdorff Measures .........--...-2-00--[1] Hausdorff Measure of Transformed Sets ........0..000 ere [II] 1-dimensional Hausdorff Measure ..........0..00 022 eee [I] Hausdorff Measure of Jordan Curves .............-00-22--

§29 Hausdorff Measures of Integral and Fractional Dimensions .......... [1] Hausdorff Measure of Integral Dimension and Lebesgue Measure

[1] Calculation of the n-dimensional Hausdorff Measure of a Unit Cube in R” 731

[I] Transformation of Hausdorff Measure of Integral Dimension. ..... . 737 [IV] Hausdorff Measure of Fractional Dimension

A Digital Expansions [I] Existence of [II] Uniqueness [10] Cardinality

...........0-2--

of Real Numbers p-digital Expansion .... 2.2.2.2... ..0.0000005 Question in p-digital Representation ............. of the Cantor Ternary Set... 0... . ee ee ee

B Measurability of Limits and Derivatives [I] Borel Measurability of Limits of aFunction

...........-2005

[11] Borel Measurability of the Derivative of aFunction

............

C Lipschitz Condition and Bounded Derivative D Uniform Integrability [I] Uniform Integrability 2... 0. [MI] Equi-integrability 2... [1] Uniform Integrability on Finite Measure Spaces

ee eee .............

E Product-measurability and Factor-measurability [I] Product-measurability and Factor-measurability of aSet .......... [II] Product-measurability and Factor-measurability ofa Function ...... F Functions of Bounded Oscillation []) Partition of Closed Boxesin R® 2.2... ee [I] Bounded OscillationinR® ........0......22 00.0000 008 [1] Bounded Oscillation on Subsets... 2... ee te te es [IV] Bounded Oscillation on 1-dimensional Closed Boxes .......... [V] Bounded Oscillation and Measurability ..............0006 [VI] Evaluation of the Total Variation of an Absolutely Continuous Function . 799

xii

Contents

Bibliography

803

Index

805

Preface to the First Edition This monograph evolved from a set of lecture notes for a course entitled Real Analysis that I taught at the University of California, Irvine. The subject of this course is the theory of measure and integration. Its prerequisite is advanced calculus. All of the necessary background material can be found, for example, in R. C. Buck’s Advanced Calculus. The

course is primarily for beginning graduate students in mathematics but the audience usually includes students from other disciplines too. The first five chapters of this book contain enough material for a one-year course. The remaining two chapters take an academic quarter to cover.

Measure is a fundamental concept in mathematics. Measures are introduced to estimate sizes of sets. Then measures are used to define integrals. Here is an outline of the book. Chapter 1 introduces the concepts of measure and measurable function. §1 defines measure as a nonnegative countably additive set function on a o-algebra of subsets of an arbitrary set. Measurable mapping from a measure space into another is then defined. §2 presents construction of a measure space by means of an outer measure. To have a concrete example of a measure space early on, the Lebesgue measure space on the real line R is introduced in §3. Subsequent developments in the rest of Chapter 1 and Chapter 2 are in the setting of a general measure space. (This is from the consideration that in the definition of a Measure and an integral with respect to a measure the algebraic and topological structure of the underlying space is irrelevant and indeed unnecessary. Topology of the space on which a measure is defined becomes relevant when one considers the regularity of the measure, that is, approximation of measurable sets by Borel sets.) §4 treats measurable functions, in

particular algebraic operations on measurable functions and pointwise limits of sequences of measurable functions. §5 shows that every measure space can be completed. §6 compares two modes of convergence of a sequence of measurable functions: convergence almost everywhere and convergence in measure. The Borel-Cantelli Lemma and its applications are presented. A unifying theorem (Theorem 6.5) is introduced from which many other convergence theorems relating the two modes of convergence are derived subsequently. These include Egoroff’s theorem on almost uniform convergence, Lebesgue’s and Riesz’s theorems.

Chapter 2 treats integration of functions on an arbitrary measure space. In §7 the Lebesgue integral, that is, an integral with respect to a measure, is defined for a bounded. real-valued measurable function on a set of finite measure. The Bounded Convergence Theorem on the commutation of integration and limiting process for a uniformly bounded.

xiv

Preface

sequence of measurable functions which converges almost everywhere on a set of finite measure is proved here. The proof is based on Egoroff’s theorem. On the Lebesgue measure space on R, comparison of the Lebesgue integral and the Riemann integral is made. §8 contains the fundamental idea of integration with respect to a measure. It is shown here that for every nonnegative extended real-valued measurable function on a measurable set the integral with respect to the measure always exists even though it may not be finite. The Monotone Convergence Theorem for an increasing sequence of nonnegative measurable functions, the most fundamental of all convergence theorems regarding commutation of integration and convergence of the sequence of integrands, is proved here. Fatou’s Lemma concerning the limit inferior of a sequence of nonnegative measurable functions is derived from the Monotone Convergence Theorem. In §9 the integral of an extended real-valued measurable function on a measurable set is then defined as the difference of the integrals of the positive and negative parts of the function provided the difference exists in the extended real number system. The generalized monotone convergence theorem for a monotone sequence of extended real-valued measurable functions, generalized Fatou’s lemma for the limit inferior and the limit superior of a sequence of extended real-valued measurable functions, and Lebesgue’s Dominated Convergence Theorem are proved here. Fatou’s Lemma and Lebesgue’s Dominated Convergence Theorem under convergence in measure are included. In §10 a signed measure is defined as an extended real-valued countably additive set function on a o-algebra and then shown to be the difference of two positive measures. In §11 the Radon-Nikodym derivative of a signed measure with respect to a positive measure is defined as a function which we integrate with respect to the latter to obtain the former. The existence of the Radon-Nikodym derivative is then proved under the assumption that the former is absolutely continuous with respect to the latter and that both are o-finite. (The fact that the Radon-Nikodym derivative is a derivative not only in name but in fact it is the derivative of a measure with respect to another is shown for Borel measures on the Euclidean space in §25.)

Chapter 3 treats the interplay between integration and differentiation on the Lebesgue measure space on R. §12 presents Lebesgue’s theorem that every real-valued increasing function on R is differentiable almost everywhere on R. The proof is based on a Vitali covering theorem. This is followed by Lebesgue’s theorem on the integral of the derivative of a real-valued increasing function on a finite closed interval in R. Functions of bounded variation are included here. §13 defines absolute continuity of a real-valued function on a

finite closed interval in R and then shows that a function is absolutely continuous if and only if it is an indefinite integral of a Lebesgue integrable function. This is followed by Lebesgue’s decomposition of a real-valued increasing function as the sum of an absolutely continuous function and a singular function. Such methods of calculating a Riemann integral in calculus as the Fundamental Theorem of Calculus, integration by parts, and change of variable of integration find their counterparts in the Lebesgue integral here. §14 treats convex functions and in particular their differentiability and absolute continuity property. Jensen’s inequality is included here.

Chapter 4 treats the L? spaces of measurable functions f with integrable | f|? for p € (0, 00) and the space L™ of essentially bounded measurable functions on a general measure

Preface

xV

space. Here Hélder’s inequality and Minkowski’s inequality are proved for p € (0, oo]. §15 introduces the Banach space and its dual. §16 treats L? spaces for p € [1, oo] as well as for p € (O, 1). §17 treats relation among the L? spaces for different values of p. The £7 spaces of sequences of numbers (a, : n € N) with }7,,cn lan|? < 00 is treated as a particular case

of L? spaces in which the underlying measure space is the counting measure space on the set N of natural numbers. The Riesz representation theorem on the L? spaces is proved in §18. §19 treats integration on a locally compact Hausdorff space. Urysohn’s Lemma on the existence of a continuous function with compact support and partition of unity, Borel and Radon measures, the Riesz representation theorem on the space of continuous functions with compact support as well as Lusin’s theorem on approximation of a measurable function by continuous functions are included here. (The placement of §19 in Chapter 4 is somewhat arbitrary.) Chapter 5 treats extension of additive set functions to measures. It starts with extension of an additive set function on an algebra to a measure in §20 and completes the theory with extension of an additive set function on a semialgebra to a measure in §21. (Semialgebra of sets is an abstraction of the aggregate of left-open and right-closed boxes in the Euclidean space R”. Its importance lies in the fact that the Cartesian product of finitely many algebras and in particular o-algebras is in general not an algebra, but only a semialgebra.) As an example of extending an additive set function on a semialgebra to a measure, the Lebesgue-Stieltjes measure determined by a real-valued increasing function on R is treated in §22, Theorems establishing the equivalence of the absolute continuity and singularity of a Lebesgue-Stieltjes measure with respect to the Lebesgue measure with the absolute continuity and singularity of the increasing function that determines the Lebesgue-Stieltjes measure are proved. As a second example of extending an additive set function on a semialgebra to a measure, the product measure on the product of finitely many measure spaces is included in §23. Tonelli’s theorem and Fubini’s theorem on the reduction of a multiple integral to iterated integrals are found here. Chapter 6 specializes in integration in the Lebesgue measure space on R”. In §24 the Lebesgue measure on R” is constructed as an extension of the notion of volumes of boxes in R” to Lebesgue measurable subsets of R”. Then it is shown that the Lebesgue measure space on R” is the completion of the n-fold product of the Lebesgue measure space on R. Regularity of the Lebesgue measure and in particular approximation of Lebesgue measurable sets by open sets leads to approximation of the integral of a measurable function by that of a continuous function. Translation invariance of the Lebesgue measure and integral and linear transformation of the Lebesgue measure and integral are treated. §25 begins with the study of the average function of a locally integrable function. Hardy-Littlewood maximal theorem and Lebesgue differentiation theorem are presented. These are followed by differentiation of a set function with respect to the Lebesgue measure, in particular differentiation of a signed Borel measure with respect to the Lebesgue measure, and density of a Lebesgue measurable set with respect to the Lebesgue measure. §26 treats change of variable of integration by differentiable transformations. Chapter 7 is an introduction to Hausdorff measures on R”. §27 defines s-dimensional Hausdorff measures on R" for s € [0, 00) and the Hausdorff dimension of a subset of R”.

xvi

Preface

§28 studies transformations of Hausdorff measures. §29 shows that a Hausdorff measure of

integral dimension is a constant multiple of the Lebesgue measure of the same dimension.

Every concept is defined precisely and every theorem is presented with a detailed and complete proof. I endeavored to present proofs that are natural and inevitable. Counterexamples are presented to show that certain conditions in the hypothesis of a theorem can not be simply dropped. References to earlier results within the text are made extensively so that the relation among the theorems as well as the line of development of the theory can be traced easily. On these grounds this book is suitable for self-study for anyone who has a good background in advanced calculus. In writing this book I am indebted to the works that I consulted. These are listed in the Bibliography. I made no attempt to give the origin of the theory and the theorems. To be consistent, I make no mention of the improvements that I made on some of the theorems. I take this opportunity to thank all the readers who found errors and suggested improvements in the various versions of the lecture notes on which this book is based.

Corona del Mar, California January, 2000

J. Yeh

Preface to the Second Edition In this new edition all chapters have been revised and additional material have been incorporated although the framework and organization of the book are unchanged. Specifically the following sections have been added: §13 [VI] Length of Rectifiable Curves §15 [VII] Baire Category Theorem [VI] Uniform Boundedness Theorem

[IX] Open Mapping Theorem

[X] Hahn-Banach Extension Theorems §16 weak convergence in L? spaces in [III] and [TV] of §16

the complete metric spaces L? for p € (0, 1) in [V] of §16 §19 [V] Signed Radon Measures [VI] Dual Space of C(X) §23 [IV.2] Convolution of L? Functions

[IV.3] Approximate Identity in Convolution Product [IV.4] Approximate Identity Relative to Pointwise Convergence

Besides these topics there are additional theorems in sections: §1, §4, §5, §8, §10, §11,

§13, §15, §16, §17, §19, §20, §21, §23, §24, §25, and §27. Also 64 problems have been

added.

To use this book as a textbook, selection of the following sections for instance makes a possible one-year course at the graduate level:

$1 to §13, §15([T] to [VI]), §16 to §21, §23(M] to [III]) It is my pleasure to thank Abel Klein for his helpful comments on the first edition of this book.

J. Yeh Corona del Mar, California

March, 2006

xvii

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Preface to the Third Edition In this edition several topics are added. Since these topics do not fit in single sections they are presented as appendices. They are: [A] Digital Expansions of Real Numbers [B] Measurability of Limits and Derivatives [C] Lipschitz Condition and Bounded Derivative [D] Uniform Integrability

[E] Product-measurability and Factor-measurability [F] Functions of Bounded Oscillation In [B], we show that if the limit of a real-valued function on R exists then it is Borel-

measurable.

In particular if a real-valued function is differentiable then the derivative is

Borel-measurable. In [C], we show that if a real-valued function satisfies a Lipschitz condition on [a, b] C R then it is differentiable a.e. on [a, b] and moreover the derivative is bounded on [a, 5].

In [D], we discuss uniform integrability and equi-integrability.

In [F], we define the notion of bounded oscillation for a real-valued function whose domain

of definition is a closed box in R”. We show that for the particular case n = 1 a function is of bounded oscillation if and only if it is a function of bounded variation. Also 93 problems have been added. There is now a total of 394 problems.

Corona del Mar, California August, 2013

xix

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List of Notations the natural numbers

ARAAANNZ -

the integers the nonnegative integers the real numbers

the complex numbers RorC the extended real number system {—co} UR U {oo}

the set of ¢ = & + in whereé,n ¢ R the n-dimensional Euclidean space

BX) Bx SBR

the collection of all subsets of a set X

Bp

(Bw), Mt, ‘L

amt!

a(€) a(€)

Miu") Ags

Jo,

Joes

Jeo.

Je

Yo» Noor Neo» Fe

D(f) RS)

{D: f u*(AN (Ei U Ea)) by the subadditivity of p* on §3(R). Thus we have 2*(A) > w*(AN (EU Ex)) + u*(AN (£1 U E2)*). On the other

hand by the subadditivity of 2* on $3 (R), the reverse of this inequality holds. Thus £1 U £2

satisfies condition (3) and is therefore a member of 9J%(*). This shows that Nt(u*) is closed under unions. We show in §2 that 29t(*) is closed under countable unions. A

collection of subsets of a nonempty set X is called a o-algebra of subsets of X if it includes X as a member, is closed under complementations and countable unions. Thus our St(jz*) is ac-algebra of subsets of R. Let us show next that jz* is additive on the o-algebra Nt(z*)

of subsets of R. Thus let £,, £2 € Dt(z*) and assume that Ej NM Ey = §. Now with E, U E, as the testing set A in the z*-measurability condition (3) which is satisfied by £1,

we have

But (EZ; U Eo)

(4)

*(E1 U Bo) = w*((E1 U Ex) 9 E1) + w*((E1 U E2) 9 Ef). Ey = Ey and (Ey U Ey) N Ef = Eo. Thus the last equality reduces to

#* (Ey U Ea) = w*(E1) + w* (Ep).

This shows that jz*, though not additive on $3(R), is additive on the subcollection Nt(u*) of PR). Now pz* is additive on MNt(*) so that we may regard it as the extension of the notion of length to sets which are members of Nt(z*). For this extension j:* to be interesting, the collection )t(j.*) must be large enough to include subsets of R that occur regularly in analysis. In §3, we show that 9)t(2*) includes all open sets in R and all subsets of R that are the results of a sequence of such set theoretic operations as union, intersection, and complementation, on the open sets.

§1 Measure on a o-algebra of Sets

§1

3

Measure on a c-algebra of Sets

[I] o-algebra of Sets Notations. We write N for both the sequence (1, 2, 3, .. .) andthe set {1, 2, 3, ...}. Whether

a sequenice or a set is meant by N should be clear from the context. Similarly we write Z for both (0, 1, —1, 2, —2, ...) and {0, 1, -1, 2, —2, ...} and Z, for both (0, 1, 2,...) and {0, 1,2,...}. Definition 1.1. Let X be an arbitrary set. A collection A of subsets of X is called an algebra (or a field) of subsets of X if it satisfies the following conditions: 1°

XeQ,

2

AEM=>

3

A,BEA>AUBEA.

AEA,

Lemma 1.2. If 2 is an algebra of subsets of a set X, then (1)

GEM,

(2)

At,..., An € A=

G3)

A BeADBANBEA,

(4)

At,..., An € A=

6)

A BeA>A\Bem.

iy Ae € A, (fy Ac € A,

Proof. (1) follows from 1° and 2° of Definition 1.1. (2) is by repeated application of 3°. Since AN

B = (A‘ U B°)*, (3) follows from 2° and 3°. (4) is by repeated application of

(3). For (5) note that A \ B = AN BS € BW by 2° and (3).

wf

Definition 1.3. An algebra A of subsets of a set X is called a o-algebra (or a o-field) if it satisfies the additional condition:

4°

(An:n EN) CA

Une

An € A

Note that applying condition 4° to the sequence (A, B, 0, 8, ...), we obtain condition

3° in Definition 1.1, Thus 3° is implied by 4°. Observe also that if an algebra 2 is a finite collection, then itis ao-algebra. This follows from the fact that when 2 is a finite collection then a countable union of members of 2 is actually a finite union of members of 2 and this finite union is a member of & by (2) of Lemma 1.2.

Lemma 1.4. If & is a o-algebra of subsets of a set X, then (0)

(An:n EN CAD (ey An € A

Proof. Note that (en 4n = (Upen AS)°- By 2°, A$ € Thus by 2°, we have (U,ey AS)" € 2.

and by 4°, nen Ag €

4

CHAPTER 1 Measure Spaces

Notations. For an arbitrary set X, let $3(X) be the collection of all subsets of X. Thus A & §B(X) is equivalent to A C X. Example 1. For an arbitrary set X, ¥3(X) satisfies conditions 1° - 3° of Definition 1.1 and condition 4° of Definition 1.3 and therefore it is a o-algebra of subsets of X. It is the greatest a-algebra of subsets of X in the sense that if 2l is a o-algebra of subsets of X and if P(X) C A thenA = F(X). Example 2. For an arbitrary set X, {4, X} is a o-algebra of subsets of X. It is the smallest

o-algebra of subsets of X in the sense that if 2% is a o-algebra of subsets of X and if Ac {B, X} then A = {B, X}. Example 3. In R?, let 9% be the collection of all rectangles of the type (a1, bi] x (a2, bz] where —oo < a; < b; < oo fori = 1, 2 with the understanding that (a;, 00] = (a, 00).

Let & be the collection of all finite unions every A € 9 is the union of finitely many, We regard § as the union of 0 members of Mis an algebra of subsets of R?. However

of members of $t. We have Ht C A since actually one, members of 9% so that A € . 9% so that @ € &. It is easily verified that & is not a o-algebra. Consider for instance,

An = (n—4,n] x ©, 1] € CM forn EN. Then U),ey An is not a finite unionof members of 9 and is thus not a member of 2.

[II] Limits of Sequences of Sets Definition 1.5. Let (A, : n € N) be a sequence of subsets of a set X. We say that (A, : n € N) is an increasing sequence and write An t if An C Anti forn € N. We say that (A, : n € N) is a decreasing sequence and write A, | if An

D> Anyi forn

ENA

sequence (A, : n € N) is called a monotone sequence if it is either an increasing sequence or a decreasing sequence. For an increasing sequence (A, :n € N), we define

qd)

lim An = U»

= {x € X:x € A, for somen € N}.

RE.

For a decreasing sequence (A, : n € N), we define (2)

bm

An = f)

An = {x €X:x

€ A, foreveryne N}.

neN

For a monotone sequence (A, :n € N),

If A, ¢, then jim An

im, A, always exists although it may be 9.

= @ if and only if A, = 9 for every n € N. If A, |, we may

have tm, An = @ even if A, #

for every n € N. Consider for example X = R and

An = (0,2) forn € N. Then n00 lim A, =. On the other hand if A, = [0, }) forn e N then A, | and

jim, A, = {0}.

In order to define a limit for an arbitrary sequence (A, : n € N) of subsets of a set X

we define first the limit inferior and the limit superior of a sequence.

§1 Measure on a o-algebra of Sets

5

Definition 1.6. We define the limit inferior and the limit superior of a sequence (A, : n € N) of subsets of a set X by setting

) liminfA, = J (7) 4x, neNk>n

(2) lim supAy ==U

4.

neNkzn

Note that (M,>n Ax : 2 € N) is an increasing sequence of subsets of X and this implies

that lim Msn At = Unew Mien Ae exists. Similarly (Usen At: 2 €N) is a decreasing sequence of subsets of X and thus

lim, Ubon Se = Onew Upon Ax exists. Thus

lim inf A, and lim sup A,, always exist although they may be 9. noo

no

Lemma 1.7. Let (A, : n € N) be a sequence of subsets of a set X. Then

(1)

liminf 4, = {x € X: x © A, for all but finitely many n € N}.

Q)

lim sup A, = {x € X: x € A, for infinitely many n € N},

@)

liming A, c lim sup Ap. n—>00

Proof. 1. Letx ¢ X. Ifx € A, for all but finitely many n ¢ N, then there exists no ¢ N such thatx € A; for all k > ap. Then x € (\gong At C nen Mion Ak = liminf An. Conversely if x € lim * inf An = Unen Mean2 Ate then x € xn, 2 Ax for some np € N thus x € A, for all k > no, that is, x € A,, for all but finitely many n € N. This proves 2. Ifx € A, for infinitely many n ¢ N, then for everyn ¢ N we havex € (Jj,, Ax thus x € Mew Upon At = lim sup Ay. Conversely ifx € limsup An = (\ye~ Uson n>00 100

and (1). and Abs

thenx € Uben Ae for every n € N. Thus for every n € N,x € Ax for some k > n. This

shows that x € A, for infinitely many n € N. This proves (2). 3. (1) and (2) imply (3). &

Definition 1.8.

Let (A,

: n € N) be an arbitrary sequence of subsets of a set X.

lim inf A, = jim sup Ap, then we say that the sequence converges and define noo

setting

> OO

jim, Ae = lim inf An

= lim sup A,,.

n00

does not exist Note that this definition of

lim

if liminf A,

If

im, An by

% limsup A, then. ‘tim, An

n>

A, contains the definition of

lim

A, for monotone

n00 n—>00 sequences in Definition 1.5 as particular cases and thus the two definitions are consistent.

Indeed if A, ft then (>, 4¢ = An for every n € N and U,en( ion At = Unen An and therefore liminf An = Unen An. On the other hand, L),, At = Uxen At for every n € Nand (yew Upon At= Uber Ax and thus lim sup An = Unen An- Similarly for An J. Note also that if (A, : n € N) is such that liminf Ay == Q and lim sup A, = @ also n00

6

CHAPTER 1 Measure Spaces

then

lim

n-00

Example.

A, = @. Let X

=

R and let a sequence

(A,

: n

€ N) of subsets of R be defined

by Ai = [0,1], 43 = [0,3], 4s = [0, 5],-.- and Ao = [0, 21, Ay = [0,4], Ag = [0, 6], .... Then lim inf An = {x € X: x

€ A, for all but finitely many n € N} = {0} and

lim sup A, = {x € X : x € A, for infinitely many n € N} = [0, 00). Thus

ny

jim, An does

notexist, The subsequence (Ap, : k € N) = (Aj, A3, As, .. .) is adecreasing sequence with jim n An, = {0} and the subsequence (A,, : k € N) = (Az, Aq, Ag, ...) is an increasing sequence with jim n An

= [0, 00).

Theorem 1.9, Let 2 be a o-algebra of subsets of a set X. For every sequence (A, :n € N) in &, the two sets limminf A, and lim supAy,, are in 2. So is im a An if it exists.

noo

Proof. For every n € N, (yo,Az € A by Lemma 1.4. Then 4° of Definition 1.3. This shows that lim inf An of Definition 1.3. Then cn lim A, exists, then lim n—>0o

noo

€ &.

nen Mon At € A by

Similarly U,., Ak € A by 4°

jon At "2eA. by Lemma 1.4. Thus limsup A, € &. If

A, = liminf A, € &. noo

.

00

[III] Generation of o-algebras Let A be an arbitrary set. If we select call {E, : a € A} a collection of sets are for instanceN = {1, 2,3,...}, Z= arbitrary set A can serve as an indexing

a set E, corresponding to each a ¢€ A, then we indexed by A. Usual examples of indexing set A (0, 1, —1,2, —2,...}, and Z, = {0,1,2,...}. An set.

Lemma 1.10. Let {2l, : a € A} be a collection of o-algebras of subsets of a set X where

A is an arbitrary indexing set. Then 1\yc4 Ua is a a-algebra of subsets of X. Similarly if

{2y : @ € A} is an arbitrary collection of algebras of subsets of X, then \yc4 Aa is an

algebra of subsets of X.

Proof. Let {M. : @ € A} be an arbitrary collection of o-algebras of subsets of X. Then Nee

A, is a collection of subsets of X. To show that it is a o-algebra we verify 1°, 2°,

and 3° in Definition 1.1 and 4° in Definition 1.3. Now X € 2%, for every a € A so that X € (ea Me verifying 1°. To verify 2°, note that if E € (\,c4 Me, then E € A, so that E° € Ay for every a € A and then E° € (\,-4 Ay. 3° is implied by 4°. To verify 4°, let (E, Unen

> n EN)

C Oye4 Me.

En € My.

Then for every a € A, we have (E, : n € N) C Ay so that

Then Unen E,€

Mees Mo.

Ot

Theorem 1.11. Let € be an arbitrary collection of subsets of a set X.

There exists the

smallest o-algebra Ag of subsets of X containing €, smallest in the sense that if U is a

o-algebra of subsets of X containing € then Ay C A. Similarly there exists the smallest algebra of subsets of X containing €.

§1 Measure on a o-algebra of Sets

7

Proof. There exists at least one o-algebra of subsets of X containing €, namely $3(X). Let (Aq

: a € A} be the collection of all o-algebras of subsets of X containing €. Then

Nacsa Mo Contains € and it is ao-algebra according to Lemma 1.10. Itis indeed the smallest o-algebra containing € since any a-algebra &% containing € is a member of {2, : a € A}

so that AD

ge, Ma.

Definition 1.12. For an arbitrary collection € of subsets of a set X, we write o (€) for the

smallest o-algebra of subsets of X containing € and call it the o-algebra generated by €.

Similarly we write a(€) for the smallest algebra of subsets of X containing € and call it

the algebra generated by €.

It follows immediately from the definition above that if €; and €2 are two collections of subsets of a set X and €; C €o, then o(€1) C o(€z). If A is a o-algebra of subsets

of X, then o (2) = 2. In particular for an arbitrary collection € of subsets of X, we have a(o(€)) =a(€).

Let f be a mapping of a set X into a set Y, that is, f is a Y-valued function defined on X. The image of X by f, f(X), is asubset of Y. Let E be an arbitrary subset of Y. E need not be a subset of f (X) and indeed E may be disjoint from f(X). The preimage of E under

the mapping f is a subset of X defined by f—!(E) = {x € X : f(x) € E}, that is, the collection of everyx € X such that f(x) € E. Thus if

EM f(X) = @ then fo)

For an arbitrary subset E of Y we have f(f—!(E)) C E. Note also that

f7W) =x, STE) = $70 \ BE) = FOO)

= 6.

FI) = X\ FI = (Fy,

I7"(Uaes Ea) = Uaes f-"(Ex)s

I" (Maes Bu) = Ques f-"(Ea). For an arbitrary collection € of subsets of Y, let f7(€)

= {f(E)

:EeE€}.

Proposition 1.13. Let f be a mapping of a set X intoa set Y. If B is ao-algebra of subsets of Y then f—'(98) is a o-algebra of subsets of X. Proof. Let us show that 2% is a c-algebra of subsets of X by showing that X € f—1(98),

f71(93) is closed under complementations in X, and f —1(98) is closed under countable

unions.

1. We have X = f—!(¥) € f—1(98) since ¥ € B. 2. Let A € f—!(93). Then A = f—!(B) for some B € 98. Since B° € %B, we have f7-\(B°) © f-1(3). On the other hand, f—1(B°) = (f-1(B))* = A®. Thus we have

Ae € f-1(8). 3. Let (A,

: 2 € N) be an arbitrary sequence in f 7108).

Then A, = f

(Bn) for

8

CHAPTER 1 Measure Spaces

some B, € % for each n € N. Thus we have

Un =U #72) = FU Be) € £1)

neN

neN

neN

since |_), Othe set B(xo, r) = {x € X : p(x, x0) 0 such that B(x, r) C E. An open ball is indeed an open set in the sense defined above. The collection of all open sets in a metric space satisfies axioms I, I, II, and IV and is thus a topology. We call this topology the metric topology of X by the metric p. A set E in a metric space (X, ) is said to be bounded if there exist x9 € X andr > 0

such that E C B(xo,r). A set E in R” is a compact set if and only if E is a bounded and closed set.

Definition 1.16. Let D be the collection of all open sets in a topological space X. We call the o-algebra o (2) the Borel o-algebra of subsets of the topological space X and we write 38x or B(X) for it. We call its members the Borel sets of the topological space. Lemma 1.17. Let € be the collection of all closed sets in a topological space (X,). o(€) =o0(D). Proof. Let E €¢ €. Then E° € O

C a(Q).

Then

Now since o(D) is a o-algebra, we have

E = (E°Y € o(). Thus € C o(D) and consequently o(€) c a (o()) = o(). By

the same sort of argument as above we have o(D) C o(€). Therefore o(€) =o (D).

©

Definition 1.18. Let (X, 9) be a topological space. A subset E of X is called a G-set if it is the intersection of countably many open sets. A subset E of X is called an F,-set if it is the union of countably many closed sets. Thus, if EF is a G-set, then E° is an F,-set, and if E is an F,-set then E° is a Gs-set. Note that every G5-set is a member of Sy. So is every F,-set. Indeed if E is a G;-set,

thenE = (),cn On where O, € O forn € N. Now O, € 0 C o(D) = Bx for every nN. Since %3y is a o-algebra, we have E = (),cy On € Bx. Let us note also that if Z is a G3-set, then there exists a sequence (O,

: n € N) of

open sets such that E = (),cy On. If we let Gz = (\eu1 Ox, then (G, :n € N) isa decreasing sequence of open sets and cy Ga = Mew On = E. Thus a Gs-set is always the limit of a decreasing sequence of open sets. Similarly if E is an F,-set, then there exists

§1 Measure on a o-algebra of Sets

11

a sequence (C,, : n € N) of closed sets such that E =

nw Cy. If we let F, = Ula

Ck,

then (F, : n € N) is an increasing sequence of closed sets and nen Fn = Une Cn = EThus an F,-set is always the limit of an increasing sequence of closed sets.

[V] Measure on a c-algebra Notations. Let R = {—oo} URU {co} and call it the extended real number system. We use the alternate notation [—oo, co] for R also. Definition 1.19. Let € be a collection of subsets of aset X. Let y be anonnegative extended

real-valued set function on €. We say that (a) y is monotone on € if y(E1) < y(E2) for Ei, Ez € € such that E, C E2,

(b) y is additive on € if y(E1 U E2) = y(E1) + y(E2) for E1, Ez € € such that E, NE. =@and E, UE. € €,

(©

y is finitely additive on € if y (U1 Ex) = Di

sequence (Ex: k =1,...,n) in € such that Ufa

y (En for every disjoint finite

E,e

€,

(d) y is countably additive on € if vy (Uncx En) = Vnen ¥ (En) for every disjoint sequence (Ey, :n € N) in € such that |),cy En € €

(©) y is subadditive on € if y(E, U Eo) < y(E1) + y(E2) for E1, Eo € € such that E,UE€&

(@ y is finitely subadditive on € if y (Uha Ex) < Di (Ex) for every finite sequence (Ex :k =1,...,n) in € such that (Z_; Ex € €, (g) y is countably subadditive on € if y (Unen En) < Yyen Y (En) for every sequence (En in €N) in € such that ncn En € €.

Note that in (c) while LJf_, Ex € € is required, it is not required that any of J2_1 Ex.

Ubi Et, ..., U2zi Ex be in €. Note also that (c) implies (b) and (f) implies (¢).

Observation 1.20. Let y be a nonnegative extended real-valued set function on a collection € of subsets ofa set X. Assume that # € € and y(@) = 0.

(a) If y is countably additive on €&, then it is finitely additive on €. (b) If y is countably subadditive on €, then it is finitely subadditive on €. Proof.

Suppose y is countably additive on €.

To show that it is finitely additive on

€, let (Ex : k = 1,...,m) be a disjoint finite sequence in € such that Ute Eye €. Consider the infinite sequence (F, : k € N) in € defined by & = E, fork = 1,...,n and F, = fork >n+1. Since# € €, (RH : k € N) is a disjoint sequence in € with

Uben Fe = Uf

Ex € €. Thus by the countable additivity of y on € and by the fact that

y(B) = 0, we have y (Ut Et) = » (Uren Fe) = Daew (Fe) = Cs y (Ex). This

proves the finite additivity of y on €. We show similarly that if y is countably subadditive on &, then it is finitely subadditive on €. Lemma 1.21, Let (£,, : n € N) be an arbitrary sequence in an algebra X of subsets of a set X. Then there exists a disjoint sequence (F, : n € N) in Q such that

12

(1)

and ®

CHAPTER 1 Measure Spaces

N

N

a=1

n=1

Ua=U"

for every N EN,

Un=U-A. neN

neN

In particular, if & is a o-algebra, then nen Fa = Unen En € &. Proof. Let F, = E, and F, = E, \ (E, U...U E,_1) forn > 2. Since 2 is an algebra, F, € A forn ¢ N. Let us prove (1) and (2) and then the disjointness of (F, : n € N). Let us prove (1) by induction. To start with, (1) is valid when N = 1 since Fy = Ej.

Next, assume that (1) is valid for some N € N, that is, Usa E, = Ux, F,,. Then we have N+1

N

N

N

N+

n=l

a=1

n=l

n=l

n=l

U me =(U%) U Frat = (U En) U (Eve \ U 2a) = U Ee

that is, (1) holds for N + 1. Thus by induction, (1) holds for every N € N. To prove (2), let x

€ Unen E,.

Then x €

E,, for some n € N and thus we have

% © ger Ex = Ufa Fe C Ujen Fn by (1). We show similarly that ifx € Ucn 7, then x € nen En. Thus we have en En = Unen Fn. This proves (2). Finally let us show that (F, : n € N) is a disjoint sequence. Consider F,, and F, where

n #m, sayn w (Uta Ex) = Dee MED, (2)

monotonicity: E,, Ez € A, Ey C FE. > w(E1) < w(E2),

G)

Ei, By € &, Ey C Ey, w(E1) < 00 = w(E2 \ Ei) = w(E2) — w(ED,

(4)

countable subadditivity: (En :n € N) CM => w (hen En) < nen (En)

and in particular

(5) finite subadditivity: (E1,..., En) CM => w (Uta Ex) < DL w(EdProof. The countable additivity of ~ on A implies its finite additivity on 2% by (a) of Observation 1.20. The finite additivity of jz on 2% implies its additivity on 2{ and then its monotonicity on 2 by (a) of Lemma 1.22. To prove (3), let £1, Hz

€ Wand

Ey

C

Ey.

Then £; and E> \ E; are two disjoint

14

CHAPTER

1 Measure Spaces

members of 2 whose union is equal to E27. Thus by the additivity of 2 on 2, we have B(E2) = B(E}) + e(E2 \ £1). If w(E1) < 00, then subtracting (£1) for both sides of the last equality, we have (£2) — (£1) = (E2 \ £1). This proves (3).

The countable additivity of jz on 2 implies its countable subadditivity on & by (b) of Lemma 1.22. This then implies the finite subadditivity of 4. on 21 by (b) of Observation 1.20. & Regarding (3) of Lemma 1.25, let us note that if 4(£,) = oo then by the monotonicity of 2 we have (Ez) = 00 also so that 4.(£2) — w(E1) is not defined.

[VI] Measures of a Sequence of Sets Let be a measure on a a-algebra & of subsets of a set X. a sequence in 2. If lim E&, exists, does lim y(E,) exist? uf jim, En) =

n>

noo

Let (EZ, : n € N) be If it does, do we have

jim, uU(E,)? The next theorem addresses this question for monotone se-

quences of measurable sets. It is based on the countable additivity of a measure. It is a fundamental theorem in that a subsequent theorem regarding the limit inferior and the limit superior of the measures of an arbitrary sequence of measurable sets as well as the monotone convergence theorem for the Lebesgue integral, Fatou’s lemma, and Lebesgue’s dominated convergence theorem are ultimately based on this theorem. Theorem 1.26. (Monotone Convergence Theorem for Sequences of Measurable Sets) Let

be a measure on a o-algebra A of subsets of a set X and let (E,

monotone sequence in Q.

:n

€ N) bea

(a)If En t, then no lim j(En) = u4( lim 00 Ey). (b) FE,

J, then

lim p(E,) = a

n—>00

HAA) < 00 such that Ey Cc A.

iim

> OO

Ex), provided that there exists a set A €

UA with

Proof. If Z,, +, then dim, En = Unen En € Wl. If Ey J, then dim, En = (hen En € A. Note also that if (E, : n € N) is a monotone sequence in 2, then (u(Z,) : 2 € N) isa monotone sequence in [0, oo] by the monotonicity of jz so that jim (En) exists in [0, oo]. 1. Suppose E,, +. Then we have u(Z,) +. Consider first the case where 2(Ey,) = 00 for some mo € N. In this case we have Aim, w(En) = oo. Since Eno C Unen En =

lim E,, we have jt( lim E,) > (Eno) = 00. Thus j4( lim E,) =0co=

noo

noo

ROO

lim (Ep).

n> 00

Consider next the case where j4(E,) < 00 for every n € N. Let Ey = G and consider a disjoint sequence (F, : n € N) in 2 defined by F,, = E, \ En-1 forn € N. We have

En = UN, Fa for every N € N and hence Unen En = Unen /n- Then we have

u( Jim En) = 2((JEn) = (Um) = a) neN

neN

neN

=o En \ Env) =O {wEn) — o(En-v)}, neN

ncN

§1 Measure on a o-algebra of Sets

15

where the third equality is by the countable additivity of yz. and the fifth equality is by (3) of Lemma 1.25. Since the sum ofa series is the limit of the sequence of partial sums we have

Do {eGin) — e(En—1)} = lim "Y7 {a Ex) — uEx-1)}

neN

k=1

= Jim {w(En) — w(Eo)} = lim (En). Thus we have j2( lim E,) = lim (E,). AO

n> 0O.

2. Suppose E, | and assume the existence of a containing set A with finite measure. Define a disjoint sequence (F, : n € N) in 2 by setting F, = Ey \ En41 form € N. Then

a

A\(\h=Ur,. neN

neN

To show this, let x € £1 \ Myen En. Thenx € EF, andx is not in every E,. Since Ey |, there exists the first set E,.+41 in the sequence not containing x. Then x € Eng \ Engt+i =

Fag C Unen Fn. This shows that Ey \ Men En C Unen Fn. Conversely ifx € Unen Fas

thenx € Fry = Eng \ Eno+1 for some no € N. Now x € En, C Ey. Sincex ¢ Engii,We

have x ¢ pen En. Thus x € E; \ Mex En. This shows that Len Fn C Ei \ Open En Therefore (1) holds. Now by (1), we have

@)

w(E1\ 7) En) =H( UF): neN

neN

Since 4 (Qnen En) < #(E1) < (A) < 00, we have by (3) of Lemma 1.25

@)

——w(Z1\ 7) Ba) =n) — 2( 1) Bn) = wy) — wf Jim 2). neN

neN

By the countable additivity of 4, we have

@)

u( U Fa) = 0 wa) = 7 wa \ Ens) neN

neN

nen

=) {oGn) — wEny1)} = fim YY fa (Ee) — a Ber)} neN

= im

k=1

{w(E1) — w(En41)} = aE) — fim, u(En+1)-

Substituting (3) and (4) in (2), we have

H(E1) — wf Jim En) = 2(Ei) — lim (En41) = #(E1) — lim jo(Ep). Subtracting (£1) € R from both sides we have a lim, En) = jim, (En).

O

16

CHAPTER 1 Measure Spaces

Remark 1.27. (b) of Theorem 1.26 has the following particular cases. Let (FE, : n € N) be

a decteasing sequence in %. Then lim (En) = a lim, E,) if any one of the following conditions is satisfied:

(a) U(X) < 00, (b) u(E1) < «0,

(c) 4(En,) < 00 for some no € N. Proof. (a) and (b) are particular cases of (b) of Theorem 1.26 in which X and FE) respectively

are the containing set A € 21 with 4(A) < oo.

To prove (c), suppose 4(En,) < 00 for some no € N. Let (F;, : 2 € N) be a decreasing sequence in 21 obtained by dropping the first no terms from (EZ, : n € N), that is, we set F,, = Engin forn € N. Lemma 1.7 implies that lim inf i= lim inf E,, and lim sup F, = lim sup EZ, and thus noo

and F,

C

lim

En, forn

F, =

noo

lim

n->00

€ N and since 4(En,)

lim »(F,) = u({ lim F,) = (

noo

n>00

E,. Now since (F, : n € N) is a decreasing sequence

noo

< 00, (b) of Theorem

1.26 applies so that

lim E,). Since (u(F,) : n € N) is a sequence obtained

noo

by dropping the first ng terms of (%(E,)

: n € N), we have im, u(F,)

= im, H(E,).

Therefore we have lim y(E,) = 2{ lim Ep). a)

n00

Let yz be a measure on a o-algebra A of subsets of a set X.

Then for an arbitrary

sequence (E, :n € N) in 2, lim inf E,, and lim sup E,, exist in 1 by Theorem 1.9 and thus

n>00

uf lim inf E,) and j.(lim sup E,,) are defined. Now (1(E,) : n € N) isasequencein [0, 00] >

and thus liminf z(E,) = lim inf w(E,) and limsupu(E,) = N00 lim pon sup 2(E,) exist in n-00 n>ook>n n>0o [0, 00]. How are j4(liminf E,) and j:(lim sup E,) related respectively to lim inf 4(E,) 7100

noo

and lim sup 4(Z,)? The next theorem addresses this question.

100

ROO

Theorem 1.28. Let 2 be a measure on a o-algebra A of subsets of a set X. (a) For an arbitrary sequence (E, :n € N) in Q, we have

()

u(liminf inf CE). n00 £,) < limn>00

(b) If there exists A € SA with u(A) < 00 such that E, C Aforn €N, then

2) (c)

@)

j(lim sup E,,) > lim sup 2(E,). n->co

both

lim E, and

ROO

n+>00

lim p(E,) exist, then

n->00

w( noo lim Eq) < n—>00 tim (En).

@ if jim En exist and if there exists A € A with w(A) < 00 such that E, C Aforn €N, then im

4

w(En) exists and

Hf lim En) = Jim w(En).

§1 Measure on a o-algebra of Sets

Proof.

1. Recall that lim inf E,

17

=

new Mon Be =

(Neen Ex : 2 € N) is an increasing sequence in %.

have y(liminf Zn) = lim

tim, Chen Ex by the fact that

Then by (a) of Theorem 1.26, we

(Myon Ee) = liminf (Myon Ze) since the limit of a

sequence, if it exists, is equal to the limit inferior of the sequence.

Since

hen Ex C En,

we have 14 (Msn Ex) < (En) for n € N by the monotonicity of 2. This then implies

lim:inf # (Men Ex) < lim inf u(E,). Continuing the chain of equalities above with this inequality, we have (1). 2. Assume that there exists A € {2 with 4(A) < oo such that E, C A forn € N. Now

limsup En = nen Uson Ex = lim Ubon Ex by the fact that (Ubon Ex

2 €N) isa

decreasing sequence in A.

Cc

n—00

Since E, C A for all m € N, we have Ubon E,

A for all

n & N. Thus we have jz(limsup Ep) = w( no9 lim Uj,RE Ex) = n>00 lim 2 (UponI Ex) by 0) n—>00 of Theorem 1.26. Now lim

(Uren 2x) = lim sup #(Usen Ex) since the limit of a se-

quence, if it exists, is equal to the limit superior of the sequence. Then by 5, Zz D En, we have 1 (Ups, Ex) = w(En). Thus lim sup (Upon Ex) = lim sup u(E,). Continuing the chain of equalities above with this jncquality, we have (2). 3. If im, E,, and jim, BA(E,) exist, then im, E,, = liminf E,, and im, BE)

lim nf s.(i,) 80 that (1) reduces to (3)

ue

=

“4. If ima, E,, exists, then im sup En = im, Ey,= lim inf E,. If there existsA € 2 with uA)

6)

< oo such that E, C "A for ne N. then by (2) and (1) we have

limsup 2(En) = #(lim sup E,} = u( lim Ep) now

Aco

noo

= 2(liminf E,,) < liminf 2(E,). n>00

n>00

Since liminf 4(Z,) < lim sup (£,) the inequalities (5) imply noo n>00

(6) Thus

(4)

liminf (En) = w{ lim En) = lim sup HE). lim ,(£,) exists and then by (6) we have ( lim E,) = lim (Ey). This proves nora n->00 ni>00

[VII] Measurable Space and Measure Space Definition 1.29. Let & be a o-algebra of subsets of a set X. The pair (X, QW) is called a measurable space. A subset E of X is said to be &-measurable if E € A. Definition 1.30. (a) if 4 is a measure on a o-algebra ‘XA of subsets of a set X, we call the triple (X, A, w) a measure space.

(b) A measure pp on a o-algebra A of subsets of a set X is called a finite measure if

18

CHAPTER 1 Measure Spaces

#(X) < 00. In this case, (X, &, 2) is called a finite measure space. (c) A measure 2 on a o-algebra & of subsets of a set X is called ao -finite measure if there exists a sequence (E, : n € N) in & such that Jen En = X and w(En) < 00 for every

n EN. In this case (X, A, ys) is called a o-finite measure space.

(d) A set D € & in an arbitrary measure space (X, A, i) is called a o-finite set if there exists a sequence (D, : n € N) in & such that Unen Dy, = D and p(D,) < 00 for every

neN.

Lemma 1.31. (a) Let (X, 2, 4) be a measure space. If D € Nis a o-finite set, then there exists an increasing sequence (F, :n € N) in Ql such that lim | F, = Dand w(F,) < oofor

1

everyn & N and there exists a disjoint sequence (G, : n € N) in'A such that |J,en Gn = D and 4(Gn) < 00 for everyn EN.

(b) If (X, &, 2) is a o-finite measure space then every D € Wis aa-finite set.

Proof. 1. Let (X, &, jz) be a measure space. Suppose D € 2 is a o-finite set. Then there exists a sequence (D, : n € N) in & such that |), 2 Dy = D and u(D,) < 00 for every n EN. Foreachn EN, let F, = Ufet D,. Then (F, : n € N) is an increasing sequence

in 2 such that jim Fr = UnenFn = Unen Dn = D and u(F,) = u(Uf-1 De) < Yhe1 H(Dx) < 00 for every n EN.

Let G, = Fy and Gy = Fy \Ufc & forn > 2. Then (Gq : n € N) is a disjoint

sequence in 2 such that L,en Ga = Unser,

= D as in the Proof of Lemma 1.21,

(G1) = w(Fi) < 00 and 4(Gn) = (Fa \ Uli Fe) S a(Fa) < 00 forn > 2. This

proves (a).

2. Let (X, &, 2) be a o-finite measure space. Then there exists a sequence (EZ, : n € N) in 2 such that LenEn = X and u(E,) < 00 for every n € N. Let D € 2. For each n EN, let Dp = DN Ep. Then (D, : n € N) is a sequence in 2 such that L),.4 Dn = D and 4(D,) < 4(E,) < 00 for every n € N. Thus D is a o-finite set. This proves (b).

Definition 1.32. Given a measure 4 on a o-algebra A of subsets of a set X. A subset E of X is called a null set with respect to the measure yz if E € Wand u(E) = 0. In this case we say also that E is a nuil set in the measure space (X, A, uz). (Note that 9 is a null set in

amy measure space but a null set in a measure space need not be 9.)

Observation 1.33. A countable union of null sets in a measure space is a null set of the measure space. Proof. Let (E, : n € N) be a sequence of null sets in a measure space (X, &, 4). Let E = UnenEn- Since & is closed under countable unions, we have E € Q&. By the countable subadditivity of 2 on &, we have u(E) < Yen #(En) = 0. Thus w(E) = 0. This shows that E is a null set in (X, 2, 2). ©

Definition 1.34. Given a measure yt on a o-algebra & of subsets of a set X. We say that the o-algebra X is complete with respect to the measure js if an arbitrary subset Eo of a null set

§1 Measure on a o-algebra of Sets

19

E with respect to us is a member of A (and consequently has (Eo) = 0 by the monotonicity of 4). When & is complete with respect to 1, we say that (X, A, 4) is a complete measure

space,

Example.

Let X = {a,b,c}. Then & = {9, {a}, {b, c}, X} is a o-algebra of subsets of

X. If we define a set function 4 on X by setting (0) = 0, w({a}) = 1, u({b,c}) = 0,

and 4.(X) = 1, then yw is a measure on 2. The set {b, c} is a null set in the measure space (X, A, 2), but its subset {5} is not a member of 2. Therefore (X, A, 2) is not a complete

measure space.

Definition 1.35. (a) Given a measurable space (X, UM). An A-measurable set E is called

an atom of the measurable space if @ and E are the only A-measurable subsets of E.

(b) Given a measure space (X, A, 4). An A-measurable set E is called an atom of the

measure space if it satisfies the following conditions :

1°

2(E) > 0,

2

Eg C £, Eg € R=

p(Eo) =0

or u(Eq) = (EZ).

Observe that if Z is an atom of (X, 20) and w(£) > 0, then E is an atom of (X, A, yz). Example. In a measurable space (X, &) where X = {a, b, c} and A = {G, {a}, {b, c}, X},

if we define a set function HX)

on 2 by setting (8) = 0, w{{a}) = 1, u({b,c}) = 2, and

= 3, then yz is a measure on &.

(X, 2, w).

The set {b, c} is an atom of the measure space

[VI] Measurable Mapping Let f be a mapping of a subset D of a set X into a set Y. We write D(f) and (f) for the domain of definition and the range of f respectively. Thus D(f=DcX,

MP) ={y oY: y = f(&) forsomex e D(P)} CY. For the image of D(f) by f we have f(D(f)) = N(f).

For an arbitrary subset E of Y we define the preimage of E under the mapping f by

f\(E) = [x eX: f(x) € E} = {x DY): f@) € E}. Note that F is an arbitrary subset of Y and need not be a subset of S8(f).

Indeed E may

be disjoint from 9%(f), in which case f—1(E) = 9. In general we have f(f—1(E)) c E. For an arbitrary collection € of subsets of Y, we let f-1(€) := {f-1(Z) : E € €}. Observation 1.36.

Given sets X and Y.

Let f be a mapping with D(f)

C X and

20

CHAPTER 1 Measure Spaces

SRCf) CY. Let E and Ey be arbitrary subsets of Y. Then

(a)

£1) =D(),

Q)

SIE) =f

@Q)

f-\(E) = (f-"(®)’ _ provided that D(f) = ¥,

4

6)

O\ =f

OMIT =DBIN\ II ®.

£7" (Use Bu) = Unea f-" (Ea), I (Muea Ea) = Maea f'(Ea).

Proposition 1.37. Given sets X and Y. Let f beamapping withD(f) C X and Rf) c Y. If 33 is ao-algebra of subsets of Y then f—'(98) is a o-algebra of subsets of the set D(f). In particular, if D(f) = X then f—) (9B) is a o-algebra of subsets of the set X. Proof. Let 93 be a o-algebra of subsets of the set Y. To show that f—'(93) is a c-algebra of subsets of the set D(/) we verify:

1°

D(f) € f-'B).

3°

(An: n © N) C f71(93) > Upen An © £713).

2

Ae f-1(B) > D(f)\A€ f7'(3).

This is done below.

1. By (1) of Observation 1.36, we have D(f) = f7'(Y) © f-1(9B) since Y € B.

2. Let A € f—1(93). Then A = f—'(B) for some B € 93. Then by (2) of Observation

1.36 we have D(f)\ A = D(f)\ f1(B)

= f-1(B*). Since B is aa c-algebra, B ¢ B

implies B° € %. Then f~!(B°) € f—1(83). This shows that D(f) \ A € f—1(98). 3. Let (A, :n € N) bea sequence in f—!(93). Then A, = f—!(B,) for some B, € 3 for each n € N. Then by (4) of Observation 1.36, we have

Un =U 6) = 7(U Be) € £1),

neN

neN

nen

since U,00 lim Fy. n—->0O

(4)

Prob. 1.2. (a) Let (A, : n € N) be a sequence of subsets ofa set X. Let (B, :n € N) bea sequence obtained by dropping finitely many entries in the sequence (A, : n € N). Show that lim inf By = lim inf An and

fim sup Ba=

imm sup A,.

Show that im, Bp, exists if

and only if jim, Ag exists andwhen ‘they exist theyae a equal. (b) Let (An: TR € N) and (B, : n € N) be two sequences of subsets of a set X such that A, = 8B, for all but finitely many n € N. Show that fim inf By, = lim inf Ap and lim sup B, = iimn sup Ay. Show that jim, B,, exists if and only’ if “iim a>00

A, exists and when

they exist they are re equal. Prob. 1.3. Let (Z,, : n € N) bea disjoint sequence of subsets of a set X. Show that im, En exists and

lim £, = 9.

n>00

Prob. 1.4. Leta € R and let (x, : n € N) be a sequence of points in R, all distinct from a, such that jim, X, = a. Show that im a {xn} exists and jima {xn} = 9 and thus

jim {xn} # {a}. Prob. 1.5. ForE c Randt

€ R, let us write E+¢ = {x +f

€R:x

€ E} and call it

the translate of E by t. Let (f, : n € N) be a strictly decreasing sequence in R such that jim, t, = Oand let E,= E +t, forn € N. Let us investigate the existence of im, En.

(a) Let E = (— 00, 0). Show that in, E, = (—00, 0]. (b) Let E = {a} where a € R. Show that Jim, E, =. (c) Let E =

[a, b] wherea, b € Randa

(d) Let E = (a, b) where a, b € Randa (e) Let E =

< b. “Show that

fim, E, = (a, 6].

< b. Show that

jim, En=

(a, b].

Q, the set of all rational numbers. Assume ‘that (i,: n € N) satisfies the

additional condition that £, is a rational number for all but finitely many n € N. Show that Jim, E,= E. OlLet E = Qas in (d) but assume that (1, : n € N) satisfies the additional condition that t,

is a rational number for infinitely many n € N and 1, is an irrational number for infinitely many n € N. Show that Jima, E,, does not exist. Prob. 1.6. The characteristic function 14 of a subset A of a set X is a function on X defined

by

1

1a@) = { 0

forx € A,

forx € A‘.

Let (A, : n € N) be a sequence of subsets of X and A be a subset of X.

24

CHAPTER

(a) Show that if jim, An = A then {b) Show that if ‘im

lim

100

1 Measure Spaces

14, = 1, on X.

14, =1, on X then

jim, Ayn = A.

Prob. 1.7. Let 2 be a o-algebra of subsets of a set X and let Y be an arbitrary subset of X. Let 3 = {ANY : A € &}. Show that % is a o-algebra of subsets of Y. Prob. 1.8. Let 21 be a collection of subsets of a set X with the following properties: 1°, XeA, 2°. A BEA>DA\B=ANB EA, Show that 2 is an algebra of subsets of the set X. Prob. 1.9. Let 2 be an algebra of subsets of a set X. Suppose 2 has the property that for every increasing sequence (A, : n € N) in 2, we have L),-w An € A.

o-algebra of subsets of the set X. Prob. 1.10.

Let (X, 2l) be a measurable space and let (Z,

sequence in % such that J, oy En = X.

Show that & is a

: n € N) be an increasing

(a) Let A, = AN Ep, that is, Uy, = {AN E, : A € A}. Show that , is a oc-algebra of subsets of E, for each n € N.

(b) Does Jen Mn = A hold? Prob. 1.11. (a) Show that if (2, : n € N) is an increasing sequence of algebras of subsets

of a set X, then _),cv 2, is an algebra of subsets of X.

(b) Show that if (1, : n € N) is a decreasing sequence of algebras of subsets of a set X,

then (cy Mn is an algebra of subsets of X.

Prob. 1.12. Let (X, 21) be a measurable space. Let us call an 2{-measurable subset E of X an atom in the measurable space (X, 2) ifE ~ @ and G and E are the only 2-measurable subsets of E. Show that if £, and E> are two distinct atoms in (X, MA) then they are disjoint.

Prob. 1.13. For an arbitrary collection € of subsets of a set X, let a(€) be the algebra generated by &, that is, the smallest algebra of subsets of X containing €, and let o(€) be the o-algebra generated by €. Prove the following statements:

@) n(ate) ==a(€), (b) o(o(€))= o(€), () a(€) Cc o(€),

(d) if € is a finite collection, then a(€) = o(€),

(©) o(a(€)) =o(€).

(Hint for (d): Use Prob. 1.18 below.)

Prob. 1.14. Let (A, : n € N) be a monotone sequence of c-algebras of subsets of a set X and let A = lim, An. {a) Show that if (ln : n € N) is a decreasing sequence then 2 is a o-algebra.

(b) Show that if (2, : n € N) is an increasing sequence then 2 is an algebra but 2 may not be a o-algebra by constructing an example. Prob. 1.15. Let € = {Aj, ..., An} be a disjoint collection of nonempty subsets of a set X

§1 Measure on a o-algebra of Sets

25

such that |_7_, Ay = X. Let ¥ be the collection of all arbitrary unions of members of €. (a) Show that ¥ = o(€), the smallest o-algebra of subsets of X containing €. (b) Show that the cardinality of o (€) is equal to 2".

Prob. 1.16. Let € = {A; : i € N} be a disjoint collection of nonempty subsets of a set X such that |_), 0 for every k € N. Define a set function yz on 2 by setting = Deen Okie. Show that us is a measure on 2.

Prob. 1.24, Let X = (0, oc) and let J = {4 : k € N} where , = (K—1,k] fork EN.

26

CHAPTER 1 Measure Spaces

Let 2% be the collection of all arbitrary unions of members of J. For every A € 2% let us define jz(A) to be the number of elements of J that constitute A. (a) Show that 2 is a o-algebra of subsets of X.

(b) Show that jz is a measure on the o-algebra 2. (c) Let (A, : 2 EN)

C 2 where A, = (7, 00) form € N.

sequence (A, : n € N) we have im, (An) ¥ H( lim

An).

Show that for the decreasing

Prob. 1.25. Let (X, 2, 4) be a o-finite measure space so that there exists a sequence (E, sn € N) in & such that JeyEn = X and 4(E,) < 00 for every n € N. Show that there exists a disjoint sequence (F, : € N) in & such that ),.j FH, = X and u(F,) < 00

for every n € N.

Prob. 1.26. Let SR be the Borel o-algebra of subsets of R, that is, the smallest o-algebra

of subsets of R containing the collection of all open sets in R. The Lebesgue measure 1,

is a measure on %$R with the property that for every interval 7 in R, 4, (7) = €(/) where £(J) is the length of 7. The Lebesgue measure jz, on SR will be constructed in §3. Here

we assume its existence and pose the following problems: (a) Construct a sequence

(E,

: n

€

N) of sets in Sg

such that lim ne

lim _j4, (E,) does not exist.

n>00

(b) Construct a sequence (E, lim

n>

: n € N) of sets in BR

E,, exists but

such that jim, ui, (Ep) exists but

E,, does not exist.

(c) Construct a sequence

(EZ,

: n

€

WN) of sets in Bp

such that both

jim, E,, and

jim, 4, (En) exist but ,( lim En) # lim y, (En).

(d) Show that for every x € R, we have {x} € Sp and yz, ({x}) = 0.

(e) Let Q be the set of all rational numbers in R. Show that Q € Sx and ,(Q) = 0. (f) Let P be the set of all irrational numbers in R. Show that P € Sp and 4, (P) = ov. (g) Construct an uncountable union of null sets that is not a null set. Prob. 1.27. Consider the measurable space (R, 93(R)) where 93(R) is the o-algebra of all

subsets of R. Let us define a set function jz on ¥3(R) by setting ~(E) for E € 93(R) to be equal to the number of elements in FE when £ is a finite set and setting 4(£) = co when E

is an infinite set.

(a) Show that yz is a measure on $3(R). (b) Show that the measure yz is not o-finite.

Prob. 1.28. Let (X, 2, 42) be a finite measure space. Let € = {E, : A € A} be a disjoint collection of members of 21 such that .(E)

a countable collection.

> 0 for every A € A. Show that € is at most

Prob. 1.29. Let X be a countably infinite set and let 2 be the o-algebra of all subsets of X. Define a set function 4 on 2 by defining for every E ¢ A WE) = { 0 if E isa finite set, oo otherwise. (a) Show that jz is additive but not countably additive on 21. (b) Show that X is the limit of an increasing sequence (E, : n € N) in MA with w(F,) = 0

§1 Measure on a o-algebra of Sets

27

for all n, but 4(X) = oo. Prob. 1.30. Let X be an arbitrary infinite set. We say that a subset A of X is cofinite if A° is a finite set. Let MA be the collection of all the finite and the cofinite subsets of the set X. (a) Show that 2 is an algebra of subsets of X. (b) Show that 2f is not a a-algebra.

Prob. 1.31,

Let X be an arbitrary uncountable set. We say that a subset A of X is co-

countable if A‘ is a countable set.

Let 21 be the collection of all the countable and the

co-countable subsets of the set X. Show that 2% is a o-algebra of subsets of X. (This offers an example that an uncountable union of members of a o-algebra is not a member of the o-algebra. Indeed, let A be a subset of X such that neither A nor A° is a

countable set so that A, A° ¢ %. Let A be givenas A = {x, € X: y eT} where I’ is an uncountable set. Then the finite set {x,} € 2 for every y < I’, but Uyerfxy} =A¢A)

Prob. 1.32. Let X be an infinite set and let 2 be the algebra of subsets of X consisting of the finite and the cofinite subsets of X (cf. Prob. 1.30). Define a set function 4 on 21 by

setting for every A € Ml: tA) = 0 if A is finite, 1 if Ais cofinite. (Note that since X is an infinite set, no subset A of X can be both finite and cofinite although

it can be neither.) (a) Show that jz is additive on the algebra 2.

(b) Show that when X is countably infinite, 2 is not countably additive on the algebra 2. (c) Show that when X is countably infinite, then X is the limit of an increasing sequence (A, :n € N) in 2 with 2(A,) = 0 for every n € N, but u(X) = 1.

(d) Show that when X is uncountable, then 4 is countably additive on the algebra 21. Prob. 1.33. Let X be an uncountable set and let 2 be the «-algebra of subsets of X consisting of the countable and the co-countable subsets of X (cf. Prob. 1.31). Define a set function

# on A by

setting for every A € A: (A) = { 0 if Ais countable, 1 if Ais co-countable.

(Note that since X is an uncountable set, no subset A of X can be both countable and

co-countable although it can be neither.) Show that yz is countably additive on 2.

Prob. 1.34. Given a measure space (X, A, 44). We say that a collection {A, : 4 € A} Cc A is almost disjoint if 41,42 € A and Ay A Az imply 4(A,, M Aj.) = 0.

(a) Show that if {A, : n € N} C (b) Show that if {A,

: n € N}

is almost disjoint then (ncn

C A is such that uf Unew An)

An) = Donen H(An)=

Donen (An)

and

H(A,) < 00 for every n € N then {A, : 2 € N} is almost disjoint. (c) Show that if we remove the condition in (b) that %(A,) < co for every n € N then the

condition (Ucn 4n) = Sonen #(An) alone does not imply that {A, : 2 € N} is almost disjoint.

Prob. 1.35. Let (X, 2, 4) be a measure space. The symmetric difference of two subsets A and B of a setX is defined by AAB

= (A \ B) U (B\ A).

28

CHAPTER 1 Measure Spaces

(a) Show that A = B if and only if AAB = 9. (b) Show that AU B = (AN B) U (AAB). (c) Prove the triangle inequality for the symmetric difference of sets, that is, for any three subsets A, B, C of aset X we have AAB

C (AAC) U

(CAB).

(d) Show that 4(AAB) < u(AAC) + 4(CAB) for any A, B,C € A. (e) Show that (A U B) = 2(AN B) + uw(AAB) for any A, Be A. (f) Show that if ~(AAB) = 0 then u(A) = p(B) for any A, Be A.

Preamble to Prob. 1.36. Let (X, &, 4) be a finite measure space. Then a function p on A x AW defined by p(A, B) = u(AAB) for A, B € A has the following properties:

1° p(A, B) € [0, #)], 2° p(A, B) = p(B, A), 3° p(A, B) < p(A, C) + pC, B).

However p need not be a metric on the set 21 since (A, B) = 0 does not imply A = B. Prob. 1.36. Let (X, UA, 4) be a finite measure space. Let a relation ~ among the members of & be defined by writing A ~ B when 4(AAB) = 0. (a) Show that ~ is an equivalence relation, that is,

1° 2 3°

A~A, A~B>B~A, A~B,B~CS>A~C.

(b) Let [A] be the equivalence class to which A belongs and let [2] be the collection of all

the equivalence classes with respect to the equivalence relation ~. Define a function p* on

[Xt] x [2t] by setting o*([A], [B]) = #(AAB) for [A], [B] € [21].

Show that p* is well defined in the sense that its definition as given above does not depend on the particular representative A and B of the equivalence classes [A] and [8B]; in other words,

A’ € [A], B’ € [B] => (A’AB’) = p(AAB).

(c) Show that p* is a metric on the set [2].

§2 Outer Measures

§2

29

Outer Measures

[1] Construction of Measure by Means of Outer Measure Definition 2.1, Let X be an arbitrary set. A set function x* defined on the a-algebra B(X) of all subsets of X is called an outer measure on X if it satisfies the following conditions : 1°

nonnegative extended real-valued: u*(E) € [0, 00] for every E € $3(X),

2°

w*(B) = 0,

3° 4°

monotonicity: E1, Ex, € $(X), E, C Eo => w*(E1) < w*(E2), countable subadditivity: (E, :n € N) C 98(X) > u*(Unen En) < Donen i (En)By definition an outer measure * is a countably subadditive set function on the o-

algebra $3(X).

If it is also additive on 98(X), that is, if it satisfies the condition that

B*(E] U Eo) = w*(E1) + w*(E) whenever £1, Ey € $3(X) and £, NM Ey = G, then according to Proposition 1.23, 4* is countably additive on $8(X) so that it is a measure on 38(X). In general an outer measure jz* does not satisfy the additivity condition on 98(X). ‘We shall show that there exists a o-algebra 2 of subsets of X, 2 C $8(X), such that when #* is restricted to W it is additive on M. Then jz* is countably additive on A by Proposition 1.23 and is thus a measure on 2. Let E € 98(X). For an arbitraryA € 98(X), we have (AN E)N (AN E*) = G and (AN E)U(AN BE) =A. Definition 2.2. Let *

be an outer measure on a set X.

We say that a set E € 9(X)

is

measurable with respect to j1* (or .*-measurable) if it satisfies the following Carathéodory condition :

(A)

= p*(AN BE) + p*(ANE’) for every A € P(X).

The set A is called a testing set in the Carathéodory condition.

We write D0t(u*) for the

collection of all u*-measurable E € 93(X).

Our next goal is to show that D0(sz*) is a o-algebra of subsets of X. Observation 2.3. The countable subadditivity of jz* implies its finite subadditivity on $3(X) by Observation 1.20.

Thus 4*(A)

< u*(AN £)+ w*(AN

E*) for any E, A € (X).

Therefore to verify the Carathéodory condition for E € 9(X), it suffices to verify that BA(A) > p*(AN E) + w*(AN E*) for every A € P(X). Lemma 2.4. Let j:* be an outer measure on a set X. Consider the collection Nt(u*) of all

B*-measurable E € $3(X). (a) If £1, Eo € Mt"),

then Ey U Eo € Mt(u*).

for E\, E, € Dt")

such that E)N E, = 0.

(b) The set function jc" is additive on IN(u"), that is, p*(E, U Eo) = w*(E1) + w*(E2)

30

CHAPTER 1 Measure Spaces

Proof. 1. Suppose £1, Ez € Wt(y*). Let A € $B(X). Since Ey € Mu") we have

qd)

u*(A) = w*(AN £1) + p*(AN ES).

With AM Ej as a testing set for Ey € 2%(u*), we have (2)

B*(AN Ef) = w*(AN Ef N Bo) + w*(AN Ef 1 E>).

Substituting (2) into (1) we have

(3)

Bwi(A) = w*(AN E}) +

(AN Ef N Ea) +

(AN EG NED).

Regarding the first two terms on the right-hand side of (3) note that

(AN £1) U (AN Ef N Ey) = AN (EU (Ef E))

=AN (E, U (Ep \ E1)) = AN (£1 U Ep). Then by the subadditivity of 4* we have

4

w(AN Ey) + w*(AN Ef 9 E2) > w*((AN Ey) UCAN Ef N E2)) = p*(AN (Ei U E2)).

We also have 2*(AN Ef M E§) = u*(AM (Ei U E2)°). Substituting this and (4) into (3), we have

p*(A) = w*(AN (Ei U Ea) + w*(AN (Ei U E2)’).

By Observation 2.3 this shows that E, U Ez satisfies the Carathéodory condition. Hence E, U Ep € Mt(p*). 2. Toprove the additivity of 2* on Dit(u*), let £1, Ez € MU(u*) and EiNE2 = G. By (a)

wehave E,\UE

€ Dt(u*). Since E, € Mt(u*), wehave u*(A) = w*(ANE1)+u*(ANE{)

for every A € 93(X). In particular, with A = £, U Ez we have

p*(E, U Ey) = u*((E, U E2) 1 £1) + w*((E1 U E2) 9 Ef). Now the disjointness of E) and £2 implies that (Z}UE2)NE, = Ey and (E\UE2)NEj = E2.

Thus *(E, U E) = p*(E1) + w*(E2).

of

The next theorem shows that for an arbitrary outer measure jz* on a set X, there exists

anon ,4*-measurable subset of X if and only if jz* is not additive on 93(X).

Theorem 2.5. Let * be an outer measure on a set X. Then the following two conditions are equivalent: @

p* is additive on Y3(X).

(ii) Every member of $3(X) is u*-measurable, that is, Dt(u*) = Y(X). Thus there exists a non *-measurable set in X if and only if * is not additive on $8(X).

§2 Outer Measures

Proof.

1.

31

Suppose y* is additive on $8(X).

Let E € 9(X).

Then for an arbitrary

A € $3(X), the two sets AN E and AN E* are disjoint members of 93(X) whose union is

equal to A so that by the additivity of * on $3(X) we have 4*(A) = p*(ANE)+p*(ANE*).

This shows that every E € 9$(X) satisfies the Carathéodory condition. Thus E € 2%(u*)

and then $8(X) Cc MNt(u*). On the other hand, since $8(X) is the collection of all subsets of X we have IN(z*) C 9(X). Therefore we have Mt(u*) = 93(X). 2. Conversely suppose $B(X) = Mt(z*). Then since pz* is additive on Mt(u*) by Lemma 2.4, j* is additive on $8(X). Example. Let X be an arbitrary set and for every E € 93(X) let 4*(E) be equal to the number of elements in E. Then ;z* satisfies conditions 1° - 4° in Definition 2.1 and is thus an outer measure on X. Moreover j* is additive on 93(X).

Lemma 2.6. Let * be an outer measure on a set X. IfE € $8(X) and w*(E) = 0, then every subset Eo of E, and in particular E itself, is a member of Wt(u*). Proof. If 4*(£) = 0, then for any subset Ep of EZ, we have 4*(E9) = Oby the monotonicity of 4*. Then for every A € 9$(X), we have u*(AN Eo)+ u*(AN ED) < w*(Ep) +4 *(A) =

z*(A) by the monotonicity of 4*. This shows that Eo satisfies the Carathéodory condition by Observation 2.3. Hence Ep € Dt(u*). a Lemma 2.7, Let * be an outer measure on a set X. If E, F € $(X) and p*(F) = 0,

then "(EU F) = "(E).

Proof. By the monotonicity and the subadditivity of z*, we have

B(E) < w(E UF) < u*(E) + w*(F) = u*(E). Therefore we have u*(E U F) = p*(E).

©

Theorem 2.8. Let jz" be an outer measure on a set X. Then the collection M(*) p*-measurable subsets of X is a o-algebra of subsets of X.

of all

Proof. 1. For any A € $3(X), we have *(ANX)+u*(ANX*) = 4*(A)+pu*(@) = w*(A). This shows that X satisfies the Carathéodory condition so that X € Dt(u*). 2. Let E € IR(u*). Then for every A € 93(X) we have

pt(A) = (ANE)

+ (ANE)

= p(AN (ED) + (ANE.

This shows that E° satisfies the Carathéodory condition so that E° € Nt(u*). 3. Let (En :n € N) C W(u*). By Observation 2.3, to show that J,,cy En € Wt(u*),

it suffices to show that for every A € 93(X), we have

ur(ay> w(An[

neN

zal) +u*(4n[ Ue]). aeN

32

CHAPTER 1 Measure Spaces

Now since E; € Dt(*), we have for every A € 93(X)

qd)

u*(A) = u*(AN £1) + (AN Ep).

We claim that for every k € N we have

@ — waH=Su(an[Ualox) +e(an[Ua]) jel i=l jal with the understanding that Ue

E; = @. Let us prove (2) by induction on k € N. Now

for k = 1, (2) reduces to (1). Next assume that (2) is valid for some k € N. Let us show

that (2) is valid for k + 1. Now with AN [Usa E;| as a testing set for Ex41 € I(u*), we have

iC»

w(aol

i

1

af) =#(4o[Un oes) +u-(an[Ue] on) - “(so[Ual nays) +a (s9[Ue)

Substituting this equality into (2) which is valid for k by our assumption, we have

way = oue(an[ Jafney) +0r(4n[U ei] zen) +0"(an[ j=1 i

i=1 1

j=l

i=l

j=l st

°

°

,

= Ye (aol Uzi] nz;) +u*(an[ Uzi] ).

j=1

ei]

j=l

This shows that (2) is valid for k + 1 under the assumption that it is valid for k. Thus by induction, (2) is valid for every k € N.

Since Uh_, Ey C Ujen Ey, we have [Ut_, Ey]° > [Ujen Ey], and then by the monotonicity of 1*, we have w*(AN[Uf-1 Ey]’) = u*(AN[Ujen £y]°)- Using this in (2), we have

r ede (ao[Ual nz) +u*(an[Uei)). jen

Since this holds for every k € N, we have jl

Bt(A) > Yura n [ UJ Bi) n a) + w(A n [ U Ej) jeN

i=1

jen

>u*(U (an [Uelo#)) +ut(an [U#) jen

(soy (Uaf're)) +#(en[Ue]) jen

§2 Outer Measures

33

where the second inequality is by the countable subadditivity of 4*. Now by Lemma 1.21

j-1

j-1

U(Ualos)=Ule[Ua)-Us

jeN

i=l

JjeN

i=1

JjeN

Therefore we have

ura) = wt(An[ Uzi]) +4*(40[ U2])JEN

jen

Thus J; 00

n—>0o

=p(lim inf Fy) < liminf (Fn) = liminf x*(Fn) =liminf lim p*(En), 71300 u*(E,) = 00 where the second equality is by the fact that lim inf Fy € Dt(u*), the second inequality is n>00

by (a) of Theorem 1.28, and the last equality is by the existence of

jim, *(E,). This and

(1) imply (2). Definition 2.12. Let :* be an outer measure on a set X. We say that s* is o-finite on J3(X) if there exists a sequence (A, : n € N) in $8(X) such that |_J, 2,

then (B,

: n € N) is a disjoint sequence in $8(X), Uncen Bn = Une An

= X, and

#*(Bn) < 2*(An) < 00 for every n € N by the monotonicity of 4* on $3(X).

In a measure space (X, A, 4), if E, F ¢ A, E C F, and u(E) = u(F) < oo, then BUF \E) = #(F)—(E) = 0. This follows from the additivity of 2 on 2. The assumption BE) < oc is to ensure that the difference j4(F) — 4(E) is defined. If * is an outer measure

on a set X, and if F € $3(X), F € Mu"), E C F, and p*(E) = u*(F) < 00, do we have z*(F \ E) = 0? In the next theorem, we show that for an outer measure z* which is regular and o-finite, this question is equivalent to the question as to whether ;z* is additive on 93(X), or equivalently according to Theorem 2.5, the question of non-existence of non pe*-measurable sets in X. Theorem 2.13.

Let * be a regular and o-finite outer measure on a set X.

following two conditions are equivalent:

Then the

§2 Outer Measures

35

@ Mu") =P). Gi) Ee PX), F € Mu"), EC F, u*(E) = w*(F) < 00 > u*(F \ E) =0. Proof. 1. To show that (i) implies (ii), assume %(yz*) = 9(X). Suppose EF and F satisfy the hypothesis of (ii). Then E and F are members of Nt(y*). Since z* is a measure on the

o-algebra Dt(.*), itis additive on M(u*). This implies that u*(F) = u*(F\ E)+p*(E). Subtracting *(Z) € R from both sides, we have w*(F \ E) = w*(F) — p*(E) = 0. This shows that (i) implies (ii). 2.

To show that (ii) implies (i), let us assume (ii). Let E be an arbitrary member of

3B(X). Since p.* is o-finite on $B(X), there exists a sequence (A, : n € N) in $8(X) such

that Unen An = X and p*(A,) < 00 foreveryn € N. Let E, = EM A, forn € N. Then (En in € N) is a sequence in $8(X) with on En = E and p*(Ey) < *(An) < 00 for

every n € N by the monotonicity of j2* on $3(X). Now since j* is a regular outer measure, there exists F, € t(u*)

such that F, > E, and p*(F,)

= p*(E,) for every n € N. By

(ii), we have u*(F;, \ E,) = 0. This implies that F, \ E, € t(u*) by Lemma 2.6. Since E, = F, \ (Fy \ Ey) and F, and F, \ E, are members of the o-algebra Dt(u*), Ey, is a

member of S)t(u*) for everyn € N. Then E = ncn En € DU(u"). This shows that every

member of 9$(X) is a member of 2)t(*). Thus (ii) implies @).

a

Definition 2.14. An outer measure y:* on a topological space X is called a Borel outer measure if By C Mi(u*). Definition 2.15. An outer measure * on a topological space X is called a Borel regular outer measure if it is a Borel outer measure on X and if for every E € 98(X) there exists F € Sy such that F > E and *(F) = p*(E).

Remark 2.16. A Borel regular outer measure 2* on a topological space X is a regular outer measure on X in the sense of Definition 2.10. This follows from the fact that we have

Bx c Mu").

[11] Metric Outer Measures Given a metric space (X, d). Let the topology on X be the metric topology by the metric d. The distance between a point x € X anda set E € 98(X) is defined by d(x, FE) = infycz d(x, y). If x € E then d(x, E) = 0 but the converse is false. If E is a closed set then d(x, E) = Oif and only ifx € E. The distance between two sets E, F € 93(X) is

defined by d(E, F) = infycz,ycr d(x, y). KEN F # BD thend(E, F) =0. FEN F =8, a(E, F) may still be equal to 0 even if E and F are closed sets. IfE is a closed set and F is a compact set, then d(E, F) = Oif and only if EN F #@. Definition 2.17. Given a metric space (X, d).

(a) Two sets E), E € $8(X) are said to be positively separated if d(E1, E2) > 0. (b) An outer measure j* on X is called a metric outer measure if for every pair of positively separated sets E,, Ex € $3(X), we have w*(E, U Ea) = u*(E1) + u*(E2).

36

CHAPTER 1 Measure Spaces

Lemma 2.18. Let j:* be a metric outer measure on a metric space (X, d). Let (A, :n € N) be an increasing sequence in 93(X) andlet A = lim Ap. If A, and A\ A,+1 are positively

n separated for everyn © N, then 4*(A) = im, B*(A,)Proof.

Since A,

+, we have *(A,)

t as

—

measure *. Since A, C A we have *(A,) lim p*(A,) < 2*(A). It remains to show

n—00

@

co by the monotonicity of the outer

< *(A) for every n € N. Thus we have

wt(A) < Jim u*(An).

Let Ap = @ and define a sequence (B, : n € N) in 93(X) by setting B, = A, \ Ay-1 for n éN. Let us show

Q

DH Bre-1) S lim w*(An),

keN

ro)

DoH" Bx) < lim u*(Ay). keN

To prove (2), let us consider w(UR vt

By-1)

=

(Us, By-1)

By_1) for an arbitrary N € N.

U Boni.

Now

we have

UP

Boy-1

Let us write C

Aon

and

Bonsi = Aon+1 \ Aow C A \ Aon. This implies N

a( L) Bort, Bat) 2 d(Aaw-1, A\ Aon) > 0 k=1

by our assumption on the sequence (A, :n € N). Then since yz* is a metric outer measure,

we have

N+1

) ur( U Bus-1) = “(qu Bu-1| u Buys

N

ut( U Bu-1) + p*(Bon41)-

k=l

Repeating the argument to :*((J¥_, Box—1) and iterating the process N times, we obtain N41

ut( U Bu-1) k=1

N+1

= Do (Bus). k=1

Since UN7! Bog_1 © Aon, we have DN? y*(Bo4_1) < u*(Aon41). Thus N+1

de Ba-) = jm, 2» H*(Bu-1) S fim w*(Aaws1) = jim, w*(An). S

=!

§2 Outer Measures

37

This proves (2). We prove (3) likewise by starting with |)"tt Box for N EN. If at least one of the two series in (2) and (3) diverges, then iim, p* (Ay)=

00 and (1)

holds. Suppose that both series converge. Then we have

4)

dH (Bn) = Yo a" (Bae-1) + ¥) w*(Bax) < 00.

neN

keN

keN

Now A= lim An = Ugen 4n = Ano U (Up2ng+1 Bn) for an arbitrary no € N. Thus by the countable subadditivity of the outer measure .*, we have

6)

BA) S B"(Ano) + DD e*(Br)n>notl

By (4) we have ng —

‘

im,

*

Laeno 41 #*(Bp)

_

*

= 0. Then p*(A)


0, we have * (Uncen En) < Cnen #*(En). This proves the countable subadditivity of z*.

Remark 2.22. The nonnegative extended real-valued set function y on the covering class 23 on which the outer measure js* is based in Theorem 2.21 satisfies no conditions other than that y(@) = 0. In particular, y is not required to be monotone on 23. Thus it is possible that we have V, W € 23 such that W C V but y(W)

> y(V). The existence of such a set

W in % has no effect on the definition of *. If such a set W is a member of a covering sequence (V, :n € N) foraset E € 93(X), then by replacing W with V we have a covering sequence of E with a possibly smaller, but never greater, sum )°,-y Y (Vn). Thus in taking infimum of }°,,cy (Vn) on the collection of all covering sequences of FE, a sequence with W as a member has no effect. Remark 2.23. Further regarding the set functions y and y* in Theorem 2.21, we have: (a) For V € 23, (V) is a one-term covering sequence in 23 for V and thus

ut(V) = int { Dyen Va): Wa i EN) CB, Unen Va > V} SV). However z*(V) = y(V) may not hold. In fact if there exist W, V € 23 such that Wc V and y(W) > y(V), then since (V) is a one-term covering sequence in 23 for W, we have

B*(W) A. By (ii) we have y*(V,) = u*(V_ NE) + u*(Vq 1 E°) for every n €N. Summing overn € N, we have

Yet) =

neN

neN

et NE) +

neN

wha NE)

=e(Umnm)+e(U nz)

=0'((U)n8)+e(Y)or) neN

neN

neN

> UX(ANE)+p(ANE), where the first inequality is by the countable subadditivity of the outer measure y* and the second inequality is by the monotonicity of 4*. According to (a) of Remark 2.23,

B*(Vn) < v(Vq). Thus Dew 7 (Vn) = nen *(Va) = BAN E) + w*(AN E*). This shows that z*(A M £) + u*(A NN E*) is a lower bound for the collection of nonnegative

extended real numbers {)7 ey 7(Vn) : (Vn: €N) C %, Unen Va > A}. Since u*(A) is the infimum, that is, the greatest lower bound, of this collection, we have y*(A)

>

B*(AN E) + w*(AN E*), The reverse inequality holds by the monotonicity of u*. Thus

(holds.

»

Problems Prob. 2.1. Let jz* be an outer measure on a set X. Show that if y2* is additive on $B(X), then it is countably additive on $8(X). Prob. 2.2. Let :* be an outer measure on a set X. Show that a non jz*-measurable subset of X exists if and only if 2* is not countably additive on $B(X). Prob. 2.3, For an arbitrary set X let us define a set function 4* on $3(X) by *(E) =

#

number of elements of EF if E is a finite set,

~~ |0o if £ is an infinite set.

(a) Show that y.* is an outer measure on X.

(b) Show that jz* is additive on $8(X), that is, z*(E, U E2) = w*(E1) + *(E2) for any

§2 Outer Measures

Al

E\, Ey € $8(X) such that £) 0 Ey = G. (c) Show that 2* is a measure on the o-algebra $3(X). (This measure is called the counting measure.)

(d) Show that Dt(z*) = 93(X), that is, every E € 93(X) is z*-measurable.

Prob. 2.4. Let X be an infinite set and let 4 be the counting measure on the o-algebra 2 of all subsets of X. Show that there exists a decreasing sequence (EZ, : n € N) in 2 such that E, | @, that is, jim, E, = @, with jim, B(E,) # 0.

42

§3

CHAPTER

1 Measure Spaces

Lebesgue Measure on R

‘We summarize here the process of constructing a measure space by means of an outer measure as developed in §2. Let X be an arbitrary set and let 27 be a covering class of subsets of X. Let y be a nonnegative extended real-valued set function on 29 such that y(B) = 0. Let us define a set function j:* on $8(X) by setting for every E € $B(X)

BE) = inf [Dew (Ye)! Vn 2 EN) CB, Upew Vn > Ef Then ,2* is an outer measure on X. If we let t(4*) be the collection of all 44*-measurable subsets of X, then Nt(y*) is a o-algebra of subsets of X. If we write yz for the restriction

of 2* to Mts"), then yz is a measure on the a-algebra Nt(z*) of subsets of X. Moreover the measure space (X, Mt(j*), 2) is a complete measure space. Specific properties of the measure space (X, D7(.*), 4) are consequences of our choice of the set function y on the

covering class 23 of subsets of X. Below we construct the Lebesgue measure space on R by the process described above.

[I] Lebesgue Outer Measure on R Definition 3.1.

Let 3, be the collection of 9 and all open intervals in R, Jo, be the collection

of @ and all intervals of the type (a, b] in R, 3¢o be the collection of @ and all intervals of the type [a, b) in R, and J, be the collection of @ and all closed intervals in R, with the understanding that (a, 00] = (a, 00) and [—o0, b} = (—00, b). Let3 = FgUBo¢ UF cqUTe,

that is, the collection of § and ail intervals in R. For an interval I in R with endpoints

a,beR a E}By Theorem 2.21, wt is an outer measure on R. We call 4} the Lebesgue outer measure on R. We write IN, for the a-algebra Wt(w*) of uf-measurable sets E € F(R) and call it the Lebesgue o-algebra of subsets of R. Members of the o-algebra 20, are called Mt, -measurable or Lebesgue measurable sets. We call (R, Mt,) the Lebesgue measurable

space. We write 1, for the restriction of p* to Dt, and call it the Lebesgue measure on R. We call (R, Mt, u,) the Lebesgue measure space on R. For some £ € 93(R), we have Drew

Ch) = 00 for any (J, : n € N) C 3p, such that

Unen fn > E sothatwe have inf | yew £n) : Un i €N) C Jo, Unen Jn > E} = 00.

Observation 3.2. Each of the four collections J,, Joc, Jeo, Tc, is a covering class of R in

the sense of Definition 2.20, and 2 is a nonnegative extended real-valued set function with

§3 Lebesgue Measure on R

43

* £(@) = 0 on each of these four collections. Thus if we define four set functions m%, m*_, m*,, and m* on $3(R) by setting for every E € $B3(R)

wa(E) = inf [Yen ln) : Gn

2 €N) C Jo, nen In > E},

us(E) = ink {Den ln): Ua im €N) C Joes Unen in DE},

ui(E) = inf [Yen €Un) : Un i © N) C eos Upen In D E}, ue(E) = inf {Yew £Un) : hin EN) C Fe, pew > E}, then each of these set functions is an outer measure on R by Theorem 2.21. Among these four, 43 is the Lebesgue outer measure 47. Actually for every E € 9(R) we have the

equalities u3(E) = 45,(E) = 42,(E) = u¢(E).

Proof. Let us show that 43(E) = u{(E) for every E € $8(R). The same kind of argument

can be used to show that 4*(Z) = 4%,(E) and u3(E) = u%,(E). Let ¢ > 0 be arbitrarily given. Let (J, : m € N) be a sequence in J, such that

Unen dn D E. Let In = Gn, by) and Jn = [aq — 2/2"), by + €/2"+"] fornn € N. Then

(Jn im € N) is a sequence in 3-, Unen Jn D E, and Dew £(Jn) = Den £Un) last equality implies by the definition of z3(£) as an infimum that u3(E) < cn Since this holds for an arbitrary sequence (J, : n € N) in 3, such that cy in have u2(E) < w5(E) + 6. Then by the arbitrariness of ¢ > 0, we have u3(E)

+ €. The &Un) +éD E, we < 3(E).

Conversely starting with an arbitrary sequence (7, : n € N) in J. given by i, = [Gn, bn] for n € N such that Unen I, > E, and defining a sequence (J, : n € N) in J, by setting

Jn = (Gn — &/2"*1, by + €/2"+) for n & N, we show by the same argument as above that

wS(E) < wi(E). Thus w3(E) = wX(E). 0

Lemma 3.3. (a) For every x € R we have {x} € 99, and u? ({x}) =0.

(b) Every countable subset of R is a null set in (R, Dt, u,)-

Proof.

1. Letx ¢ R.

For every ¢ > 0, we have x € (x — 2,x +8)

€ J, and thus the

sequence (J, :n € N) = ((x —e,x +e), 9,9,...) in J, is a covering sequence for {x}.

This implies that we (x)

< Saen

Un)

= 2¢.

By the arbitrariness of ¢ > 0, we have

u* ({x}) = 0. By Lemma 2.6, this implies {x} € t(u*) = Mt. 2. A singleton in R is a null set in (IR, t,, 4,) by (a). A countable subset of R is then a countable union of null sets and is therefore a null set by Observation 1.33.

&

Lemma 3.4. 5 = £ on J, that is, ws (1) = £(1) for every interval I inR. Proof. 1. Consider the case where J is a finite closed interval given by I = [a, b] where a,b € R,a 0, we have J = [a,b] C (a —e,b +) € Jy. Thus

((a@ —¢,b + £), 9,9, ...) is a covering sequence in 3, for I. This implies that we have

HED

< £((a —e,b4+ €)) + £8) + €(@) + --- = £2) + 2e. Since this holds for every

& > 0, we have

@)

ur) £(Z) and (2) holds. Thus consider the case where every member in

the covering sequence is a finite interval. Let us drop any member in the covering sequence that is disjoint from J and drop also any member that is contained in another member of the sequence. The resulting sequence (J, : n € N) is a covering sequence for J with

YMren’On) S Donen £Un). Since (J, : 2 € N) is an open cover of the compact set J, it has

a finite subcover. Renumber the members of the sequence if necessary so that J C Ux, Jy where J, =

(ay, by) for k = 1,...,N and a, < az < --: < ay.

Nowifa; = a; for

some i # j, then either J; C J; or J; C J; which contradicts the fact that none in the sequence (Jz : k = 1,...,N) is contained in another in the sequence. Thus we have aj < @ @ — a1) + @ — a2) +--+ (Gy — ay_-1) + (bn — ay)

= by —a, = b-a=£). Thus

) ncn £Un)

>

€(1).

This proves (2).

According to (2), £(I) is a lower bound

for [Donen £Un) : Gn: €N) C30, Une n > I}. Since *(Z) is the infimum of this collection of nonnegative extended real numbers,

we() = £0).

we have BED

>

£(1).

Therefore

2. If Z is a finite open interval given by J = (a, b), then by the monotonicity and finite

subadditivity of wy and by (a) of Lemma 3.3, we have

u(G@. 5) < uz (a,b) < wz (lal) + uE(G@, 5) + ue({b}) = Hi (G@.)). Thus 4:* (a, b)) = 24 (La, b]) = £([a, 6]) = £((@, b)), where the second equality is by our

result in 1.

§3 Lebesgue Measure on R

45

3. If I is a finite interval of the type 7 = (a, b], then since (a, b] = (a, b) U {b} and

Ht ((6}) = 0 by (@) of Lemma 3.3, we have y*((a, b]) = n((a, 6) by Lemma 2.7. Thus uk (a, b]) = £((a, b)) = £((a, b]). Similarly for a finite interval J of the type I = [a, b).

4. Let I be an infinite interval. Consider for instance J = (a, 00) where a € R. For any

n € N such thatn > a, we have (a, 00) > (a,n) and y:*((a, 00) > w*((a,n)) =n-a,

by the monotonicity of j.* and by our result in 2. Since this holds for every n > a, we

have j4* (a, 00)) = 00 = £((a, 00)). Similar arguments apply to infinite intervals of other types.

&

Proposition 3.5. The Carathéodory condition for the 47 -measurability for E € $(R)

@ uh(A) = wt(AN £) + u(AN E9) for every A BR) is equivalent to the condition @ ui)

= wE( NE) + wR

E*) for every I € Ip.

Proof. Since ;4* is an outer measure based on the set function £ on the covering class J, of subsets of R, the equivalence of (i) and (ii) is a particular case of Theorem 2.24. Lemma 3.6. 3 C 9t,, that is, every interval in R is a Dt,-measurable set.

Proof. 1. Let us show first that every open interval in R is 9)t, -measurable. By Definition 3.1, wf-measurability and 99t, -measurability of a subset E of R are synonymous. Thus according to Proposition 3.5, a subset E of R is 9), -measurable if and only if it satisfies the condition (1)

BEC) = BECO E)+ BEC NE’) for every I € Jy. Consider an infinite open interval (a, 00) in R where a € R.

Then for an arbitrary

I €73, we have

T=INR=I1N{@, 0) UG, oo)*} = {1 Now I

(, 00)} U {7

G, 00)*}.

(a, 00) is either an interval or 9 and similarly I M (a, 00)° is an interval or 9. Since

IN (a, 00) and IN (a, 00) are disjoint and their union is equal to I, we have

2)

£(D) = €(L.0 @, 00) + €(EN (a, 00)').

According to Lemma 3.4, we have £(I) = u*(I), £(19 (@, 00)) = u*(I £(1. @, 00)*) = w* (TN (@, 00)*), Substituting these into (2), we have

(3)

Ga, 00)), and

BD) = wt (ING, 00)) + 2k(EN (@, 00)*) for every I € 3p.

This verifies condition (1) for (a, 00) and shows that (a, 00) is Dt, -measurable.

By similar argument, we show that an infinite open interval (—oo, 6), where b € R,

is DJt,-measurable.

If a,b

€ Randa

< b, then R =

(—oo, b) U (a, 00).

Then since

(—oo, b) € M,, (a, 00) € Mt, and Mt, is a o-algebra of subsets of R, we have R € Mt,.

46

CHAPTER 1 Measure Spaces

This shows that every infinite open interval in R is 29%, -measurable. An arbitrary finite open interval is given by (a, b) where a, b c R anda < b. Then (a, b) = (—00, b) N (a4, 00). Since (—0o, b) € Mt, (a4, 00) € Mt, and Mt, is a o-algebra of subsets of R, we have

(a, b) € MM,. Thus we have shown that every open interval in R is Dt, -measurable.

2. LetJ be an interval in R which is not an open interval. Then J = {a}UZ, J = TU{b},

or J = {a} U TU {6}, where J is an open interval, a, b € R, and {J, {a}, {b}} is a disjoint collection.

Now J € St, by our result above and {a}, {6} € Mt, by Lemma 3.3. Then

since Mt, is a o-algebra of subsets of R, we have J € Mt,.

a

A set E in a topological space X is said to be dense in X if for every non-empty open set O in X we have EM O # @. In particular if E is a dense subset of R then every open interval (a, 6) in R contains infinitely many points of E. Indeed implies that (a, 8) contains a point x, € EZ. Then the subinterval a point x2 € E and the subinterval (a, x2) of (a, x1) contains a Thus (a, 8) contains an infinite sequence (x, : n € N) of distinct

the denseness of E in R (a, x1) of (a, 8) contains point x3 € E and so on. points in E.

Observation 3.7. If E is a null set in (R, 9,,, 12,), then E° is a dense subset of R. Proof. For every open interval J inR, wehave 1, (I) > 0. ThusifE ¢ 9, andy, (Z) = 0,

then by the monotonicity of 2, on t,, E cannot contain any open interval as a subset. This implies that E° M I 4 9 for every open interval J and therefore E°isdenseinR. Examples.

(a) If Q is the set of all rational numbers and P is the set of all irrational

numbers, then Q is a null set in (R, 9t,, 4,) and P « Mt, with u,(P) =o. (b) For every interval J in R, we have 7 Proof.

1.

P ¢ Mt, with u,(1

P) = e(J).

The set Q of all rational numbers is a countable set and thus a null set in

(R, M,, u,) by (b) of Lemma 3.3. The set P of all irrational numbers, being the com-

plement of Q in R, is a member of 9%,.

,(P) + #,(Q)

1, (P) = 00.

= 4,(R).

Since u,(Q)

By the additivity of 2, on Dt,, we have

= 0 and w,(R)

= £(R) = 0, we have

2. Let I be an interval in R. By Lemma 3.6, 7 is a member of Qt, . This implies that

I

P and 7M Q are members of It,.

Since P and Q are disjoint and their union is R,

INP and IN Q are disjoint and their union is 7. By the additivity of 4, on 2%,, we have #2

NP)

+ 4,29

Q) = w,(D

= £CD by Lemma 3.4.

Since z,(Q)

= 0, we have

#,(1 N Q) = 0 by the monotonicity of 4, on Mt,. Thus z,(7 N P) = £(7). 3. We append here a direct proof for u*(Q) = 0. Since there are only countably many rational numbers, we can represent Q = {x, : n € N}. For an arbitrary e > 0, let &

é

Ih = (% — Sat % + Set)

forn EN.

Then I, € 3p and £(I,) = 3 forn € N and moreover ncn Jn > Q. Now

Nem@= ee

neN

neN

§3 Lebesgue Measure on R

Thus

47

ui (Q) = int { 174):Grin EN) CI, ncN

Since this holds for every « > 0, we have wt (Q)=0.

> O} 0 for every nonempty open set O in R.

(RB, 9, 4,) has no atoms. (BR, 9, w,) (IR, Dt,, 4,)

is translation invariant. is positively homogeneous.

Non it, -measurable sets exist. The Lebesgue outer measure wt is a Borel regular outer measure.

Property (i) is immediate. Indeed since the measure space (R, 9, , 14,,) is constructed

by means of an outer measure 27 it is a complete measure space according to Theorem 2.9. ‘We prove the rest of the properties below. Theorem 3.8.

measure space.

The Lebesgue measure space (R, Ut,,,)

is a o-finite, but not finite,

Proof. Note that 2, (R) = £(R) = oo by Lemma 3.4. This shows that (R, 9, ,) is not a finite measure space. Let us show that (R, 9t,, 4,) is a o-finite measure space. Let E,

= (—n,n) forn

¢ N.

Then E,

€ ®t, forn

€ N and

0. Proof.

Every nonempty open set O in R contains an open interval 7.

,(0) > 4,(D =t0) > 0. 0

Then we have

Proposition 3.11. Regarding atoms in (R, 90t,) and atoms in (R, Mt, w,), we have: (a) In the Lebesgue measurable space (R, 2,), the only atoms are the singletons.

48

CHAPTER 1 Measure Spaces

(b) The Lebesgue measure space (R, Mt. H) has no atoms. Indeed if E € IN, and i, (E) > 0, then for every & > 0 there exists Ey C E, Eo € M,, with 4, (Eo) € (0, é]. Proof. 1. For every x € R, we have {x} € 9t, by (a) of Lemma 3.3. The only subsets of

{x} are 8 and {x} and hence @ and {x} are the only 99%, -measurable subsets of E. Thus {x}

is an atom.

Now let E € 99t, and suppose E contains at least two points. Let x € E. Then {x} € Mt,. But {x} # O and {x} # E. Thus £ is not an atom of the measurable space {R, Dt,). This proves (a).

2. For e > O arbitrarily given, let J, = (( — 1)e, ne] forn € Z. Then (, :n € Z)isa

disjoint sequence in J,

with ), 0. Thus there exists some mp € Z such

that 44, (En,) > O. Since En, = EM Ing C Ing, we have pt,(Enp) < Un) monotonicity of 4, . Thus we have jz, (E,,) € (0, €]. This proves (b). ©

= & by the

Definition 3.12. Let X be a linear space over the field of scalars R. (a) For E C X and xo € X, we write

E+xo={x+x0:x€ E} ={yeX:y=x+4x9 for somex € E} and call the set the xq-translate of E.

{b) Fora € R, we write

aE = {ax:x e E}={ye¢X:y =ax for somex ¢ E} and call it the dilation of E by factor a. (c) Fora collection € of subsets of X,x € X, anda € Rwewrite€+x

anda€é= {aE: Ee €}.

Observation 3.13. For translates of sets the following equalities hold:

(E +41) + x2 = E+ (x1 +9), (E+x)° = E°+x, E, CE, > Ei,

+xCE.+x.

For translates of collections of sets we have

(Une Eu) +* = Unea Ea +2), (Nea Ea) +¥ = Nac Za +2). For dilations of sets we have

B@E) = @B)E, (aE) = aE‘.

= {E+x

Be

e}

§3 Lebesgue Measure on R

49

Definition 3.14. Given a measure space (X, A, 4) where X is a linear space.

(a) We say that o-algebra & is translation invariant if for every E ¢ Mand x € X we have E+xee (b) We say that the measure 1 is translation invariant if & is translation invariant and if for everyE ¢ Mandx

€ X we have w(E +x) = w(E).

(c) We call (X, &M, 41) a translation invariant measure space if both XM and us are translation invariant.

Example.

As an example of a o-algebra of subsets of R that is not translation invariant,

let I, = (2 —1,n] forn e€ Z and let 2 be the collection of # and all countable unions of

members of the collection {J, : m € Z}. It is easily verified that 2% is a o-algebra of subsets

of R. 2% is not translation invariant since J, € 2 but J, + 4 = (n—},n+ 4] is nota member of 2.

Lemma 3.15. (Translation Invariance of the Lebesgue Outer Measure) For every E € $B(R) and x € R, we have pF (E + x) = wf (EZ). Proof. Note that for every J € 3, andx € R, we have +x € J, and £44 + x) = £(1). Let E € 93(R). Take an arbitrary sequence (J, : n € N) in J, such that U,on D E. For an arbitrary x € R, (7, +x :n € N) is a sequence in J, with £(2, + x) = €(J,) and

Unen(n+*) = (Unen in) +x 3 E+x. Thus Yen 2h) = new £Un +x) = ut (E+z)

by the fact that (J, + x : n € N) is a covering sequence for E + x and by the definition of wy (E +.) as an infimum. Since this holds for an arbitrary covering sequence (J, : n € N) for E and since 2* (E) is the infimum of Donen £Un), we have uE(E) > wi (E +x) for any E € P(R) andx ER.

Applying this result to Z + x and its translate (E + x) — x = E, we have the reverse

inequality u7(E + x) > u(E.). Therefore u*(E +x) = wi(E). Theorem 3.16. measure space we have E+x Then DM, +x Proof.

&

(Translation Invariance of the Lebesgue Measure Space) The Lebesgue (R, M.,,, 12,) is translation invariant, that is, for every E € M, andx ER € Mt, and w,(E +x) = u,(E). LetNM, +x:= {E+x:E € wt}. = Mt, for everyx € R.

Let E € St, and x ¢€ R. To show that E + x € 20,, we show that for every

A € $P(R), we have u*(AN(E+x)) + wt(AN(E +3)*) = wt (A). By Lemma 3.15

Br(AN (E+x)) + ut(AN(E+2)’) = wt ({AN(E +%)}— x) + wh({AN (E+ x)}- x)

= pt ((A-x)ME) + ut((A—x) NE’) = uh(A—x) = ur*(A), where the second equality is by Observation 3.13, the third equality is by the Carathéodory condition satisfied by our E az}.

By (1) we have

(a17.) = lalé(Lh,).

Observe also that

Unen 2h, DAE

& o(Upen th) DE

& Unen thn 2 E.

Substituting these in (3), we have

4)

wt@E)= lain { Ynen £(4%) > Unin €N) C3o, Unen dn D Eh.

Let Jn = 14, forn € N. Then J, € 13, = 3, by (2). Substituting this into (4), we have

HE (@E) = [alin { DexIn) : In i EN) C Joy pen Jn > Ef = |elur (2),

§3 Lebesgue Measure on R

51

by Definition 3.1 for we (£). This completes the proof.

Theorem 3.18. (Positive Homogeneity of the Lebesgue Measure Space) Consider the measure space (R, Wt, , 4,). LetE € MN, ° ER, anda, (a) fa =0, thenaE € Wt, and p,(@E) =

:= (aE: E € Mt}.

(b) Ifa #0, then aE € Mt, and pw, (aE) = jal, CB. (c) Ifa #0, thena IN, = Mt.

Proof. 1. If « = 0, then aE = {0} € OM, and pz, (WE) = pz, ({0}) = 0 by Lemma 3.3.

2. Let us assume that« # 0. Let E € 9t,. To show thataZ € Mt, , we show that for every A € 93(R), we have

qd)

H(A) = BE(AN@E) + wi (AN @E)’*).

Now with 1A as a testing set for E € 20t,, we have

@

ut (24) = ut (Anz) + uy (LANE).

By Lemma 3.17 and Observation 3.13, we have

us (24) = ag,

ut (LANE) = wt (SAN dak) = butane),

Hy (GAN EY) = uy (GAN Gok’) = ut (AN @E)’). Substituting these equalities in (2) and multiplying the resulting equation by |a|, we obtain (1). This shows that a

€ 29%, . Then by Lemma 3.17, we have

H,(@E) = wi @E) = |a|wF(E) = lou, (E). 3. Foreverya € R, wehaveawE € Mt, by (a) and (b) and thus aM,

C M,.

Ifa £0,

then 2 € R and thus 197, c Mt,. Then M, = 19, = a1 Mt, c oM,. Therefore we have at,

C Dt, C aM, and thenaMt, = Mt, fora #0.

o

[MM] Existence of Non-Lebesgue Measurable Sets Consider the interval [0, 1) in R. Define addition modulo 1 of x, y € [0, 1) by

°

x+y=

x+y

ifx+y1.

The operation + takes a pair of numbers in [0, 1) toa number iin [0, 1). Itiis commutative,

that is, x ty Ec

(0, l) andy

=y

+ x, and associative, that is, (x $ ») teaxt

e€ [0, 1), let E+

y = {ze [o, 1):z=x

this set the y-translate ofE modulo 1.

4

forsomex

Cy + 2). For € E}, and call

52 Lemma 3.19.

CHAPTER 1 Measure Spaces LetE c [0, 1). fE € Dt, thenfor every y € [0, 1), we have E + ye MM,

and 4,(E + y) = 4, (E). Proof. Let E € 2, andy € [0, 1). Then [0, 1 — y) and [1 — y, 1) are two disjoint subsets of [0, 1) whose union is [0, 1). Define two disjoint subsets of E by FE, = EM [0,1—-y) and EF, = EN[1—y, 1). Then £) UE, = E. Also Ej, Ey € Mt,. Thus

qd)

H,(E) = pw, (Fi) + 4, (£2).

Now E + y=E\ +

Q)

€ Mt, by the translation invariance of (R, 9, 4,) and we have

by (E1 + y) = wy (E1 + 9) = 1, (Ed).

Ontheotherhand, Ey + y = Exty—le OM, by the translation invariance of (R, t,,, 44,) and we have

8)

by (Ba + y) = a, (Ea + y — 1) = 2, (E).

Now

(4)

E+ y=(1UR)+y= (Ei +y)U (Ee +9).

The disjointness of E; and E> implies that of FE, + yand E> } y. Since E; + yand E> + y are members of 9Jt, as we showed above, we have E + y € 9t,. Then by (4), (2), (3), and

°

°

°

(1), we have 2, (E + y) = 4, (21 + y) + #, (22 + y) = Bw,(£1) + 4, (E2) = 4, (E).

0

Theorem 3.20. The interval [0, 1) in R contains a non-Lebesgue measurable set.

Proof. For x, y € [0, 1), let us say that x and y are equivalent and write x ~ y ifx —yisa rational number. Then x~X x~ysy~x, XY

YMZSXWE

Thus ~ is an equivalence relation among elements of [0, 1). Let {EZ} be the collection of

the equivalence classes with respect to this equivalence relation. {F,} is a disjoint collection

and (J, Za = [0, 1). Any two elements of [0, 1) which are in one equivalence class differ

by a rational number and any two elements of [0, 1) which are not in the same equivalence class differ by an irrational number. Let P be a subset of [0, 1) constructed by picking an element from each Ey. (According to the Axiom of Choice, given a nonempty collection of nonempty sets it is possible to select

an element from each set.) Let {r, : 2 € Z,} be an enumeration of the rational numbers in

[0, 1) with ro = 0. Let

P,y=P+rq

forn €Zy.

§3 Lebesgue Measure on R

53

Note that Pp = P + ro= P collection, that is, wm

+ 0 = P. Let us show first that {P, : n € Z+} is a disjoint

mné€Zi,m~n=> Py N Pr =H.

Assume the contrary.

Then P,

P,

we have x = pm + Tm for some Pm for some p,

€ P. Then py

°

~ @ and there exists x € Py 1 Pp. € P. Similarly

+ Tm = Pr

+ Tn.

Since x € Pu

since x € Py, we have x = pp + Tn Since pm

+ Ym is either pm + rm or

Pm +m — 1 and pr + rp is either py, + rp OF Pn + fn — 1, the difference py — py isa

rational number. Then pm, pn € Eq for some o E,. Thus py = Pr-

eA. But P contains only one element of

Then the equality Pm + Tm = Pn + Tn, Which we established above,

is reduced to Dm + Tm = Pm + r,. This implies r,, = r, and then m = n, a contradiction. This proves (1). Let us show next that

Q)

U & =(0,1).

neZ,

Since P, C [0, 1) form € Z, we have Unezy. P,, C [0, 1). To prove the reverse inclusion we show that if x € [0, 1) thenx € P, for somen € Z,. Now if x € [0, 1), thenx € Ey

for some a. Since P contains one element from each equivalent class, there exists p € P such that p € Ey. Thus x and p differ by a rational number. Since both x and p are in [0, 1), their difference is a rational number in [0, 1), that is, their difference is 7, for some n € Z,. Nowifx > p, thenx = p+/, so thatx € P,. On the other hand, if x < p, thenp = x +7, so thatx = p — ry. Let ry, = 1 — ry, a rational number in [0, 1). Then x=ptrn—l=p

+ Tm so thatx € P,,. Thus in any case x is in P, for some n € Z,.

This shows that [0, 1) c Une, P,, and proves (2). Finally to show that P ¢ S)t,, assume the contrary, that is, P € 9%, . Then by Lemma

3.19, Py = P + rq € Mt, and w,(P,) = w,(P). By (2) and (1) we have

@)

1 = ,(00,D)=1( U a) = 0 aad = a). neZ,

neZy

neZy

Since P € Dt,, u,(P) is defined and y, (P) > 0. If uz, (P)= 0, then Ynez, “,(P)= so that (3) reduces to 1 = 0, a contradiction. If 4, (P) > 0, then DineZy B,(P) = oo so that (3) reduces to 1 = 00, a contradiction. This shows that P is notin 9Jt,. a

[TV] Regularity of Lebesgue Outer Measure Lemma 3.21. (Borel Regularity of the Lebesgue Outer Measure) The Lebesgue outer measure ue on R has the following properties: (a) For every E € $(R) and € > 0, there exists an open set O in R such that O D E and

HI(E) < N(O) < URE) +e.

54

CHAPTER 1 Measure Spaces

(Note that the strict inequalities u*(E) < u%(O) and 47(O) < 4} (E) + may not hold.) (b) For every E € $8(R), there exists a Gs-set G in R such that G D E and

Bx (G) = ut(E). (©) The Lebesgue outer measure jy is a Borel regular outer measure. Proof. 1. Let E € B(R) and let ¢ > 0. If ~f(Z) = oo then with the open setR > E we have HIE) < #7 (R) =oO=o+e= ui (E) +e. Thus (a) is trivially true in this case. Now consider the case 4*(E) < oo. By the definition of #z* (Z) as an infimum there

exists a sequence (J, : n € N) in 3, such that Lon J, > E and w?(E) = Donen l(n)
E, and by the monotonicity and the countable subadditivity of wh, we have

uI(E) = wt(0) = ut(

neN

hb) = ute) = Yen) < wt) +6. neN

neN

(Note that for E € $8(R) such that wt (E) = o, it is impossible to have an open set O in R

such that O > E and thestrict inequality w¥(E) < 43(O) < u}(E) +6, thatis, 00 < 00.) 2, LetE € 98(R). Fore = 1 forn é N, there exists an open set O, > E such that

BEE) < w8(O,) < w*(E)+1 by (a). Let @ = Myex On. Then GisaG5-setandG > E. Since G C O, for every n € N, we have wi (E) < ui(G) < HE (On) < Hui (E) + 1, by the

monotonicity of 4*. Since this holds for every n € N, we have u*(E) < w*(G) < uz (E) and therefore *(E) = n3(G). 3. Since Br C Mt, = Wt(u7) by Theorem 3.9, ef is a Borel outer measure on R. A G,-set is a member of BR. Thus wy is a Borel regular outer measure by (b).

Theorem 3.22. For E € $8(R), the following conditions are all equivalent : @ Eem,. (ii) For every € > 0, there exists an open set O > E with ui(O \ E) 0, there exists a closed set C C E with uy (E\C)

(v)

There exists an Fz-set F C E with u3(E \ F) =0.

(ii), (ii) > (ii, and Gii) => (i). This establishes ( (i). Finally we show that (iv)=> (v) and (v) > @. (Note that proving (iv) > () is actually superfluous.)

1. @) => Gi). Assume that £ satisfies (i). According to Lemma 3.21, for every ¢ > 0 there exists an open set O > E with w7(E) < ut(O) < ut (E) +e. Since E € Mt, with O as a testing set in the Carathéodory condition satisfied by Z, we have w*(E) + & >=

ui(O) = ui(ON E) + wi(O 0 E*) = pi(£) + wi(O \ £). If wi(E) < oo, then

subtracting 47 (Z) from the last inequality we have uf (O \ E) < &. If #7 (Z) = on, let E, = EN(n—1,n]

forn

€ Z.

Then (EZ, : n € Z) is a disjoint sequence in

t, with

Unez En = E and y,(En) < #,((1—1,n]) = 1. Applying the result above to £, for each

§3 Lebesgue Measure on R

55

n € Z, for every ¢ > 0 we have an open set O, > E,, such that zp, (On \ En) < 3712-Fle, If we let O = Unez O,, then O is an open set, O D> E, and

)

o\e=(U%)\(U%)=(U%)o(Um

-Ufo.n(Us)'}-Ulon(U4)] neZ

ne

e

neZ

neZ

neZ

neZ

< Jn \ En). neZ

Then we have

HE(O\ BE) < wt ((JOn\ End) < un neZ

nezZ

\ En)

ey? cheap le 1,42, “4a 32h 3 on 327 30 3 =e This shows that E satisfies (ii).

2. (ii) > Gi). Assume that E satisfies (ii). Then fore = 4, n &N, there exists an

open set O, > E with u*(O, \ E) < 3. Let G = yey On, a Go-set containing E. Now

G C O, implies p¥(G \ E) < wi(On \ E) < } for everyn € N. Thus uF (G \ £) = 0.

This shows that E satisfies (iii),

3. (ii) > G@). Assume that E satisfies (iii). Then there exists a G3-set G D E with

ut(G \ E) = 0. Now u*(G \ E) = 0 implies that G \ E € 97t(u*) = Mt, according to

Lemma 2.6. SinceE C G, we haveE = G \ (G \ E). Then the fact that G and G \ E are in 2, implies that E is in Mt,. 4. @ = (iv). Assume that E satisfies (i). Then E € Mt, and hence E° € Mt,. As we showed in 1, 2, and 3, (i), (ii), and (iii) are all equivalent. Thus applying (ii) to our

E* € Mt, for every « > 0 we have an open for any two sets A and B, we have A\ B = O\ E*° = E \ O° and then ui (E \ O°) < &. OD E*. If we let C = O°, then we have (iv) 5. (iv) > G@). Assume that E

set O D E* with u7(O \ E°) < &. Now AN BS = BC (A‘)* = Bo \ A‘. Thus Now O° is a closed set and O° C E since for E.

satisfies (iv). Then forevery ¢ > 0, there exists a closed set

Cc Ewith p*(E\C) < e. Now E\C = C°\ E* so that u*(C°\ E°) < ¢. AlsoC CE

implies C° > E°. Let O = C*. Then O

is an open set, O D E° and pi(O \ E°) < «.

This shows that E° satisfies (ii). Since (ii) is equivalent to (i), we have E° € DM,. Then E € Mt,. Thus E satisfies (i). 6. (iv) = (v). Assume that E satisfies (iv). Then for every n € N, there exists a closed

set C, C E with u*(E\ Cy) < i Let C = Unen Cn, an Fo-set contained in E. Now EDC DC, sothat E\C C E\C, and us(E\C) < ut(E\ Cr) < } for every n eN.

Thus ue (E\C) =0. This shows that £ satisfies (v). 7. (vy) => @. Assume that E satisfies (v). Then there exists an F,-set F C E with

uy (E\ F) =0. Now uf (E \ F) = 0 implies that E \ F ¢ Dt(ut) = Mt, by Lemma 2.6.

56

CHAPTER 1 Measure Spaces

Since F Cc E, we have E = F U(E \ F). Since F is an F,-set, F € Sp E € Mt,. This shows that E satisfies (i).

C Mt,.

Thus

Remark 3.23. According to (b) of Theorem 3.21, for every E € 98(R) there exists a Gs-set G D E with 4*(G) = ut (£). This does not imply that u*(G\ Z) = 0. IndeedifE ¢ Mt, and wi (E) < 00 then wi (G \ E) > 0. On the other hand, ifE € 9, and z*(Z) < 00 then uF(G \ E) = 0. Proof. Let E € B(R) and let G be a Gz-set such that G D and pi (G) = uz (E). Since G D E we have G = EU (G \ E) and then by the subadditivity of the outer measure 7 we have ue (G) < ue (E) + wy (G\ E£). If ut (E) = ue (G) < oo, then subtracting this

real number from the last inequality we have 0 < yi (G \ E). Now if ui(G \ E) = 0 then G \ E € Dt(ut) = Mt, by Lemma 2.6. Since E C G, we have E = G \ (G \ E). Then G € Mt, and G \ E € Mt, imply that E € Mt,. Thus u*(G \ E) = 0 implies that E€ Mt,. Then E ¢ Mt, imlplies w*(G \ E) # 0 and hence w*(G \ E) > 0. Returning to the equality G = EU(G\ £), if E € Dt, then G\E € QM, also and then HE(G) = wE(E) + wiG \ E) by the additivity of u* on Dt,. If uF(Z) = ut (G) < 00, then subtracting this real number from the last inequality we have u7(G\\ E)=0. © Next we derive from Theorem 3.22 some approximation theorems for Lebesgue measurable sets by open sets in terms of symmetric differences. The symmetric difference of two sets A and B is defined by AAB = (A \ B) U(B \ A). Theorem 3.24. Let E € 98(R). Then E € Mt, if and only if for every e > O there exists an open set V with w*(EAV) 0 there exists an open set V > Ewith u*(V \ £) E, wehave V\E = EAV. Thus p*(EAV) 0, there exists an open set V with ut (EAV) < e. To show that E

€ 9}t,, we show that £ satisfies (ii) of Theorem 3.22.

Let ¢ >

0 be

arbitrarily given. By our assumption there exists an open set V with w*(EAV) < 5. Consider the set E \\ V C EAV. By Lemma 3.21, there exists an open set W > E\ V with p*(W) < ui(E\V) +5 < ui(EAV) +3 < 26. If we let O = W UV, then O is an open set containing E and

O\ E=(WUYV)\EC WU(V\ EB) C WU(EAYV) so that

2 1 HE(O\ E) < wf) + wy (EAV) < 36+ 36

Thus £ satisfies (ii) of Theorem 3.22 and therefore E € 20t,.

8.

©

Theorem 3.25. Let E € Mt,. If 4,(E) < 00, then for every « > O, there exist finitely many disjoint finite open intervals J,,..., Jm such that 11, (E Amy

Jn)

0 there exists an open set O > E such that

(a)

1,(0\ E) < 5.

If 2, (E) < 00, then by the additivity of 2, on St, we have

@)

#,(O) = 4, (E) + 4,(0 \ E) < 00.

Now 4*(O) = (0)

< 00. Thus by the definition of 7*(O) as an infimum, there exists

a sequence (J, : n € N) in 3, such that Ucn J, D O and

)

Yl) < ui(0) + 5 N

ee) < 5.

By the triangle inequality for symmetric differences, that is, AAB C (AAC) U (CAB) for any three sets A, B, and C, we have

eam e40)u(0af U m})u({U m}4{ Ua): n=1

neN

neN

n=1

By the monotonicity and finite subadditivity of 1, we have

(5)

u,(EA U) In) < m,(EAO) + u, (0A) h) n=1

N

néeN

+m({U m}o{ U4])neN

Since E Cc O, we have EAO

©

n=1

= O \ E and then by (1) we have

1, (EAO) = 1,(0\ E) < 5.

Since O C nen Jn. we have OA nen In = (Upen In) \ O. Then we have

7)

1,(04 Uh) =m ({ Um} \ 0) =H. Um) - #00) neN

neN

neN

7 (la) — (0) + Un) — (0) < 5

neN

neN

358

CHAPTER 1 Measure Spaces

where the second equality is by (3) of Lemma 1.25 which is applicable since j4,(0) < 00 by (2) and the last equality is by (3). Finally since Ue Tn C Unen In we have

{UJa{Un}={Ua} {Us} < Um and then

®

N

m({U»fa{U m}) s.( Ue) = ata) = Dee) N

n>N

n>N

Using (6) - (8) in (5), we have 1, (EA UN, In) < €. Notethat

en £(Un) < 00 according

to (3) and this implies that J, is a finite interval for every n € N. Let us note that since {i, :n=1,..., N} is a finite collection of finite open intervals, Ua 7, is the union of

finitely many disjoint finite open intervals Jj,...,Jm.

Remark 3.26. Theorem 3.25 is not valid without the assumption that 4,(EZ) < oo. For

example, let E = Uncen En where E, = (n—},n) forn ¢ N. Then E € St, and #, (EB) = oo. Let {h,..., Iv} be an arbitrary finite disjoint collection of finite open intervals in R. Then there exists No €¢ N such that UN, I, C (—No, No). This implies N

N

n=1

n=1

EALJhDE\ J RDE\(CNoM)=

Le.

n>No+1

Thus we have

"

uz (ZA J i) > w( n=1

U

n>Not1

En) =>

n>Notl

1

“z=

n>Notl

3 =O

Since every open interval in R has a positive Lebesgue measure, every Lebesgue measurable set containing an open interval has a positive Lebesgue measure. The converse is false, that is, a Lebesgue measurable set with a positive Lebesgue measure need not contain an open interval. For instance the set of all irrational numbers contained in (0, 1) has Lebesgue measure equal to 1 but this set contains no open interval. The next theorem shows that if E € 2M, and y,(E) > 0, then there exists a finite open interval J such that 4, (EZ as close to jz, (1) as we wish.

Theorem 3.27. Let E € 90, with ,(E) finite open interval I such that a 4,1)

J) is

> 0. Then for every a € (0, 1) there exists a

< u,(END

< w,(D.

Proof. Consider first the case where 4,(E) € (0, 00). Leta € (0,1). Then i =1+7 with n > 0. Since yz, (EZ) € (0, 00), we have n 4, (£) € (0, 00). By Lemma 3.21, there exists an open set O > E such that

qd)

H,(0) sw, (E) +04, (E) =

1

+ mu, (EZ) = ght (2) < 00.

§3 Lebesgue Measure on R

59

Now since O is an open set in R, it is the union of a disjoint sequence (J, : n € N) of open intervals in R. Then 4,(0) = nen #, Un). Since E C O, we have

Hy (E) = #, (E00) =m, (EN

neN

bh) =

EN hy).

neN

Using these expressions for jz, (O) and y, (Z) in (1), we have

1

(2)

Det) $5 yoo (EN hn).

neN

neN

Therefore there exists at least one no € N such that 4,(n,) < du, (E Ing). Since Bt, (O) < 00, J, is a finite open interval for everyn ¢ N. Let I = Jy). Then / is a finite

open interval with op, (I) < (EN 2) < a, (0).

Now consider the case 4, (EZ) = oo. By the o-finiteness of the Lebesgue measure space

(R, Dt, 4,), there exists a IN, measurable subset Eo of E with 4, (Ep) < (0, 00). Then

by our result above for E € 9%, with 2, (E) € (0, 00) , there exists a finite open interval T such that a 2, () < w,(EoND 0, then by Theorem 3.27 for every a € (0, 1) there exists a finite open interval J, such that

a)

ett, la) < 1, (E 0 Ta).

Let Jo = ( - 5 by, da), $ 4, (a). Let us show that for sufficiently large a € (0, 1) we have Jag C ACE). Let z € Jy be arbitrarily chosen. We show that z ¢ A(£) by showing

that z = x — y for some x, y € E. Now (EN iy) U [Enh) +z] C i,UC, +2). Since

z € Jy, the set Jy U (Jy +2) is an open interval with 2, (Ly U (la +z) < {1+ $}u, (la). Thus we have

Qa,

(EN ta) U[EN Ta) + z]) < H(Ta U Ua +2) = {1+ 5} ola).

By the translation invariance of (R, 90, 4,), we have 4, ((E

Now if we assume that EM I, and (EM I,) + z are disjoint, then

(3)

Ia) +z) = uw, (EN hy).

4, (EMI) U[EN dy) +2) = (EN Ie) + 2, (EN Ia) +2) =2y, (EN Ty) > 204, (la),

60

CHAPTER 1 Measure Spaces

by (1). Thus, if we assume that E 9 I, and (EZ M I,) + z are disjoint then by (2) and (3)

4

da E, O € Or}. Proof. Let E ¢ $(R). If O € Op and O D E, then by the monotonicity of ut and by the fact that O € 20, we have w*(E) < wf(O) = u,(O). Thus we have HE(E)

< inf {u,(O)

: O D E, O € Og}.

To prove the reverse inequality, lete > 0 be

arbitrarily given. According to Lemma 3.21, there exists O € Op such that O D E and 4,(O) < wZ(E) +6. Thus inf {u,(0) : O > E,O € Or} < wi (E) +6. From the arbitrariness of ¢ > 0, we have inf {4,(O):

O > E,O

€ Ox} < By(E).

'

Definition 3.31. The Lebesgue inner measure of E € Y(R) is defined by

Hs, (E) = sup {u,(C): CC E,C € Ex}. (Note that 9 C E and 4 € €p so that the collection of all closed sets contained in E is

nonempty.)

Observation 3.32. For every C € Ep, we have pi»,,(C) = 4, (C).

Proof. This is an immediate consequence of Definition 3.31. We say that a topological space X is a o-compact space if there exists a sequence (K, : n € N) of compact sets in X such that |), 0 there exists Co € Eg such

that Co C E and

1, (Co) > a — 8.

Let (K, : n € N) be an increasing sequence in Rg such that We have im, 1,(Kn) = ,(Co). Then

lim Ky = 100

nen Kn = Co.

p> sup Hz (Kn) = tim, (Kn) = (Co) > @ — 8. ne!

Then by the arbitrariness of e > 0, we have 8 > a. Next consider the case a = oo. In this case, for every M > 0 there exists Cy € Ex such that Cy C E and 4, (Cy) > M. Let (K, : n € N) be an increasing sequence in Rg

such that lim, Kn = Unen Kn = Cu. Then

jim Hz (Kn) = p2, (Cy)

so that

M < u,(Cmu) = jim, Hy (Kn) = sup Mz (Kn) < B. By the arbitrariness of M > 0, we have B = co=a.

Bf

Theorem 3.34. The Lebesgue inner measure [y,, on R has the following properties: 1°

nonnegative extended real-valued : 14,,(E) € [0, 00] for every E € BR),

2°

pe, @) = 0,

3°

monotonicity: E,, Ex € $8(R), Ey C Ez => ts,,(E1) < bx, (E2),

4°

countable superadditivity : (Ey, :n EN) C PR),

disjoint > tee (Upen En) >

Dnen a, (En),

5°

translation invariance: E € $8(R) andx ER => ey, (E +x) = bs, (2),

6°

positive homogeneity: E € SB(R) anda € R => py,,(@E) = lal, (E).

62

CHAPTER 1 Measure Spaces

Proof. Properties 1° - 3° are immediate from Definition 3.31. Property 5° follows from

the translation invariance of (IR, 99, 4,) applied to the closed sets C in Definition 3.31. Similarly 6° follows from the positive homogeneity of (R, Vt, 44,). To prove 4°, let (E,

: 2

€ N) be a

disjoint sequence in $(R).

If ts, (En)

=

co for some no € N, then we have ts,,(Upew En) 2 H+,, (Eno) = 00 by 3° so that Bet (Upen En) = 00 = Yen He,,(En). Thus consider the case where j1+,,(En) < 00 for every n € N. Let N € N be arbitrarily fixed. Let ¢ > 0. Now for every n € N, since bex,,(En)

< 00 there exists C,

€ €p such that C, C Ey, and 4,(Cy)

> bs,, (En) — 7

by Definition 3.31. Since (EZ, : n € N) is a disjoint sequence, (C, : n € N) is a disjoint sequence, For the closed set UN Cy contained in Ua En, we have by 3° , Observation 3.32, and the finite additivity of 4,

N

N

n=1

n=l

= Doar (Ca) > So ae, (En) —€. By the arbitrariness of e > 0, we have is,,(\U_; En) = My Me, (En). Then by 3°, we have jis,.(Unen En) = 2M Hx, (En). Since this holds for every N ¢ N, we have

Bs (Uyen En) 2 nen H,z(En). This proves 4°. a

Remark 3.35. (a) Needless to say condition 4° in Theorem 3.34 does not hold without the assumption of the disjointness of (EZ, : n € N). For instance if we let E} = Ez = [0, 1]

and E3 = @ forn > 3, then fi4,,(Uen En) = Hs,,(E1) = 1, but Dery Me, (En) = 2. (b) Let us observe also that the superadditivity of 44,, on 93(R) implies its finite superad-

ditivity on $B(R), that is, if (Z, : nm = 1,..., N) is disjoint finite sequence in $3(R) then

Hm (Uper En) = Dopt Han En)-

Observation 3.36. For every E € $3(R), we have ,,,(E)

< ue (E).

Proof. Let E € $3(R). Let Co be an arbitrary closed set in R such that Cp C E and let Op be an arbitrary open set inR such that Op D E. Then Cy C Op so that jz, (Co) < 4,(Oo). Since this holds for an arbitrary Op € OR such that Op > E, we have

(Co) E, O € Op} = ut (E) by Proposition 3.30. Since this holds for an arbitrary Cp € @p such that Co C E, we have sup {u,(C); C C E,C € Eg} < ur(E), that is, us, (E) < BE(E).

1

The next theorem is a criterion for the Lebesgue measurability of a subset of IR in terms of the Lebesgue inner measure. Theorem 3.37. Let E ¢ $3(R).

(@) FE € M,, then 1s, (E) = 13 (E).

§3 Lebesgue Measure on R

63

) f ps,,(E) = ur (E), then E € 0, provided that BE(E) < 00. Proof. 1. Suppose E € $t,. Let I, = (#—1,n] and E, = EN|, forn € Z. Consider the disjoint sequence (EZ, :n € Z) in Mt, with 4, (En) < 00 for everyn € Zand ,-7 En = E. By (iv) of Theorem 3.22, for an arbitrary ¢ > 0 there exists C, € €p such that C, C E,

and 41, (En \ Cn) < 37127Mle for every n € Z. Since ,(Cn) < 4,(En) < 00, we have

H, (En) —

(Cn) = B, (En \ Cy) so that

My (En) < (Ca)

le

+ 32m

le Ssup {u,(C):C C Ey, C € Ep} + 32Ii le

=H, (En) + 320°

Then we have

UNE) = u,(E) = (UL) Bn) = nez

nezZ

as En)

0, we have HE(E)


bs, (E) by Observation 3.36. Therefore 2», (E) = 4% (E). This proves (a). 2. Conversely suppose f,,(Z) = *(E) and 4*(E) < oo. To show that E € Dt,

it suffices to show that for every ¢ > 0 there exists O € Op such that O > E and wy (O \ E) < & according to (ii) of Theorem 3.22. Now according to Proposition 3.30 we

have y* (EZ) = inf {u,(O) : O > E, O € Op}. Thus there exists O € Op such that O D

E and p,(0) < uzi(E) + 5. Since 4,4,,(£) = sup {u,(C) : C C E,C € €g} and since Bx, (E) < HIE) < 00 there exists C € @p such that C C E and py, (E) — 5 < p,(C).

Since C C E and thus O \ E c O\ C, we have

ui(O\ E) $ uk(0\C) =H,(0 \C) = 2, (0) — 4,(C)

0 there exists C € €g such that C C E and

Hx,,(E) — & < 1, (C) < bs, (E). (b) There exists an F,-set F in R such that F C E and p,(F) = ps, (E). Proof. 1. We have i.,,(Z) < #3(E) < 00 by Observation 3.36. It follows then from the definition of 2,,,(£) that for every ¢ > 0 there exists C € €g such that C C E and bex,,(E) — € < t,(C). By Observation 3.32 and by the monotonicity of u,,,, we have

1 (C) = ba, (C) < bs, (CE). This proves (a). 2.

(Ch)

For every n € N there exists C,

€ €g such that C,

C

E and py,,(£) — i
and 4, (F) = bs,,(E1 U E2) by Lemma 3.39. By Lemma 3.21, there exists a G3-set G > Ez

such that 4, (G) = wi (Ez). Now F c (F\G)UG Thus we have

and F\G

C (E, UE) \GC

&}.

Hs, (E1 U Ea) = 4, (F) Su, ((F \G) UG) bs, (E1) + wy (E2). The next measure and the Lebesgue always equal

theorem says that if EF is a Lebesgue measurable set in R with finite Lebesgue if we decompose the set into two parts in an arbitrary way, then the sum of inner measure of one part and the Lebesgue outer measure of the other part is to the Lebesgue measure of the set.

Theorem 3.41. Let E ¢ B(R). If E €¢ Mt, and w,(E)

we have jix,,(AM E) + wi(A° Proof.

For an arbitrary A

E) = pw, (E).

€ $3(R), AN

< 0, then for every A € BR),

E and ACN E are two disjoint sets whose

union is E so that u*((A MZ) U (A°N E)) = w*(E) < 00. Thus by (a) and (b) of Proposition 3.40, we have jtx,,(Z) < bs, (AN E)+ HI (Ae NE)< ue (E). But E € Mt, and this implies 4,,,(Z) = “#7(E) = p(B) according to (a) of Theorem 3.37. Thus

Ba, (AME) + pe(ASNE) = uw, (EZ).

&

Lemma 3.42. Leté € (0, 1) be an irrational number and let

A={n+mé:m,néZ}, B={n+mé:

m,n €Z,

n even},

C={n+mé

: m,n € Z,

n odd}.

Then A= BUC, BNC =8%, C = B +1 (that is, the translate of B by 1 € R),0€ B, and A, B, and C are ail countable dense subsets ofR. Proof. Clearly A, B, and C are all countable subsets of R. It remains to show that they are

dense in R. Let us observe that if m’, m’, n’, n’” © Z and (n’, m’) ¢ (n", m”) then

69)

ntm'é £n" + mE.

To prove this, suppose (n’, m’) 4 (n”, m”) but n’ + m’& =n" + mE, that is,

(2) Now since (n’, m’) then by (2) we have the fact that& is an then by (2) we have

ni —n" = (n" —m’)é. 4 (n,m) we have either m! 4 m" or n’ # n” or both. If m’ 4 m” & = (n’ — n”)/(m” — m’) which is a rational number. This contradicts irrational number. On the other hand, if n’ 4 n” then n’ — n” ¢ 0 and m” — m’ #4 0 so that € = (n' — n”)/(m” — m’), contradicting the fact

that& is an irrational number. This proves (1).

66

CHAPTER 1 Measure Spaces To show that A is dense in R, we show that every open interval 7 in R contains at least

one point of A. Now for everyi ¢ N, there exists a unique; € Z such thatn; +7& e [0, 1). Letx = mj +ig € Afori e N. Letk € N be so large that } < (1). Consider the collection {x; : i = 1,...,4 +1} C A. Note that x1,..., x,41 are all distinct by (1). Let xj, = 1,...,%+ 1, be enumerated in increasing order and let the result be labeled as 0 < yi < --- < yey < 1. Nowiff yjii — yj > } for all j = 1,...,k, then yi41 —yi

= R- i =

Thus yj41 — yj
7/1 A contains infinitely many points of A. This contradicts A(Eo) N A = {0}, or@. Thus if Eo € 9M, then 2, (Eo) = 0.

3. Suppose Eo € Dt,. Then 4,(EZo) = 0 as we showed in 2. By the translation invariance of QR, Mt, Hy), we have Ep + ay, € SM, and yt, (Eo + aq) = w, (Eo) = 0 for

ne€ Z,.

Since nex

(Eo + an) = R and {Zp + a, : n € Z4} is a disjoint collection,

we have 1, (R) = 4, (Unez, (Eo + 4n)) = Dez, Hi (Eo + dn) = 0, a contradiction. Therefore Ey ¢ t,.

By means of the non-9Jt, -measurable set Eo in Proposition 3.43 above, we construct

next a set M € 98(R) such that both yz»,,(M@) = 0 and y,,,(M°) = 0. Then we have Bex,, (4) + ps, (M°) = 0 # ps,,(R). This shows in particular that the Lebesgue inner measure j,,,, though superadditive, is not an additive set function on $8(R).

Theorem 3.44.

There exists a set M € 98(R) with the following properties:

(1)

a.)

= ba, (M9) = 0,

(2)

Ha, (MOE) = fs, (MEO E)=0 for every E € BR),

3)

wi(MNE)

= wi(M° NE) =4,(E£) for every E € Mt, with 4,(E) < 00.

Proof. 1. Let A, B, and C be the three subsets of R defined in Lemma 3.42 and let Eg be

the non-T, -measurable subset of IR defined in Proposition 3.43. Let M = Eo+ B. By the definition of jzz,,, to show that y,,, (M4) = 0, it suffices to show that for every closed set F C M we have 4, (F) = 0. We show that actually for every F € 9)t, such that F Cc M, we have :,(F) = 0. Let A(F) be the difference set of F as in Definition 3.28. Since F € Mt, to show that 4, (F) = 0 it suffices to show that A(¥) contains no open intervals

in R according to Theorem 3.29. Since C is a dense subset of R by Lemma 3.42, every open interval inR contains some points of C. Thus to show that A(F) contains no open intervals, it suffices to show that A(F)NC = 9. Toshow A(F)NC = G, letz € ACF). Thenz = z1—z2

where 21, 272 € F C M = E+B so thatz = (41 +1)— (42 +52) = (41 — x2) + (1 — D2) with x1, x2 € Eo and bi, by € B. Now if x; = xg, then z = b; — b2 € B and thus z ¢ C.

On the other hand if x1 # xz, then by the definition of Eo in Proposition 3.43, we have

%1 —X2 ¢ A. Now x1 — x2 = z+ (bz — by). Since bj, bp € B, we havebp —b, EC BCA. Ifz € A, then since A is closed under addition, we have x; — x2 € A, a contradiction. Thus

z ¢@ A. Then since C C A, we have z ¢ C. Thus in any case, ifz € A(F) thenz ¢ C. This shows that A(F) 9 C = @ and proves 4,,,(M) = 0. To show that j»,, (M°) = 0, recall that we showed in 1 of the Proof of Proposition 3.43

68

CHAPTER 1 Measure Spaces

that Ep + A = R. Thus we have

=R\M=(Eo+A)\(Eo+B) =Eop + C =(Eo + B+) =MH+1. By the translation invariance of j,,, on $3(R) by 5° of Theorem 3.34, we have the equalities Bex, (M°) = bs,,(M + 1) = Uts,,(M) = 0. This completes the proof of (1). 2. (2) follows from (1) by the monotonicity of z,,, on $3(R). 3. Let E € Mt, and u,(E) < oo. According to Theorem 3.41, for every A

PBR), we have pis,,(A NE) + wi(A°N E) = p,(E).

have 0 + wi(M ME) = j,(E) by (2) and similarly with the choice A O+ we (ME NE) = yw, (E) by (b). This proves (3). a

Theorem 3.45. measurable set. Proof.

Every 2%,-measurable set E with ,(E)

Let M be as in Theorem 3.44.

€

With the choice A = M° we

>

= M we have

0 contains a non-DN,-

Then by (2) and (3) of Theorem 3.44, we have

Hx,, (MN E) = Oand wi (MME) = 4, (E) > Oand therefore u.,, (MOE) A ut (MN E). This shows that MN

E ¢ 90t, by (a) of Theorem 3.37.

u

Problems Prob. 3.1. Let a decreasing sequence (E,:n € N) C St, in the Lebesgue measure space (R, Dt, 4.) be given by E, = [n, 00) forn € N.

(a) Find

lim £, and u,( lim E,).

R00

(b) Find ‘lim 2, (Ep).

100

noo

Prob. 3.2. Consider a sequence (E, :n € N) C Mt, defined by E, = [0, 1) U[n,n + 1) when n is odd and E, = [0, 1) U [n,n + 2) when n is even. (a) Show that im, E,, exists and find im, En. (b) Show that "im.DH, (E,) does not exist.

Prob. 3.3. For each of the following sequences (EZ, :n € N) Cc M,, (a) show that im, E,, exists and find im, En,

(b) show that jim, #2, (E,) exists and lim, qd (2)

Q)

(En) # By (im, En).

E, = [0, 1) U [n,n +1) forn € N, E,, = [0, 1) U [n, 2n) forn € N,

E, = [n,n + 1) forn € N.

Prob. 3.4. Let 3 be the collection of 9 and all finite open intervals in R. Define an outer

measure j:* on R by setting for each E € J3(R)

u*(E) = inf { Dren ln)! (hh

Show that j2* = 23 defined in Observation 3.2.

2 EN) CT, nen in DE}.

§3 Lebesgue Measure on R

69

Prob. 3.5. Let € be a disjoint collection of members of 9Jt, in the Lebesgue measure space

(R, Dt,, 4,). Show that if 2,(E) > 0 for every E € €, then the collection € is at most

countable.

Prob. 3.6. For E € Dt, with 4, (E) < oo, define a real-valued function g, on R by setting @;(%) = 4,(EN(-00,x]) forx eR. (a) Show that ¢, is an increasing function on R.

(b) Show that

lim g,(x) =O and lim 9,(x) = p,(E). x00 x->00

(c) Show that g, satisfies the Lipschitz condition on R, that is,

lee") — @,(@")| < |x’

x"|

forx’,x" ER.

(d) Show that g, is uniformly continuous on R. Prob. 3.7. Let E € Mt, with 4,(Z) < oo. Show that for every a € (0, 1), there exists a subset E, of E such that Ey € 9M, and u, (Eq) = au, (E).

(Hint: Use the continuity of the function g, in Prob. 3.6.) (Thus the Lebesgue measure space (R, Mt, Hz) not only does not have any atoms but in fact every Lebesgue-measurable set with finite measure has a measurable subset with an arbitrarily designated fraction of the measure. See Prob. 3.8 for E € Vt, with uz, (EZ) = 00.)

Prob. 3.8. Let E € 20, with 4, (£) = oo. Show that for every 4 € [0, 00), there exists a subset E, of E such that E, € Mt, and u,(E,) =A. Prob. 3.9. Let E c R and we (E) = 0. Show that E£° is a dense subset of R, that is, every

non-empty open set O in R contains some points of E°, in other words no non-empty open set O in R can be disjoint from E°. Prob. 3.10. Consider the measure space (R, Sp, ,)(a) Show that if FE €¢ BR and? € R, then E+¢ € Bpand Let

BR+t:={E+t:

wp, (EZ +t) = pw, (E).

E € BR}. Show that Be +t = Bp for every s ¢ R.

(b) Show that if Z € Sy anda € R, thenaE € Bp and pz, (@E) = |al|u, (E).

Leta BR

:= {aE : E € BR}. Show that By

= BR for everya € R such thata 4 0.

Prob. 3.11. Let E and F be two subsets of R. Suppose E C aF for some a € R. Show

that w*(E) < |a|u3(F).

Prob. 3.12. Let f be a real-valued function on an open interval J = (a, b) C R such that the derivative f’ exists and | f’(x)| < y forx € J where y is a nonnegative real number. (a) Show that for every Ig = (ao, bp) C (a, b) the image f (ip) is either a singleton or an

interval in R and

#,(f Co) < ¥ 4, (io).

(b) Show that for an arbitrary subset E C I we have

ur (f(E)) < y uf (E). Prob. 3.13. From the interval [0, 1] remove the open middle third (4, 2). Remove the open middle third from each of the two remaining intervals [0, 4] and [3 1]. This leaves us four

70

CHAPTER 1 Measure Spaces

closed intervals. Remove the open middle third from each of the four. Continue this process of removal indefinitely. The resulting set 7 is called the Cantor ternary set. (a) Show (b) Show (c) Show dense, or

that T is a Borel set and that 2, (7) = 0 and thus that T is nowhere dense. non-dense, if the interior

indeed T is a compact set in R. T is a null set in the measure space (R, Br, 4,). (A set A in a topological space (X, 9) is called nowhere of its closure is empty, that is,(a)° =.)

(e) Show and [0, 1] (Thus the countable

that T is an uncountable set and in fact a one-to-one correspondence between T can be established. Cantor ternary set is an example of a null set in (R, Mt,, 2) which is an unset.)

(d) Show that T is a perfect set. (A set A in a topological space (X, 9) is called perfect if A = A’ where A’ is the derived set of A, that is, the set consisting of all the limit points of the set A. A set A is a closed set if and only if A > A’. A point ina set A is called an isolated point of A if it is contained in an open set which contains no other point of A. Thus aset A is perfect if and only if it is closed and has no isolated point in it.)

Prob. 3.14. Construct a collection {E,,, :

€ N, k € N} of subsets of R such that

U (1)keN Bat) #

neN

keN

(LUneN Ene):

Prob. 3.15. Let Q be the set of all rational numbers in R. For an arbitrary ¢ > 0, construct an open set O inR such that O > Q and H3(O) 1—«.

Prob. 3.18. Consider the measure space (IR, 99, ,) -

(a) Show that every null set V in (R, 9t,, 4,) is a subset of a Gs set G which is itself a null set in (R, Mt, ,) «

(b) Show that every E € St, can be written as E = G \ N where Gis a G; set and Nisa

null set in (R, Mt, 4,,) contained in G. (c) Show that every EZ € Dt, can be written as E = F U N where F is an F, set and N is anull set in (R, Mt, 4,) disjoint from F. Prob. 3.19. Let E C R, EF € 2%,, and 4, (EZ) < oo. Show that for every ¢ > 0 there exists acompact setC C E such that u,(E\C) 00

a0

§3 Lebesgue Measure on R

71

Prob. 3.21. (a) Consider the “ur -measurability condition on E € §8(R): (1)

ECA) = Bi(A NE)+ BHA

1 E*) for every A € $B(R).

Show that this condition is equivalent to the following condition: (2)

ad

= BENE)

+ eink)

for everyI € Jy.

(b) Let J be the collection of all open intervals in R with rational endpoints. This is a countable subcollection of J,. Show that condition (2) is equivalent to the following

apparently weaker condition : @) BEG) = BEG NE) + we

0 E*) for every J € 3.

Prob. 3.22. Prove the following statements: (a) Let A C R. If there exists e9 € (0, 1) such that ui(A NI)

interval J then 47 (A) = 0.

< eo£(Z) for every open

tb) Ltt E CR. Tf uy (E) > 0, then for every € € (0, 1) there exists a finite open interval

I, such that

ef(I,) < w(E Nb).

(c) Let E CR.

that

If F(Z)

> 0, then there exists a finite open interval Jp and 9 > 0 such

(ENID+A)N(EN Ip) #G for every h € R such that (d) For E CR, wedefine

AE

AE contains an open interval.

= {z

|h| < dp.

€ RR:z =x — y wherex,y € E}. If ui (Z) > 0, then

Prob. 3.23. For E C R, we define E+ E ={z¢R:z=x + where x, y € E}. We say thatE is symmetric with respect to 0 € Rif —E = E. Show that if E is symmetric and “uz(E) > Othen E + E contains an open

interval.

Prob. 3.24. For E Cc R, we define E+ E = {z¢R:z=x-+y wherex, y € E}. Show that if HN(E) > O thenE + E contains an open interval. Prob. 3.25. ForZ, F c R, wedefine E+F

= {z ¢ R: z=x+y

where x € E andy € F}

and E—F = {z¢R:z=x-—y wherex € E andy € F}. Prove the following statements: (a) Tf pi(E 1 F) > 0, then £ + F contains an open

(b) If zF

interval.

(EM F) > 0, thenEZ — F contains an open interval.

Prob. 3.26. Show that there exist two null sets E and F in the Lebesgue measure space R, Mt,, w,) such that E+

F =R.

Prob. 3.27. Let @ be the set of all rational numbers in R. Show that for any x € R we have jim, 194:(¢) # 1g9(x) and indeed lim 1942(*) does not exist.

me

are

Prob. 3.28. Let J be a collection of disjoint intervals in R. Show that J is a countable collection.

72

§4

CHAPTER 1 Measure Spaces

Measurable Functions

[1] Measurability of Functions Notations. For a set D we write {x € D : ---} for the subset of D consisting of those points x in D satisfying one or more conditions represented by -.-. Note that {x ¢ D:---}=6 when the defining conditions --- are not satisfied by any x € D. We also abbreviate {xe D:.---}as{D: ---} when there is no ambiguity. Thus for example, if f is an extended real-valued function defined on a subset D of aset X anda € R, then {x € D: f(x) < a} is the subset of D consisting of all x € D at which f(x) < @ holds. We also use the abbreviation {D : f < a}.

Definition 4.1. Let (X, A) be an arbitrary measurable space and let D € XM. An extended real-valued function f defined on D is said to be 2l-measurable on D if it satisfies the

condition that {x € D: f(x) a} © & for everyw € R. Let us observe that

()

Oj,

(2)

a Oand f(x) 4 00, we have

{D(5)+ 5 > Of= {B(G): F > O}\ {D(G) : F = co} em.

2, Let a > 0. Then by (2) we have

{D(F)+ 5 > oe} = {D(G) 10 < F a}= (DG): >, f > O}U{DG):5 >a, f 0 > a. Thus we have (69)

{D(F): ¢ >

f > 0} = {D(F): fF > Ope a.

Now a < 0 is equivalent to 4q ~< 0. Then the condition f < 2 < Oimplies f < 0. Then by (2) we have

©

{DG):4>aF aif and only if f,(x) > @ for some n € N. However

lim f,(@) =a n>00

f,(x) = @ for somen € N.

Consider for instance a strictly increasing sequence ( fax) ine N) such that f,(x) < a for everyn € Nand lim f,(x) =a. Wehave lim f,(x) > a but f,(x) < @ for every

N nen.

n>00

100

Theorem 4.22. Let (X, A) be a measurable space and let (f, : n € N) be a sequence of

extended real-valued UA-measurable functions on a set D € A. (a) The functions Jin we max f,, inf f,, sup f,, liminf f,, and limsup f, are

- measurable on'D.

n=1..N

_

neN

neN

neN

neN

(b) Let D, = {D: jim, fn €R}. Then D, € A and im, fn is A-measurable on De. Proof. 1. Let us show the &-measurability of natin

f, on D. Leta € Randx € D.

Then min {fi(x),... , fv(@)} a}em

n=1

by (ii) of Lemma 4.4. This proves the 2{-measurability of nay 2. Let us show that inf fri is 2-measurable on D. inf. Sn(x) is the greatest lower bound of { Are) nel

fn on D.

Let a € Randx

e€ D.

Now

ine N}. If the greatest lower bound of

{fn(x) : n € N} is less than o then o is not a lower bound of { f,(x) : n € N} so that

n(x) < @ for somen € N. Conversely if f,,(x) < a forsomen € N then since the greatest lower bound of { ine) ine N} does not exceed f;, (x), itis less than a. Therefore we have inf fn) < a if and only if f(x) < @ for somen € N. Thus we have

{D: int fa n fz 1m € N) isan increasing

100 noo sequence. By our result above, infy>, f; is M-measurable on D for every n ¢ N. Then by Theorem 4.21, imni { intron fe} is 21-measurable on D. This proves the 2{-measurability

of lim inf fron D. The A-measurability of lim sup f, on D is proved likewise. n—>00

5. To prove (b), note that since De={D: liminf f, = lim sup Jn} and since lim inf Sn and iim n sup Jn are two A-measurable functions on D, we have De € & by (1) of Theorem 4.16. The! QA-measurability of lim inf J, on D implies its A-measurability on D, by (a) of Lemma 4.7. Then since

lim

n>00

/, =liminf f, on D,,

n—00

lim

n>00

jf, is A-measurable on D,.

Theorem 4.23. Let (X, 21) be a measurable space and let (f, :n € N) be a sequence of

extended real-valued U-measurable functions on a set D € LU. Let @

De= {D : im, fn € R},

2)

De={D: limfn €R},

@)

Doo = {D: jim, fa = oo},

(4) ()

Do ={D: jim, fa = —oo}, Dre ={D: lim, fa does not exist},

so that D, and Dy, are disjoint and Dz U Dae = D; De, Doo, and D_oo are disjoint and D_-U Deo Doo = De. Then De, Dc, Doo, Doo, Dne € A and lim f, is A-measurable

n00

on each of De, De, Doo, and D_oo. Proof.

By (b) of Theorem 4.22, D,

€ 2 and

tim, Fn is A-measurable on D,.

Then

De, € A and Deo, D_oo € A by (b) and (a) of Corollary 4.5 respectively. By (a) of Lemma 47, lim

Fn is A-measurable on D,, Doo, and D_o. Finally Dye = D\ De € A.

Proposition 4.24. Let (X, 2) be a measurable space and let (f, : n € N) be a sequence of extended real-valued {1-measurable functions on a set D € UA. Let D, be the subset of D on which the sequence converges and let Doo and D_oo be the subsets of D on which the

84 Measurable Functions

85

limit of the sequence is equal to oo and —o0 respectively. Then

(1)

De=() U (){2: liwse - ful < 3},

Q)

Doo =

(6)

D-wo=()

meN NeN peN

U (iP: free 2m),

meN NeN peN

VU (td:

meéN NEN peN

Proof. 1. sequence, lén — Gn'| For a

fren < —mh-

A sequence of real numbers (a, : n € N) converges if and only if it is a Cauchy that is, it satisfies the condition that for every ¢ > 0 there exists N € N such that < € forn,n’ > N. sequence of extended real numbers (a, : x € N) to converge, it is necessary that

there exists N ¢ N such that a, € R form > N. This condition implies that a, — a, exists for n,n’ > N. Thus a sequence of extended real numbers (,, : n € N) converges if and only if for every¢ > 0 there exists N ¢ N such that |a, — o,| < ¢ forn,n’ > N. Now for each x € D, (f,(x) : n € N) is a sequence of extended real numbers. Thus

(fn(x) : n € N) converges Ve

> 0, 4N € N such that | f(x) — fy'(x)| < ¢ forn,n’ > N

Ve

> 0,4N € N such that | fy+p(x) — fy(x)| < ¢ forall pe N

m for every p € N. From this we have (2). Similarly for (3). Notations. For two extended real-valued functions f and g defined on a set D, we write

fAg =wmin{f, g} and f v g = max{f, g}, thatis, (f A g)(x) = min{f(x), g@)} and (f ¥ g)(x) = max { f(x), g(x)} for x € D.

Definition 4.25. Let f be an extended real-valued function on a set D. The positive part St, the negative part f—, and the absolute | f| of f are nonnegative extended real-valued

86

CHAPTER 1 Measure Spaces

functions on D defined for x € D by setting

f*(@) = Ff V 0)@) = max{ f(x), 0}, f-@) = -CFf A0)(@) = — min{F (x), 0}, IF1@) = |F@)INote that at any x € D, at least one of max{ f(x), 0} and min{ f (x), 0} is equal to0 and

hence at least one of f+(x) and f~(x) is equal to 0 so that the difference f+ (x) — f-(x) is always defined and moreover

F*@) — f7@) = max(f(), 0} + min{ f(x), 0} = f(x). We have also, considering the case f(x) > 0 and the case f(x) < 0,

FTG) + f-@) = max{ f(x), O} — min{ f(x), 0} = 1F

| = 1F1@).

Proposition 4.26. Let f be an extended real-valued A-measurable function ona set D € A where 2 is a a-algebra of subsets of set X. Then ft, f—, and |f| are A-measurable functions on D. Proof. The X-measurability of f+ and f~ is from Theorem 4.22. The &-measurability of

|f| is from Theorem 4.12.

[V] Continuity and Borel and Lebesgue Measurability of Functions on

R

An extended real-valued function f on a subset D of Ris said to be continuous at xy € D if f(xo) € R and if for every « > 0 there exists § > 0 such that | f(x) — f(xo)| < ¢ for

every x € (x9 — 6, x9 +6) D. We say that f is continuous on D if f is continuous at every x € D. It follows then that if f is continuous on D, then its restriction to a subset Do of D is continuous on Do. Let D, and D2 be two disjoint subsets of R. Let f, and f2 be continuous on D, and D2

respectively. If we define a function f on D, U Dz by setting fix)

forx € Di,

fa)= | fax) forx € Dp, then f may not be continuous on D, U D2 and in fact it may be discontinuous at every x

€

D, UD).

(Thus if we merge two continuous functions, the result may not be a

continuous function. Compare this with (b) of Lemma 4.7 according to which the result of merging countably many measurable functions is a measurable function.)

Example. Let Q and P be respectively the set of all rational numbers and the set of all irrational numbers in R. We have QM P = Gand QU P = R. Let fj be defined on G

84 Measurable Functions

87

by setting fi(x) = 1 for every x € Q, and let fo be defined on P by setting fo(x) = 0 for every x € P. Since f; is constant on Q, it is continuous on Q. Similarly fo is continuous on P. If we define a function f on Q U P = R by setting

rom

fi@)=1

forxe Q,

fo(x) =0

forxe

P,

then f is discontinuous at every x € R. Since f, is constant on Q € Bp, fi is BR-measurable on Q by Observation 4.8 and

similarly 2 is S$y-measurable on P. Then f is Sp-measurable on Q U P = R by (b) of Lemma 4.7. Actually the 23p-measurability of f on R can be shown directly by observing that for every a € R we have

{x eR: f(@) f(D). Then the composite function g o f is a real-valued IN, -measurable function on D.

88

CHAPTER 1 Measure Spaces

Proof. A real-valued continuous function g on aset E € SR is a BR-measurable function

on E by (a) of Theorem 4.27. Thus the Corollary is a particular case of Theorem 4.28.

Definition 4.30. Given a measure space (R, A, 4). Let f be an extended real-valued function on a set D € SM. We say that f is continuous a.e. on D (or to be more precise,

(A, 4)-a.e.

on D), if there exists a null set N in (R, A, 2) such that N C D and f is

continuous at everyx € D\ N.

Theorem 4.31.

Let f be an extended real-valued function on a set D € M,.

If f is

continuous (ON,, {4,)-a.e. on D, then f is IM, -measurable on D. Proof.

Since f is continuous a.e.

on D, there exists a null set N in (R, 90, , 4,) such

thatN Cc D and f is continuous at every x € D\ N. The restriction fi of f to D\ N is continuous at every x € D \ N and thus 9%, -measurable on D \ N by (b) of Theorem 4.27. On the other hand the restriction fo of f to N is S)t,-measurable on N since every function on a null set of a complete measure space is measurable on the null set by (a) of Observation 4.20. The 99, -measurability of f; on D \ N and the t, -measurability of fo on N imply the 9t, -measurability of f on (D \ N) UN

= D by (b) of Lemma 4.7.

In general if D € SBR or D € M, and f is a continuous function on D, the image of Dby f, f(D), may not be in Bp or Yt,. However we have the following special case. Proposition 4.32.

Let f be a real-valued function with both D(f) and ACf) in Br.

Suppose f is a homeomorphism (that is, f is continuous and one-to-one and its inverse function is also continuous). Then for every &R-measurable subset B of D(f), we have

F(B) € Bp.

Proof. Rg)

Let g be the inverse function of f. = Df)

€ BR.

Then we have D(g)

Since g is continuous on D(g)

€ BR,

= R(f)

€ Bp and

g is a BR-measurable

function on D(g) by (a) of Theorem 4.27. Then g is a By/BR-measurable mapping from R to R. Let B C D(f) and B € BR. Since f and g are one-to-one, we have f(B) = g7'(B). Since g is a S3z/BR-measurable mapping, we have g~!(B) € Br. This shows that f(B) ¢ Br.

of

[VI] Cantor Ternary Set and Cantor-Lebesgue Function [V1.1] Construction of Cantor Ternary Set Let T = [0, 1].

Let I,,1 = (3, 3), the open middle third of Ty. Let G1 = 1,1 and let T; = To \ G1.

Let 12,1 and Ib,2 be the open middle thirds of the two closed intervals constituting 7). Let G2 be the union of 41,1; 1,1, 42,2 and let 7, = Tp \ Go. Let 13,1, 4,2, 13,3, and 3,4 be the open middle thirds of the four closed intervals constituting 7). Let G3 be the union of 41,1; 1,1, 2; 151, 5,2, 3,3, b,4 and Tz = Tp \ Ga, and

so on. In general let Jj,1, ..-, J,2-1 be the open middle thirds of the 2*—! closed intervals

constituting 7,1. Let Gy be the union of 1,1; ...; Ie1,.--, Ayae-1 and T, = Tp \ Gx. Gy

84 Measurable Functions

89

is the union of 2° + 2! + 2? +... + 2-1 = 2* — 1 disjoint open intervals contained in To

and (G; : k € N) is an increasing sequence of open sets contained in Tp. 7; is the union of 2* disjoint closed intervals contained in 7 and (7% : k € N) is a decreasing sequence of closed sets contained in Jp. Let

G= jim Ge=

Ge

and T =T\G.

We call T the Cantor ternary set. Note that TG

= 6, TUG

= [0, 1] and

T=™\G=MnG =mn(UG) =mn(() G4) keN

keN

= (oN GD = (\o\ Gd) = (He = jimTe. =

cy

keN

=

keN

=

ite

keN

Theorem 4.33. The Cantor ternary set T has the following properties: (a) T

is a null set in the Borel measure space (R, Br, H,).

(b) G = [0, 1] \ T is a union of countably many disjoint open intervals in R; G is dense in [0, 1], and 2, (G) = 1. (c) T is an uncountable set. Indeed the cardinality of T is equal to ¢, the continuum. (d) T is a compact set inR.

(e) T is a perfect set in R, that is, T is identical with the set of all its limit points. (f) T is nowhere dense in R, that is, the interior of its closure, (Ty, Proof.

1. Since T is a closed set, T € BR.

contained in [0,1] and

is an empty set.

Since (7; : n € N) is a decreasing sequence

lim 7 = T, we have 4,(T) = u,( lim T) =

k->00

k00

lim y, (Tk).

k->00

Now #, (Jo) = 1, u,(T) = 2, (DB) = (2), and so on and in general 4, (Ti) = @y* for k € N. Thus y,(T) = jim, (2) = 0. This shows that T is a null set in (R, Bp, /4,). 2. G = [0,1] \ T is a union of countably many disjoint open intervals in R. Being a null set in (R, Br, j4,) by (a), T is a null set in (R, M,,, 4,). Then by Observation 3.6, T® is dense in R. Since G = [0,1] T°, the denseness of T° in R implies the denseness of G in [0, 1]. Since G U T = [0, 1] and px, (T) = 0, the additivity of u, on Bp implies

that 4, (G) = 4, (10, 11) — 4, (7) = 1.

3. T is the result of indefinitely iterated process of deleting open intervals from [0, 1]. In the first step an open interval is deleted from [0, 1], leaving 2 disjoint closed intervals. In the second step an open interval is deleted from each of the 2 disjoint closed intervals, leaving

2? disjoint closed intervals. In the third step an open interval is deleted from each of the 27

disjoint closed intervals, leaving 2° disjoint closed intervals and so on indefinitely. The two endpoints of each of the 2* disjoint closed intervals in the k-th step of deletion are never deleted in subsequent steps of deletion and are thus elements of T. Thus the cardinality of T is at least equal to 280 = ¢, the continuum. Since T c [0, 1] and the cardinality of [0, 1]

is equal to c, the cardinality of T is equal to c. 4. T is a bounded closed set and is thus a compact set in R.

90

CHAPTER 1 Measure Spaces

5. Let 7’ be the set of all limit points of T. Since T is a closed set, we have T > T’. It remains to show that T Cc T’. To show that every xo € T is a limit point of 7, we show that (xp — 6, xp + 6) contains at least one point of T other than xo itself for every 5 > 0. Now since xp € T and T = (\zen Te, we have xp € Ty for everyk € N. Given6 > 0, let k € N be so large that x < 8. Since xp € 7; and since 7; is the disjoint union of vd

closed intervals each with length e the one closed interval among the 2* constituting T that contains xo is contained in (xp — 6, xp + 6). Thus (xo — 6, xp + 6) contains points of

T other than xp. Thus T c T’. This shows that T = T’. 6.

Since T is a closed set, we have T = T and then (ry

=

T°.

To show that

T° = @, assume the contrary. Then T° is a non-empty open set in R so that wz, (T°) > 0 by Observation 3.10. This contradicts the fact that jz, (7) = 0. Therefore T° = and then

(TY =9.

0

If J is an open interval in R then x, (7) > 0. Aset E € Mt, with u,(£)

> 0 need

not contain an open interval. The set of all irrational numbers in an interval is such a set. Below we present a closed set F in R with 2, (F) > 0 and containing no open interval. Its

construction resembles that of the Cantor ternary set.

Example. Let @ € (0,1). We have ) cx ha = a. From the closed interval [0, 1]

delete an open interval in the center with length ha. From each of the two resulting closed intervals delete an open interval in the center with length bye From each of the resulting closed intervals delete an open interval in the center with length aye and so on. If we let G be the union of all the deleted open intervals, then G is an open set contained in [0, 1] and 44,(G) = dor + pot pete =a. The set F = [0, 1] \ G is a closed set and 2, (F) = 1—a > 0. The fact that F contains no open intervals can be shown as follows.

After n-th step in the process of deleting open intervals, we have deleted an open set Gy,

with u,(Gn) = {+ +---+ w]e = {1- phe. We are left with a closed set Fr consisting of 2" disjoint closed intervals each with length [1 —{1— +}a] < 4. Note that F = [0, 1]\ G = [0, 1]

(Unen Gn)’ = 10, 19 (Mew Fn) = New Fn. Now let

be an open interval. Then8 := yz, (1) > 0. Letn € N be so large that x < B. Since F, is the union of disjoint closed intervals of length less than de I cannot be contained in F,.

Then J cannot be contained in F = (),cx Fn-

[VL2] Cantor-Lebesgue Function

The open set G, defined in the construction of the Cantor set, is the union of the disjoint open intervals: 11,1; 2,1, 2,2; 13,1, 2, 6,3, Ba; -. +5 Teas +++ 5 Ig,ae-15 «., With length 2, j) =

84 Measurable Functions

91

1/3* for j =1,...,2*-! andk € N. Letus define a real-valued function t on G by setting

to(x) =

No Ql Nal ys

age gee

forx in I,,1,

*

vee

OE

forx in 1,

4,2 respectively,

forxin bi, 52, 5,3, 13,4 respectively, forx in Ii,1,..., Tg,2¢-1 respectively,

Thus defined, z is an increasing function on G. If x’ and x” are two points in G and if the distance between the two is less than ® then the difference between +(x’) and r(x”) does not exceed ¥ Thus for every ¢ > 0, ifk € Nis so large that 4 |to(x’) — to(x”)| < x 00

increasing on [0, 1].

To show that t(0) = 0, consider the following sequence of intervals in the construction

of T:

hy

=

G. 2), bi

=

(d. 3). Bai

be the midpoint of i,1, that is, x,

=

ae

definition of t on G, we have t(xx) = *

70) =t( lim xg) = k->00

=

(#:

fork

%), woody € N.

Then

=

(#. 3).

tim, xk = 0.

see

Let xy

From our

Then by the continuity of t at x = 0, we have

lim r(xg) = lim x = 0. We show similarly that r(1) = 1. um

k->00

k->00

We use the Cantor ternary set and the Cantor-Lebesgue function to construct : (a) a strictly increasing continuous function transforming a null set with Lebesgue measure 0 into a set with positive Lebesgue measure, (Proposition 4.37), (b) a continuous function transforming a Lebesgue measurable set into a non-Lebesgue measurable set, (Proposition 4.38),

92 (©)

CHAPTER 1 Measure Spaces a 9%,-measurable function f and a Mt, -measurable set E such that f~1(E) is not

Mt, -measurable, (Proposition 4.39), (d) a Lebesgue measurable set which is not Borel measurable, (Proposition 4.40).

Proposition 4.35. Let X be an arbitrary topological space and Y be a Hausdorff space. Let f be a continuous mapping of a compact set K in X into Y. If f is a one-to-one mapping, then f is a homeomorphism of K to f(K). Proof. The compactness of K and the continuity of f imply that f(K) is a compact set in Y. Consider the relative topology on K derived from the topology of X and the relative topology on f(K) derived from the topology of Y. If C is a closed set in K then C isa compact set in K so that f(C) is a compact set in f(K).

Since Y is a Hausdorff space,

F(K) is a Hausdorff space and this implies that the compact set f(C) in f(K) is a closed

setin f(K). Thus f(C) is a closed set in f(K) for every closed set C in K. This implies

that f(V) is an open set in f(K) for every open set V in K. Now since f is a one-to-one mapping, an inverse mapping g exists. For every open set V in K, g~1(V) = f(V) which is an open set in f(K). This shows that g is a continuous mapping of f(X) to K. Therefore f is ahomeomorphism of K to f(K).

Lemma 4.36. Let @ = t +2 where t is the Cantor-Lebesgue function on [0, 1] and: is the identity mapping of [0, 1], that is, (x) = x for x € [0,1]. Then @ is a homeomorphism of (0, 1] onto [0, 2]. Furthermore both y and its inverse function w are strictly increasing functions on their respective domains of definition. Proof. Since both r and « are real-valued, continuous, and increasing on [0, 1], so is g =t+u. Since 1(0) = 0, 7(1) = 1, £0) = O, and i(1) = 1, we have gO) = 0

and g(1) = 2. By the Intermediate Value Theorem for continuous functions, for every

a € (0, 2) there existsx € (0, 1) such that g(x) = a. Thus ¢([0, 1]) = [0, 2].

Since t is increasing and : is strictly increasing on [0, 1], 9 = + ++ is strictly increasing on [0,1]. Thus g maps [0, 1] one-to-one onto [0,2]. Then by Proposition 4.35, g is a homeomorphism of [0, 1] to [0, 2]. The inverse function y of g is defined on [0, 2] and maps [0, 2] one-to-one onto [0, 1]. Since ¢ is increasing on [0, 1], ¥ is increasing on [0, 2].

Since yy is one-to-one, it is strictly increasing.

Proposition 4.37. There exists a strictly increasing continuous function f on [0,1] and a compact set K Cc

[0, 1] such that K is a null set in (R, Bp, 2,) and p, (f(K)) =1

Proof. We show that the function g = t +: in Lemma 4.36 and the Cantor ternary set T are such function and set. Let G be the open set defined in the construction of T. G is the union of countably many disjoint open intervals. Let (J, : 2 € N) be an enumeration of the open intervals constituting G, Let c, be the value of the Cantor-Lebesgue function on J, forn

€ N.

Then g(J,)

=

(© +7)(n)

=

Jn t+ cn, the c,-translate of J,.

Thus

0G) = 9 (nen Jn) = Unew(Jn +n) € BR. Form # n, if the open interval J, is to the left of the open interval J,,, then cy, < cm since 7 is an increasing function. Thus J, + cp

and Jm + Cm temain disjoint. Then by the countable additivity of 42, and by the translation

84 Measurable Functions

93

invariance of (R, Dt, , 4,), we have

#1, (G)) = Dt,Ga + en) = nen

> a, Jn) = wy, (G) = 1.

neN

Now

e(T) = 9((0, 11\ G) = g([0, 11) \ eG) = [0, 2] \ eG) € Br and hence jz, (e(T)) = 14, ([0, 21) — 2, (@(G)) = 2 — 1 = 1. Thus for the Cantor ternary

set T, which is a compact setin R and anull setin (R, Bp, j2,) and for the strictly increasing

continuous function

on [0, 1], we have #,(9(T)) =1.

Proposition 4.38. There exist a set E € 90, and a continuous function f on an interval containing E such that f (E) ¢ ,.

Proof. Let T be the Cantor ternary set and let the function g and y be as in Lemma 4.36.

‘We showed in the Proof of Proposition 4.37 that p(T) € SR and pu, (e(T)) > 0. Then by

Theorem 3.45, there existsaset A C g(T) suchthatA ¢ SJt,. Now w(A) C ¥(¢(T)) =T.

Since T is a null set in the complete measure space (R, Mt, , j2,), its subset ¥-(A) is a null set in (R, Mt,, 4,). Let E = (A) and f = g. Then E C [0,1], E € Mt,, fisa continuous function on [0, 1] and f(E) = @oy)(A)=AEMI,.

u

If f is a real-valued 99, -measurable function on a set D € Mt,, then f isa Mt, /Brmeasurable mapping by (a) of Theorem 4.6 so that f 1B) € Mt, for every B c Br.

However for E € 9%,, we may not have f—!(E) € 9,. An example is given in the next Proposition. Proposition 4.39. There exists a real-valued IN, -measurable function f for which we have

f1E) ¢ Mt, for some E = M,.

Proof. Let T, A, gy and y be as in Proposition 4.38. We have A C 9(T), A ¢ Dt, and (A)

€ Mt,.

Now

y is a real-valued continuous function on [0, 2] and hence a St, -

measurable function on [0, 2]. Since y is a one-to-one mapping, we have y—! (¥(A)) = A ¢ OM,. Let f = w and E = y(A).

Then f is a real-valued St, -measurable function

on [0, 2], E € Mt, and f-'(Z) =p" (w(4)) =A EM.

Proposition 4.40. There exists E € Wt, such that E ¢ Br. Proof. Let T, A, g and % be as in Proposition 4.38. We have A C 9(T), A ¢ Mt, and ¥(A) € 9,. Let us show that (A) ¢ 3p. Now since g is a homeomorphism of [0, 1] onto [0, 2], for every S_-measurable subset B of [0, 1] we have y(B)

Proposition 4.32. Thus if (A)

€ BR

by

€ Sp, then g(y(A)) € Br. But y(y(A)) = A ¢ M,

and consequently y{y-(A)) ¢ 3p. This shows that y(A) ¢ Bp.

94

CHAPTER 1 Measure Spaces

Problems Prob. 4.1. Given a measurable space (X, 2). Let f be an extended real-valued function on aset D € &. Let Q be the collection of all rational numbers. (a) Show that if {x ¢ D: f(x)

n={

1 for (x, y) € (0, 2) x ©, 1) withx = y eZ,

0

otherwise on (0, 1) x (0, 1).

96

CHAPTER 1 Measure Spaces

For eachy € (0, 1), define afunction f, on (0, 1) by setting f(x) = F(x, y) forx € (0, 1). Consider the uncountable collection of functions {f, : y € (0, 1)}. For each y € E,

fyplx) =

y=

1

whenx=y,

10 whenx € 0, 1)\ {yh

and for eachy € (0, 1) \ EZ, we have

fy(x) = 0 when x € (0, 1). Thus f, is a 9)t,-measurable function on (0, 1) for each y € (0, 1). But

suPyc0,0 Sx) =

1

whenxe &£,

9 whenx € (0, 1)\E.

Since E ¢ 01,, the function sup,y0 Prob. 4.20. Let F be as in Prob. 4.18. Show that if yo = c or d, the subset of D on which the function lim F(., y) exists is a 29, -measurable set and the function lim F(-, y) is

y>90 ®t, -measurable on the subset.

y>yo

Prob. 4.21. Let f be a real-valued function on R. Consider the derivative of f atx defined by

€¢ R

(Df) = jim £G+-F@) if the limit exists inR are

(We say that f is differentiable at x € R only if (Df)(x) € R.)

Show that if f is continuous on R then D(Df)

D(Df).

€ %3p and Df is Sg-measurable on

Prob. 4.22. Let F be a real-valued continuous function on [a, b] x [c, d]. Show that the function f defined on [a, b] by f(x) = fi F(x, y) dy for x € [a, b] is a BR-measurable function on [a, 5].

Prob. 4.23. Let (X, &, 2) be a measure space and let f be an extended real-valued 2-

84 Measurable Functions

97

measurable function on X. For A € R, let E, = {x € X: f(x) > A} and define a function gy on R

by setting g(a) = u(E,) fora € R.

(a) Show that ¢ is a nonnegative extended real-valued decreasing function on R. (b) Show that if ~(X) = M > 0 then g is bounded above by M on R.

(c) Show that if p(49) < oo for some Ao € R, then ¢ is right-continuous on [Ap, 00). (d) Show that if g(Ag) < oo for some Ag € R, there may be s € (so, 00) at which ¢ is not left-continuous. Prob. 4.24. Let (X, U, jz) be a measure space and let f be an extended real-valued Ameasurable function on X. For A € R, let F, = {x € X : f(x) > A} and define a function

y on Rby setting 4(A) = #(F;) fords € RB. (a) Show that y is a nonnegative extended real-valued decreasing function on R.

(b) Show that if 4(X) = M > 0 then y is bounded above by M on R. (c) Show that if ¥(A9) < co for some Ag € R, then y is left-continuous on (Ag, 00). (d) Show that if ¥(Ao) < 00 for some Ao € R, there may be s € [s9, 00) at which w is not

right-continuous.

Prob. 4.25. Let (X, 2, 4) be a measure space. Let (f, : 2 € N) be an increasing sequence

of extended real-valued %-measurable functions on X and let f = AER, let £, = lim E, = E and

n00

{X: f, > A} forn lim p(£,) = h(E).

€ N and let E

=

{X

Aim, fa on X.

: f

>

For

A}. Show that

n00

Prob. 4.26. Let (X, 2, 2) be a measure space. Let (f, : n € N) be an increasing sequence of extended real-valued {{-measurable functions on X and let f = lim | fn on X. For n AER, let F, = {X: fy > A} form € N and let F = {X : f > A}. Construct (f, :m € N) such that

lim

n00

Prob. 4.27.

F, ~ F and

lim p(F,) < u(F).

n00

Let us call a sequence (a, : n € N) of extended real numbers eventually

monotone if there exists mo

increasing or decreasing.

€ N such that (a,

: n

>

no) is monotone, that is, either

Let (X, SA, 42) be a measure space. Let (f, : n € N) be a sequence of extended real-valued M-measurable functions on X such that (f,(x) : # € N) is eventually monotone for every

xeEX.

Letf = lim. fn on X. Fora € Rlet E, = {X: f, > a} form € N and let nt

E={X:f >a}. (a) Show that in, E, = E.

(b) Show that if (X, 2, jz) is a finite measure space then lim n-

(En) = w(E).

Prob. 4.28. Let (jf, : n € N) and f be extended real-valued functions defined on a set D such that Jim, Sa(x) = f(x) for every x € D. (a) Fora ER, lett FE = {xe D: f(x) > a}and EF, = {x CE: Show that jim, E,, exists and moreover iim, E,=E., (b) Fora € R, let EF = {x € D: f(x) a} forne N. f,)

a}

n

(1)

w{D:

a}.

Q

w{D: f MW. Therefore A= A. w Let (X, &, 2) be the completion of a measure space (X, A, 14). Since A Cc A, a Ameasurable function f onaset D € 2 C MA may not be A-measurable on D. The following theorem shows that there exists a null set N in (X, &, 4) such that f is 2-measurable on

D\N.

Theorem 5.6. Let (x

A, B) be the completion of a measure space (X, X, 2). Let f be an

extended real-valued M-measurable function on a set D € M C A. Then there exist a null

set N in (X, A, ) and an extended real-valued A-measurable function g on D such that

f =g on D\N. (In other words, if f is an extended real-valued U-measurable function

ona set D € &, then there exists a null set N in (X, A, 2) such that f is A-measurable on

D\N.)

Proof.

Let {r, : » € N} be the collection of all rational numbers.

For every n € N,

let E, = {D: f < ra}. By the &-measurability off on D, we have E, € A. Then

E, = A, UC, where A, € & and C,, is a subset ofa null set B, in (X, A, w). If we let N = Unen Bn, then N is a null set in (X, &, w). Define an extended real-valued function

102

CHAPTER 1 Measure Spaces

g on D

by setting

a) =

f(x)

forxe D\N,

0

forx € N.

It remains to show that g is {-measurable on D. Now for every n € N, we have

qd)

{D:g0on D, then f, ydy © [0, oo].

4° Ify 0on D, then f, gdu = 0 if and only if u{D : g £0} =0. 8 Ife >0omD, EC D,andE € &, then f, gdp < fygvdu. Proof. 1. 1°, 2°, 3°, 4°, and 5° are immediate from Definition 7.4. 2. Let us prove 6°. Let ¢ = )77_, aj1p, be the canonical representation of g. Case 1. Consider the case that a; # Ofori = 1,...,”. In this case we have

{D:g9 #0} =D.

qa)

Thus if p{D : 9 # 0} < oo then 4(D) < 00 and then f, 9 du € R by 5°, that is, g is #-integrable on D.

Conversely if g is -integrable on D then f, ¢du = S°7_, aj4(D;) € R. This implies

that ~(D;) < co fori = 1,..., and then u(D) = wh (D;) < oo. Then by (1) we have p{D : p #0} < co. Case 2. Consider the case that one of {a1, ..., dn} is equal to 0, say a1 = 0. In this case we have

Q)

{D: 9 #0} = D2 U---UDp.

Note that a, = 0 implies a4(D,)

on the product 0 - 00. Thus if u{D

: g #

0}


0 on D implies that a; > Ofori = 1,...,. LetE C D and £& ¢€ 2. Then the restriction of y to E is a simple function on E and its canonical representation is given by wha @;lp,ng. Then since p(D; 0 E) < 2(D;) and a; > 0, we

have

[odu = Dd aim(Dr NE)< Dam(Di) = [oan This proves 8°. Observation 7.6. Let g be a simple function on a set D € Ql in a measure space (X, 2, jz). Let {Ei, ..., Ey} be a disjoint collection in & such that Ui E; = D. Theng is a simple function on £; forj = 1,..., k and if g is 4 semi-integrable on D then so is g on E; for

j=1,...,kand furthermore fy ody = Diy Se, 94h.

Proof. Since E; 0 on D, then gy + 2 is jz semi-integrable on D and moreover

87 Bounded Function on Set of Finite Measure

135

Spolgitmidu = fyoidut fypmdu.

(g) If at least one of 1 and ¢ is p-integrable on D, then @ +

2 is

semi-integrable on

D and moreover we have fy{gi + gr} du = fygidut fy grdu. In particular, if both g and gz are -integrable on D, then so isp) + 92.

(hb) FD, Dz € A, Di ND2 = Band D, UD2 = D, then fy gdp = fy, gdutSp, 9du. Proof. 1. To prove (a), suppose g1 = ¢2 on D \ E where E is a subset of D which is a

null set in (X, 2, 2). By Observation 7.6, we have fy gi du = Jove gidut froidu and fy gidu = Joyg 24h + frg2dp. On D\ E we have y = ¢ so that we have Joye vide = Joye 2 dh.

Since u(E) = 0, we have f, gi du = 0 and f,

1° of Observation 7.5. Thus we have fy gidu = fy gdp.

du

= 0 by

2. (b), (c), (d), and (e) are immediate from Definition 7.4.

3. To prove (f), let g, = 7%, aj1z, andg = jai bj1p, be the canonical representations of g, and g2 respectively. If we let G;,; = E;M F;, then we have a disjoint collection

(Gij:#=1,...,m5 f= 1,...,n} in A with UP, UP) Gj = D. We have

[eran = Damen = Dal Pues] = anes) i=1

i=1

j=l

i=1 j=l

and similarly [nau

n = Yaw) j=l

” = ba] j=l

m Due. “i=l

min = YY 4j4Gi,))i=1 j=l

On the other hand since g + g2 assumes the value a; + b; on the set G;,; fori =1,...,m and j = 1,...,, we have by the Comment made after Definition 7.4

[ {eit ebde => YG +5) uG,;) i=l j=l

= VarvG.)+ Yb i=1 j=l

=|

D

odut

4, To prove (g), let y1 = 7

i=1 j=l

eG.)

f gdp. D

a1z, and gy, = °j_1 bj1p, be the canonical repre-

sentations of y; and g2 respectively. Let G;,; = Ej 1 F; as in 3. We have the equalities

Sp vide = Yih D5a1

a (Gis) and fy pode = SHE 1 P71 bju(G;,;). Let us show

that if at least one of gy and ¢g» is j4-integrable on D, then the simple function gy; + g2 is 2 semi-integrable on D. Let us assume that g is -integrable on D. Since g; + g assumes the value a; + b; on G;,;, to show that 1 + ¢2 is 4 semi-integrable on D we

show that 7721 )ja1(@ + b;)(Gi,;) exists in R.

Suppose among the m x n sum-

mands in }7y2y DVja1 Gi + bj) (Gi,;), we have (a; + b;)u(Gi,;) = 00 for some (i, j) and (a; + be)u(Gz,2)

=

—oo for some (k, £).

Now

since g. is -integrable on D and

136

CHAPTER 2 The Lebesgue Integral

thus b;2(G;,;) and beu(G;,2) are finite, we have a; 2(G;,;) = 00 and azp(Gi,2) = —00 and then a;4(E;) = co and ag(Ex) = —oo. This is impossible since gy is jz semiintegrable on D and ia a; 4(E;) exists in R. This shows that it is impossible that one

of the m x nm summands assumes the value oo and another the value —oo. an Lia

Therefore

+ b;)u(Gi,;) exists in R.

5. (h) is a particular case of Observation 7.6.

[11] Integration of Bounded Functions on Sets of Finite Measure Given a measure space (X, U, jz). Let D be a 2-measurable set with 4(D) < oo. Then every simple function on D is js-integrable on D by 5° of Observation 7.5. Let f be a bounded real-valued function on D, say | f(x)| < M for x € D for some constant M > 0. Let g and ¥ be simple functions on D such that g(x) < f(x) and f(x) < W(x) forx € D. (Such simple functions always exist. For instance g(x) = —M and (x) = M forx € D

will do.) Now sinceg < ¢ on D, we have f, gdu < fp v du by (b) of Lemma 7.7.

Observation 7.8. Let f be a bounded real-valued function on aset D € 2 with 4(D) < co in a measure space (X, A, 2). Suppose f(x) € [M1, M2] for some constants M;, Mz € R

such that M, < M2. Then we have

Miu(D) < o0}=|J{D:h-g > }} keN

so that

WMD: g #h} }}=0

then z{D

for every k € N,

: g # h} = 0 andg = hae. on D.

To prove (3), letk € N be fixed.

Since

On S85 f Sh < Wy forn EN, wehave{D:h—g > }}Cc{D:¥n—o,> }} and then u{D :h-—g > 3}

1

1)

pH{Dih-s2;}

1

1

}}=0 and then multiplying byk we have u{D :h — g > }} = 0. This proves (3). If f is a bounded real-valued 2{-measurable function on a set D € A with w(D)

< co

in a measure space (X, 2, 14), then supper fy gy du = inf sey fy ¥ du € R by Theorem 7.9 and Observation 7.8. This is the basis of the following definition. Definition 7.10.

Given a measure space (X, A, 4).

Let f be a bounded real-valued A-

measurable function on a set D € A with 4(D) < 00. We define the Lebesgue integral of Ff on D with respect to u by

ffD feomas) = opesf f JD oau= fev int [ Jp yaw er. In other words, with the callection ® of all simple functions g on D such that p < f we let

[ F(e)u(ds) = sup{ I gdp:ge a}. We say that f is Lebesgue integrable on D with respect to js or simply jz-integrable on D.

87 Bounded Function on Set of Finite Measure

139

If D € & with u(D) < oo then a simple function f defined on D is also a bounded real-valued 21-measurable function on a set with finite measure so that f,, f du is defined by both Definitions 7.4 and 7.10. Thus we need to verify that the two definitions of f, f du

are consistent for such a function. Let us write J(f) for the integral of f on D in Definition

7.10, reserving the notation f,, f dy. for the integral of a simple function f as in Definition

7.4, Thus I(f) = sup{ f, du: y € &}. Now if f is itself a simple function on D, then f € © and moreover for every p € © we have g < f so that fp gdu < fp f du by (b) of Lemma 7.7. Thus I(f) = sup { fy ¢du: 9 € ®} = J, f du. This shows the consistency of Definitions 7.10 and 7.4.

Lemma 7.11. Given a measure space (X, A, 4). Let f; and fo be bounded real-valued %&M-measurable functions on a set D € A with u(D) < oo. If f, = fo ae. on D, then

Sy fidu =p frodu.

Proof. For i = 1 and 2, let ; be the collection of all simple functions g on D such that Gi < fy. By Definition 7.10, we have

qd)

[fdu=on{ [dui

Let us show that corresponding to every v1 €

eo}.

there exists g €

2 such that f, gi du =

to gadu. Now since f; = fo ae. on D, there exists a null set Dg in (X, A, ~) such that Do C Dand fi = fo on D\ Do. Since f| and f2 are bounded on D, there exists M > 0 such that fi(x), fo(x) € [—-M, M] for x € D. Let us define a simple function gz on D by

setting g(x) = gi(x) forx € D \ Do and g(x) = —M forx € Dp. Then g so that g2 € 2.

By (h) of Lemma

[ nau= f D

=

D\Do

D\Do

radu

f

adu= f

vdut

f

edu=f

Do

Do

Thus corresponding to every g, € © there exists g, € Then the collection of real numbers { to gidu:

g

collection { f, y2 du : go € 2} and therefore we have @

< f, on D

7.7 and 1° of Observation 7.5, we have

D\Do

D

gidu+0

ody.

2 such that fy gidu = fy grdu.

€ o,} is a subcollection of the

sup{ f ordu:ore oi} 0. Now

GB)

[erau-

sup f edu =c sup {2 f eau}

D

oxcf =c

JD

sup

osef

1

exef

—gdz=c

sf

JD

JD

sup

JDC

= cow f edu=c f

LC

glesf

D

1

—gpdp

IDE

f dp,

where the third equality is by (e) of Lemma 7.7. This proves (1) for the case ¢ > 0. When c¢ = —1, we have

@4)4)

pe —fldu= ae

du=— vit, inf }wp f edu { |[eauod

gs-f

=— fs-eJp inf [odu=-intPe)f oan =—

fan,

where the third equality is by (e) of Lemma 7.7 and the last equality is by Definition 7.10. This proves (1) for the case ¢ = —1. When c < 0, recalling the results above for the cases ¢=—landc > 0, we have

©

ferdu=f (-te)sau=-f tirau=—tei f tau=e f rau.

This proves (1) for the case c < 0. With (3), (4), and (5), we have (1).

2. To prove (2), let g; and gz be simple functions on D such that g; < f; and g2 < fo. We have 91 + 92 < fi + f2 on D. By (g) of Lemma 7.7, we have

| adn

f

D

D

godu= [ {yitg}du< D

sup

f gdp,

9sfithsD

since gy + ¢2 is one simple function on D such that g; + 2 < fi + fo. Then we have

f gidu+ D

sup | @du


inf [ du, Athy De a

that is,

O)

[nau f frdu> fit fdu. D

D

D

By (6) and (7), we have (2).

Lemma 7.13. Given a measure space (X, A, 2). Let f and g be bounded real-valued &M-measurable functions on a set D € A with wD) < oo. Then

a

fsemD=f sans f dy.

If |f| < M on D where M

> 0, then

If D raul < f D [fldu }} keN

and then

0< w(D1) j)keN

Thus there exist ky € N such that u{D : f > 3} > 0. Let us define a simple functiong

on D by setting

1

9%)=

—

forxe{D: f > gh,

0

forxe D\{D: f > x}.

Then yg < f on D. Thus we have

[rane

fodu=fufpire

p}>o.

This contradicts the assumption that tb f du = 0. Therefore we have f = 0 ae. on D. 3. Consider the general case that f is a bounded real-valued %-measurable function on aset D € A with w(D) < oo, f > Oae. on Dand fy f du = 0. Let us show that f =0 ae. on D. Since f > 0 ae. on D, there exists a null set E in (X, A, 42) such that E c

f = 0on D\ E. Then by (h) of Lemma 7.7 and 1° of Observation 7.5 we have

o=frau=fisdu+f) tana ff saw.

D and

87 Bounded Function on Set of Finite Measure

143

Thus we have f > 0on D\ Eand fry, fdu=0. This implies that f = 0 a.c.on D\ E according to 2. Thus there exists a null set F in (X, 2, %) such that F C D\ E and f =0 on (D\ £)\ F = D\ (EU F). As the union of two null sets, E U F is a null set in (X, A, wz). Thus f = 0 ac. on D. This completes the proof of (a).

4. To prove (b), note that if f < g ae. on D then g — f > Oae. on D. If in addition

we have f, fdu = fy gdu, then f,{g— f}du = 0 by Lemma 7.12. Then by (a), we

have g — f = Oa. on D, thatis, f =gae.onD. Lemma 7.15.

Given a measure space (X, 2%, 4).

measurable function on a set D

€ & with w(D)

©

Let f be a bounded real-valued A< oo.

Let (Dy, : n € N) be a disjoint

sequence in Xi such that ex Dn = D. Then fy fd = Vaen tp, f au.

Proof. Let ¢ be an arbitrary simple function on D such that y < f. Lety = )7?_, aj1z, be its canonical representation. Let gy, be the simple function on D, which is the restriction of @ to Dy. Its canonical representation is given by g, = ye a;1z,np,- Note that for each i=1,..., p,{E; 1D,

in € N} is a

disjoint collection in & with Unen(Ei 0 Da) = Ei

so that 2(E;) = doen HCE; 1 Dy). Now

[edu = awe) =a i=1

ace: .D»)

i=1

[E« BEN vo} = Lf néN

s

xf

Li=1

neN

nds

fd,

where the third equality is by the fact that if }>, 1 1,n, ---: ) neN @p,n are convergent series

of real numbers then ) ,en {On +--- +p, T= Ynen On +--+ Donen %p,n, and the last inequality is from the fact that 9, vis just one simple function on D, such that g, < f

so that fr

Ondu < UPp 0 such that

|fn(x)| < M for all x € Dandn EN. Since f is also bounded on D, we assumeM > 0 has been so chosen that | f(x)| < M for all x € D. Now since (f, : 2 € N) converges to f a.e. on D and since 4(D) < 00, according to Theorem 6.12 (Egoroff), for every n > 0 there exists a &-measurable subsetE of D with (EZ) < 4 such that (f, : n € N) converges to f uniformly on D\ E, that is, for every ¢ > 0 there exists N € N such that | f,(x) — f(x)| N. Then forn > N, we have by Lemma 7.15

[iii sian= fl te flan fife Fld N, we have limsup f;, |f, — f|du < eu(D) + 2Mn. Then since this holds for every n > 0 and¢ > 0, we have lim sup tb lf — fldu = 0. Now Sp lfn — fldp > 0 for every n € N so that lim inf tb if,

f\du > 0. Then we have

o0o

we have lima So lfn — fap =. This proves (1). To derive (2) from (1), note that by Lemma 7.12 and Lemma 7.13, we have

@)

[ff man—f rau|=|f im saul < fim sian. D

D

D

D

Applying (1) to the last member of (3), we have lim \fp fpdu— fy fdu| =0. Now for a sequence (a, : n € N) of real numbers

jim, |a,| = 0 implies jim, da = 0. Thus we

have lim, {fp fr du — fp f du} = 0 and then jim, Ipfrdu=fyfdu.

a

87 Bounded Function on Set of Finite Measure

145

Given a sequence of extended real numbers (a, : n € N). If there exists a real number

a such that every subsequence (a,, : k € N) has a subsequence (am, :Le N) converging to a, then the sequence (a, : m € N) converges to a. (This can be proved as follows. Suppose that the sequence (a, : n € N) does not converge to a. Then for some ¢9 > 0 we have a, ¢ (a — £9, a + &9) for infintely many n ¢ N and thus there exists a subsequence (Gn, : k € N) such that a,, ¢ (@ — £0,4 + £9) for every kK € N. This subsequence

(an, : & € N) cannot have any subsequence (aq,, : £ € N) converging to a. This contradicts the assumption that every subsequence (a, : k € N) has a subsequence (a,,, : £ € N)

converging to a. Therefore (a, : n € N) must converge to a.) Using this property of a sequence of real numbers, we can replace the convergence a.e. condition on the sequence of functions in Theorem 7.16 by the weaker condition of convergence in measure. Recall that since 4(D) < oo, convergence a.e. on D implies convergence on D in measure according to Theorem 6.22 (Lebesgue). Corollary 7.17. (Bounded Convergence Theorem under Convergence in Measure) if the condition that (f, : n € N) converges to f a.e. on D in Theorem 7.16 is replaced by the weaker condition that (f, : n € N) converges to f on D in measure, the conclusions (1) and (2) are still valid.

Proof. Assume that (jf, : » € N) converges to f on D in measure. We are to show that im, fo lfn — f| dy = 0. Consider the sequence of real numbers (a, : n € N) defined by a, = fol

— f|dy for n

€ N. Thus we are to show that lim Gn = 0. Take an arbitrary

a

subsequence (ap, : k € N). Consider the sequence of functions (f,, : k € N). Since (f; :

n €N) converges to f on D in measure, the subsequence (f,, : k € N) converges to f on D in measure too. Then by Theorem 6.23 (Riesz), there exists a subsequence ( Sz, 1 8 € N) which converges to f a.e.on D. Thus by Theorem 7.16, we have jim tb fog, —Fl du = 0, that is, the subsequence (am,

OO

fe N) of the arbitrary subsequence (a,, : k € N) of

(a, : n € N) converges to 0. Therefore the sequence (a, : n € N) converges to 0, that is,

dim Joli - fldu =0. 0

[111] Riemann Integrability Let J = [a,b] CR. Let P = {xo,..., xn} wherea = x9 < --- < X_ = b and consider

a partition of J into subintervals J, = [xg—1, x4] fork = 1,...,n. We call xo, ..., x, the

partition points in P. The mesh of P is defined by |P| = max{€() :k = 1,...,n}. LetPB be the collection of all partitions of J. If P, P2 € 98 and P; C Po, then we say that P2 is

arefinement of P;.

Definition 7.18. Let f be a bounded real-valued function on I. For P € 38 given by P = [x0,...,%n}, let Ie = [xe_1, x4] and& € hy be arbitrarily chosen fork =1,...,n. The Riemann sum of f corresponding to P and the selection § = (& :k =1,...,n) is

146

CHAPTER 2 The Lebesgue Integral

defined by

S(F9,6) => FEE). k=1

We say that f is Riemann integrable on I if there exists J € R such that for every e > 0 there exists 6 > 0 such that

|S(4P,&) —J| 0. With P

9B givenby P = {xo,..., xn}, let Te = [xe-1, xe],

my = infzer, f(x) and My = sup, cy, f(x) fork =1,...,n. The lower and upper Darboux

sums of f corresponding to a partition P are defined by

(ly

S(f,P) =

> mL)

k=1

and S(f,P) = >> Mee). k=1

Note that

(2)

—ME(D < S(f,P) < S(f,P) < Me),

and

(3)

Py C Po => S(f, Pi) < S(F, Pa) and S(f, P1) = S(f, Pr).

Let

a

s(f)= se S(F.P) and S(f) = jnf S(F.9).

We call S| ( f )and s( Ff ) the lower and upper Darboux integrals of f on I. Clearly

6)

—MLD < S(f) < S(f) < Mew).

Lemma 7.20, Let f be a bounded real-valued function on I = [a, b]. Then S(f) = S(f) if and only if for every & > 0 there exists Po € YB such that

@

S(f, Po) — S(f, Po) 0 there exists Pp € $Y such that (1) holds. Then

5(f) = jit, 3(F. 9) < 5(f, 70) = 5(F Po) - S(V 0) + 5(F. 7) 0, we have 5(f) < S(f). On the other hand, S(f) > S(f) for any bounded real-valued function f on I. Therefore 5(f) = S(f). 2. Conversely suppose S(f) = S(f). Since S(f) = suppeg S(f, 9), for every s > 0 there exists P, € $8 such that

S(f, P1) > S(F) — §, and similarly there exists Pz € $8 such that

S(f. P2) < S(f) + 5. If we let Po = P) UP2 € P, then S(f, Pi) < S(f, Po) and S(f, Pa) > S(f, Po) by (3) in Definition 7.19 so that

S(F, Po) — S(f Po) < S(F, Pa) — SCF, Pi) < (S(4) + 5} - (S(F) - 5} =

0

Proposition 7.21. Let f be a bounded real-valued function on I = [a, b]. Then we have

S(f) = S(f) ifand only if for every « > 0 there exists8 > 0 such that a

S(f,P)-S(4,P) 0 there exists § > 0 such that (1) holds. Take any Pp € 3

such that |Po| < 8. Then S(f, Po) — S(f, Po) < e so that S(f) = S(f) by Lemma 7.20. 2. Conversely suppose S(f) = S(f). Let ¢ > 0 be arbitrarily given. Then by Lemma

7.20, there exists P* € 98 such that

Q)

S(f, P*) — SUF, P") < 8.

Let P* = {ao,..., an} where a = ap < ++» < ay = b. Since f is bounded on I, there existsM > O such that | f(x)| < M forx € I. Let5 = gz. To verify that our 6 satisfies (1), let P € 9 be such that |P| < 5. Let P = {xo,...,.x,} where a = xg {S(F, Px) — SCF Pe-1)} k=l

0 be so small that U(xg, 8) C ISim,k(m)" Then

©

F*@%o)= lim sup f< 80 y049,8)

sup

UGo.dm)

f
0 such that

©

f*(%o) S_

Since

sup f < f*(@o) +8. UGa.dnI

jim, |Px|= 0 and since x9 € I?im, k(n) for some k(m)

€ N for every m € N, there

exists Mt: > 0 such that form > M we have (7)

Inkem) C U (xg, 8) NT.

By (5), (7), and (6), we have for m > M

F* (x0) < Vn(%o) = sup fs Jr.)

sup

U(@.8)NE

f < f*(xo) +e.

This proves (4). Being a countable set, E € Sy and then J \ E € Bp. Since y», is a BR-measurable function on [, it is 23n-measurable on J \ E by (a) of Lemma 4.7. Then (4) implies that

f* is a Bp-measurable function on I \ E by Theorem 4.22. Since E is a countable set, any extended real-valued function on E is Sp-measurable on E. (If g is an extended realvalued function on a countable set £ in R, then for any w € R the set {x € E: g(x) 0. Then by (3), we have |%_(x)| < 2M for x € J and m e€ N. According to (4), we have

lim m= moo

f* on J \ E, that is, a.e. on Z, Thus by Theorem 7.16 (Bounded Convergence

Theorem), we have

0°)

. sim, [ vmau, _= f f° * dn,.

With (8) and (9) we have f, f*du, = S(f). The equality f, f,du, = S(f) is proved similarly.

Theorem 7.27.

Let f be a bounded real-valued function on I = [a, b]. If f is Riemann

integrable on I, then f is INt,-measurable and Lebesgue integrable on I with respect to

My, and moreover f° f@)dx = f, f du,. Proof.

If f is Riemann integrable on J, then the lower and upper Darboux integrals of

f on J are equal, that is, S(f) = S(f), by Theorem 7.22. Then by Lemma 7.26, we have f, fadu, = f, f*du,. Since f, and f* are Bg-measurable on J, they are Mt,-

measurable on I. Since f, < f* on I, the equality of the Lebesgue integrals of f, and f* implies that f, = f*, (t,, 4,)-a.e. on I by (b) of Lemma 7.14. Then since f, < f < f* on I by Observation 7.25, we have f = f, = f*, (OM, w,)-a.e. on J. The completeness

of the measure space (R, Vt, , z,) then implies that f is $9t, -measurable on J according to (b) of Observation 4.20, Now that f is Jt, measurable on J andf = f*, (Dt, #,)-a.c.0n

I, wehavef, f du, = f, f* du, by Lemma 7.11. But f, f*du, = S(f) = C f@)dx

by Lemma 7.26 and Theorem 7.22. Thus f? f(x)dx =f, fdu,.

Theorem 7.28. Let f be a bounded real-valued function on I = [a, b] and let E be the set of all points of discontinuity of f in I. Then

Ff is Riemann integrable on I

Of. = f*, (MN, u,)-ae. onl

u,(E) =0. Proof.

1.

If f is Riemann integrable on 7, then as we saw in the Proof of Theorem

7.27, fx = f*, (Ot, #,)-ae. on I.

Conversely if f. = f*, (,, w,)-a.e. on J, then

Sy fxdu, = f, f* du, by Lemma 7.11. Then we have S(f) = S(f) by Lemma 7.26. But this implies the Riemann integrability of f on J by Theorem 7.22.

2. By Observation 7.25, f is continuous at xo € J if and only if f,(%0) = f*(xo). Thus

fe = f*, (OM,, 4,)-2.e. on J if and only if z,(Z) = 0.

o

154

CHAPTER 2 The Lebesgue Integral

Example. Let Q be the set of all rational numbers and P be the set of all irrational numbers in R. Let f be a bounded real-valued function on [0, 1] defined by f(x) = Oif x € [0,1] Q and f(x) = lifx € [0,1] P. The set £ of all points of discontinuity of f in [0,1]. Then E = [0,1] and %,(Z) = 1. Thus by Theorem 7.28, f is not Riemann integrable on [0, 1]. On the other hand since [0,1]N Q, [0,1] MN P € Br, f

is a simple function on (R, %$g) and its Lebesgue integral with respect to 2, is given by

Soy f du, = 0+ #, (10,119 Q) +1- 2, (10, 11M P) = 1. Problems

Prob. 7.1. Let (X, 2, 2) be a measure space. Let f be areal-valued M-measurable function on aset D € 2l with 4(D) < oo. Show that for everye > 0 there exists a set FE € 2 such that FE C D, p(E) < « and f is bounded on D \ E.

Prob. 7.2. Let f be an extended real-valued 2{-measurable function on a set D € Aina measure space (X, 2, 44). For M,, Mz € R, Mi

Mz be defined by

g@) =

M,

if f@) Mo.

< Mb, let the truncation off at M; and

£@) if f@) €[M1, Mol,

Show that g is 2l-measurable on D.

Prob. 7.3. Let f be a bounded real-valued 2{-measurable function on a set D € Aina measure space (X, 2, jz). (a) Show that there exists an increasing sequence of simple functions on D, (gy, :n € N),

such that g, + f uniformly on D.

(b) Show that there exists a decreasing sequence of simple functions on D, (w, : n € N),

such that ¥, | f uniformly on D.

Prob. 7.4. Let f be a real-valued 2-measurable function on a set D € 21 in a measure space (X, A, 44). Suppose f is not bounded on D. (a) Show that it is not possible that for an arbitrary e > 0 there exists a simple function g on D such that |o(x) — f(x)| < ¢ forallx € D.

(b) Show that there cannot be any sequence of simple function which converges to f uniformly on D. Prob. 7.5. Given a measure space (X, A, 4). Letus call a function g an elementary function if it satisfies the following conditions:

1°

2°

Dw) eA,

gis M-measurable on D(y),

3° assumes only countably many real values, that is, 9%(y) is a countable subset of R. Let f be a real-valued &%{-measurable function on a set D € A. (a) Show such that (b) Show such that

that there exists an increasing sequence of elementary functions on D, (gy, : n € N), y, + f uniformly on D. that there exists a decreasing sequence of elementary functions on D, (%, : n € N), ¥, | f uniformly on D.

87 Bounded Function on Set of Finite Measure

155

Prob. 7.6. Given a measure space (X, 2, 4). Let g be a nonnegative elementary function onaset D € A with u(D)

assumed by

< oo. Let {oj : i € N} be the set of nonnegative real values

and let D; = {x € D : y(x) = a;} fori € N. We call the expression

= Yap, ieN

the canonical representation of gy. We define

[eau = ¥ e:2(D;) € [0, oo]. ieN

Let f be a nonnegative real-valued function on D. Let ® be the collection of all nonnegative elementary functions g such that 0
s fy v du, where p

and 7% are simple functions on D. The integral of f thus defined can be computed as the limit of a sequence of integrals of simple functions. Prob. 7.9. Let f be a bounded real-valued A-measurable function on a set D € 2% with HAD) < 00 satisfying f(x) € (-M, M] forx € D for some M For each n € N, let

> 0.

Dra= {x € D: f(x) € (&-1)4,4")]} fork=—n+1,---,n

Define simple functions g, and yw, on D by setting and

Gn = Dhani - DT

dys

156

CHAPTER 2 The Lebesgue Integral

Show that

; Un = Vike nt

im, pond

=

ke “Da

.

fp f du and Jim, Ip ¥ndu

= fy f dp.

Prob. 7.10. Let f be a bounded real-valued &%-measurable function on a set D € A with BAD) < 00 satisfying f(x) € (-M, M] forx € D for some M > 0. Let P = (yo, .-., Yp) be a partition of [—M, M], that is, -M = yo < ---< yp = M. Let I, = [ye-1, yx] fork = 1,..., p and let [P| = max;—1,..., p Ck).

Let Ey = {x € D: f(x) € (ye-1, yx]} fork =1,..., p.

Let g(P) and ¥(P) be simple functions on D defined by g(P)

¥OP) = Dy Ye Ley.

=

yey Ye-1 > 1g, and

Let (P, : n € N) be a sequence of partitions such that lim [P,| = 0. n

Show that im, {pePndu = fy f du and im, Ip ¥ Pn) du = fp f dp. Prob. 7.11. Consider a sequence of functions (f, : n € N) defined on [0, 1] by setting

nx I(x) = Tame?

forx € [0, 1].

(a) Show that (f, : 2 € N) is a uniformly bounded sequence on [0, 1] and evaluate .

nx

7. Joy 1tae

(b) Show that (f, : n € N) does not converge uniformly on [0, 1]. (Recall that for the Riemann integral, uniform convergence of (f, : n € N) on [0, 1] implies

dim, Jo fae)ax = fg lim fax) dx.)

Prob. 7.12. Given a measure space (X, 2, 2). Let (f, : 2 € N) be a sequence of extended. real-valued 2{-measurable functions on a set D € &% with u(D) < oo and let f be an

extended real-valued 2{-measurable function on D that is real-valued a.e. on D. Consider the following two conditions on (f, : » € N) and f:

Pim forte an = 0 1°

(f, :n €N) converges to f in measure on D.

Show that 1° and 2° are equivalent, that is, 1° 2°. Prob. 7.13. Let (X, &, 1) be afinite measure space, that is, 4.(X) < co. Consider extended real-valued &{-measurable functions that are real-valued a.e. on X. Let us declare two such functions equal if they are equal a.e. on X. Let ® be the collection of all such functions.

“t

If al du for fee. ohe= f Foe

(a) Show that p is a metric on &. (b) Show that @ is complete with respect to the metric p.

(c) Let (f, : m € N) be a sequence in @ and let f € &. Show that lim n

pCa, f) = 0if

and only if (f, : 2 € N) converges to f in measure onD.

Prob. 7.14. Given a measure space (X, 2, 2). Let (f, : n € N) be a uniformly bounded

sequence of real-valued 2{-measurable functions and let f be a bounded real-valued 2measurable function on D € A with ~(D) < 00.

87 Bounded Function on Set of Finite Measure

157

Show thatif f, 5 f on D then im, folin—fldu =O and lim, Sy ind = fy f du. This is the Bounded Convergence Theorem under Convergence in Measure. direct proof without using the subsequence argument as in Corollary 7.17.

Construct a

Prob. 7.15. Let Q be the collection of all rational numbers in (0, co). Let each x € Q be

expressed as 2 where p and g are positive integers without common factors other than 1. Define a function f on [0, 1] by setting 1s

f@)=4?

- ifxe[0,1]N

0

|

Qandx=

ifxe[0,1\Q.

Show that f is Riemann integrable on [0, 1] and evaluate h F(x) dx.

Prob. 7.16. Let f be an extended real-valued function on R. Define two extended realvalued functions y and y on R by letting g(x)= liminf f() and ¥(x)=

tim n sup fo)

for x € R. Show that g and ¥ are 33y-measurable on R. (Note that f itself is not* assumed to be BR-measurable on R.) Prob. 7.17. Let f be an extended real-valued function on R. Define a subset of R by

={xeR: iim £0) exists in R} and define an extended real-valued function A on D by setting

AG) = lim f(y) forx € D. Show that D € Sx and A is %p-measurable on D.

Prob. 7.18. Let (X, ) be a metric space with a metric topology by the metric p. Forx € X and r > 0 the open ball with center x and radius r is defined by setting

Bix,r) = {y € X: p(x, y) 0, let D, = {x € D: w(x) > &}. show that D, is a closed subset of D in relative topology of D, that is, D, = F M D where F is aclosed set in X.

(d) Let A be the subset of D consisting of all discontinuity points of f and let B be the subset of D consisting of all continuity points of f. Show that A = F M D where F is an Fy-set in X and B = GN D where G is a G5-set in X.

Prob. 7.19.

Let Q be the set of rational numbers in R and let F be the set of irrational

numbers in R.

158

CHAPTER 2 The Lebesgue Integral

(a) Show that Q is a an F,-set and P is a G;-set. (b) Show that Q cannot be a Gs-set and P cannot be an F,-set.

Prob. 7.20. (a) Construct a real-valued function f on R such that f is discontinuous at every rational x € R and continuous at every irrational x € R. (b) Show that it is impossible for a real-valued function f on R to be continuous at every rational x € R and discontinuous at every irrational x € R.

§8 Integration of Nonnegative Functions

§8

159

Integration of Nonnegative Functions

[I] Lebesgue Integral of Nonnegative Functions Definition 8.1. Given a measure space (X, 2, 2). Let f be a nonnegative extended real-valued A-measurable function on a set D € U. We define the Lebesgue integral of f on D with respect to p by

f fdz= D

sup | gdy (0,e co],

O 0}, we have

LCE) > 0. Now since f > Oon E, wehave E = Unen En where En = {D: fz 1}. The sequence (E,: n € N) is an increasing sequence in 2 so that jim, En= By Theorem

1.26, we have

no € N such that 2(E,,)

go = tylEn + 0-

im, H(E,)

=

w(E)

>

Unen En= E.

0. This 3 implies that there exists

> 0. Let us define a nonnegative simple function go on D by

1p\z,,: Then 0 < gp < f on Dand

| fduw=

sup | gdp =f

D

O 0,

which contradicts the assumption f,, f du = 0. 3. To prove (c), let Do be a M&l-measurable subset of D. Corresponding to each nonnegative simple function yp on Do such that 0 < yp < f on Dp, let us define a nonnegative

simple function g; on D by

_

go{x)

forx € Do,

1) = 1) 9

tore D\ Do.

Then 0 < gi < f on D and fy, odu = fy yi du. Thus we have

sup

[

O 0 ae.

on D.

Thus

(Do) = 0. Now Do U Dy + D so that tim j(Dp U Dy) = u(D). If w(D) > 0, then there exists ky € N such that ~(Dp U Dy)

> O°

Then by (c) we have f,, f du > Jo, fdu>

> 0. Since 4(Do) = 0, we have 4(Dj,) > 0.

EH(Dia)

> 0. This is a contradiction. Thus

BD) =

5. Let us prove (¢). Fori = 1 and 2, let &; be the collection of all nonnegative simple

functions g on D such that 0 < g < fj. Since fi < f2 on D, ifa simple function ¢ is such 0 < yg < f) on D, then 0 < gy < f2 on D also. Thus 9; C ©. Then that

{ [eauiver}c{

f eau:ve

or}

§8 Integration of Nonnegative Functions

161

so that

[ fau=s{ D

6.

If fi =

f oduie coi} 0, we have feraumef

ran.

Iff is nonnegative real-valued, then for c = 0 we have cf = 0 on D and fr cfdp =0

but cf, f du does not exist when f,, f dus = 00.

Proof. For c > 0, the equality is proved in the same way as in Lemma 7.12 for bounded real-valued measurable functions on a set of finite measure. © In Lemma 7.12, we proved the equality f,{fi + faldu = fp fidut fp fodp for two bounded real-valued measurable functions f; and f2 on a set D with ~(D) < oo. By using the fact that f, f du = suppo[ aman an] =>o f edu= Yo van, i= i=l

neN

neN © i=l

neN

a

neN

162

CHAPTER 2 The Lebesgue Integral

where the fifth equality is by the fact that for series of nonnegative extended real numbers Donen Lm

--->

Linen @pin We have Donen {on

pret

Xn}

=

Vnen@ia

tee

+

Donen %p,n- This proves the countable additivity of v on 2{ and completes the proof that v is a measure on 2. & Theorem 8.5. (Monotone Convergence Theorem) Given a measure space (X, Mt, 4). Let (fn: n € N) be an increasing sequence of nonnegative extended real-valued A-measurable

functions on a set D € Nand let f = lim. fy on D. Then lim. Ip frdus=fyfdu. Proof. D,

1.

Since (f, : n € N) is an increasing sequence of nonnegative functions on

im, Fn(x) exists in [0, 00] for every x € D and thus f =

extended real-valued function on D.

im, Jn is a nonnegative

f is 2-measurable on D by Theorem 4.22.

Since

fn < f on D, we have fy frdu < fy f dy for every n € N by (e) of Lemma 8.2. Also Sn ~ Snsrimplies f, fr du < fy fa+iduand thus (fp, fr du : n € N) isan increasing se-

quence of nonnegative extended real numbers bounded above by {,, f dj and consequently

jim fp frdu s fy fan.

2. Let us prove the reverse inequality tim. tp indu> fpf du. Let be an arbitrary

nonnegative simple function on D such that 0 < g < f. Witha € (0, 1) arbitrarily fixed, we have 0 < ag < g < f on D. Let us definea sequence (E, : n € N) of subsets of D by

qa

E, = {x€D: fx(x) = ag(x)}

forn EN.

Since f, and ag are %A-measurable on D, we have E, Sn

< Savi implies E,

C

€ 2 by Theorem 4.16.

Eni1 for every n € N so that (Z,

Now

: n € N) is an increasing

sequence in 2, Since E, C D for every n € N, we have (),cy En C D. Let us show that actually cn En = D. To show this, we show that ifx € D then x € E, for somen € N. Letx € D. If f(x) = 0, then since 0 < yg < f we have g(x) = Oalso. Since0 < f, < f, we have f,(x) = 0 for every n € N. Thus f,(x) = 0 = ag(x) for every n € N so that x € E, for every n € N. On the other hand, if f(x) > 0 then since 0 < g < f and @ € (0,1) we have f(x) > ag(x). Since f,(x) t f(x), there exists N(x) € N such that Fuoy(x) = ag(x). Thus x € Ey). This completes the proof that ifx € D thenx € E,

for some n € N. Therefore D C Uncen En so that D = nen EnLet us define a set function v on & by setting v(A) = ta gdp for A € A. Then by Proposition 8.4, v is a measure on (X, 2). Now

@

[pane D

f

fadu > f

En

En

apdu =a f

En

gd =av(Ey),

where the second inequality is from (1). Since (Z, : n € N) is an increasing sequence, we have im, E, = Unen E,, = D. Then by Theorem 1.26 (Monotone Convergence), we

have

lim v(E,) = v( lim E,) = v(D). Lettingn > oo in (2), we have

n00

n00

lim

n->00

[

'D

ind

> oa

lim

no

v(E,) =av(D).

§8 Integration of Nonnegative Functions

163

Since this holds for every a € (0, 1), letting @ + 1 we have

jim, [ frau > oD) = f au. Since this holds for an arbitrary nonnegative simple function g on D such that0 < 9 < f, we have

tim,

n-v00 D

fadu

>

sup

Os N so that lim, fx) ne

N so that

Fe | f'™(@)

i Jim, f [] (x)

== im” i

= F(x); if f(x) = 00, then f(x) = n for every ==o= =

imi i f(x) =— and similarly if = —oo, then

—_—

— FH = —n for every n € N so that lim /'

” FB,

(x)

=

=

1 lim

—_

n==-

—

—oo ==

f(x).

In

particular if f is nonnegative then f™] + f on D. Lemma 8.16. Given a measure space (X, A, 2). Let f be a y-integrable nonnegative extended real-valued A-measurable on a set D € UW. Then for every e > 0, there exists a constant M > 0 such that

fir- 1 au= frau frau 0, we have 0 < f™1 < f on D so that the jz-integrability of f on D implies that of £4 by (e) of Lemma 8.2. Then f,{f— fl} du = fy f du—fy fl dy by Corollary 8.9.

For n € N, consider f™!, the truncation of f at level n. Then (f!! : n € N) is an

increasing sequence of nonnegative real-valued 2{-measurable functions such that f"] + f

on D. Thus by Theorem 8.5, we have tim, Ip f™ du = fy f dp. Since f, f dp < ov, for every ¢ > 0 there exists N ¢ N such that f, fdu—e < fy fl dy forn > N. Thus Ipfdu-fpf™du N so that fz,(x)

= f(x) for everyn

> N.

Hence

lim fg,(x)

100 x € D. In particular if f is nonnegative then fz, + f on D.

= f(x) for every

Lemma 8.18. Given a measure space (X, 2, 2). Let f be a js-integrable nonnegative extended real-valued W-measurable function on a a-finite set D € A. Then for every & > 0, there exists a A-measurable subset E of D with 4(E) < 00 such that

qd)

[r-seau=

fo ran—f

fedu 0 such that

Q)

fur — fort-mwi} dm, = [iran

Proof.

1.

For any E

C

D such that E

- [ font-m,m\ dit, < 8.

€ A, we have O

< fg


0 there exists N € N such that we have

Spf 4u— fp fe, du N. LetE = Ey and we have the inequality in (1).

2. When (X, 2, z) = (R, Mt, u,) and D € M,, we let E, = DM [-n,n] for

n €N.

Then (E, : n € N) is an increasing sequence in 2,

such that jim, En

= Dand

#4, (En) < 00 for every n € N. Then (2) follows by the same argument asin1.

The next Theorem shows that if f is a nonnegative extended real-valued 2{-measurable

function on a o-finite setD € QM and if f, f du < ov, then f, f du. can be approximated

by the integral of a bounded nonnegative real-valued 2-measurable function which vanishes outside of a set with finite measure. Theorem 8.19. Given a measure space (X, U, 2). Let f be a p-integrable nonnegative extended real-valued A-measurable function on ao-finite set D € A. Then for everye > 0, there exist a constant M

that

> 0 and a &X-measurable subset E of D with u(E)

[itr-0"

@

1 au=f

tan

f

< 0 such

(78), du 0 such that @

[lr-

(f°) praca}

au, = [iran

= fF)

pean

dp, O and E € & such that E C D, we have 0 < (fl), < fll < f on D. Thus the j:-integrability of f on D implies that of fl

and (fl),

Corollary 8.9, we have J, {f — (fM4l),} du = fy fdu— fy (fF), du.

Then by

Let ¢ > 0. The y-integrability of f on D implies according to Lemma 8.16 that there exists M > 0 such that

@)

[iran fray 0 however small, if we take n € N so large that 3 < 6 then

H, (En) = 5 . Problems

Prob. 8.1. Let (X, &, 2) be an arbitrary measure space and let D € SA with 4(D) > 0. Let

Ff be a nonnegative extended real-valued 2-measurable function on D. Show that if f > 0 on D then fy, f du > 0. Prob. 8.2. Let (X, 2, 4) be an arbitrary measure space.

Show that if there exists a non-

negative extended real-valued &-measurable function f on X such that f > 0 on X and ty f dp < oo then (X, A, 4) is a o-finite measure space.

Prob. 8.3. Let (X, A, 2) be a measure space and let € be a collection of disjoint members

of 2% with positive 4 measure. Show that if (X, A, jz) is a o-finite measure space then the collection € is at most countable. Prob. 8.4. Given a measure space (X, &, jz). Let f be a nonnegative extended real-valued

&M-measurable function on a set D € A with ~(D) < oo. Suppose f > 0 a.e. on D. (a) Show that for every 6 > 0 there exists 7 > O such that for every 2-measurable subset E of D with n(E)

> 6, we have f, fdy

> 7, (that is, if F € A, E Cc D, and w(£) is

bounded below by a positive real number then f,. f dit is bounded below by a positive real number). (b) Show that (a) does not hold without the assumption 4(D) < 00. Prob. 8.5.

Given a measure space (X, A, ~) with 4(X)

< oo.

Let f be a nonnegative

extended real-valued 2-measurable function on X such that f > 0, w-ae. on X. (E, : n € N) be a sequence in & such that jim, Ja, f du =0. Prove the following:

(@) lim u(En) = 0.

Let

174 {b) {c)

CHAPTER 2 The Lebesgue Integral lim (E,,) = 0 does not hold without the assumption that f > 0, j-a.e. on X. 7) jm, (E,)

= 0 does not hold without the assumption that 4(X) < oo.

Prob. 8.6. Given a measure space (X, Ml, 2). Let f and g be two nonnegative extended real-valued 2-measurable function on a set D € 2 such that f < ga.e.on D.

(a) Show that if f,, f du = fp gdu < co, then f = g ae. on D.

(b) Show by constructing a counter-example that if the conditions f, f du f (> & 4




ieN

lim ¢y,i-

jim,

Tot

Prob. 8.13. Given a measure space (X, 2, 4). Let f be a nonnegative extended real-valued. -measurable function on a set D € 2 with w(D)

< oo. Let D, = {x €E D: f(x) > n}

forn € Z,. Show that f, f du < 00 if and only if Prez, BDz) < &.

Prob. 8.14. Given a measure space (X, Ql, 4). Let f be a bounded nonnegative real-valued -measurable function on a set D € A. Let D, = {x € D: f(x) > n)} forn € N. Show

that if f, f du < oo then DY cy H(Dn) < 00.

Prob. 8.15. Let (X, &, #2) be a measure space with u(X)

< co. Let f be a nonnegative

extended real-valued -measurable function on X that is finite z-a.e. on X. Show that f in 4-integrable on X if and only if

>

2ufxe Xs f(x) > 2"} < 00.

nek,

Prob. 8.16. Let (X, 2, 2) be an arbitrary measure space. Let f be a bounded nonnegative real-valued A{-measurable function on X. Show that f is jz-integrable on X if and only if

D

Fulx ex: fe)> 2} 00 2

folhr — f\? du < &, for every n € N for some fixed p € (0, 00).

Show that the sequence (f, : 2 € N) has a subsequence (f,, : k € N) which converges to f ae. on D.

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CHAPTER 2 The Lebesgue Integral

Prob. 8.19. Given a measure space (X, &, jz). Let (f, : n € N) and f be extended realvalued 21-measurable functions on a set D € & and assume that f is real-valued a.e. on D.

Suppose lim n

tb |fn — f |? du = 0 for some fixed p € (0, 00). Show that the sequence

(fn 11 € N) converges to f on D in measure. Prob. 8.20. Let (X, &, 4) be a measure space and let f be an extended real-valued Ameasurable function on X such that f If? du < oo for some p € (0, 00). Show that

lim APu{X: |f| =a} =0.

Ao oO

Prob. 8.21. Let (X, U, 44) be a o-finite measure space. Let f be an extended real-valued

A-measurable function on X. Show that for every p € (0, 00) we have Pdp= fo” [isdn

pol

: fl >a} a). wf{X:

Prob. 8.22. Let (X, A, 2) be an arbitrary measure space and let f be anonnegative extended real-valued 2{-measurable function on X. Consider {t € [0, oo] : u(f-1({#})) > O}, a subset of the range of the function f. Show that if f is 4-integrable on X then the set is a countable set. Prob. 8.23. Let (X, 2, jz) be an arbitrary measure space and let f be a jz-integrable nonnegative extended real-valued 2{-measurable function on X. Let us define two nonnegative extended real-valued functions g and h on [0, co) by

a) =u{x eX: f(x) >t}

fort € [0, 00),

h(t) = p{x eX: f(x) =t}

fort € [0, 00).

Show that

(a)g t} fort € [0, 00). Prove

©

fi ran= f [0, 1 BO ACAD = [10, ttt eX LO) > tha.

(b) Define a function h on [0, 00) by h(¢) = w{x € X: f(x) = t} fort € [0, 00). Prove 2)

[reu=f

me mca) =f

sa

eX: f(x) > thud).

§9 Integration of Measurable Functions

§9

177

Integration of Measurable Functions

[1] Lebesgue Integral of Measurable Functions If f is an extended real-valued %A-measurable function on a set D € Min a measure space

(X, A, w), thenf = f+—f- where f+ and f~ are the positive and negative parts of f, that

is, f+ = max {f, 0} and f— = —min{f, 0}. Since f+ and f— are nonnegative extended real-valued 2-measurable functions on D, fy ft dy and f,, f~ du exist in [0, 00] by

Definition 8.1, but f, ft du — fy f~ du may not exist in R. Definition 9.1.

Given a measure space (X, A, 4).

Let f be an extended real-valued

&A-measurable function on a set D € A. If fy f* du — fp f— du exists in R, then we say

that f is Lebesgue semi-integrable on D with respect to 2, or simply jt semi-integrable on

D, and define fy f du = fy ft du— fy f- du. We say that f is Lebesgue integrable on D with respect to js, or simply integrable on D, only when f, f du ER.

For the class of bounded measurable functions on a set with finite measure and for the class of nonnegative extended real-valued measurable functions the Lebesgue integral has been defined by Definition 7.10 and by Definition 8.1 respectively. Let us reconcile these definitions with Definition 9.1 for the class of extended real-valued measurable functions which includes the last two classes of functions as particular cases.

To do this, let Jo(f),

I(f), and J(f) be the integrals of f in the sense of Definition 7.10, Definition 8.1, and Definition 9.1 respectively. If f is a nonnegative extended real-valued A{-measurable function on a set D € A, then

f=ft—f- where f+ = fand f~ =0. Then J(f) = 1(f*)-1(f-) = 1(f)-1@) =

I(f). Thus Definition 8.1 is consistent with Definition 9.1. If f is a bounded real-valued M-measurable function on a set D € M with ~(D) < oo, then

In(f) = Io f* — f-) = ln f*) — (fF) = sup f

du — sup f

gsft/D

=

sup

gsf-

[edu-

O 2. Then {D,

a disjoint collection in 2 with (g_, Dy = En forn € N and en

: n € N} is

Dn = Unen En = D.

Thus by (b) and (a) we have

[rau-> J, fee=aed ff te neN

= lim

noo

Jp

Dy

fau=

tim, f

SF dp.

This proves (c).

For the integral of a measurable function defined a.e. on a measurable set, the following definition is convenient. Definition 9.11.

Given a measure space (X, 2A, w). Let f be an extended real-valued

A-measurable function only on D \ N where D, N € &, N C D, and N

is @ null set in

(X, A, w). Suppose f is p semi-integrable on D\ N. We write f, f dufor f, f du where fis an extended real-valued A-measurable function on D defined by a

f

onD\N,

O

omN.

Note that fy Fdu = foyy faut fy Fae = fovy f du +0 by (@) of Lemma 9.10.)

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CHAPTER 2 The Lebesgue Integral

Lemma 9.12. Given a measure space (X, A, 4).

Let f be an extended real-valued A-

measurable function on a set D € A.

(a) If f is semi-integrable on D, then so is — f and [,(—f) dp =— tb f du. (b) If f is 2 semi-integrable on D, then so is cf for everyc # Oand fy cf du=e fy f dp. (© Fff is w-integrable on D, then Of is defined a.e. on D and [,0- f du = 0-fp f dp.

Proof. (a) follows from the fact that (—f)+ = f- and (—f)~- = f+. (b) follows from (a) and Lemma 8.3.

real-valued ae. on D.

(c) follows from the fact that if f is js-integrable on D, then f is

It is usually shown that if f and g are both z-integrable, then so is f + g and moreover Spf + etdu = fy fdu+t fp gdp. Here we show that the equality holds even when f

and g are only jz semi-integrable as long as the sum f,, f du + fp g du exists in R.

Observation 9.13. Let a,b € R be such that a + b exists in R. Then max {a + b, 0}
0, then max {a + b, 0} < max {b, 0} max {a, 0} + max {6,0}. If a > 0 andb > O also, then max{a + 6,0} = a+b

max {a, 0} + max {b, 0}.

Theorem 9.14. Given ameasure space (X, U, 4). Let f and g be two extended real-valued

MA-measurable functions on a set D € A. Suppose f and g are both

semi-integrable on

D. fp fdut fypgdu exists in R, then (a) f + ¢ is defined a.e. on D,

(b) f + g is 4 semi-integrable on D,

© folf +etdu= fy fdut fogau.

In particular if both f and g are p-integrable on D then f + g is -integrable on D and

SAf +eldu= fp fdutfygdu.

Proof. 1. Suppose f, f du+ fy g ap exists in R. Then either

@

[rau

f eau € (—00, oo],

[raw

[eau € [—c0, 00).

or (2)

Let us consider case (1). Case (2) can be treated likewise. Now if we assume (1), then we have

Q)

[tran f etaver, oo],

§9 Integration of Measurable Functions

183

and

@

[ frau, f g7 dys € [0, 00).

By (a) of Lemma 8.2, (4) implies that f~, g~ € [0, 00) a.e. on D. Then f, g € (—o0, 00]

ae. on D so that f + g is defined and f + g € (—0o, co] ae. on D.

2. Let us show that f + g is 2 semi-integrable on D. By Observation 9.13, we have

| (f+)

ae. on D,

0. Now since both f and g are real-

valued on D, we have E = pen Ex where Ey = {x € D: g(x) — f(x) > f}. Then 0 < wW(E)

< Dten HER).

Thus there exists ko € N such that 2(E;,)

> 0 so that

Jag{g -— ftdu = joHEm) > 0. Now fe {g— fidu = fy, sdu— fr, fdu by Corollary 9.15. Thus we have Sex gdp > Sing fdut EHEw)

> Jeg f du. This is a

contradiction. Thus 2(Z) = 0. Similarly z(F) = 0. This shows that f = g ae. on D. Now consider the general case that f and g are two j.-integrable extended real-valued A-measurable functions on D. The y-integrability of f and g implies that f and g are real-valued a.e. on D by (f) of Observation 9.2. Thus there exists a null set Do contained in D such that f and g are real-valued on D \ Do. Let two functions f and 2 be defined

§9 Integration of Measurable Functions

185

by setting f = = fon D\ Doand f = 0 on Do, and similarly g = g on D \ Do and g = 0

on Do. Then f and 7 are real-valued on Dand f, fdu= fefap= fggdu= tpap

for every E € 2 such that E C D so that by our result above we have f = Z ae. on D, Then since f = f a.c. on D and g = # ac. on D, we have f = g ae. on D. 2. Suppose D € Mis a c-finite set and f and g are yz semi-integrable on D. Suppose

Saf de

= fy dp for every E € & such that E C D. The o-finiteness of D implies

according to Lemma 1.31 that there exists a disjoint sequence (A, : n € N) in 2

such that

Unen 4n = D and (An) < 00 for every n € N. Since a countable union of null sets is a null set, to show that f = g a.e. on D, it suffices to show that f = g a.e. on A, for every

n € N. Now suppose the statement that f = g a.e. on A, is false for some m € N. Then at least one of the two sets E = {A,: f < g} and F = {A,: f > g} has a positive measure. Suppose 2(£) > 0. Now £is the union of three mutually disjoint -measurable sets given

by

E® = {A,:—00 < f 0. For brevity, let EX = E® |, ,. Then foslg — f}dp > RulE) > 0. Since —mo < f < g < £9 on E* and p(E*) < u(A,) < 00, f and g are y-integrable on

E* and then fy.{g — f}du = fueg du — {yx f due by Corollary 9.15. This then implies

See 8 4b — fre f dp = EH")

> Oso that f,.gau > fxs f du which contradicts the

assumption that f, f du = f, g dy for every E € 2 such that E Cc D.

Consider next the case (E™) > 0. Now E® = | Jpey ES where E® = {A,:—00 < f 0. Forbrevity

let us write E* = EP. Then since g = co on E* and 44(E*) > 0 we have f,.. gd = 00. On the other hand, fp. f du < €o2(E*) < £ou(An) < 00. Thus fy. fdu # fp. edu,

contradicting the assumption that f, f dj = f, g dy for every E © & such that E C D. Finally consider next the case u(E®) > 0. Now E® = yen EQ where ES) = {An :f=—0,g>

—m}.

ThenO < p(E®)< >, en H(ES’) so that there exists mo € Nsuch that u(E@) > 0. For brevity let us write E* = EF. Then fy. f du= —ooand f,. gdp = —mou(E*) > —00

since u(E*) < (A) < oo. Thus fy. f du # fre gdp. This is a contradiction.

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CHAPTER 2 The Lebesgue Integral

Thus we have shown that the assumption (EZ) > 0 leads to a contradiction. Similarly the assumption j(F') > 0 leads to a contradiction. Therefore f = g a.e. on A,, for every

néNandhence f=

gae.onD.

Let us observe that (b) of Proposition 9.16 is not valid without the a-finiteness of the set D e€&. As acounter-example, consider a measure space (X, 2, 4) where X is an arbitrary nonempty set, 2 = {B, X}, and w(@) = 0 and z(X) = oo. Let f = 1 on X andg = 2

on X. Then f, f du =0 = fygduand fy fdu = co = fy gdu so that f and g are semi-integrable on X. We have f, f du = fy g du for every E € A. But f(x) 4 g(x) for any x € X.

[II] Convergence Theorems The Monotone Convergence Theorem (Theorem 8.5) for an increasing sequence of nonnegative extended real-valued functions is extended next for a monotone sequence of extended.

real-valued functions. Theorem

9.17. (Generalized Monotone

Convergence

Theorem) Let (X, 2, jz) be an

arbitrary measure space. Let (f, :n € N) be a monotone sequence of extended real-valued U-measurable functions on a set D € Uandlet f = lim fy. n>00 (a) If (fi. : n € N) is an increasing sequence and there exists a js-integrable extended real-valued U-measurable function g such that f, > g on D for everyn € N, then Sn, n EN, andf are p semi-integrable on D and dim, fo fndu = fp f dp. (b) if (fn in € N) is @ decreasing sequence and there exists a -integrable extended realvalued A-measurable function g such that f, < g on D for everyn EN, then f,,n € N,

andf are

semi-integrable on D and lim, fp fndu= fof du.

Proof. 1. Assume the hypothesis in (a). Since g < f, and g is y-integrable on D, f, is p-semi-integrable on D by Lemma 9.5. Similarly for f. By the z-integrability of g on D, there exists a subset E of D which is a null set in (X, 2, 2) such that g is real-valued on D\ E by (f of Observation 9.2. Then f, — g is defined on D \ E. Now (f, — g:n €N)

is an increasing sequence of nonnegative extended real-valued 2-measurable functions on D\ E and thus by Theorem 8.5 (Monotone Convergence Theorem), we have

dim, ff

lim

n-siau= fr

2» —ghdp=

-ahan

—ghdp.

By the ji-integrability of g, we have Jove gdu éR. Thus Jove fadpe— Jove gdy exists

in R and this implies that Sovel fn —g}du= Sone fad — Jove g dp by Corollary 9.15. Similarly we have Sovelf —g}du= Jove fdp— Joye 8 ah. Therefore we have lim

n>00

Joye Sn

du- [

p\E®

and thus dim, Jove irdp = Joye f du. Definition 9.11.

d=

D\E f

du — [

De®

du

Then we have dim, to fndu = fp f du by

§9 Integration of Measurable Functions

187

2. To prove (b), assume the hypothesis in (b). By Lemma 9.5, f,, € N, and f are p semi-integrable on D. Now (—f, : n € N) is an increasing sequence and — f, > —g on D for every n € N. Since —g is jz-integrable, (a) implies im, {pCi du = fyo(-fdp and thus Jim, tp fpdu=fofdu.

a

Remark 9.18. (a) In Theorem 9.17, if some entry in the monotone sequence (f, : n € N),

say fno for some no € N, is j1-integrable, then f,,. can serve as the bounding ,:-integrable function g for the monotone sequence (fy : n > no). Thus im. fy frau = fy fdu. (b) Theorem 9.17 does not hold without the existence of the jz-integrable lower bounding function g for an increasing sequence (f, : n € N) or the js-integrable upper bounding

function g for a decreasing sequence (f, : n € N). For example, in (IR, 9t,, j2,), let h=

-} on Rfor n € Nand let f = 0 on R. Then (f, : 1 € N) is an increasing sequence

and im, fn = f on. We have f, f, du, = —00 for everyn € N while f, f du, =0.

Fatou’s Lemma (Theorem 8.13) fora sequence of nonnegative extended real-valued measurable functions is extended to a sequence of extended real-valued measurable functions as follows. Theorem 9.19. (Generalized Fatou’s Lemma) Given a measure space (X, A, 1). (fn

im

Let

€ N) be a sequence of extended real-valued A-measurable functions on a set

Dem. (a) If there exists a t-integrable extended real-valued U-measurable function g such that Su = g on D for everyn €N, then fy, n € N, and lim inf Sn are ut semi-integrable on 7 Dand

a

[imine fay s timing ff dy. D

noo

n> 00

'D

In particular if f = lim | Sn exists a.e. on D, then f is pz semi-integrable on D and 1

@)

ff frau stiming f fan.

(b) If there exists a s-integrable extended real-valued A-measurable function g such that Sn timsnp ff, du. noo

n>co

D

In particular if f = im, Fn exists a.e. on D, then f is semi-integrable on D and

«)

[fanz timsup f fav.

188

CHAPTER 2 The Lebesgue Integral

Proof. 1. To prove (1), assume the hypothesis in (a). Since f;, > g on D for every n € N, we have lim inf fn > gon D. Then by Lemma 9.5, f,, n € N, and lim inf fn are pt 7 7

semi-integrable on D. Now liminf f, = lim inf f; and (inf fj : 2 € N) is an increasing noo

nO

k>n

ken

sequence of extended real-valued 2%{-measurable functions on D satisfying jaf te = gon D for every n € N. Theorem), we have

2

Thus by (a) of Theorem 9.17 (Generalized Monotone Convergence

Jf simi du = ff iit fed = Jim, fat fc = liminf

| int fed

00

Lemma 9.5. For the sequence (— f, : nm € N), we have —fn> —g on D for everyn EN and —g is yz-integrable on D. Applying (1) to the sequence (— f, : n € N), we have

6

[ lim inf (— fy) dy. < limint [ (fa) du = liming (— [ frau).

Now lim inf(—a,) = — lim sup a, for an arbitrary sequence (a, : n € N) in R. Thus no

noo

[ lim inf (—f,)du = — [ lim sup fa dy timint (- f feau) = —timanp ff dy. Using these two equalities in (5), we obtain (3). If f = im, Jn exists a.e. on D, then we

have f = lim sup fn ae.

on D. Then fy fdu= fy lim sup f, di by Definition 9.11.

Substituting this ¢ equality in (3), we have (4).

The Dominated Convergence Theorem which we prove next contains the Bounded Convergence Theorem (Theorem 7.16) as a particular case. Theorem 9.20. (Lebesgue’s Dominated Convergence Theorem) Given a measure space {X, A, pw). Let (f, :n € N) be a sequence of extended real-valued A-measurable functions ona set D € 2 such that |f,| < g on D for everyn € N for some p-integrable nonnegative

§9 Integration of Measurable Functions

189

extended real-valued M-measurable function g on D. if f = jim, Sn exists a.e. on D, then f is p-integrable on D and furthermore

(1)

jim, f frdp= [sau

and

@

dim, [fe -fidu =o.

Proof. 1. Since | f,| < g on D, we have —g < f, < g on D for everyn € N. Sinceg is 4-integrable on D, —g is also yz-integrable on D. Thus by Theorem 9.19 (Generalized Fatou’s Lemma), f,,

€ N, and f are yz semi-integrable on D and

3)

fiA fan stimint imint f[| tea,

as well as

@

ff fanz timsup ff, ay. D

no

D

Since liminf fp frdu < lim sup fp fad, (3) and (4) imply timint f fady =limsup f noo

Thus

im, Sp fadu =

'D

n—>00

'D

fy f du € R. This proves

fedu=

ff fan. 'D

(1).

2. Let us show that f is not only 4 semi-integrable but in fact jz-integrable on D, that

is, fp fdu € R. Since —g < f, < g on D, we have — fygdu < fy fadu < fogdu for every n € N. Thus f, f du= jim, Jp finde €[- edu, fpedu] and therefore we have f, fd eR. f=

3. To prove (2), we apply (1) to the sequence (|,— f|: € N) as follows. Since jim, fn ae. on D, there exists a null set EZ, in (X, 2, 2) contained in D such that

f=

"tim, fn on D\ Ej. Since f is -integrable on D, f is real-valued a.e. on D by

of Observation 9.2 so that there exists a null set Ez in (X, 2, 4) contained in D such that f is real-valued on D \ E2. Let E = E, U Ep, a null set in (X, A, wz) contained in

D. Now

f is real-valued on D \ E so that f, — f is defined on D \ E forn € N. Since

—g < fp < g on D forn

€ N and since lim

fn = f on D\ E, wehave —g < f < gon

1 D\ E. Then |fa— fl < |fal+lfl < 2g on D\ E. Since f = jim, fx on D\ E, we have Jim, fa — f =00n D\ E and consequently lim, | fn — f|=0o0n D\ E. Thus applying (1) to the sequence (| f, — f| : 2 € N) which is dominated by the y-integrable function 2g

and converges to 0 on D \ E, we have im, Joye lin — fldu = fpyg 0du =0. Then by Definition 9.11, we have im

Solfr—fldu=0.

&

190

CHAPTER 2 The Lebesgue Integral

Theorem 9.20 still holds if the condition that | f,| < g on D for every n € Nis replaced by the condition that | f,| < g ae. on D for each n ¢ N. In this case we redefine f, to

be equal to 0 on the null set on which | f,|< g does not hold. The redefined functions f, now satisfy | hil, a; : n ¢ N) in R, and lim supa, is the limit of the decreasing sequence n—>00

(supy>n az

7m € N) in R. It follows then that if lim inf a, < cforsomec € R, thena, c, then a, > ¢ for infinitely many n € N. n> 00

Theorem 9.21. (Fatou’s Lemma under Convergence in Measure) Given a measure space (X, A, yw). Let (f, :n € N) be a sequence of extended real-valued A-measurable functions on a set D € & converging in measure to an extended real-valued A-measurable function

Ff which is real-valued a.e. on D. (a) If fa = g on D for every n é€ N for some p-integrable extended real-valued 2M-measurable function g on D, then fy, n € N, and f are all semi-integrable on Dand

a

[tau stimint f f.an.

(b) if fa


limsup f frdu. D

noo

D

Proof. Assume the hypothesis in (a). Since f, > g on D, the j-integrability of g on D implies the z semi-integrability of f, on D, thatis, f, fa du exists in R, foreveryn € N by Lemma 9.5. The convergence in measure of (f, : n € N) to f on D implies the existence

of a subsequence (f,, : k € N) which converges to f a.e. on D by Theorem 6.24. Then f > g ae. on D and this implies the » semi-integrability of f on D by Lemma 9.5. Thus

we have proved the existence of the integrals f,, f, du forn € Nand f,, f dy. To prove the inequality (1), let a, = f, fadu forn

show a

< lim inf @, let us assume the contrary.

€ Nanda

= f, fdy.

Then there exists c €

To

R such that

a>ec> lim inf ap. This implies that a, < c for infinitely many n € N and thus there 200

exists a subsequence (a, : k € N) such that a,, < ¢ for every k € N. The convergence in measure of (f, :n € N) to f on D implies the convergence in measure of the subsequence

§9 Integration of Measurable Functions

191

(fu, : & € N) to f on D. This then implies the existence of a subsequence Cu,

:€eN)

which converges to f a.c. on D by Theorem 6.24. Then by (2) of Theorem 9.19, we have fofaus lim inf fo Fri, Ui, that is, a < lim inf apy, But a,, 0, there exists a constant M > 0 such that

[frau frau) < fir $01an 0, we have | f!4| < || on D. Then the j-integrability of f on D implies that of f!1. Thus by (c) of Corollary 9.15 and (c) of Observation 9.2, we have

[itauff 700ay

fu - fay < f If — £94) dy. D

D

For every n € N, we have |f — fl] < |f| + |f™l| < 2/f|. Also Jim D implies im, \f -—

|

=00n

= fon

D. Applying Theorem 9.20 (Lebesgue’s Dominated

Convergence Theorem) to the sequence (| f — f'*]| : n € N) of functions dominated by the #-integrable function 2| f|, we have

fir- —f"\du= sana fl timflim |fif -—f™riandu =0.

lim noo

Thus for every ¢ > 0, there exists M € N such that fy |f —fIdu 0, there exist a A-measurable subset E of D with w(E) < 00 such that

@

[rau

ff teaul < fir

feldu 0 such that

@ Proof.

| [ Faw, - [ font-man dit, For every E

C

D such that E

0 there exists M € N such that jim, So lf —fe,|du < eforn > M. WithE = Ey, we have the inequality in (1). When

(X, % 2) = (R, Mt, u,), we let E, = DM [-n, n]. Then (2) follows from (1).

§9 Integration of Measurable Functions

193

Theorem 9.25. Given a measure space (X, A, 2). Let D € A be ac-finite set. Let f bea B-integrable extended real-valued A-measurable on D. Then for every ¢ > 0 there exist a constant M > 0 and a U-measurable subset E of D with (E)

eau| < fir - ()glau 0 such that

@)

| [ fdu, - f (f°)

sean dite < [ LF — (7!) ov aan te 0, there exists § > 0 such that

[frau 0 there exists a continuous real-valued function g on R such that te |f —gldu, interval.

< &. Moreover g can be so chosen that it vanishes outside of a finite closed

Proof. Let ¢ > 0 be arbitrarily given. By (2) of Theorem 9.25, there exists M > 0 such

that

a

[le — (aoa ts
A(C) and moreover A(E) > —e for every E € A such that E c C’. Suppose for some e > Ono such subset C’ of C exists.

Then there exists E, € & such that Z,} C C and A(#£i) of 4 we have A(EZi) + A(C \ £1) = A(C). Since A(C)

< —e. By the finite additivity € R and since £; and C \ Ej

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CHAPTER 2 The Lebesgue Integral

are {-measurable subsets of C, we have A(E1), A(C \ £1)

MC \ Ey) = AC) — ACE1) > ACC) since —A(E1)

€ R by Lemma

10.5.

Thus

> & > 0. Now since C \ FE, € &,

C\ E, CC and ACC \ E}) > ACC), by the assumed non-existence of the set C’ there exists

E, € 2% such that EF, C C \ E, and A(E2) < —e. Thus continuing, we have a disjoint sequence (E, : n € N) of 2-measurable subsets of C with A(E,) < —e for every n € N. Let E = Unen En. By the countable additivity ofA, we have A(E) = }> en A(En) = —00. But the fact that E Cc C and A(C) € R implies that 4(£) ¢ R by Lemma 10.5. This is a

contradiction. This proves that for every ¢ > 0 there exists a 2l-measurable subset C’ of C such that A(C’) > ACC) and ACE) > —e for every 2l-measurable subset E of C’.

2. Let us show that if C € 2% and A(C) € R, then there exists a positive set A C C with (A) > A(C). Let us define inductively a decreasing sequence (C,, : n € N) in & with M(C,) € R forn € N as follows. Let C; = C. Suppose we have selected C1, ..., Cy-1By our result in 1 there exists C, C C,_1 such that C, € & with A(C,) > A(Ca_-1)

and A(E) > -} for every A-measurable subset E of C,. With the decreasing sequence

of sets (C, : n € N) in 2 thus selected, let A = (cy Cn. Since (C, : n € N)isa decreasing sequence and since 4(C;) = A(C) € R, we have A(A) = fim (Cn) by (b) of Theorem 10.8. Since (A(C,) : n € N) is an increasing sequence of real numbers and since 4(C1) = A(C), we have tim A(Cn) > A(C). Therefore 4(A) > A(C). To show that A is

a positive set, let Z be an arbitrary 2-measurable subset of A. Then E C A= (\,cw Cn so that E C C, for every n € N. This implies that A(EZ) > -} for every n € N and thus (E) > 0. This shows that A is a positive set. 3. Leta = sup{A(E) : E € A}. Let us show that there exists a positive set A for A such that A(A) = a. Since A(@) = 0, we have a € [0,00]. Then we can select (C, : n € N) in & such that A(C,,) € R for every n € N and A(C,,) t a. Then by our result in 2, for every n € N there exists a positive set A, for 4 such that A, C C, and

M(An) = A(Cy). Let A = U,en 4n € 2. By the definition ofa we have A(A) < a. Let

us show that actually 4(A) = a. Let F, = Jf_; Az for n € N. Then F, is a positive set for i by (a) of Lemma 10.11. Now (F, : n € N) is an increasing sequence in 2 with

Unen Fn = Upew 4n = A so that A(A) = A lim, F,) = lim ACFn) by (a) of Theorem

10.8. On the other hand F,, = (F, \ An) U Ap and the finite additivity of 4 on 2{ implies ACF) = ACFn \ An) + ACAn). Since F, \ Ay is a 2l-measurable subset of the positive set F,, for 4, F, \ Ap is a positive set forA by (a) of Observation 10.10. Thus A(F, \ An) > 0. This implies A(F,,) > A(A,). Thus we have

MA) = lim A(F,) > limsupA(A,) > limsupA(C,) = lim 4(C,) = a. n—>00

n>+00

Therefore we have A(A) a@ = (A) < ©.)

Now

let B = A°.

=

a.

(Since A(E)

n>00

€

noo

[—oo, 00) for every E

We have AN B = @ and AU B = X.

€ 21, we have

To show that {A, B} is

a Hahn decomposition for (X, 2, 4), it remains to show that B is a negative set for A.

Let E € Aand E C B. If A(E) > O, then by the finite additivity of 1 on 2 we have MAU E) = 1(A) + A(E) > ACA) = a, contradicting the definition of a. Thus ACE) < 0. This shows that B is a negative set for 4. Thus we have proved the existence of a Hahn decomposition for a signed measure space (X, 2, 4) in which A(Z) € [—00, 00) for every

Ee.

§10 Signed Measures

221

4. Consider a signed measure space (X, 2%, A) in which A(E) € (—oo, co] for every E € &. Then —A is a signed measure on (X, 2f) such that —A(EZ) € [—o0, 00) for every

E € &. Thus by our result above there exists a Hahn decomposition {A, B} for the signed

measure space (X, &, —A). Then {B, A} is a Hahn decomposition for (X, A, A).

5. Let us prove the uniqueness of a Hahn decomposition. Suppose {A, B} and {A’, B’} are two Hahn decompositions of a signed measure space (X, A, A). Since A is a positive set for A, its 2{-measurable subset A \ A’ is a positive set for A by (a) of Observation 10.10. On the other hand A \ A’ = AM(A’)° = AN B’ which is a 2-measurable subset of a negative set B’ for A and hence a negative set for 4. by (a) of Observation 10.10. Thus A \ A’ is both a positive set and a negative set for A and hence a null set for 4 by (b) of Observation 10.10. Similarly A’ \ A is a null set for 4. Then AAA’ = (A \ A’) U(A’ \ A) is a null set forA by Lemma 10.11. Similarly BAB’ is a null set ford. Corollary 10.15. In an arbitrary signed measure space (X, A, A), a maximal positive set and a maximal negative set for . always exist. Indeed if {A, B} is a Hahn decomposition for (X, A, a), then A is a maximal positive set and B is a maximal negative set for }. Proof. Let {A, B} be a Hahn decomposition for (X, 21, 4). Let A’ be an arbitrary positive

set for 4. Then A’ \ A, being a 2-measurable subset of a positive set A’, is a positive set for A by (a) of Observation 10.10. On the other hand, A’ \ A = A’M A° = A'N B being a &A-measurable subset of a negative set B is a negative set for A by (a) of Observation 10.10. Thus A’ \ A is both a positive set and a negative set for 1. and is therefore a null set for 1 by (b) of Observation 10.10. This shows that A is a maximal positive set for 1. Similarly B is a maximal negative set for A. & Definition 10.16. (Mutual Singularity of Two Signed Measures) Two signed measures }1 and )2 on a measurable space (X, 2M) are said to be mutually singular and we write 4, L dz if there exist two A-measurable sets C, and Cz such that C} C2 =, Cy UC2 = X, C1 is a null set for 42 and C2 is a null set for 1. Example. In Definition 10.16 above one of the two sets C; and C2 can be 8. Then if A, and

Az are two signed measures on a measurable space (X, 21) and A, is identically vanishing,

that is, A1(Z) = 0 for every E € &, then we have A; 1 A2. Indeed with C; = @ and Co = X we have C1} N Co = G, C1 UCo = X and C; = Dis a null set for Az and Cz = X is a null set for Ay.

Definition 10.17. (Jordan Decomposition of a Signed Measure) Given a signed measure space (X, Ui, A). If there exist two positive measures p and v, at least one of which is finite, on the measurable space (X, 2M) such that 4 1 v andi Jordan decomposition of i.

= p —v, then {y, v} is called a

(Note that the requirement that at least one of the two positive measures jz and v is finite is

to ensure that yz — v is defined.)

A Hahn decomposition and a Jordan decomposition of a signed measure space are

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CHAPTER 2 The Lebesgue Integral

related. The existence of one implies that of the other as the next proposition shows. Proposition 10.18. Let (X, 2, 4) be a signed measure space. {a) If {A, B} is a Hahn decomposition of (X, 2, 4) and if we define two set functions

and

von A by u.(E) = (ENA) and v(E) = —A(EN B) forE € &, then {y, v} is a Jordan decomposition for i.

(b) If {z, v} is a Jordan decomposition of } and C, and C2 are two A-measurable subsets

ofX such that C; Cz = B, Cy U C2 = X, Cy is a null set for v and Cz is a null set for H, then {C,, C2} is a Hahn decomposition of (X, 2, A).

Proof. 1. Let us prove (a). Every %-measurable subset of the positive set A for A is a positive set for A and every {-measurable subset of the negative set B for A is a negative set for 4 by Observation 10.10. From this it follows that 4 and v are positive measures on (X, 2). Since MB) € (—oo, 0]. finite additivity of E € Aso that 4 = null set for 4. Now

4 cannot assume both oo and —oo, we have either A(A) € [0, 00) or Thus at least one of yu and v is a finite positive measure. Also by the 4, we have A(E) = A(E NA) + ACE NB) = u(E) — v(E) for every — v. To show that yz 1 v, we show that A is a null set for v and Bisa v(A) = —A(AN B) = —A(@) = 0 so that A is a null set for the positive

measure v. Similarly 4(B) = A(B NM A) = AB) = 0 so that B is a null set for the positive measure jt. This shows that

2 | v. Therefore {j2, v} is a Jordan decomposition for A.

2. Let us prove (b). We show that C; is a positive set for 1 and C2 is a negative set for A. Now since {2, v} is a Jordan decomposition for (X, 2, 4), we have A(E) = 2(E) — v(E) for every E € 2. To show that C is a positive set for A, let E € Mand E C C. Since C; is a null set for the positive measure v, its subset E is a null set for v. Thus v(E) = 0 and consequently 4(Z) = #(E) > 0. This shows that C; is a positive set for A. To show that Cz

is anegative set for A, let EF € Mand FE C Cy. Since C2 is a null set for the positive measure #, its subset E is a null set for z. Thus 4(£) = 0 and consequently A(E) = —v(E) This shows that C2 is a negative set for A.

< 0.

Observation 10.19. Given a signed measure space (X, 2, 4). Let C, C’ € A be such that

CAC’ is a null set for A. Then for every E € 21, we have AE NC) =AENC’).

Proof. SinceC = (CNC’) U(C\ C’), wehave ENC = {EN(CNC)}IU{EN(C\C)} and thus AE NC) = A(EN(CNC))+A(EN(C\ C)). Since CAC’ is a null set for A, its subset C \ C’ is a null set for A and then so is its subset EM (C \ C’).

Thus

A(EN(C\C’)) = O and hence A(E NC) = A(EN(CNC’). Interchanging the roles ofC and C’, we have ACE NC’) = A(EN(C/NC)). Thus we have MENC)=A(ENC’).

©

Lemma 10.20. Let (X, U, 4) be a signed measure space. Let {A, B} and {A', B’} be two Hahn decompositions of (X, %, 4). Then MEN A) = A(EN A’) and ENB)

for every E €

= MENB’)

Proof. By the uniqueness of the Hahn decomposition (Theorem 10.14), AAA’ and BAB’ are null sets for 4. Then the Lemma follows from Observation 10.19.

§10 Signed Measures

223

Theorem 10.21. (Jordan Decomposition Theorem) Given a signed measure space (X, A, A). A Jordan decomposition for (X, 2%, d) exists and is unique, that is, there exists a unique pair {2, v} of positive measures on (X, A), at least one of which is finite, such that p 1 v and }. = yt — v. Moreover with an arbitrary Hahn decomposition {A, B} of (X, &, i), if we define two set functions p. and v on A by setting

u(E) =A(EN A)

and

v(E)=-A(ENB)

forE eM,

then {, v} is a Jordan decomposition for (X, A, 4). Proof. 1. By Theorem 10.14, a Hahn decomposition {A, B} of (X, 2, 4) exists. According

to (a) of Proposition 10.18, if we define two set functions jz and v on & by w(E) = ACENA)

and v(Z) = —A(EN B) for E € A, then {y, v} is a Jordan decomposition for 4. Therefore a Jordan decomposition for (X, 2, 4) exists.

2. To prove the uniqueness of the Jordan decomposition, let {4z, v} be the Jordan decomposition for (X, 2, A) defined in 1 and let {j’, v’} be a Jordan decomposition for (X, 21, 4). Then there exist A’, B’ € 21 such that A’

B’ = G, A’U B’ = X, A’ is a null set for v’ and

B’ is a null set for 2’. By (b) of Proposition 10.18, {A’, B’} is a Hahn decomposition for (X, A, A). Then by Lemma 10.20, we have A(ENA) = CENA’) andA(ENB) = ACENB’) for every £ € 2{. Then for an arbitrary E € A, we have

M(E) = MENA) =A(EN A’) =p(ENA)—v(ENA’) sincedA =p’! —v' = p'(E( A’) since A’ and E 2 A’ are null sets for v’ =K(EN A) +p(ENB’) since B’ and EQ B’ are null sets fory’ = p’(E) _ by the finite additivity of 2’. Thus z = yp’ on 2. Similarly v = v’ on A. This proves the uniqueness of the Jordan decomposition as well as the expressions (1) for 4 and v. Definition 10.22. (Total Variation Measure) Let p and v be the unique positive measures on # L vandd = p — v. Let us call and v the A+ for uw and i— for v. The total variation of i

(1)

Given a signed measure space (X, Xi, i). &, at least one of which is finite, such that positive and negative parts of4 and write is a positive measure |A| on X defined by

[A] =at(B)40(E) forE eA.

The positive and negative parts, 4+ and \~, are also called the positive and negative variations of .. For every E € U, we have

(2)

[A|G2) = 44 (EZ) +4°(E) 2 ATE) - 7)

= ADL

The next proposition explains the terminology ‘total variation’ for |A|. Proposition 10.23. Given a signed measure space (X, ‘A, A). For E € A, let Qp = sup {Sta |ACEe)| : {E1,..., En} CM,

disjoint, Ry

Ey = E,n € N}.

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CHAPTER 2 The Lebesgue Integral

Then |A|(E) = ag. Proof. Let E ¢€ SA. Then for any finite disjoint collection {£;,..., Zn} in A such that User Ex = E, we have

Yad k=1

=o at ed —a-(Ed s k=1

k=1

flat Ed + a (Eb|}

=) A") +2°d} = VA k=1

=] 1@.

k=1

Thus we have ag |ACE1)| + |ACE2)| = |ACE 0 A)| + [ACE 9 BD]. If we define (EZ) = ACE M A) and v(E) = —A(E NB) for E € A, then {y, v} is a Jordan decomposition for (X, 2, 4) by (a) of Proposition 10.18. By the uniqueness of the Jordan decomposition (Theorem 10.21), we have At(E) = ACE N A) and A~(Z) = —A(E 1 B).

Now since A isa positive set for A, its subset EA is a positive set for 4 by (a) of Observation

10.10 so that |A(E M A)| = ACE M.A) = At(E). Similarly since B is a negative set for A so is its subset EM B so that A(E NM B) < 0. Then |A(E NN B)| = —ACE NM B) = A (EB). Thus

we have wg > At(E) + 47(E) =| A|(Z). Therefore we have |A|(Z) =o. The next proposition is the converse of Proposition 10.3. signed measure can be represented as an integral.

It shows that an arbitrary

Proposition 10.24. Let (X, 2,1) be a signed measure space. Let {A, B} be a Hahn decomposition of (X, A, 4) and let |A| be the total variation of 4. Then we have

ae) = f {1a —1g}d]A| for Ee &. E Proof. Let {A, B} be a Hahn decomposition of (X, 2, 4). By Theorem 10.21, the positive variation At and the negative variation A~ of the signed measure A are given by

a

MCF) =A(ANF)

and 27(F)=—A(BNF)

for F € Wl.

Then for every E € &, we have

[fta-tohatal =f taneatal - f toned lal =1 [can 8-110 2) =AT(AN EB) +4 (ANE) —At(BN E) —A- (BNE)

= MANE) —- MBN AN E)—- MAN BNE) +ABNE) =MANE)+A(BN E) =A(B),

§10 Signed Measures

225

where the third equality is by [A] = 4+ +-4-, the fourth equality is by (1), the fifth equality is by AN B = @, and the sixth equality isby AUB=X. The next proposition relates the positive variation, the negative variation, and the total variation of a signed measure obtained by integrating a function to the positive part, the negative part, and the absolute value of the function. Proposition 10.25. Let f be a jy semi-integrable extended real-valued A-measurable function on a measure space (X, A, jz). Let us define a signed measure i and three positive measures 41, [22, and 3 on A by setting

ae)= f fan, me) = f fan,

wat)= [fa wt)= f ifldn, for E

€ &

Then 1,

po, and 3

variation, and the total variation of i.

are respectively the positive variation, the negative

Proof. Note that A(E) = fy f du = fy ft du — fy f- du = wi(E) — w2(E) for every Ee€MAsothatA = uw, — po. If welet A = {X: f > O} and B = {X:f

< 0}, then

A,Be A ANB=8, AUB =X, p(B) = 0, and y2(A) = 0 so thatzy 1 jo. This shows that {j1, 42} is a Jordan decomposition of 2, and jz; and j22 are the positive variation and the negative variation of A. Then for every E € A, we have

wtE) =f ifidn= fi rtawt [du = ni(B) + vB) =| 1. This shows that j23 is the total variation of A. Observation 10.26. Given a signed measure space (X, 2, A).

(a) A set E € & is a null set for A if and only if E is a null set for |A|. (b) If aset E € A is a null set for A, then Z is a null set for both At and A~.

Proof. 1. Let us prove (a). According to Theorem 10.21, with an arbitrary Hahn decomposition {A, B} of (X, 21, A), the positive and negative parts of A, A+ and A~, are given

by (

AT(E) =MENA)

and

47-(E)=—-MENB)

forEe A.

Then we have

(2)

[A| (2) =at(2) +47 (2) = MEN A)-A(ENB)

forE € Wl.

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CHAPTER 2 The Lebesgue Integral

If E is a null set for A, then so are its subsets EZ

A and E M B by (a) of Observation 10.10

so that |A|(E) = 0 by (2) and this shows that E is a null set for the positive measure |A|. Conversely if E is a null set for [A], then for every 2-measurable subset Eg of E we have |A(Eo}| | Aa] L Az > ag L] Ao] ] Aa] Lae o] Aa] LI el,

(b) [Aa] L Aa Af LaAganday 1 Ao.

Proof. 1. The equivalence A; Ll Az 0 implies that cA+ and cA~ are measures on

(X,%).

Since a+ and A~ are the positive and negative parts of the signed measure A,

we have At | A7 on (X, &) and this implies that cat | cA~ on (X, QM). Then by the

§10 Signed Measures

231

uniqueness of Jordan decomposition of a signed measure (Theorem 10.21), the equality (2) implies that (cA)+ = cat and (cA)~ = cA-. Substituting these in (1), we have

@)

[racn=f raor)- f rae) =e f raxt— ff parr} =e f

rar.

This proves the theorem for the case c > 0.

2. Consider the case c = —1, Now —A = —{4+ —A7} = a7 —A7. Since A~ and

At are measures on (X, 2{) and A~ L At on (X, &), uniqueness of Jordan decomposition

implies that (—4)+ = A~ and (—A)~ = A*, Substituting these in (1) for c = —1, we have

(4) [racn=f rar-f raxr=-{f rar

f

par-}=- f rar.

This proves the theorem for the case c = —1. 3. Consider the case c < 0. Now ifc < 0, then c = —|c|. Thus we have

(5)

[ racry= f ra(—ten)=— f rater) =-lel f fanme fi ras,

where the second equality is by (4) and the third equality is by (3). This proves the theorem. for the casec => >

I, f du =—oo > Sa fdp € (—00, 00] >

=>

fi fdweR. fi fdueR.

Ja, f du # —00.

Jaf du #0. fy. f du € (—00, oo]. Ja, £ 4. € [—00, 00).

Prob. 10.2. Let (X, 2, 4) be a signed measure space. Let f be an extended real-valued A-measurable function on a set A € 2%. Let Ag € Wand Ap Cc A.

(a) Show that if f, f dd exists in R, then so does Say faa.

(b) Show that if f, f dA exists in R, then so does f,, f dd. Prob. 10.3. Given a measure space (X, M, 44). Let f be a yz semi-integrable extended real-valued A{-measurable function on X. Define a signed measure A on (X, &) by setting ME) = fy, f du for E eX. (a) Find a Hahn decomposition of (X, &, 4) in terms of f.

(b) Find a Jordan decomposition of 4 in terms of f. (c) Find the total variation [| of 4 in terms of f.

Prob. 10.4. Let (X, A, 4) be a signed measure space and let {A+, 4—} be a Jordan decom-

position of A. Let E € 2. Prove the following statements: At(E) = sup {A(Ep) : Eo C E, Eo € 2, A” (E) = — inf {A(Eo) : Eg C E, Eo € A}. Prob. 10.5.

Let (X, 2,4) be a signed measure space.

regarding the infinity of the signed measure A:

Prove the following statements

§10 Signed Measures

233

(a) If there exists F € 21 such that A(E) = oo then A(X) = oo.

(b) If there exists Z € A such that A(E) = —oo then 4(X) = —o0. Prob. 10.6. Given a (a) Show that if £ € set for A with A(Eo) (a) Show that if Z € set for A with A(Eo)

signed measure space (X, 2, A). 2 and A(£) > 0, then there exists a subset Ep of E which is a positive > A(E). Mand A(E) < 0, then there exists a subset Eo of E which is a negative < A(E).

Prob. 10.7, Let (X, 2, 4) be a signed measure space. Prove the following statements: (a) Let A be a positive set for 4. If A° is a negative set, then A is a maximal positive set. (b) If A is a maximal positive set for 1, then A‘ is a negative set for A.

(c) Let A be a positive set for 4. Then A is a maximal positive set if and only if A‘ is a negative set for A. (a’) Let B be a negative set for A. If B° is a positive set, then B is a maximal negative set. (b’) If B is a maximal negative set for A, then B° is a positive set for A. (c’) Let B be a negative set for 4. Then B is a maximal negative set if and only if B° is a positive set for A. (d) A set A is a maximal positive set for 4 if and only if A° is a maximal negative set for A. (d’) A set B is a maximal negative set for A if and only if B° is a maximal positive set for 2.

Prob. 10.8. Consider the measure space (X, &, 4) = ([0, 2x7], Mt, M0, 27], w,). Define

asigned measure A on ([0, 27], St, N[0, 2z]) by setting A(E) = tr sinx 44, (dx) for every

E € 9, N[0, 27].

(a) Let C = [42, 3x]. Show that C is a negative set for A.

(b) Show that C is not a maximal negative set for 1. (c) Let ¢ > 0 be arbitrarily given. Find a St, © [0, 27]-measurable subset C’ of C such

that A(C’) > A(C) and A(E) > —e for every Mt, M [0, 2 ]-measurable subset E of C’. Prob. 10.9. Given a signed measure space (X, 21, 4). For E € A, let and

ag =sup { f) |ACEz)| : {E1,..., En} C A, disjoint , Z_, Ex = E,n € N},

Be = sup { nen |A(En)| : {En in EN} CA, disjoint, Ucn En = E}.

Observe that [4|(Z) = eg according to Proposition 10.23. Show that ag = Br. Prob. 10.10. Let 42 and v be two positive measures on a measurable space (X, 2f). Suppose p 1 v. Show that if p is a positive measure on (X, 2%) and if p is nontrivial in the sense that p(X) > 0, then (uz + ) L (v + p) does not hold. Prob. 10.11. Let 42 and v be two positive measures on a measurable space (X, 2). Suppose for every ¢ > 0, there exists E € A such that u(#Z) < ¢ and v(E*) < s. Show thatz L v.

Prob. 10.12. Let (X, 2, 44) be a o-finite measure space and let f be a nonnegative extended. real-valued 2%-measurable function on X such that f


0 such that cu < v on A, not even

when y is a finite positive measure.

Example 1. Let A and B be two non-empty subset of a set X such that AN B = 9 and AUB=X.

functions

Then Now does cv
0 may be, we have cu(A) = c > 0 = v(A). Thus there not exist c > 0 such that cz < v on 2. Similarly there does not exist c > 0 such that pon A.

Example 2. Consider the measure space (X, 2, 2) = ([0, 1], 9t, M[0, 11, u,). Let f be a function on [0, 1] defined by

F@)

x)=

|x sin 1|

forx € (0, 1],

| 0

forx =0.

Let us define a set function v on 2 by setting v(E) = tr f dp, for E € €. Then visa positive measure on (X, 20) by Proposition 8.12. Now since lim f(x) =0, foreveryc > 0 x!

there exists a € (0, 1] such that f(x) < 5 forx € [0, a]. Then we have

v({0, al) = |

¢

[0,a]

fdp, < glx (10, al) < cu, ([0, a]).

This shows that there does not exist c > 0 such that cz, < v on 2. Notations. For a measurable space (X, 2t) and for aset EF € A, we writeANE A € QM}, that is, the collection of all 2{-measurable subsets of E.

={ANE:

Lemma 11.8. Let js be a finite positive measure and v be a positive measure on a measurable space (X, A). Suppose y and v are not mutually singular. Then there exists c > 0 and E € A with p(E) > 0 such that cu 0. Then since 4(A)

< )>,cy #(An) there exists no € N such that 4(A,,)

> 0.

Now {Ano; Brg} being a Hahn decomposition of (X, 2, v — ae), the set Ay, is a positive

set for v — ab. Let us select E = Aj, and c = i. Then z(Z) > 0 andc > 0 and since E is a positive set for the signed measure v — cyz we have (v — cjt)(Eo) = 0, that is,

cp(Eo) < v(Ep) for every Eg CANE. Let us note and if 4;(£) + measure on (X, or when one of

w

that if Ay and Az are two signed measures on a measurable space (X, 2) A2(E) exists in R for every E € &, then Ay + Az is defined and is a signed 2). This is the case when one of A; and A does not assume the value co 4; and 42 does not assume the value —oco.

Lemma 11.9, (a) Let 41 and Az be signed measures and yt be a positive measure on a measurable space (X, 2). IfA1 K pt, 42 K pw, and dy + Az is defined, then {A +42} & yw. (b) Let 1, A2, and d be signed measures on a measurable space (X, QM). fA, Ld, d2 LA,

and h1 + Aq is defined, then (41 +42} 12.

Proof. (a) is immediate. Let us prove (b). If Ay L A, then there exist Ai, By € AiN B, = @, Ai U By = X, A; is a null set forA and B; is a null set for A,. Ag A, then there exist A>, Bp € 2 such that Az M By = @, A> U Bp = X, A? for 4 and B, is a null set for 42. Consider Ay U Ag and (Ay U A2)° = AGN A5

2% such that Similarly if is anull set = BLN Bo.

Since A; and Ap are null sets for A, Ai U Az is a null set for 4 by Lemma 10.11. Since

B, is a null set for (1, so is its 2{-measurable subset Bj M Bo. Similarly By M Bp is a null set for Az. Let E be an arbitrary 2{-measurable subset of Bj M Ba. Then A1(Z) = 0 and A2(E) = O so that {Ay + Ag}(E) = A1(E) + A2(Z) = 0. Thus B, N B is a null set for Aa +A. Therefore Aj U Az is a null set forA and its complement B; M Bz is a null set for 41 +A. This shows that {Ay +Ag} LA.

Theorem 11.10. (Existence of Lebesgue Decomposition)(Lebesgue-Radon-Nikodym)

Let jz be a a-finite positive measure and i. be a a-finite signed measure on a measurable

space (X, &). Then there exist two signed measures hg and i, on (X, A) such that rg < , As L 4 = hg +A and moreover ig is defined by 4g(E) = Spf duforE € UA where f is a ut semi-integrable extended real-valued A-measurable function on X. Proof. 1. Consider the case that both jz and 2 are finite positive measures. We prove Theorem 11.10 by constructing a js semi-integrable extended real-valued 2{-measurable functionf onX such that if we define A,(E) = te Ff dp for E € Mand define Ay = A—Ag

thenAg < wanda, 1 p. Let ® be a collection of nonnegative extended real-valued %-measurable functions g on X defined by

®={g:9200nX and f, ody < A(E) for everyE € Mt}. Since the identically vanishing function on X is such a function,

4 9. Let us observe

that if gi, g2 € ®, then max {¢1, y2} € ®. Indeed if we let A = {x € X : gi (x) > g2(x)},

§11 Absolute Continuity of a Measure

239

then for every E € 2 we have

[ E maxtor ontdu = f ENAmax fo endu+ f E\Amax {91, va}dy -f

ENA

odu

f

E\A

gradu

< MEM A)+A(E \ A) = ACE). This shows that max {y1, 2} € ©. It follows by induction that for every finite collection {g1,..-, Pn} C P, we have max {gi,..., Gn} € ®. Let

a= sup f ged

JX

gdp.

Let us show that there exists f € © such that [ tau=a. Note that since fyodu

< A(X) for everyg € &, we have 0 < @ < A(X)

< oo. By the

definition ofa there exists a sequence (yg, : n € N) in ® such that Jim, tk Gn dp = a. Let Ff = S8up,en Gn. To show f € ®, let y, = max {9}, ..., On} form € N. Then (y, : n € N)

is an increasing sequence in &. Now since %» > @ for every n € N, we have sup, cy Yn = SUP, cn Yn = f. On the other hand, 4, = max {g1,..., Ga} < sup,cn @n = f for every

n € N so that sup,ew Wn < f. Thus f = sup,cy ¥n- Then since (¥, : n € N) is an increasing sequence, we have y, t+ f on X. Then by the Monotone Convergence Theorem (Theorem 8.5) and by the fact that tr Wndu

a

< 0(E) for every E € UA, we have

[i rau= jim, ff vndu < 06).

This shows that f € ®. To show that f, f dj = or, note that from the fact that #, > 9, on X, we have fy fdu = im, fe tndu >= lim, Sx Gn du =e, On the other hand, since

f €@wehave fy fdu 0 on X, Aq is a positive measure on 2. Since Ag(X) = Ie fdp

=a

< 00,

Aq is a finite measure. For every E € 21 with u(E) = 0, we have te f dy = 0 and thus Aa & pt. By (1) and (2), Ag(Z) < A(E) for every E € Mand thusA > A, on A. Since both 4 and A, are finite measures, if we let Ay = A — Ag then A, is defined and is a finite positive

Measure on A. Let us show that A, set Eg € & with (Zo)

4. Suppose not. Then by Lemma 11.8, there exist a

> 0 and a constant ¢ > 0 such that A, > cu on AN

Ep. Consider

240

CHAPTER 2 The Lebesgue Integral

the nonnegative extended real-valued {-measurable function f + clz, on X. Let us show that this function is a member of &. Now for every E € &, we have

[ir +eteohau = [sau + ements) < ha(E) + As(E 1 Eo) S Ag(E) + As(E) = ACE). This shows that f+clz, € . Now fy{f+clz,} du = oa+cu(Eo) > a since

(Ep) > 0.

This contradicts the definition of a. Thus we have A, L py. 2. Next consider the case that both yz and A are a-finite positive measures on (X, 2).

Now the o-finiteness of jz implies that there exists a disjoint sequence (Fj, : m € N) in 2& such that _),,cn Fm = X with (F,)

< 00 for every m € N

and similarly the o-finiteness

of J implies that there exists a disjoint sequence (G, : n € N) in & such that L),ny Gn = X with 4(Gy) < oo foreveryn € N. Then {Fn OG,

: m EN, n

€ N} is a disjoint collection

in A, pen Unen Fin 1 Ga = X with w(Fn 1 Ga) < 00 and ACF N Ga)

< 00 for

every m € Nandn EN. Let (X, : n € N) be a renumbering of the countable collection {Fim 1G, im € N,n € N}. Then (X, : n € N) is a disjoint sequence in A, Unen Xn, =X with 4(X,) < oo and A(X;,) < 00 for everyn € N. Let fi, and A, be the restrictions of 2 and A to the measurable space (X;,, 2 X,,) for n €N. Then py and A, are finite positive measures on the measurable space (Xn, AN Xn).

Thus by our result in 1 we have

(3)

An = Anja t+ Anes Anja K ns Ans Ln

@

dna E) = [ Fadi, for E €LNXm

on (Xn, AN Xp),

E

where f, is a nonnegative extended real-valued 2 M X,-measurable function on Xp. Let us define two set functions 4, and A, on 2 by setting

6)

Aa(E) =) Ana(EN Xn) for E € M,

©

s(E) = }ans(EOX_)

neN

neN

for E € Q.

It follows immediately that A, and A, are two positive measures on (X, 2). To show the countable additivity of 4, on 2 for instance, let (E; : k € N) be a disjoint sequence in 2. Then we have

da(U Be) = Dana LU Be Xn) = 7 Yo nalEe Xn) keN

neN

keN

= oan a(Ex keNacN

neNkeN

Xn) = Dra E)keN

It remains to show that for the two positive measures Ag and A, on (X, 2) defined

respectively by (5) and (6) we have (7)

A=hatAs;

Aa K M5 Ay Le

on (X, MW).

§11 Absolute Continuity of a Measure

241

Now for every E € 2 we have

AE) = DO MEN Xn) = So an(E neN

Xp)

neN

= > {An,alE M Xn) + An,s(EM Xn)} neN

= Doan clEN Xn) +) dns(E neN

neN

Xn)

= ha(E) + As(E) where the third equality is by (3) and the last equality is by (5) and (6). This shows that we have A = Ag +A, on (X, 2). a null set in (X, A, uw). Then w(E) = 0 Let us show that Ag < u on (X, 2). Let E be and this implies z,(E 9 Xn) = w(E X,) = 0 for everyn € N. Since Ana K fin On

(Xn, Xn 1 A) by (3), we have Ana(E OM Xn) = 0 for every n € N. Then by (5) we have Ag(E) = VnenAna(E 1 X,) = 0. Thus E is a null set in (X, A, 4,2). This shows that

Aa & won (X, A).

Let us show that A, 1 yz on (X, 21). By (3) we have Ans -L My on (Xn, AN Xz) for everyn € N. Thus there exist A,, B, € AM X, such that A, 1 B, = GB, A, UB, = Xp,

B,, is a null set for ,,5, and A, is a null set for zn. Let A= Ucn An and B= ey Ba. Then A, B € 2. Since (X, : n € N) is a disjoint sequence and A, and B,, are disjoint subsets of X, for every n € N, {An, By: n € N} is a disjoint collection. Then we have

ANB= (U*) (Us) =U (4.0 (U»)) ~Us=4, neN

AUB=(U4n)U(U Be) = Un UB) =U Xn =X. neN

neN

neN

neN

Now (A) = Den M(AN Xn) = Donen Hn(An) = 0 since A, is a null set for ,. Also by (6) we have As(B) = Yo en Ans(BM Xn) = Docu An,s(Bn) = 0 since By is a null set for An,s. This shows that A, L 4 on (X, 2). This completes the proof of (7). Let us show that for the positive measure 1, on (X, 2) defined by (5) there exists a

nonnegative extended real-valued {{-measurable function f on X such that

(8)

Aa(E) = [ fdu

forE eA.

With the nonnegative extended real-valued 2 1 X,,-measurable function /, on X,, defined in (4) for n 1 yu by (6) of Lemma 11.9.

Let f = fi — fo. Now since A+ is a finite signed measure and At = AS + A+, Ar

is a finite signed measure. Then by (11) we have iz fidu = F(X) € BR so that f; is real-valued j-a.e. on X. Thus f = fi — f2 is defined yz-a.e. on X and is an extended

real-valued {{-measurable function. Then for every E € 2% we have

[rau= [in-man= f fiau- ff ran =AP(E) — 25 (EZ) = aE)

§11 Absolute Continuity of a Measure

243

where the third equality is by (11) and (12) and the last equality is by (13). This completes the proof that A = Ag +As, Aq K ft, Ay L pe and Ag(E) = Se f du for E € Mt wheref

is an extended real-valued 2{-measurable function on X for the case that A* is a finite and 47” is aa-finite positive measure. The case that A+ is a o-finite and A~ is a finite positive measure is treated similarly. = Observation 11.11. Let jz be a positive measure and A be a signed measure on a measurable space (X, 2).

(2) b) ©

A Oand 42 > 0, A-ae, on X.

{c) Show that plve Prob. 11.10.

%.2

=0,d-00. on X

Given a measurable space (X, 2).

For A € 2, let us write 2M

A for the

a-algebra of subsets of A consisting of subsets of A of the type EN A where E € 2. Prove the following theorem related to the Lebesgue Decomposition Theorem:

Theorem. Let jz and v be o-finite positive measures on a measurable space (X, 2M). There

exist A, B € Asuchthat ANB = 9, AUB = X, p ~ von and yz | v on the measurable space (B, 212 B).

the measurable space (A, ANA)

Prob. 11.11. Prove the following theorem related to the Lebesgue Decomposition Theorem: Theorem. Let jz and v be o-finite positive measures on a measurable space (X, 2M). There

exist a nonnegative extended real-valued {{-measurable function f on X and a set Ag € A with (Ag) = 0 such that v(E) = fy f du + v(EM Ap) for every E € 2.

Chapter 3 Differentiation and Integration §12

Monotone Functions and Functions of Bounded Variation

[1] The Derivative To define the right-derivative and the left-derivative of an extended real-valued function defined on an interval in R let us review the definition of the right limit and the left limit of a function. Let f be an extended real-valued function on (a, 5).

1. If there exists a € R such that for every¢ > Othere exists § > 0 such that | f(x)-—a| 0 there exists 5 > 0 such that f(x) > M forx € (a, a + 6) then we

say

that oo is the right hand limit of f as x | a and write lim f(@) = oo. If for every M > 0 xia there exists

> 0 such that f(x)

< —M

forx

€ (a,a +4) then we say that —oo is the

right hand limit of f as x | a and write lim f@) = —00. Ya

2. If limyjo f (x) exists in R, then for every sequence (x, : 2 € N) in (a, b) such that lim x, =a wehave lim f(x,) = lim f(x). noo

n>00

xla

Conversely if for every sequence (x, : n € N) in (a, b) such that im, Xn = @ the limit im, f Xn) exists in R

then the limit is the same for all such sequences (x, : » € N) in

(a, b) and moreover iim f(x) exists and lim f@)=

tim, S Gn).

3. The left limit of f as x + b, that is, as x tends to b from the left, is defined likewise. Thus for instance lim F(x) = B € Rif for every ¢ > 0 there exists § > 0 such that *

|f@) — B| < ¢ for x € (b — 4, b).

257

258

Chapter 3 Differentiation and Integration

4. Let f be an extended real-valued function on [a, b]. Let x € [a, b) and f(x) € R. We

define the right-derivative of f at x as follows. Let 6 > 0 be so small thatx +6 < b. Define a function g on (0, 8) by setting

a) = fern — f@)

forh € (0, 8).

The right-derivative of f at x is defined by

(D, fY) = kim a), provided the limit exists in R. Definition 12.1. Let f be an extended real-valued function on [a, b].

(a) fx € [a,b), f(x) € R, and

pm SOTA —FO) (Dr f)(%) = jim a

exists in R, then we call (Dy f)(x) the right-derivative of f at x and say that f is right semi-differentiable at x. We say that f is right-differentiable atx only if (D, f)(x) € R. (b) ifx € (a, bl, f(x) € R, and

mn SHA) — FO) (Def) = him a exists in R, then we call (De f)(x) the left-derivative of f at x and say that f is left semi-differentiable at x. We say that f is left-differentiable at x only if (Def)(x) € R. (c) fx © (a,b), f(x) € R, and both (Dz f)(x) and (D, f)(x) exist in IR and are equal, we let (Df)(x), or as an alternate notation f'(x), be the common value, Thus

+h) — #@) = Ofla) = jim: 0 FETY= FA) h We call f'(x) the derivative of f at x and say that f is semi-differentiable at x. We say that f is differentiable at x only when f'(x) € R. (d) Let us agree to say that f' exists on [a, b] if D, f exists on [a, b), Def exists on (a, b],

and Def = Dyf on (a, b).

If a real-valued function f on (a, b) is differentiable at some x €

(a,b), then f is

continuous at x. The converse is false. In fact there exists a continuous function on R which is not differentiable at any x € IR. To show such an example, let us prove a necessary condition for differentiability.

Proposition 12.2. Let f be a real-valued function on (a,b). Suppose f is differentiable at some x & (a,b).

Then for any two sequences (a, : n € N) and (by, : n € N) such that

a, t x andb, | x asn— 00, we have

tim 10) = F@n) _

n>00 = bn — Oy,

f'@).

§12 Monotonicity and Bounded Variation

259

Proof. Forn € N, let A, = (6, — x)/(bn — ay). Then we have A, € [0, 1] and 1 —A,

(& — an) / (bn — Gn). Now

@

f'@) =Anf'@) +

=

- ad f’@

and

2

Fn) — F@n) _ Fn) — FR) | FG) — Flan) bn — Gn

bn — Gn

_ 3,1

bn — Gn

-f@) +d —a yf 2 by — xX

= FG@n) X—4G,

Subtracting (1) from (2), we have

FOr) = FCW) p(y) = dyn + (1 — An) Bas by — Qn where

b

hg = LOD=FO _ pre) and By =< LOA FO) _ prep, by — Xx

X—@,

By the differentiability of f at x, we have

jim, A, = O and im, B, = 0. Then since

(Aan : 2 € N) and (1 — A, : n € N) are bounded sequences of real numbers, we have jim An An =Oand jm d — A,) B, = 0. Therefore we have

lim

nO

This completes the proof.

{2

= Fn) _ re} -0.

nm — Gy

um

Example. (A Continuous Nowhere Differentiable Function on R) Let ¢ be a real-valued function on R defined by setting g(x) = |x| forx € [—1, 1] and then setting g(x +2) = g(x) for x € R. Define a sequence (yg, : n € N) of functions on R by setting g,(x) = o(4"x)

forx € Randn EN. Let

fH = > ()"e.@)

forx eR.

neZy

Then f is continuous at every x € R and not differentiable at any x € R. Proof.

The function g is continuous and periodic with period 2 and |g] < 1 on R.

The

series of functions in the definition of f converges uniformly on R by the Weierstrass Test in Calculus and thus f is a continuous function on R. Let us show that f is not differentiable at an arbitraryx € R. Fora fixed m € Z1, there exists k € Z such that k m, we have 4"b, — 4"am, = 4°4-™ which

is an even integer so that |p(4"b,,) — y(4"a,,)|

= 0 by the fact that y is periodic with

period 2. Next ifn = m, then 4a, and 4b», are integers and 4"b», — 4”a», = 1 so that |p(4" bp) — 9(4"am)| = 1. Finally ifn < m, then 4"a, and 4"b,, are not integers and

furthermore no integers lie between them so that by the definition of g and by the fact that Vb

— Fay, = 4°, we have |9(4"bm,) — 9(4"am)| = 4*—”. Summarizing these results,

we have

0

Gn ©m) — Gn(Gm)| = |p 4" bm) — 94" am)| =

forn > m,

_

4°-™"

forn

(Gyan = tn\Gn)

= | » (5)

ee

fale) |.

Applying the inequality ||| — |8|| < la — A| fora, B € R, we have

a —_

le

m

eal - LG) im)

—

Pm

mol

=() ee LG) ee 3

=3" Thus we have

lim, |f Gn)

™

1

a

ama

3

"

1

am

n=0

2

nm

Gn(bm) — fn(Gn)| bm — Gm

=" ui

5

7

0

2°

— f(@m)\/lom — am|

= 0.

Since ay, t¢ x and by

| x as

m — od, this implies that f is not differentiable atx by Proposition 12.2. If f is a differentiable function on [a, 5], then the real-valued function f’ need not be continuous on [a, #]. (For example, the function f(x) = x? sin 1/x for x € [—1, 1] \ {0} and f (0) = 0 is differentiable on [—1, 1] but f’ is discontinuous at x = 0.) Nevertheless

f', like a continuous function, has the intermediate value property.

Theorem

12.3.

(Intermediate Value Theorem for Derivatives) Let f be a real-valued

function on [a, b]. If f is differentiable on [a, b] and if f'(x1) # f! (x2) for some x1, x2 € [a, b], x1 < x2, then for every c € R between f(x) and f'(x2) there exists xo € (x1, x2) such that f'(x9) = c. Proof.

Let us assume f’(x;} < f’(x2).

Let g(x) = f(x) — cx forx € [a,b].

Then

g is a real-valued continuous function on [a, b] and in particular on [x1, x2] so that it assumes a minimum on [x1, x2] at some point in [%1, x2], that is, there exists xq € [x1, x2] such that g(xo) < g(x) for x € [x1, x2]. Let us show that actually xo € (x1, x2). Now

g'(x) = f'(x) — ¢ so that g’(x1) < 0 < g’(x2). Since

er) = (Drea) = lim © (1

+A) — 8G)
O such that g(x; + hk) — g(x1) < O fork e€ (0,5). Thus there exists & € (x1, x2) such that g(&) — g(x1) < 0, thatis, g(&) < g(x1). Thus xo # x1. Similarly

7

_

_

8 (x2) = (Deg) (2) = hn

gGa +h) — G2) _ lim §@2) — g@2 — A) > h

ho

h

0

so that there exists § > 0 such that g(x2) — g(x. — hk) > Oforhé e€ (0,8). Then there exists € € (x1, x2) such that g(x2) — g(&) > 0, that is, g(x2) > g(€). This shows that Xo % X2. Therefore xq € (x1, x2). Then since g is differentiable at x9 and since g assumes a minimum at xo, we have g’(xo) = 0. Then f’(xo} =c. Example. If we define a real-valued function f on [—1, 1] by setting f(x) = 0 for x € [-1,0) and f(x) = 1 forx e€ [0,1], then f does not have the intermediate value

property so that by Theorem 12.3, f is not the derivative of any function on [—1, 1].

Regarding the measurability of the derivatives of a measurable function, we have the following Proposition 12.4, Let f be a real-valued IN, -measurable function on [a,b]. Let E be a IN, -measurable subset of [a,b].

If D; f exists on E, then D, f is a IN,-measurable

function on E. In particular, if D, f exists on a.e. on [a, b] then D, f is a Dt, -measurable function on [a, b]. The same holds for Dzf and f’. Proof. Let us extend the definition of f to R by setting f = 0 on R \ [a, b]. Let us define

a sequence of real-valued function on R by setting gn(x) = n{f (x ++) — f(x)} for

x ER. The Mt, -measurability of f on R implies that of g, by Theorem 9.31 (Translation

Invariance). iim

Let D, = {R : Jim, gn exists in R}.

By Theorem 4.23, D, € 90t, and

8n is Mt, -measurable on D,. Now if (D, f)(x) exists for some x € [a, 5), then

F@+D-F@) _ sing fet I __ Tim, gn(x). (Dr f(x) = lima FE+M-FO) h ~~ 1-00 Thus if D, f exists on a 99t,-measurable subset E of [a, b), then E C D, and moreover

Df

= lim 1

8x on E so that D, f is 2%, -measurable on E.

w

We shall show in §13 that if f is an absolutely continuous function on [a, 5] then f’ exists a.e, on [a, b] and Jab) f' du, = f(b) — f@. We shall show also that if f is a real-

valued function on [@, b] and f’ exists and is finite everywhere on [@, b] and is jz, -integrable

on [a, 5], then fi,4 f’du, = f)— f@.

To define the Dini derivates let us review the definition of the right limit inferior, the right limit superior, the left limit inferior, and the left limit superior of an extended real-valued function defined on an interval in R. Let f be an extended real-valued function on (a, 5).

262

Chapter 3 Differentiation and Integration 1. The right limit inferior and the right limit superior of f as x | a are defined by

lim inf f@) = Lin int, F@x) and

lim sup f(x) =lim ala

sup

f(x).

810 (a,a+8)

Let us note that | int, yf (x) increases as 6 | 0 so that bm aint, yf («) existsin R. Similarly sup

Gat)

f(x) desrenres

as 6 | 0 so that lim

and lim sup f (x) always exist.

Moreover

x{a

sup

fe) pane in R.

540 (@,a+8)

Thus lim inf f(x)

lim F(x) exists if and only if lint f@ Ra

=

lim sup f (x) and when this is the case we have lim F(x)= liminf f (x) = limsup f(x). xa xa x{a 2. There exists a decreasing sequence (x, : n € N) in (a,b) such that x, | a@ and jim, Ff Gn) = im inf f(x). Similarly there exists a decreasing sequence (x, : n € N) in xa

(a, b) such that x, | a and

lim

f(x,) = limsup f(x).

100

xa

( Let us prove the first of the two statements for instance. Let y = Lim inf f(x) ER, that

is, y = fm inf, f). Since wait (i

y Fi increases as 6 | 0, for an axbitrary sequence

:k EN) of positive numbers och that 5¢ | 0 we have

fi, ith, FO = Ym inf SO = Select x, € (a,a + 6;) such that att, yF@) since we have have dim, lim jm

F(x)

= y.

int inf

f@)


a@ and

jm

Xp =a.

Then we can select a subsequence (x;, : 2 € N) of (x, : & € N) such that 3Xk, ~) a. Let (%p in EN) be the sequence (x; :n € N). Then (x,: m € N) is a decreasing sequence in

(a, b) such that x, | a and

lim f (xn) = lim f(xz,) = lim fax) =v.)

n->00

n->00

k—00

3. The left limit inferior and the left limit superior of f as x ¢ b are defined by

liminf f@) = fim int FG x) and

Tim sup {@O= im

SUP

f(@).

4, Let f be an extended real-valued function on [a, 5]. Let x € [a, 6) and f(x) € R.

We define the lower-right Dini derivate of f at x as follows. Let 5 > 0 be so small that x +6 < b. Define a function g on (0, 6) by setting

g) =

f@ +h) — FR) h

forh € (0, 8).

§12 Monotonicity and Bounded Variation

263

The lower-right Dini derivate of f at x is defined by (D1 f)(x) = lim inf ath). Definition 12.5. Let f be an extended real-valued function on [a, b]. (a) Ifx € [a, b) and f (x) € R, we define

Ds Nee) = tig FEAM= IO) my gg LE+N= IO oe h

(Dt f)(x) = lim sup 2

ALO

840 he (0,8)

+) — fF) =lim

h

sup FG@ +h) — FR)

540 he(0,8)

(b) Ifx € (a, b] and f(x) € R, we define

D_H(0) =tipjge FEFD= FE) yg h

h

840 he(—8,0)

h

FE*M= IO) op h

(D~ fe) = lim sup LE FO A~LO _ iy sgyp Ato

We call (D, f)(x),

(Dt f)(x),

h

(D_f)(x),

FEF“ LO oR

840 ne(—8,0)

and (D7 f)(x)

eR.

h

respectively the lower-right,

upper-right, lower-left, and upper-left Dini derivates of f at x.

Note that the four Dini derivates always exist in R and (D_f)(x)

(Dif)

< (Dt f)@).

f is right semi-differentiable at x if and only if (D; f)(x) = (Dt f)()

Ff is right-differentiable at x if and only if (D4 f)(x) = (D* f)(x) € R.

< (D7 f)(x) and

_

ER.

Ff is left semi-differentiable at x if and only if (D_f)(x) = (D7 f)(x) ¢ R. Ff is left-differentiable at x if and only if (D_ f)(x) = (D7 f)(x) ER.

f is semi-differentiable at x if and only if the four Dini derivates of f at x are all equal. F is differentiable at x if and only if the four Dini derivates of f at x are all finite and equal.

[M1] Differentiability of Monotone Functions A real-valued function f on an interval J in R is called an increasing (or non-decreasing) function if f(x1) < f(x2) for every pair x1, x2 € I such that x; < x2, and a decreasing (or non-increasing) function if f (1) > f(@2). If f is a decreasing function, then —f is an

increasing function and thus we need not treat decreasing functions separately. A function is called a monotone function if it is either an increasing function or a decreasing function. Let f be a real-valued increasing function on an open interval J in R. Let a be an arbitrary point in J. Since f(x) t asx t¢ a and since f(x} < f(a) forx € I such that x 0, the

264

Chapter 3 Differentiation and Integration

Theorem 12.6. Let f be a real-valued increasing function on an open interval I in IR. Then Ff has at most countably many points of discontinuity in I. Proof. Let a; and a2 be two points of discontinuity of f in J and assume a; < a2. Consider

the two open intervals J (a1) = (f(a1—), f(ai+)) and J(@) = (f(@2—), f(@4). To show that J (a1) and J (az) are disjoint, let us take an arbitrary point xp € (a1, @2). From

the fact that f is increasing on J, we have f(aj+) < f(xo) < f(a.—). Thus the two open intervals J (a1) and J (az) are disjoint.

Let A be the set of all points of discontinuity of f in 7 and consider the collection of

the open intervals {J(@) : a € A} where J(a) = (f(a—), f(@t)).

From J(@) select

arbitrarily a rational number rz. Since {/(a) : a € A} is a collection of pairwise disjoint

sets, {rz : a € A} is acollection of distinct rational numbers. Since there are only countably many rational numbers, A is at most a countable set.

If f is a real-valued increasing function on an open interval J in R, then f is continuous except possibly at countably many points in J. Since continuity of f at x € I implies F(x)

= f(x+), we have f(x)

= f(x+)

except at countably many x € J.

Thus if we

define a real-valued function g on I by setting g(x) = f(x+) forx € I, then g = f except at countably many points in J. Since f is an increasing function, we have f(x1+) < f(x2+)

for any x1, x2 € I such that x; < x2. Thus g is an increasing function on 7. Furthermore g

is right-continuous on J. (This can be shown as follows. Leta € I be arbitrarily chosen. By the definition of (a+), for every e > 0 there exists

> 0 such that f(x) — f(a+) < efor

x € (a,a+8). Then f(x+)— f(a+)

a, forx eR.

{neN: &, 0 be arbitrarily given. Now since Donen @, < 00, there exists N € N such that Dev a, < é. Let > 0 beso small that the interval (xo, x9 +4) does not contain any point in the finite collection {&, ..., £7}. Then f(x)— f(xo) < D,.y On 0, since LneN Erg) a, = B < oo, there exists a finite subcollection No of N(é,,) such that nen a, > B —&. Since Np is a finite collection, there exists x < En, such thatx > & forn € No. Then f(x) > neNno G, > B —e. This shows that f(&,,—) = lims +6, F(x) = B—«. Then by the arbitrariness of ¢ > 0 we have f (&,.—) > 8. Therefore

FGnp—) = B by (2). Substituting this in (1), we have f (Eno) — f (Gna—) = Ong:

To show the continuity of f at every xo € R \ E, let N(xo) = {2 EN: & < xo} and B = Yrenta) 2a. Then f(x) = Drencay) %n = 8. By the same argument as above, we have f (xo—) = £ and thus f (xo) = f(xo—), proving the continuity of f atxp. Example 2. In Example 1, let us assume that EZ

= @. For x € R, let [x] be the greatest

integer not exceeding x. Let f be the function in Example 1 and let g be defined by

g(x) = f(x) +[]

forx eR.

Then the function g has the following properties: 1° gis areal-valued right-continuous increasing function on R. 2° EU Zis the set of points of discontinuity of g. The jump of g at &, is equal to a, for every n € N and the jump of g at an integer is equal to 1. 3°

lim

7-00

g(x) = —ooand

lim g(x) = 00.

x00

Example 3. Consider a finite open interval (a, b) in R. The real-valued function ¢ defined

on (a, b) by (x) = 52,

— a) — § for x € (a, b) is a homeomorphism of (4, b) onto

(-%. 3). The function y-(y) = tany for y € (—3, 3) is a homeomorphism of (—3, 3)

onto R and thus the function # = o g is a homeomorphism of (a, b) onto R. With the function g in Example 2, leth = go. Then h is a real-valued right-continuous increasing function on (a, b) with lim, 2 k(x) = —oo and lim,+, (x) = oo, and the set of points of discontinuity is a dense subset of (a, b).

266

Chapter 3 Differentiation and Integration We show next that a real-valued increasing function f is differentiable a.e. on [a, b].

For this we require the Vitali Covering Theorem.

Definition 12.7, Let E be a subset of R. A collection G of finite closed intervals is called a Vitali cover of E if for every x € E andé > 0 there exists I € % such thatx € I and £1)

< &. (Singletons are not admissible as finite closed intervals.)

The collection of all finite closed intervals in R is a Vitali cover for any subset of IR. For an arbitrary subset E of R, the collection {[x — 1x + 4] in © N,x € E} is an example

of Vitali cover of E. So is the collection {[x,x + }]:n €N,x € E}.

Observation 12.8. Let 23 be a Vitali cover of a subset E of R. For an arbitrary open set O such that O > E, let 239 be the subcollection of 23 consisting of all members of 23 which are contained in O. Then 83g

is a Vitali cover of E.

Proof. To show that 2% is a Vitali cover of Z, we show that for every x € E ands > 0 there exists J € 239 such that x € I and £(1) < ©. Nowifx ¢ E, thensinceE C O andO

is an open set there exists 5 > O such that (x —6, x+)

C O. Since &% is a Vitali cover of E,

there exists J € QJ such that x € J and £(1) < min (46, e}. ThenJ C (x —8,x+8) CO so that J € 23.

Also £(Z) < e. This shows that 279 is a Vitali cover of E.

a

Theorem 12.9, (Vitali Covering Theorem) Let %3 be a Vitali cover of an arbitrary subset E of R. Then there exists a countable disjoint subcollection {I, :n € N} of © such that

wo

ut(E\ Uh) =0. neN

if a (E) < 00, then the countable disjoint collection {I, : n € N} can be so chosen that

Q)

Yeh) < 00,

neN

and for every ¢ > 0 there exists N ¢ N such that

e))

N

ut(E\Un) E such that (E) < u,(O) < ui(E) +1 < oo. Let Bo be the subcollection of 2G consisting of those members of 23 which are contained in O. By Observation 12.8, Wo is a Vitali cover of E. Let ly € MQ be arbitrarily chosen. If WME \ 4) = 0, then the collection {J,} satisfies conditions (1), (2), and (3). If wy (E \ lh) > 0, then we select a disjoint subcollection {J : n € N} of 33g inductively as follows. Suppose for some n € N

we have selected a disjoint subcollection {11,..., I,} of Wo. If u* (E\Uger &) = 0, then

the collection {f1, ..., Jn} satisfies conditions (1), (2), and (3). If u*(E \ Uz &) > 0, then we select I,41 € Zo

as follows.

§12 Monotonicity and Bounded Variation

267

For brevity let n

(4)

Fr=(Jh&

and

0, =O\ Fy.

k=1

Then F, is a closed set in R and O, = O / F; is an open set in R and moreover we have

(5)

F,10O,=9

and

F,U0O,=0.

Let (6)

Wo, ={l€ Bo: 1 C On}.

Since 239 is a Vitali cover of E, it is certainly a Vitali cover of the subset E \ F, of E. Since E \ F, C O\ Fy = On, the collection 9,

by Lemma 12.8. In particular, Bo, 4 8. Let (7)

defined by (6) is a Vitali cover of E \ F,

d, = sup {£(1) : I € Go,}.

Since Wo, # O and £(Z) > 0 for every J € Wo,, we have d, > 0. On the other hand, for every I € Bo, we have I C O, C O s0 that £(2) < u,(O) < oo and then dy < 2,(O) < oo. Let us select Jn41 € Wo, such that

®

€Un41) > Adn.

Since I,41 C On, we have 419

F, = 0. Thus {4,..., Jn, J241} is a disjoint collection

in Go. If u*( Uztt ix) = 0, then conditions (1), (2), and (3) are satisfied by our collection {Ti,.. +> Jn Ingi} and we are done. If wet (Ugtt I,) > 0, then we repeat the selection process above to select J,42 € Wo,,,. If this selection process does not terminate in finitely many steps then we have a disjoint collection {7, : n € N} in Bo satisfying condition (8). By the disjointness of the collection {Z, : n € N} we have

Yt) = Ya,te) = 4,(U fh) < #,(0) < 00

neN

neN

neN

so that condition (2) is satisfied. Let us show that the collection {J, : n € N} satisfies (1). Let F = nen ‘n. If wf(E \ F) = 0, then we have (1). Suppose pf(E \ F) > 0. For eachn € N, let J, be aclosed interval with the same midpoint as J, but with £(J,) = 5£(,)-

Then Yen £m) = 5 Donen £Un) < 00 and this implies im, Dnzp £Un) = 0. Thus

(9)

jim, m(U In)

im, Len) =0.

If we show that

(10)

E\FC\|J4A forevery peN, n>p

268

Chapter 3 Differentiation and Integration

then 2*(E \ F) < #,(Un>p Jn) for every p € N and then by (9) we have * : ut(E\ F) < Pne(U In) —=0

and condition (1) is satisfied. Thus to prove (1) it suffices to prove (10).

Let p € N be arbitrarily fixed. Now since u7(E \ F) > 0, we have E \ F 4 G. Let x € E \ F be arbitrarily chosen. We show that there exists Z* € 28g such that

(11)

xé€I*C Jq

forsomeg > p.

Then we have x < (J,., Jn so that E\ F C U,>, Jn proving (10). Let us prove the existence of such I* € %39. Nowsincex € E\ F, wehavex ¢ F, sothatx € O\F, = On for every n € N. Then x € (),,¢y On and in particularx € O,. Now since Wo is a Vitali cover of E, it is a Vitali cover of the subset {x} of Z.

Now since O, is an open set and

{x} C Op, Wo, = {I € Wo : I Cc Oz} is a Vitali cover of {x} by Observation 12.8. Thus there exists I* € Mo, such that x € I* C Op.

It remains to show that J* C J, for some q > p. For this let us note first that since x € O, = O\F, and F, > I, wehavex ¢ I, for everyn € N. Thensincex € I*, we have

I* # I, for every n € N. Let us show that there existsN € N such that I*M Fy 4 6. Now

by (2) we have Den £Cn)

< oo so that fim, £Un) = 0. This implies that im, d, =0

by (8) so that for sufficiently large N

€ N we have dy

I* ¢ Woy and hence I* ¢ Ow by (6) so that I*N Fy Let g = min {n EN: I* 0 F, # OB}. Now since I* (F, : n € N) is an increasing sequence, this implies have I* 1 F, # @ but I* M F,_1 = G. This implies last inclusion and (6), (7) and (8) imply £(1*) < dg_1

# C g 7*
p. By the definition of g, we N I, # @ and I* C Og_1. The 2£(I,). Now since I*N I, 4 B

and since the closed interval J, has the same midpoint as J, and £(J,)= 5£(I,), we have

I* C Jq. Then since g > p we have I* C (>, Jn. Sincex € I*, we have x € U,,> JnBy the arbitrariness of x € E \ F, we have E\ F Cc Un>p Jn. This proves (10) Y and completes the proof that our disjoint collection {Z,: n € N} satisfies condition (1). Let us prove (3). Since }° cn &Un) < 00, for every ¢ > 0 there exists N € N such that

Yyewi1 £Un) < €. Now we have Fy = F\ Unswii dn = FO (Upsaii Jn) 80 that FE = F°U (Un=wH1 I,). Thus we have

En =(ENF)U(EN

LU) bh) ce\P/U(

a>N4+1

LU 4).

n>N+1

Since uf (E \ F) = 0 by (1), we have

wEN FR) N4+1

This proves (3) for our disjoint collection {Z, :n € N} in Wo. 2. Let E be an arbitrary subset of R. Let Z, = EM (m,n +1) forn e Z, the collection

of all integers. Since

9 is a Vitali cover of E, 93 is a Vitali cover of E, for everyn € Z. Let

§12 Monotonicity and Bounded Variation

269

Gant) be the subcollection of 2 consisting of those members of 2 which are contained in the open interval (n,n + 1). According to Observation 12.8, 3,41)

is a Vitali cover

of E,. Since 4*(En) < 00, by our result in 1 there exists a countable disjoint subcollection

Una: & € N} of Goan41) such that u*(En \ Uren Inz) = 0. Now {hye 2k € Nin € Z}

is a countable disjoint subcollection of 83 such that

Ue\UU be=U(@\UU me) ¢U (Ea\U it):

neZ

neZkeN

and

neZ

neZkeN

neZ

keN

ot(U%\UU ma) < Dat(Fo\ Ua) =0. neZ

neZkeN

nezZ

keN

Now E C { Uneg En} U Z so that

E\UUmec{Ue\UUsafuz. neN keN

nceN

neNkeN

Then we have

ut(z\ UU hha) < ut( Ua\UU ha) +p) =0. neNkeN

neN

neNkeN

This proves (1).

Theorem 12.10. (H. Lebesgue) Let f be a real-valued increasing function on [a, b]. Then the derivative f' exists and is nonnegative on (a, b)\ E where E isa null set in (R, Mu 1) contained in (a,b).

Furthermore f' is 3%,-measurable and 1, -integrable on (a,b) \ E

with Sia,b) f' du, < f()-—f@.

Inparticular f' is real-valued, that is f is differentiable,

(Mt, 4;,)-a.e. on [a, b]. Proof.

1.

Since f is an increasing function, if f’(x) exists at some x

€ (a, 5) then

Ff’) = 0. To show that f’ exists a.c. on (a, b), we show that the four Dini derivates are equal a.c. on (a, b). Since D_f < D~f and Dif < Dt f, it suffices to show that

D-f 0.

Show that

Prob. 12.7. According to Prob. 12.6, continuity of f on [a, 6] and existence and boundedness of the derivative f’ on (a, b) imply that f is a BV function. These conditions are sufficient but not necessary for f to be a BV function. To show this, find a real-valued function f on [a, b] satisfying the following conditions: 1° f is continuous on [a, 5]. 2° f' exists on (a, b). 3° fis not bounded on (a, 5).

4° f € BV([a, b)).

Prob. 12.8. Let f(x) = sinx forx € [0, 27].

(a) Show that f € BV([0, 27]) moreover we have yon (f) =4. (b) Find the total variation function vy of f on [0, 277]. Prob. 12.9.

Consider the interval [0,1] in R.

Forn

e€ N, let b,

=

i and let a,

=

AGn + by+1). Let f be a real-valued function on [0, 1] such that f(0) = 0, f(a,) = 0 for everyn EN, f(b,) = 1 foreveryn € N and f is linear on [a,, by] and [bp+41, ap] for every

nen.

Thus defined, f is bounded on [0, 1], continuous at every x € (0, 1] and discontinuous at

x = 0. Prove the following statements:

(a) For every & € (0, 1), we have f € BV([eo, 1]) and VIC) < 2(%0 — 1) where no € N such that a f

7 1 _ 1 Pld + x prlec(Ena) = f Ufa, + ia Ha(E)-

Since this holds for every n € N, we have BECF(E))

Theorem for the particular case that f’ is bounded on E.

< te \f'ldu,.

This proves the

For the general case, let E, = {x € E : |f'(x)| € In —1,n)} forn € N. Then {E, :n € Nh} is adisjoint collectionin Mt, andE = (_), cy En- By ourresult above, we have

ut(f(En)) < fr, Lf" du, for every n € N. Now f(E) = f(Upen 2n) = Upen f(En)Then by the countable subadditivity of u* we have

e(F()s ek(fEn) = > [ If'ldu, = [ If'ldpy. neN

This completes the proof.

neN

um

[I] Banach-Zarecki Criterion for Absolute Continuity Definition 13.6. Let f be a real-valued function on [a, b]. We say that f satisfies Lusin’s Condition (N) on [a, b] if for every subset E of [a, b] which is a null set in (R, 90,, u,), the set f (E) is also a null set in (R, Dt, u,). Lemma 13.7. [f a real-valued function f is absolutely continuous on [a, b), then for every & > O there exists 8 > 0 such that for every countable collection {Ian blineN } of non-overlapping closed intervals contained in [a, b] such that )°,.-j(bn —Gn) < 5, we have

Ynen fbn) — FG@a)| < € and new {SUP jay,041 f — inffao f} 0.

By the absolute continuity of f on [a,b], there exists 6 > 0 such

that for any finite collection of non-overlapping closed intervals {[az, bg] : k = 1,...,n}

contained in [a, b] with )-7_ (bg — ae) < 6, we have )7_, If be)— f(ae)| < 3. Take an arbitrary countable collection {Ian, bl ine N} of non-overlapping closed intervals contained in [a, b] such that Daenthn — a) < 8. Then for every N € N, the collection {[ax, b:] :k = 1,..., N}is a finite collection of non-overlapping closed intervals contained

in [a, b] with VP

— ax) < 8 so that yw

\f Ge) — f @x)| < §. Since this holds for

every N EN, wehave F nen lf (bn) — f@n)|= lim FP |x) — Flaw)! cn (sup; f — inf, f} < 2. Now

by Lemma 3.21, there exists an open set O containing FE and contained in (a, b) such that z,(O) < ui(E) +6=65. Leto= Unen(@ns by) where {(an,b,) : n € N} is

a disjoint collection of open intervals. If we let J, = [@n,b,], then {J, : n € N} is a countable collection of non-overlapping closed intervals contained in [a, b] such that Dnen£Un) = Dnen (bn — Gm) = (0)

e.

< 8. Then Donen { SUP ian ,Bal f —infpe,,o,] f}
0, we have u*(f(E)) = 0. This shows that f satisfies Condition (N) on [a, 5]. 2. Conversely suppose f satisfies conditions 1°, 2°, and 3°.

Let us show that f is

absolutely continuous on [a, b]. Now since f is a function of bounded variation on [a, b], its derivative f’ exists a.e. on [a, b], is 2M,,-measurable, and yz, -integrable on [a, b] by

Theorem 12.21. By Theorem 9.26 (Absolute Uniform Continuity of the Integral), the j,integrability of f’ implies that for an arbitrary ¢ > 0 there exists 8 > 0 such that wD

[ \f'ldu, 0 ae. on [a, b]. Now for x € [a, 5]

ro=f san=f ir- rau, + f Ia,

[a,x

fl du, = Bax) + Fae)

and therefore

4(x) = Hix) + Fix) = Hix) + fla) > fx) Then since

for ae. x € [a, BI.

{n] + f on [a, b], we have Fj > f ae. on [a, b]. This proves (7).

By (7) we have Stax Fodu, > Stax f dy, for x € [a,b]. This and (6) imply Sax Fidu, = Sax fd, for x € [a,b]. This proves (3) for the case that f is a #2, integrable nonnegative extended real-valued 9)t, -measurable function on [a, 5].

5. Finally let f be an arbitrary j, -integrable extended real-valued

Jt, -measurable

function on [@, 5]. We let f = f+ — f~ and apply the result in 4 to f+ and f~. Then Fy

exists a.e. on [a, b] and is 99%, -measurable and 2, -integrable on [a, b] and Fj = f ae. on Ca |

If f is an absolutely continuous real-valued function on [a, b] then the derivative f’ exists a.c. on [@, b] according to Corollary 13.3.

Lemma 13.16. Let f be an absolutely continuous real-valued function on [a, b]. If f’ =0 a.e. on [a, b] then f is constant on [a, b). Proof.

To show that f is constant on [a, b] we show that for every c € (a, b] we have

f£@© = $@. Now since f’ = 0 a. on [a, b], we have f’ = 0 ae. on (a,c). Thus there

exists a null set F in (IR, 90, ,) contained in (a, c) such that f’ = 0 on E = (a,c)\ F. We have E € 9, with w,(E) = u,((a,¢)) =¢—a < 00. Lete > Oand 9 > 0 be arbitrarily given. Let 5 > 0 be a positive number corresponding

to our ¢ > 0 in Definition 13.1 for the absolute continuity of f on [a,c].

x € E we have f’(x) = 0 so that there exists kh, > 0 such that

q)

Now at every

|f@e +h) — f(x)| < qh for everyh € (0, hz).

Then the collection of closed intervals 3 = {[x,x +h]:h € (0,h,),x € E} isa Vitali

cover of E. Since 44, (E) < 00, by the Vitali Covering Theorem (Theorem 12.9) there exists

§13 Absolutely Continuous Functions

295

a finite disjoint subcollection {J : k = 1,..., N} of 2 such that

N

@)

m(E\U&) 0 be chosen so small that x +h, < c. Let & = [xx, yx] with x, € EC (a,c) fork =1,..., N andlet{ :k =1,..., N} be so numbered that

6)}

yore a {F@e) — Fen} — > [F Ge) — Fad} + 68 k=l

f=1

o|F@d) — F@)| + 55 |F@n - F@)| + 68. k=1

é=1

Since ((ag, by) : k € N) is a disjoint sequence with ()pen(ax, be) = Vi, Usen(ce, de) is

a disjoint sequence with |),.4(ce, de) = V2 and since V, M V2 = , we have (az, by) 1 (ce, dg) = @ for any k, 2 € N. Let Po be the partition of [a, b] with a,, by :k =1,...,m and cg, dg: £=1,..., as the partition points. Then m

n

k=1

é=1

V2(F, Po) = >> (Fe — F@)| + >> |F(be) — F@)|. Thus

[

[a,b]

\fldu, < Ve(F, Po) + 6 < Ve(F) + 6e.

Since this holds for an arbitrary ¢ > 0, we have Sab lfldu,

< ve (F).

This completes

the proof that V2(F) = Sas |fl4erTheorem 13.22. Let f be a real-valued absolutely continuous function on [a, b). Then

v2(f) = [ Polat Proof. If f is a real-valued absolutely continuous function on [a, b], then by Corollary 13.3 the derivative f’ exists (Mt, , #,)-a.e. on [a, b] and is 9, -measurable and j2, -integrable

on [a, b] and moreover by Theorem 13.17 we have f@) =[

f Oma)

+ f(a)

forx € [a,b].

Then by Theorem 13.21, we have V2(f) = fi, a) |f’@| 4, (1). [V] Calculation of the Lebesgue Integral by Means of the Derivative [V.1] The Fundamental Theorem of Calculus Let us consider the problem of recapturing a function by integrating its derivative. According to the Fundamental Theorem of Calculus, if a real-valued function f is differ-

entiable everywhere on [a, b] and its derivative f’ is Riemann integrable on [a, b], then

f f'(z)dx = f(b) — f@).

According to Theorem 7.28, f’ is Riemann integrable on

[a, 5] if and only if f’ is bounded and continuous a.e. on [a, b]. In this way, the Fundamental Theorem of Calculus relies on the continuity of f’. In contrast to this, if f is an

§13 Absolutely Continuous Functions

301

absolutely continuous function on [a, b], then according to Theorem 13.17, f’ exists a.c. and is j2, -integrable on [a, b] with Jia.b) Ff’ dw, = f(b) — f@. In the next two theorems

we show that if, instead of assuming the absolutely continuity of f, we assume that f is differentiable everywhere on [a, b] and f’ is bounded on [a, b] (as in Theorem 13.23 be-

low), or if f is continuous on [a, b], f’ exists a.e. on [a, b], f’ is 4,-integrable on [a, b], and the upper-right Dini derivate D* f > —oo on [a, b) (as in Theorem 13.26 below), then

Fran f dtu, = FO) — Fla).

Theorem 13.23. (a) if a real-valued function f is differentiable everywhere on [a, b] and i

is a bounded function on [a,b], then f is absolutely continuous on [a,b] and thus

Son fan, = FO) - f@.

(b) If a real-valued function f on [a, b] satisfies the Lipschitz condition on [a, b], that is,

there exists a constant M > 0 such that | f(x’) — f(x")| < M|x’ — x” | for every x',x" € [a, b], then f is absolutely continuous on [a, b] so that Seat f' dp, = f)-

f@.

Proof. 1. The existence of f’ on [a, b] € 9, implies that f’ is Mt, -measurable on [a, 5] by Proposition 12.4. Suppose | f’| < M on [a, 5] for some constant M > 0. Then by the Mean Value Theorem we have | f(x”) — f(x)| = lf)" — x’)| < M(x" — x’) where

& € (x’, x") forany x’, x” € [a, b] such that x’ < x”. Givene > 0,let8 € (0,i) Then for any finite collection of non-overlapping closed intervals {Iae, by]: k =1,...,n}

contained

Mé

fidp,

in [a, b] with 7_,(bx — ay) < 8, we have 7_, |f (bx) — Fa) < MD F_1 Oe — a) <
0on [a,b]. For an arbitrary ¢ > 0, let § € (0, §). Then for any finite collection

{[ag, by] :

= 1,...,n} of non-overlapping closed intervals contained in [a, b] such that

Diya (be — 94) < 8, we have D_1 |f (be) — f(ax)| < M Ri (e — a4) < ME < 8.

This shows that f is absolutely continuous on [a,b]. Then by Theorem 13.17, we have Sab f'du,

=fo®)-f@.

a

Regarding the next lemma, let us note that while a real-valued increasing function on [a, b] has at most countably many points of discontinuity, at a point of continuity of f we may have f’ = oo. For example, if f(x) = ./x for x € [0,1] and f(x) = —./|x| for x €[-1,0), then f is continuous atx = 0 and f’(0) = oo. Lemma 13.24. Let E C [a, b] be a null set in (R, t,, 2,).

Then for every e > O, there

exists an absolutely continuous increasing function f on [a, b] such that f(b) — f(a) 0a. on [a, b) and D+ (f —G,) > —oo everywhere on [a, ), and similarly for D+(H,, — f). Now for any two real-valued functions g and g2

on a set D, we have supp {v1 — ¢2} > supp 1 — SUPp ¢2. Thus for everyx € [a, b) we have

6)

Dt (f — Ga)(x) = (D* f)(x) — (Dt Ga) ().

By (4) and (1) we have

©

(D*G,)(x) = lim 110sup 22@ +9 t — Gn@) = lim sup ; oF

Jixxt4]

8nd,

< dit =n, t

By (5) and (6) and by the fact that D+ f > —oo on [a, b), we have D*(f —Gn)(x) > —00

for every x € [a, b). By Theorem 13.15, Gi, exists and G/, = gp a.e. on [a, b] and hence D+tGn = gn ae. on [a, b). Thus by (5) and (1), we have Dt(f — Gn) > f’ — gn > Oon [a, b) \ E.

This verifies that f — G, satisfies the hypothesis of (b) of Lemma 13.25 and

therefore f — G,, is an increasing function on [a, b]. Then for every x € [a, b], we have

(f — Ga)@) = (fF — Gn)(a), that is, f(x) — f@

recalling (3), we have

o)

= fiax Bn dH,. Letting n > oo and

f@)-f@z [ Fain, Fors 6 [aD

Similarly for every x € [a, b) we have

(8)

D* (Hy — f(x) = (D* Hy)(x) — (Dt A))

306

Chapter 3 Differentiation and Integration

and by (1) we have

9)

(D* Hy) (x) = lim sup

Hy(x +1) — Bn)

110

t 1 = lim sup — [ hyadp, oF Joe xte] 1 > yom

=-n”,

From (8) and (9) and the fact that D+ f > —oo on [a, b), we have Dt (H, — f)(x) > —00 for all x € [a, b) and D*+(H, — f) > hy, — f’ = 0 on [a, b) \ E and thus by (b) of Lemma

13.25, H,, — f is an increasing function on [a, b]. From this we derive by the same argument as above that

€ a,b f@-f@< [ din, for

(10)

With (7) and (10), we have f(x) — f(a) = Stax f' dp, forx € [a, b]. Now that f is an

indefinite integral of the 2, -integrable function f’, f is absolutely continuous on [a, b] by Theorem 13.15. Corollary 13.27. Let f be a real-valued continuous and increasing function on [a, b]. If D+ f > —0o on [a, b), then f is absolutely continuous on [a, b). Proof. If f is a real-valued increasing function on [a, b], then by Theorem 12.10, f’ exists

a.e. on [a,b] and 0 < Sia.2 f'dp,

< f(b) — f@

< co so that f’ is 2, -integrable on

[a, b]. If we assume that Dtf > —oo on [a, b), then the conditions in Theorem 13.26 are all satisfied and thus f is absolutely continuous on [@, 5].

Letg be a real-valued continuous function on [a, 5]. If f is also a real-valued continuous function on [a,b] and if f =

g ae.

on [a,], then f =

g on [a,b].

(Indeed the set

E = {[a, 6]: f # g} is a mull set in (R, 9, ,) and thus E° is a dense subset of R by

Observation 3.6. Then forevery x € [a, b], there exists asequence (x, : n € N)in[a, BINES such that jim, xn = x. By the continuity of f and g, we have jim, fG@n) = f(x) and jim, g(%n)

= g(x).

Since x, € [a,b] M E°, we have f(x,)

= g(x) for every n € N.

Then f(x) = g(x).) Let g be a real-valued continuous function on [@, b]. Let k be a real-valued function on [a, b] such that k = g a.e. on [a, b]. We show in the next Corollary that if # is the derivative of some differentiable continuous function on [a, b], then h = g on [a, b].

Corollary 13.28. Let h and g be two real-valued functions on [a, b]. Suppose

1° 2° 3° Then

h=gae. on[a, b], h= f' for some differentiable continuous function f on [a, b], g is continuous on [a, b]. h = g on [a, b].

§13 Absolutely Continuous Functions

307

Proof. Condition 3° implies that g is jz, -integrable on [a, b]. Then by 1°, # is w, -integrable on [a, 6]. By 2°, the continuous function f is differentiable on [a, b] and its derivative h is

4, -integrable on [a, b]. Since f’ = h we have Dtf = f’ =h > —oo on [a, b). Thus f satisfies all the conditions in Theorem 13.26 so that by Theorem 13.26 and 1° we have f@-f@=

[a,x]

haw, =

[

[a,x]

gdp,

forevery x «€ [a,b].

This shows that f is an indefinite integral ofg. Then 3° implies that f’ = g on [a, b] by Lemma 13.14. Therefore h = g on [a,b] by 2°. Example. Let # be a real-valued function on [—1, 1] defined by A(x) = 1 for x = 0 and h(x) = Oforx € [—1, 1]\ {0}, and letg = Oon[—1, 1]. We haveh = g ae. on [—1, 1]. If h were the derivative of some real-valued continuous function on [—1, 1], then by Corollary 13.28, kh would be equal to g on [—1, 1] contradicting the fact that k(0} £ g(0). Therefore his not the derivative of any function on [—1, 1]. [V.2] Integration by Parts

If two real-valued functions f and g are differentiable on [@, b] and the derivatives f’ and g’ are Riemann integrable on [a, b], then fg’ and f’g are Riemann integrable on [a, b] and

JP Fade’ @)dx t+ f2 f(x) dx = f()g() — f@g(a). This is the Integration by

Parts Formula for the Riemann integral. A corresponding formula for the Lebesgue integral may be stated as follows. Theorem 13.29. Let f and g be two real-valued absolutely continuous functions on (a, b]. Then fg! and f'g are .,-integrable on [a, b] and for every x € [a, b] we have [a,x]

fa! du, +f

78 du, = f(x)g) — f@g@).

Proof. The absolute continuity of f and g implies that of fg. Thus by Theorem 13.17, the derivatives f’, g’ and (fg)’ exist a.e. on [a, b] and are St, -measurable and jz, -integrable on [a, 5], and

a)

[

(fay du, = f(x)g@) — f@s@)

forx € [a, b]. Now (fg)' = fg’ + f’g wherever both f’ and g’ exist, that is, a.e. on [a, b]. Then for every x € [a, b] we have

(2)

[a,x ]

fe du, =f [a,x] ife'+ F'ehdu,.

Since g’ is j1,-integrable on [a, b] and since f is absolutely continuous on [a, b] and is

hence bounded on [a, b], fg’ is 4, -integrable on [a, b]. Similarly for f’g. Thus for every x € [a, b] we have by Proposition 9.14

(3)

[a,x]

(fa' + f'shdu, =|

18 de

+f

{feat

308

Chapter 3 Differentiation and Integration

With (1), (2), and (3), we are done. Corollary 13.30. Let f and g be two real-valued absolutely continuous functions on [a, b] and let F and G be arbitrary indefinite integrals of f and g on [a, b] respectively. Then Fg and fG are j.,-integrable on [a, b] and for every x € [a, b] we have

[ream+f tedu, = FWGe)—- FOG. [a,x] [a,x] Proof. If f and g are absolutely continuous functions on [a,b], then f and g are p,integrable on [a, 6]. If F and G are arbitrary indefinite integrals of f and g respectively on [a, b], then F and G are absolutely continuous on [a, b] and furthermore F’ = f and G’ = g ae. on [a, b] by Theorem 13.15. Applying Theorem 13.29 to the absolutely continuous functions F and G, we have the Corollary. [V.3] Change of Variables A standard Change of Variable Theorem for the Riemann integral states that if f is a realvalued continuous function on [@, 8] and if g is a real-valued increasing function with continuous derivative g’ on [a, 5] and g(a) = o and g(b) = 8, then we have the equality

fF £0) dy = °F 0 g)(x)g’(x) dx. Note that the continuity of g’ on [a, 5] implies that g is absolutely continuous on [a, b]. In Theorem 13.32 below, we prove a corresponding Change of Variable Theorem for the Lebesgue integral which does not require the continuity of g’.

Lemma 13.31. Let g be an absolutely continuous increasing function on [a, b] which is not

constant on [a, b] and let g([a, b]) = [a, B]. Let D = {[a, b] : g’ > 0}. (a) IfG C [a, B] isa Gp-set, then w,(G) = S-@ g' dp.

(b) FE C [a, 6), E € Mt, u,(E) =0, then g(E)N D € Mt, u, (gE) ND) =0. (c) FE

C [a, B)] and E € M,, then ge

(E) NDeM,,

[a, b] and p, (E) = Jta,b1 1-19" du,

1,-1(2)8"

is Mt, -measurable on

(Let us note that since [a, b] is a compact set and g is a continuous mapping, a(Ia, b))

is a compact set. The continuity of g also implies that g([a, b]) is an interval by the

Intermediate Value Theorem. Thus g([a, b]) is a finite closed interval which we denote

by [a, 8].

Since g is an increasing function, we have g(a) = a and g(b) = 8. The fact

that g is an increasing function implies also that the derivative g’ exists and g’ > 0 ae. on [a, b], g’ is $t, measurable and jz, -integrable on [a, b] by Theorem 12.10. The absolute

continuity ofg implies that Jie

a du, = g(b) — g(a) = B — @ by Theorem 13.17. The

Ot, -measurability of g implies that g~'(E) € MM, for Z ¢ Br, but for E € Mt, we may not have g (BE)

€ Dt, and 1,-1(z) may not be a Mt, -measurable function. Part (c) of the

Lemma asserts that 1,-1;z)g’ is a Proof.

Mt, -measurable function.)

1. To prove (a), as a particular case of a G3-set consider first an open interval

§13 Absolutely Continuous Functions

309

contained in [w, 6], say (a9, Bo) C [a, B]. Since g isa continuous mapping, g~!((ao, Bo)) is an open set contained in [a, b]. The monotonicity of g implies that the open set g—! (@o, Bo)) is an open interval, rather than a union of two or more disjoint open intervals. The continuity

of g implies that g~1({ao}) is a closed set contained in [a, b] and similarly g—({Ao}) is a closed set contained in [a, b]. Let a9 = max g1({ao}) and bp = min g~1({fo}). By the

continuity of g we have g(a) = 0 and g(bo) = fo and thus g~!{(ao, Bo)) = (ao, bo). Now

w

#,((@o; Bo)) = Bo — %0 = g(bo) — g(a) = | -f

(aosbo)

g'du, -f

8-*((ao.60))

where the third equality is by Theorem 13.17. Next, let O be an open set contained in [a, 8].

ge’ du, g' dp,

Then O

=

LJneny(@n, Bn) where

((@n, Bn) : 2 € N) is a disjoint sequence of open intervals contained in [a, 6]. Then

2 4,(0)= im (@n, A) =o [ “Geanay® nen

=

Jessen)

[ ve (eangy 8 aM

neN

' edu,

= |

deo”

g du,

where the second equality is by (1) and the third equality is by the disjointness of the

sequence (g~!((an, Bn)) : 2 € N) implied by the disjointness of ((n, Bn) : n € N).

If G is a G5-set, then G = ( 0, the equality of the measures of the two

sets implies Jetemnn, aan, = Sten, 8 du,. Using this in (4), we have

mE=f

8 (END,

sian =fg-(E)ND ad,

= [pow 1,-1z)1pg' du L -f ap) 1,-1¢z)8' du,.L © @

§13 Absolutely Continuous Functions

311

Theorem 13.32. Let f be an extended real-valued IN, -measurable function on [a, B). If g is an absolutely continuous increasing function on [a, b] with g(a) = a and g(b) = B, then

@

[e,A]

£O) uy, dy) = [

[a,b]

CF © g)(x)g'(x) m, x)

in the sense that the existence of one of the two integrals implies that of the other and the equality of the two. Proof. Consider the case f = 1g where E is a 9Jt, -measurable subset of [a, 8]. Then by (c) of Lemma 13.31, we have

fo[a, 8] fama fo[e, 8) tedu,=02)= fo[2,6] tyres! dn, a, = | [a,b] 1e(e(2))9') u, (dx) = [

(a,b)

(x) uw, (dx). (Cf 0 g)()g'

This shows that (1) holds for this particular case. Then (1) holds when f is a nonnegative

simple function on [@, 8] by our result above and by the linearity of the integrals with respect to the integrands. If f is a nonnegative extended real-valued 9)t, -measurable function on [a, 8], then by Lemma 8.6, there exists an increasing sequence of nonnegative simple

functions (g, : m € N) such that g, ¢ f on [a, 8]. Since g is absolutely continuous and

increasing on [a, b], g’ exists and g’ > 0 ae. on [a, b]. Define 2’ = 0 on the exceptional null set so thatg’ > Oon [a, b]. Then we have (y,0g)g’ + (f og)’ on[a, 6]. By Theorem

8.5 (Monotone Convergence Theorem), we have lim Jive, pend, lim no

= Jie, a) f du, and

Sa,py(@n°8)8' dp, = Jiao (Ff °8)8' du, Since g, is anonnegative simple function,

we have fra, Gn dit, = fia,n(Gnog)s’ du, foreveryn EN. Thus we have fra, 4) fdu,= Si a,6\(F © 8)g' du. This shows that (1) holds when f is a nonnegative extended real-valued Dt, -measurable function on [@, 8]. When f is an extended real-valued 2, -measurable function on [a, 8], we decompose f as f = f+ — f~ and apply the result above to ft and f~. ©

[VI] Length of Rectifiable Curves Definition 13.33. Let f,,..., fm be real-valued functions on [a, b). Consider a mapping

x of [a, b] into R™ defined by x(t) = (fi), .-.. fu(t)) € R™ fort € [a,b]. We call the subsetC = {x(¢) : t € [a, b]} of R” a curve in R™ defined by the functions f,.... fm-

Definition 13.34. Let C be a curve in R™ defined by m real-valued functions fi, ..., fm

on [a,b]. LetP = {a = to < --- < t, = b} be a partition of (a, b] and let Bqy be the

collection of all partitions of [a,b]. Consider the points x(to), ...,x(f,) € C and the line segments x(fo), X(f1), ...,%(tn-1), X(tn) in R™. Let

(1)

P(P) = x(to), (1) UU

XCn-1), (fn).

312

Chapter 3 Differentiation and Integration

We call P(P) the polygonal line inscribed to the curve C by the partition P of [a, b]. The

length of the polygonal line P(P), written £(PO)), is given by

Q

e(P@)) = YeeiDxe

->

k=1

Sito — file). i=1

We define the length of the curve C by setting

Q)

£(C)=

sup £(P(P)).

PeBan

We say that the curve C is rectifiable if £(C) < 00. Observation 13.35, For ai, ...,@,, € R, we have

max |a;| janax, lal 1AdP. ka

1

By Observation 13.35 we have for each fixedk = 1,...,n

max, |Akl= os

Saye < 14) i=1

i=1

§13 Absolutely Continuous Functions

313

and then summing over k = 1, ..., n and recalling (1) we have

@)

>amax |Aj| < £(P(P))< > {Diath

Now for each fixed i = 1,...,m, we have

Vif?) = Suiew - feaol= Dial se max, |A;| < £(P(?)) so that we have

(3)

max

i=1,....m

Ve (fi,?) < e(PO)).

On the other hand, from (2) we have

@)

m

n

Pm) < > { yo Au} = 3°> {li — F@-DI} = k=1

i=l

i=1 k=l

m

i=l

V2(4. 9).

With (3) and (4) the proof is complete.

Theorem 13.37. Let C be a curve in R™ defined by m real-valued functions fi, ..., fn on [a, B]. Then C is rectifiable, that is £(C) < 00, if and only if fi, ..., fm € BV([a, 6). Proof. By Lemma 13.36, for any partition P of [a, b] we have

2x, Vi (fis < (PO) ?)

Y vel f. 2) i=l

Then

q)

oe

japax, Va P (fi, P) < op > (PO) = sp So

b P(f;, P).

Nowhi V2(fi, P) < D1 Suppem,, Ve (fi,P) so that (2)

sup

,

ye (fi,?) , sup

Peas ja

PEBas jay PERad

V2(fi, P) = y,

sup V2(fi, P) = Y¥2(0)

=1 PEBap

By (1) and (2) and recalling that £(C) = suppeg,, £(P(P)) by Definition 13.34, we have

(6)

£(C) < >) v2 (fi). i=1

314

Chapter 3 Differentiation and Integration

Now if fi,...fm € BV([a, b]), then V2(f;) < 00 fori = 1,...,m and £(C) < 00 by (2). Conversely if £(C) = suppesy, , £(P(P)) < 00, then (1) implies be P) < 00. onan, Vi(fi,

oe

This implies that we have suppeg,, , V2(fi, P) < cofori = 1,...,m, thatis, V2(f;) < co

fori=1,...,msothat fi,..., fm € BV([a, b]).

©

Let C be a curve in R™ defined by m real-valued functions f{,..., fm on [a,b].

In

Theorem 15.35, we showed that £(C) < 00 if and only if fi,.... fm € BV([a,6]). We show below that if fi,..., fi, are absolutely continuous on [a, 5] then we have

“=f

Dean,

—

Proposition 13.38.

AY2

Let f be a real-valued function which is differentiable (that is, the

derivative f' exists and is finite) (Dt, ,)-a.e. on [a, b]. Forn €N, let

hezath

qd)

a"

fork =0,1,...,2"

and let

Jn = nti thd) fork =1,...,2".

@

Define a sequence of functions (gy : n € N) by setting (3)

galt) =

Lobb) — Fmt)

oes

0

fort =tiajsk=1,...,2".

tak — bnk-1

Tyas = My..."

Then {4)

3 _ gt im, 8.=f ae. on[a, b].

Proof. According to Proposition 12.2, if f is a real-valued function on (a, b) and differentiable at some ft € (a, 6), then for any two sequences (a, : m € N) and (6, : n € N) in {a, b) such that a, ¢ t and b, | ¢ we have

6)

ay= pe, — FG) fO= in LO)

Suppose f is differentiable a.e.

on [a,b].

Then there exists a null set NW in the measure

space (IR, 0t,, 14,) such that N C [a, b] and f is differentiable at every¢ € [a, b]\ N. Let Qn = {tnx

k =0,1,...,2"} forn € N and let Q =

ncn Qn.

Then @ is a countable

§13 Absolutely Continuous Functions

315

subset of [@, b] and is therefore a null set in (R, Nt, #,). Now NU Q C

[a, 6] is anull set

in (R, 9, 2,). Consider [a, b] \{N U Q}. To prove (4) we show that im, ao=f'o

for every¢ € [a, b] \ {NW U Q}. Let Sn = (Jn 2K =1,...,2"} form € N. Then J, is a disjoint collection of open

intervals for each n € N. Moreover if n’,n” € Nandn’

[a,b] a

i=l

CA? du,. »

Proof. If f, € BV([a, b]) then by Theorem 12.21 the derivative f/ exists (O0t,,, ,,)-a.c.

on [a, b] and f/ is 99t, -measurable and jz, -integrable on [a, b]. The jz, -integrability of f/ implies that ff is finite, that is, f; is differentiable, (Mt,, ,)-a.c. on [a, 5]. For n € N, let

()

hecat+kea” fork =0,1,...,2%,

(2)

Ink = Onk-1 tng)

QB)

P, = {a=tho 0 defined with

some p € (0, 00) by setting

re= {4

x? sin! forx € (0, dl, for x = 0.

Show that f is absolutely continuous on [0, 5] if and only if p € (1, 00). Prob. 13.21.

Find an absolutely continuous function on [a, b] that does not satisfy any

Lipschitz Condition on [a, 5].

§13 Absolutely Continuous Functions

321

Prob. 13.22. (If f is a real-valued absolutely continuous function on [a, b] then according to Theorem 13.18, f’ exists z,-a.e. on [a, b] and f’ is Jt, -measurable and 4, -integrable on [a, b] and f’ is finite 2, -a.e. on [a, b].) Let E C [a, b] be an arbitrary non-empty null set in the measure space (R, Wt, , 2). Show

that there exists a real-valued absolutely continuous increasing function f on [a, b] such that f’(x) = 00 for every x € E. Prob. 13.23. (The Cantor-Lebesgue function t on [0, 1] is a real-valued continuous and increasing function with derivative t’ = 0 a.e. on [0, 1].)

Show that there exists a real-valued continuous and strictly increasing function f with detivative f’ = 0 ae. on [0, 1].

Prob. 13.24. Let f beareal-valued function on [a, 6]. Suppose f ([a, b]) C [c, d]. Let g be

areal-valued function on [c, d]. Suppose f satisfies a Lipschitz Condition with coefficient M, > 0 on [a, b] and g satisfies a Lipschitz Condition with coefficient Mz > 0 on [c, d]. Show that g o f satisfies a Lipschitz Condition with coefficient MzM, > 0 on [a, b].

Prob. 13.25.

Let f be a real-valued absolutely continuous function on [a,b].

Suppose

Prob. 13.26.

Let f be a real-valued absolutely continuous function on [a,b].

Suppose

f(Ia, b)) C [c, d]. Let g be areal-valued function on [c, d] satisfying a Lipschitz Condition. with coefficient M > 0 on [c, d]. Show that g o f is absolutely continuous on [a, b]. fi(Ia, b)) C [c, d]. Let g be a real-valued absolutely continuous function on [c, d]. Assume

further that f is increasing on [a, b]. Show that g o f is absolutely continuous on [a, b]. Prob. 13.27. Let f and g be two real-valued functions defined on [0, 1] by and

f@) = J _|

«={ 3

forx € [0, 1],

2 1 x*|sin;|

forx € ©, 1],

* forx =0.

(a) Show that f and g are absolutely continuous on [0, 1]. (b) Show that g o f is absolutely continuous on [0, 1].

(c) Show that f o g is not absolutely continuous on [0, 1]. (Hint for (c): Show that f o g is not of bounded variation on [0, 1].) Prob. 13.28. Let f be an absolutely continuous function on [a@, b] and g be an absolutely continuous function on [c, d]. Assume that f ([a, b]) C [c, d]. Show that go f is absolutely continuous on [a, 5] if and only if g o f is of bounded variation on [a, 5].

Prob. 13.29. For p € (1, 00), let _ |

re)={

x? sin,

0

*

forx € 0,5],

forx =0.

(Thus defined, f is absolutely continuous on [0, b] by Prob. 13.20.)

(a) Show that f’ exists on [0, 6] and f’ is continuous on (0, 5]. (b) Show that f’ is 2, -integrable on [0, 5]. (©) Show that when p € (2, 00), f’ is continuous on [0, 5]. (d) Show that when p € (1, 2], limz jo f ’(x) does not exist.

322 (e) (f) (g) (h)

Chapter 3 Differentiation and Integration Show Show Show Show

that that that that

when when when when

Prob. 13.30, Let

p p p p

€ € € €

[2, (1, [2, (1,

acon

00), f’ is bounded on [0, 5]. 2), f’ is not bounded on [0, 5]. 00), f satisfies a Lipschitz Condition on [0, 5]. 2), f does not satisfy any Lipschitz Condition on [0, 5].

% forne

_ J x2cos% forx €

se={ 4

*

(0.61

for x = 0.

(0,

d],

(a) Show that f’ exists on [0, 5]. (b) Show that f is absolutely continuous on [a, 5] and f’ is ,-integrable on [a, b] for every a € (0, b]. (c) Show that f is not absolutely continuous on [0, 5] and f’ is not jz, -integrable on [0, 5].

(int: With a = (2n + 1)” and By = Qn)? forn € N, compute fig) f’ dit.) Prob. 13.31. LetD C [a, bland D € Mt, with z,(D)

> 0. Show thatforevery 4 € [0, 1],

there exists c € [a, b] such that u, (DN [a, cl) =4- 4, (D).

(Hint: Consider the Lebesgue integral of 1pnja,xj for x € [a, 5].) Prob. 13.32. Let D C R, D € M,, and 4, (D) € (0, 00). Show that for everyA € (0, 1),

there exists

€ R such that 4, (DM (—00, &]) = 4-4, (D).

Prob. 13.33. Definition.

Let E C BR, E € Mt,

and I be a finite interval in R.

We call

(er yy hy, (E nT ) the average density of E over the interval I. For an arbitrary x € R, the density of E at x is defined by

pE@) = Him ag, (EN@—h,x+h)).

Show that PEC) -{

1 0

forac. x € E, forae. xe E°.

§14 Convex Functions

§14

323

Convex Functions

[I] Continuity and Differentiability of a Convex Function Geometrically a convex function is a real-valued function on an interval Z in R such that for any two points x1 < x2 in J the chord joining the two points (1, fi (x1) and (xa, f(x2)) on

the graph of f is on or above the graph of f on the interval [x1, x2]. An analytic definition of a convex function is given as follows.

Definition 14.1. A real-valued function f on an interval I in R is said to be convex on I if for any x1, x2 € I and i € [0, 1], we have f (Ax; + (1—A)x2) < Af (x1) + (1 — A) fa). Let us call this condition the convexity condition. The equivalence of this definition to the geometric definition will be proved below. We shall show that a convex function f on an open interval J is continuous on I, differentiable everywhere except at countably many points in 7, the derivative is an increasing function on the subset of J on which it exists, and furthermore f is absolutely continuous on every finite closed interval contained in J. Observation 14.2. With x1, x2. € R such that x1 < defined by g(A) = Axi + (1 — A)x2 for A € [0, 1] [0, 1], mapping [0, 1] one-to-one and onto [x1, x2] immediate from the fact that g(A) = (x1 — x2)A +

x2, the real-valued function ¢ on [0, 1] is continuous and strictly decreasing on with ¢(0) = x2 and g(1) = x1. This is x2 for A € [0, 1].

Lemma 14.3. (a) Let f be a real-valued function on an interval I in R. For x1, x2 € I such that x, < x2, and’. € [0,1], let& = Ax, + (1 — A)x2. Then the following conditions are

all equivalent:

@

FE) SAF) + A — A) Fa).

w

sos2OFBt poy + So i = $6)

@) _ (iv)

f@< vasa x) + f(a). f@)— Feu) _< fon) ra — F@) foré # x1, Eom

x.

(b) A real-valued function f on an interval I in R is a convex function if and only if for any %1,%2 € I such that x1 < x2 the chord joining the two points (x1, fi (1) and (x2, f(x2))

on the graph of f is on or above the graph of f on the interval [x1, x2].

Proof. 1. To show the equivalence of (i), (ii), and (iii), we show that the right sides of (i), i), and (iii) are all equal. Note that by € = Axy + (1 —A)x2 = (x1 — x2)A + x0, we have

324

Chapter 3 Differentiation and Integration

= {x2 — E}/{x2 — xi} and 1 — A = {& — x1}/{x2 — x1}. Then

Af Gy) +0 -A)f@2) = 2A pq

4

10

This shows that the right side of (i) is equal to that of (ii). Next, writing x2 — € = (x2 — x1) + (1 — &), we have by simple computation

x2—€ fa) + §-x1 f(y) = f@2) - ep x2 — X41

x2

x2-

& — x1) + fn.

This shows that the right side of (ii) is equal to that of (iii). To show the equivalence of (ii) and (iv), we multiply both sides of (ii) by x2 — x1 to

obtain (x2 — +1) FE) < (2 — 8) fF) + & — 1) f 2). Adding —(€ — x1) f &), we have (x2 — x f ©) — € — x) £@) = G2 — HOF G1) + E — x1 Ff G2) — f E)} so that we have (2 — EMF E) — FD} Ss E — xf G2) — f(D}. For§ F x1, x2, by dividing by the

factors x2 — & and & — x, we obtain (iv). Conversely starting with (iv) and retracing the calculation above, we have (ii) for € # x1, x2. But (ii) holds trivially for € = x; and for & = x2. This proves the equivalence of (ii) and (iv).

2. Let f be a real-valued function on an interval J in R. Let x1, x2 € 7 and x; < x9.

The equation of the chord through the two points (x1, f(x1)) and (x2, f(x2)) is given by

a) = Pedx2— LOD

mi) + Far) for € bx, m2).

According to (a), the convexity condition (i) is equivalent to condition (iii). Therefore f is convex on J if and only if f(€) < g(€). o The next Proposition gives some fundamental inequalities for convex functions. Continuity and differentiability of a convex function are consequences of these inequalities. Proposition 14.4. Let f be a convex function on an interval I in}. Then for any u,v, w eI such that u < v < w, we have

FM-f/M < LMA FH. v-4u

w—u

~

Fw) — FO) wv

Proof. Since v € (u, w), there exists A € (0, 1) such that vy = Au + (1 — A)w. Solving this equation forA, we haveA = {w — v}/{w — u} and then 1 — A = {v — u}/{w — u}. By the

convexity condition satisfied by f, we have

fv) SAf@) + I-A FQ) = and then

qd)

(w — u) fF) < (w— vf) + & — «) Fw).

§14 Convex Functions

325

To derive the first inequality of the Proposition, let us write (1) as

(w— uw) fF) < {(w —#) —(v—o)} fO + Then we have

— wf).

(w—u){f@) — F@} < @-—w{F@) - F@}-

Cross-multiplying, we have the first inequality of the Proposition. To derive the second inequality of the Proposition, rewrite (1) as

(w — u) f(v) < (w — v) f@) + {(w — 4) — (w—v)} Fw) and then

(w —u){ fv) — fw)} < @w — v){ fF — f(w)}.

Cross-multiplying and then multiplying by —1, we have the second inequality of the Proposition. & Theorem 14.5. Let f be a convex function on an open interval I in R. Then we have: (a) f is both left- and right-differentiable at every xo € I, that is,

im

(

(De f)(@o) = lim atx

(2)

[O-~ FEO) op xX—XO

(D; f) (a0) = x4xp tim FOO FOxX—X%O

*

CQ

Furthermore for any x1, X2 € I such that x, < x2, we have

3)

2)— fF) (D, fr(e,) < FP= FOV < D, prow). %2— *1

(b) f is continuous on I.

(©) Def < Df on I.

(d) Dz f and D, f are real-valued increasing functions on I.

(e) f is differentiable, that is, the derivative f' exists and is finite, everywhere except at countably many points in I. If f! exists at x1, x2 € I, x1 < x2, then f'(x1) < f’(x2). Proof. 1. To prove the right-differentiability of f at any xo € 7, let x1,x2 € J be such that x9 < x1 < x2. Identifying xo, x1, x2 respectively with u, v, w in the first inequality in

Proposition A.4, we have {f(x1) — f (*o)}/{*1 — xo} < (f @2) — f(%o)}/{x2 — x0}. This

shows that { f(x) — f(xo)}/{x — xo} | as x | xo so that the right-derivative of f at xo,

that is, (D, f) (xo) = limyyxo{ f (x) — f (xo)}/{x — xo} exists in R and moreover we have (D, f)(xo) < {f @&) — f(xo)}/ {x — xo} for any x > xo. This proves the first inequality in (3). Letu, x € I be such that u < x9 < x. Identifying u, xo, x respectively with u, v, win

Proposition A.4, we have {f (xo) — f(#)}/(%o — 4} < {f(%) — fo) }/{x — xo}. Thus for

x > xo, (f (x) — f (x0) }/{x —xo} is bounded below by the constant { f (xo) — f ()}/{xo—4}. Therefore we have (D, f)(xo) = limzyay{f() — f(xo)}/{x — xo} € R and f is rightdifferentiable at xo.

326

x2

Chapter 3 Differentiation and Integration Similarly, to prove the left-differentiability of f at xo, let x1,x2 € Z be such that < x1 < xo. Identifying x2, x1, x9 respectively with u, v, w in the second inequality

in Proposition 14.4, we have {f (xo) — f (2)}/{xo — x2} = {f@o) — f@D}/ {x0 — x}. Thus { f (xo) — f(x)}/{%0 — x} t as x t xo.

Therefore the left-derivative (Dg f)(xo)

=

limy taf f(x) — f @o)}/{x — xo} exists in R and (De f)(x0) >= {Ff (o) — f @)1/ {x0 — x} for any x < xo, proving the second inequality in (3).

Next let x,w

€ I be such that

x < xq < w. Identifying x, x9, w respectively with u, v, w in Proposition 14.4, we have

{f@o) — f@)/ {x0 — x} < {f(w) — f@o)}/{w — xo}. Thus for x < xo, the quotient {fGo0) — £@)}/(o — x} is bounded above by the constant {f(w) — f(xo)}/{w — xo}

and therefore (Dz f)(xo)=

differentiability of f at xo.

limztuy{ f(*)— f(*o)}/{x — xo} € R.

This proves the left-

2. To prove (b), letxp € I. By (1), f is left-differentiable at xo so that f is left-continuous at xo. By (2) f is right-continuous at x9. Therefore f is continuous at xo.

3. To prove (c), let x9 € J. Let x’,x” € I be such that x’ < xp < x”. Identifying x’, xo, x" respectively with x, v, win Proposition 14.4, we have { f (xo)— f(x)}/{xo—x"}
0: (L()|’ < Mllvl forall ¢ Vv} € [0, 00).

(b) For the particular case (W, K’, | - i) = (K’ ,K’,|- } a linear mapping L of V into K’ is called a K'-valued bounded linear functional, abbreviated as b.Lf, if there exists M > 0

such that

|L(v)| < Mllv||

foralive V.

We write %3{V, K’} for the collection of all K'-valued bounded linear functionals on the

normed linear space (V,K, | - |l). For L € S3{V, K’} we define

Ell yyy = int {M > 0: |L(v)| < Mllvll for all v € V} € [0, 00). VK’ We show next that the infimum of all bounds of a bounded linear mapping is itself a bound for the linear mapping.

Proposition 15.30. Let (V, K, | -|I) and (W, K’, | - |’) be two normed linear spaces where either K = K’ or K = R and K’ = C. Let L be a bounded linear mapping of V into W. Then

IL@)I < IL lywllvl| for every v € V.

If L is not identically vanishing, that is, if there exists v € V such that L(v) # 0 € W, then

HE lly > 0

Proof. 1. By the definition of ||Z||, as the infimumof all bounds for L, for every k ¢ N there exists a bound Mj for L such that |Lllyy < Me < Lllyy + ie Since M; is a bound

for L, we have for every v € V

IL@)I! < Mellvll < {IL yy + g}lvllSince this holds for every k < N, we have ||Z(v)||’
0.

o

354

Chapter 4 The Classical Banach Spaces

[VI.3] Equivalence of Continuity and Boundedness of a Linear Mapping

Proposition 15,31. Let (V, K, | ||) and (W, K’, | - ||’) be two normed linear spaces where

eitherIK = K’ or K = R and K’ = C. Let L be a linear mapping of V into W. Then L is a bounded linear mapping if and only if L is continuous at 0 € V.

Proof. 1. Suppose L is a bounded linear mapping. To show that L is continuous at 0 € V, we show that for every ¢ > 0 there exists 5 > O such that ||Z(v) — L(0)||/ < e for every v € V such that ||v — O|| < 4, that is, ||Z(v)||’ < ¢ for Now according to Proposition 15.30 we have ||L(v)||’ < lZlly,yv = 0 then ||L(@)|' = 0, that is, L(v) = 0 € W constant mapping and hence continuous on V. If ||L||,y

every v € V such that ||v|| < 6. IZ llywllvll for every v < V. If for every v € V so that Lisa > 0 then for an arbitrarily given

e>Oletsé= (ILllyv) ‘e- Then for v € V such that ||v|| < 6 we have

ILO < UL llywlloll < [Llly75 = WLllyp (ILllyy)

-1

& =

so that Z is continuous at 0 € V. 2. Conversely suppose L is continuous at 0 € V. Then for every « > 0 there exists 5 > Osuch that ||L(v)||’ < ¢ for every v € V with |u|] < 3.

Now let v ¢ V and v #0 € V. For the element aha” € V we have laprl =$ N. Then forn > N we have lim sup ||Zn(v) — Lm(v)|ly,y S & Substituting this into (4), we have forn > N m—>00

Ln(v) — LOI! s ellvll and then

Zn —Lllyw =

This shows that jim, Ln — Lily

sup

veS1(0)

[|Ln(v) — L(e)|’ < .

w =9.

In particular since (W, K’, | - II’) := (K’, K’, | - |) is a Banach space, the linear space

58{V, K’} over the field K’ of all K’-valued bounded linear functionals on V is a Banach space by our result above. [VL.6] Dual of a Normed Linear Space

Let (V, KK, | - [l) and (W, K’, | - ||’) be two normed linear spaces where either K = K’ or

K = Rand K’ = C. Consider the collection 3{V, W} of all bounded linear mappings of V into W. In Definition 15.29, we defined a function | - ||y,y on the linear space B{V, W} over the field of scalars K’ by setting for every L € B{V, W}

Lllyy = inf {M > 0: |L@)I < Mull forall v € V}. We showed in Theorem 15.37 that || - ll,» is a norm on the linear space 3{V, W}.

showed in Theorem 15.36 that if we let S1(0) = {v € V: ||ul] = 1}, then

WLllyv=

sup IL(v)Il.

ves, (0)

We

360

Chapter 4 The Classical Banach Spaces

‘We showed in Theorem 15.38 that if (W, K’, |l- ) is a Banach space, that is, W is complete with respect to the norm | - ||’, then $8{V, W} is a Banach space with respect to the norm

I Ilyw-

Let us consider the particular case that (W, K’, | - |I’) = (IK, K, | - |}, a Banach space. Thus we consider the linear space 23{V, K} over the field of scalars K of all K-valued bounded

linear functionals on V. Let us write V* for this linear space 93{V, K}. If for every f € V* we define

If lle = int {M > 0: | f(v)| < Mllvl| for all v € V},

then | - ||. is anorm on V* and V* is a Banach space with respect to the norm | - ||». As in the general case, if we let $1(0) = {v € V: |lv|| = 1}, then

Iflle=

sup |f()I-

veS;(0)

Definition 15.39. Let (V, K, | - ||) be a normed linear space where K = R orK = C. Let V* be the linear space over the field of scalars K of all K-valued bounded linear functionals

on V. Let | - ||. be a norm on V* defined for every f € V* by

If lle = inf {M > 0: |f(@)| < Mull forall v € V}. We call the Banach space (v*, K, ||- lls) the dual normed linear space of the normed linear

(V.K, Il- II).

In a normed linear space (V, K, | - ||) and its dual space (V*,K, | - lle) if we let 510) = {v € V: |lvll = 1}, then as we showed above we have || fle = supyes,@ Lf) for every f < V*. We show in Proposition that if we let S¥(0) = {f € V*: Ilflle = 1}, then we have ||v|| = SUP rest) lf (@)| for every v € V.

[VII] Baire Category Theorem Definition 15.40. Let (X, d) be a metric space. The diameter of a subset E of X is defined

by diam(E) = sup, ycx d(x, y).

Theorem 15.41. (Cantor Intersection Theorem) Let (X, d) be a complete metric space and let (F, : n € N) be a decreasing sequence of non-empty closed sets in X. if lim diam(F,) = 0, then there exist x9 € X such that Nex Fn = {xo}. 1

Proof. 1. Let us show that (),,1 Fn contains at most one point. Suppose a, b € (J)/,cn Fn

anda # b. Then§ := d(a,b) > 0. Now a,b € (\,en Fn implies that a,b € F, for everyn € N. Then diam(F,) = SUP, yeF, d(x, y) > d(a,b) = 8 for every n € N. This contradicts the fact that lim diam(F,) = 0. Therefore a = b. This shows that ),,cn Fn

n> contains at most one point.

§15 Normed Linear Spaces

2.

361

Let us show that (), 0 there exists N & N such that d(x, x,) < ¢ form,n > N. Now since

jim, diam(F,,) = 0, there exists

N e& N such that d(F,) < ¢ forn > N. Letm,n > N. We may assume without loss of generality that m > n > N. Then Fy C F, so that %m,x, € F,. This implies that A(Xm, Xn) = SUP, yer, a(x, y)} = diam(F,) < ¢. This shows that (x, :n € N) is a Cauchy

sequence in X with respect to d. Then by the completeness of the metric space (X, d), there

exists x9 € X such that

jim, (Xn, Xo) = 0. It remains to show that xp €

hen F,,. For

every k € N, consider the sequence (x, : n > k) and its range Ey = {x, : n > k}. Since (F, :n € N) is adecreasing sequence and since x, € F, foreveryn ¢ N, we have Ex C Fy.

Now the convergence of (x, : m > k) to xo implies thatxp € Ey. Then xp € Ey C Fi = since F;, is a closed set. Thus x9 € Fy for every k € N and therefore xo € Qhen F,.

Wl

Definition 15.42. Let (X, d) be a metric space and let E be a subset of X. (a) We say that E is dense in X if E = X. (b) We say that E is nowhere dense in X, or non-dense in X, if(£)°

=.

Remark 15.43. A set E in a metric space (X, d) cannot be both dense and non-dense in

X. In fact, if E is dense in X then E = X so that (E)° = X° = X # @ so that E is not

non-dense in X.

Proposition 15.44, Let (X, d) be a metric space and let E be a subset of X. (a) E is dense in X if and only if for every non-empty open set G in X, we have GN E (b) E is dense in X if and only if for every open ball B in X, we have BN E # ©.

# ©.

Proof. For an arbitrary set E in X let us write E’ for the set of all limit points of E. Then

E=EUuE'.

1. Let us prove (a). Suppose E is dense in X, that is, E = X. Let G be a non-empty open set in X.

Since E = X, G contains a pointx

€ E = EU E’.

Ifx

€ E, then we have

GNE #9. Ifx & E’, then the open set G containing x must contain at least one point of E other than x so that GN E # B. Conversely suppose that for every non-empty open set G in X we have GN E # @. Let xeéX. Ifx e€ Ethenx € E. Ifx ¢ E, take an arbitrary open set G containing x. Since GNE # G, G contains a point ofEF distinct from x. Thus x is a limit point of E. This shows that if x ¢ EF thenx € E’ C EUE’ = E, Therefore every x € X is contained in E. This shows that E = X, that is, E is dense in X. 2. Let us prove (b). If £ is dense in X, then by (a) for every non-empty open set G in X we have GN E # @ so that in particular for every open ball B inX we have BNE # @. Conversely suppose for every open ball B in X we have BN E # @. Let G be an arbitrary non-empty open set in X. Then G contains an open ball B. Since BN E # @ and since G > B, we have GN E #9. Then

E is dense in X by (a).

362

Chapter 4 The Classical Banach Spaces

Proposition 15.45. Let (X, d) be a metric space and let E be a subset of X. {a) E is non-dense in X if and only if every non-empty open set G in X contains anon-empty open set Go such that Gg N E = @.

(b) E is non-dense in X if and only if every open ball B in X contains an open ball Bo such that By NE

= 6.

Proof. 1. Let us prove (a). Suppose £ is non-dense in X, that is, (E)° = @. Let G be an

arbitrary non-empty open set in X. Then G ¢ E for otherwise we have (E)° > G # 9, a contradiction. Now G ¢ E implies thatG\ E #4 @. LetGp = G\E # G. We

have Go E

Go = G\ E

=

@ and thus Gp NE

=

@. To show that Go is an open set, note that

= GN (E)’. Since G and (E)° are open sets, Go is an open set. Thus we

have shown that every non-empty open set G in X contains a non-empty open set Gp such that Gp NE = 9. Conversely suppose every non-empty open set G in X contains a non-empty open set

Go such that GoM E = @. To show that F is anon-dense set in X, assume the contrary, that

is, (E)° #9. Then G := (E)” is a non-empty open set. Let Go be an arbitrary non-empty open set contained in G. Then Gp C G = (EY

x in Go is either in E or in E’ or in both.

C E = EU E’, Thus an arbitrary point

Ifx € E, then Gg N E

4 @. If on the other

hand x € E’, then Go is an open set containing a limit point x of E so that it must contain a point of E and then we have GoM E ¥ @. Thus in any case for any non-empty open set Go contained in G we have Go N E ¥ @, a contradiction.

2. (b) follows from (a) by the fact that an open ball is a non-empty open set and every non-empty open set contains an open ball. If E is non-dense in X, then since every open ball B is anon-empty open set, B contains a non-empty open set Go such that Go MN E = @ by (a). Since Go is a non-empty open set, it contains an open ball Bp. Then Go M E = 9 implies By N E = G.

Conversely suppose every open ball B in X contains an open ball Bo such that Bg E = B. Let G be an arbitrary non-empty open set. Then G contains an open ball B and B contains an open ball Bo such that By 1 E = @. Thus every non-empty open set G contains a non-empty open set Go := Bo such that Go N E = @. Then by (a), £ is a non-dense set in X. a Definition 15.46. Let (X, d) be a metric space. A subset E of X is called a set of the first category if E is a countable union of non-dense sets in X. A subset E of X which is not a Set of the first category is called a set of the second category. Theorem 15.47. (Baire Category Theorem) Let (X, d) be a complete metric space. Then X is a set of the second category. Proof. To show that X is a set of the second category, we show that X is not a countable union of non-dense sets in X. We show this by showing that for any countable collection of non-dense sets in X, {E, : n € N}, we have cn

such thatx ¢ E, for every n € N.

En # X, that is, there exists x € X

Let us write B,(x) = {y € X : d(x, y) < r} and C,(x) = {y € X : d(x, y) 00 7

Since (X, d) is a complete metric space, this implies that there exists x € X such that hen Cra/2n) = {x} by Theorem 15.41. Now x € C,,/2(%n) C By, in) for every n € N. Then since B,,,(%,) M En = 9 for every n € N, we havex ¢ E, for every n € N. Therefore

x €Unen En and U,e~ En FX.

[VI] Uniform Boundedness Theorems Theorem 15.48. Let (X, d) be a complete metric space. Let ¥ be a collection of real-valued continuous functions on X with the property that for every x € X there exists M, > 0 such

that

@

LF) < My forall f e F.

Then there exist a non-empty open set O in X anda constant M > 0 such that (2)

|f@)| M,, we have | f(x)| < &, for all f € F so that x € Ej,. Thus every x € X is in Ey for some k € N so that we have

©

x=. keN

Let us show that E; is a closed set in X for every k ¢ N. Let f € F¥. Since f is areal-valued continuous function on X, | f| is areal-valued continuous function on X. Then Ey, ¢ as given

by (4) is the preimage of the closed set [0, &] in R under the continuous mapping | f|. Thus

Ex, 7 is aclosed setin X, Then E; as given by (5) is the intersection of a collection of closed

sets in X and is therefore a closed set in X. Since X is a complete metric space, according to Theorem 15.47 (Baire Category Theorem) X is a set of the second category, that is, X is not a countable union of non-dense sets. Then (6) implies that there exist ky € N such that

Ejy is not anon-dense set, thatis, (Ej,)° # B. Since Ey, is a closed set we have Ej, = Ey.

Thus we have Ey, #G. Let O = Ei and M

forallx €¢ Oandall fe.

a

= kg. Then by (3) we have | f (x}| < ko = M

Theorem 15.49. (Uniform Boundedness Theorem) Let (V, K, | - ||) be @ Banach space and (W, K’, | - |I') be a normed linear space where either K = K' or K = R and K' = C.

Consider the collection 3B{V, W} of all bounded linear mappings of V into W and let Bo{V, W} be a subcollection of B{V, W} such that for every v € V there exists M,

such that

(a)

> 0

IL@)|I' 0 such that

Q)

ILllyy ||w|' is arealvalued continuous function. (This is an immediate consequence of Observation 15.2.) Thus as a composition of two continuous mappings, L and | - ||’, fr is a real-valued continuous

function on V. Let¥ = {f, : L € So{V, W}}. This collection of nonnegative continuous

functions on V has the property that for each v € V we have fr(v) = ||L(v)||’ < M, for

§15 Normed Linear Spaces

365

some M, > 0 by (1). Thus by Theorem 15.48 there exist a non-empty open set O in V and aconstant M > 0 such that

fi)

0 and B,(wo) for an open ball in W with center at wo € W and radius r > 0, that is,

Arvo) = {v € V = lv — voll H — H where the second set inclusion is by Observation 15.55. By Observation 15.54, H — H is an open set in W. Since H — H contains 0 € W there exists r > 0 such that

B,(0) C H — HC L(Azn(0)). Then by Lemma 15.56 we have

B,(0) C L{Aang(0)) C E(Aang(0)). By Observation 15.54 we have

Brjtag(0) = BHO) CL Aan(O)) = L( 5 Atno®)) = L(Ax 0).

§15 Normed Linear Spaces

If we 2. open there

371

leta = 47 then we have (1). For the open set9 in V we have L() = , an open set in W. Let G be a non-empty set in V. To show that Z(G) is an open set in W we show that for every w € L(G) exists b > 0 such that

(2)

B,(w) C L(G).

Let w € L(G). Then there exists v € G such that w = L(v). Since v € G and G is an open set in V there exists r > 0 such that A,;(v) C G, that is, vy ++ A;(0) C G by Observation

15.53, Then L(G) > L(v + A-(0)) = L(v) + L(A; (0)). By (1) there exists a > 0 such

that B,(0) C L(A1(0)) and thus

Bar (0) = rBa(0) C rL(A1(0)) = L(r.A1(0)) = L(A-) by Observation 15.53. Thus L(G) > L(v) + Bar(0) = Bor(L(v)). Then since w = L(v) we have B,,-(w) C L(G). If we let b = ar then we have (2).

Corollary 15.58. Let (V, K, | - ||) and (W, K, | - |I') be two Banach spaces over the same

field of scalars K. Let L be a one-to-one continuous linear mapping of V onto W. the inverse mapping of L is a continuous mapping of W onto V.

Then

Proof. Let A be the inverse mapping of L. Now A maps W one-to-one onto V. Let us show that A is a linear mapping. Let wi, w2 € W and o1,a@2 € K. Since L maps V one-to-one onto W there exist unique v1, v2 € V such that L(vy) = w1 and L(v2) = wo, that is, vy) = A(w1) and vz = A(w2). Then

A(@iwy + a2w2) = A(wi LQ) + o2L(v2)) = A(L@iv1 + o202)) = av + @202 = a A(wi) + 2A (w2). This shows that A is a linear mapping of W into V. To show that A is a continuous mapping we show that for every open set G in V, A~!(G) is an open set in W.

open setin W.

But A(G)

= L(G) and according to Theorem 15.57, L(G) is an

Definition 15.59. Let X and Y be two sets and let T be a mapping of a subset D of X into

Y. The subset of X x Y defined by {(x, T(x)) € X x Y : x € D} is called the graph of the

mapping T.

Proposition 15.60. Given two normed linear spaces (V, K, | - ||) and (W, K, | - ||’) over the same field of scalars K. (a) Let us define addition and scalar multiplication on the product space V x W by

C2)

1, wi) + (v2, we) = (41 + 02, wi + we) for (v1, w1), (v2, w2) € Vx W,

(2)

y@,w)=(yu, yw) for (v,w) € V x W andy €K.

372

Chapter 4 The Classical Banach Spaces Then V x W

is a linear space over the field of scalars K.

(b) Let us define a function | « ||” on V x W by setting

@)

cv, wll” = [lull + wll’ for (v, w) € V x W.

Then | - ||” is a norm on the linear space V x W. (c) V x W is complete with respect to the norm | - ||" if and only if V and W are complete with respect to the norms || - || and || - ||’ respectively.

@ If (V.K, ||-|l) and (W, K, | |’) are Banach spaces then (V x W, K, | - ||") is aBanach space.

Proof. 1. It is readily verified that, with addition and scalar multiplication defined by (1) and (2), V x W satisfies conditions 1° - 9° of Definition 15.1. Thus V x W is a linear space

over the field K.

2. Similarly it is easily verified that | - ||” defined by (3) satisfies conditions 1° - 4° of

Definition 15.1. Thus | - ||” is a norm on the linear space V x W.

3. Let us note that for (v1, w1), (v2, w2) € V x W we have by (1) and (3)

4

Wr, wi) — a, we) II” =

1 — 92, wi — we) Il” = lot — vel] + wr — wal’.

Thus we have

(5)

ivr — val, wa — wall’ < (or, wr) — (ve, wa)ll” = vy — vei] + wi — wall’.

Now consider arbitrary sequences (v, : » € N) in V, (w, : n € N) in W, and ((up, wy) :

nN) in V x W. It follows from (5) that ((u,, w,) :

€ N) is a Cauchy sequence with

respect to the norm | - ||” if and only if (vu, : 2 € N) is a Cauchy sequence with respect to the norm | - || and (w, : 2 € N) is a Cauchy sequence with respect to the norm || - ||’. Suppose V is complete with respect to || - || and W is complete with respect to || - ||’.

To show that V x W is complete with respect to | - ||”, let (Un, Wn) : 2 € N) be a Cauchy

sequence in V x W with respect to || - ||. Then as sequence in V with respect to || - || and (w, : ~ respect to ||: ||’. By the completeness of V with W with respect to | -||’ there exist v ¢ V and w

lim. ma

||w, — w|| = 0. Then by (4) we have

we noted above (vu, : n € N) is a Cauchy € N) is a Cauchy sequence in W with respect to | - || and the completeness of € W such that jim, ll’n — v|| = 0 and

Him, (vn, wa) — (v, w) I” = ima {vq — vl + [lon — wI"} = 0.

This shows that V x W is complete with respect to | - ||. Conversely supposeV x W is complete with respect to | - ||”. To show that V is complete with respect to || - || and W is complete with respect to || - ||’, let (uv; : 2 € N) be a Cauchy sequence in V with respect to || - || and (w, : 2 € N) be a Cauchy sequence in W with respect to || - ||’, Then as we noted above (Qn, Wa) in E N) is a Cauchy sequence in V x W

with respect to | - ||”. By the completeness of V x W with respect to | - ||” there exists (v, w) € V x W such that lim, Gn, Wa) — (v, w) ||” = 0. Then by (5) we have

lim00 [|v — vi] < n-00 Tim |I(va, wa) — (v, w) ||" = 0.

§15 Normed Linear Spaces

373

This shows that V is complete with respect to | - ||. Similarly W is complete with respect

to Il - Il’.

4. If (V, K, | - ||) and (W, K, | - |’) are Banach spaces then V is complete with respect

to the norm || - || and W is complete with respect to the norm || - ||/. Then by (c), V x W is

complete with respect to the norm | - ||’, that is, (V x W, K, | - ||") isa Banach space.

Theorem 15.61. (Closed Graph Theorem) Let (V, K, ||-||) and (W, K, |l-|I’) betwo Banach spaces over the same field of scalars K. Let L be a linear mapping of V into W. If the graph

of L isa closed set in the Banach space (V x W,K, | - ||") where ||(v, w)Il” = [lvl] + llwll’ for (v, w) € V x W, then L is a continuous mapping of V into W.

Proof. Let G be the graph of L, that is, G = {(v, L(v)) € Vx W: v € V}. Then Gisa linear subspace of V x W. Indeed if (v1, L(v1)), (v2, L(v2)) € G and v1, » € Kthen by (1) and (2) of Proposition B.9 we have

vi(er, L(v1)) + vo(v2, L(v2)) = (viv + yov2, NLE@1) + L002) = (nv + yu, Lui + y202)) € G. If we assume that G is a closed set in the Banach space (V x W, K, ||- ||”) then (G, K, [I- II’) is a Banach space by Proposition 15.10. Let us define two mappings P, and P, of G C V x W

into V and W respectively by

Pi(v, L@)) =o

for (v, L(v)) € G,

Po{v, L(v)) = L(v)

for (v, Lv) € G.

Clearly P; and P, are linear mappings. Moreover we have

| Pi(v, £@))] = loll s doll + ILO) = |, £@))]", | Pa(v, L(v))| = ILO) < llvll + ILO = |v, L@) ||”. This shows that the constant 1 is a bound for the linear mapping P, and for the linear mapping P2. Thus P; and P2 are bounded linear mappings. This implies that P; and P, are continuous linear mappings by Theorem 15.32. Now P, is a one-to-one mapping of the Banach space G onto the Banach space V. Thus the continuity of ZL implies that its inverse mapping Q is a continuous mapping of V onto G by Corollary 15.58. Now L = Pz o Q. Then the continuity of the mapping @ of V onto G and the continuity of the mapping P, of G into W implies that L is a continuous mapping of VintoW. [X] Hahn-Banach Extension Theorems [X.1] Hahn-Banach Extension Theorem for Real Linear Spaces Let P beaset. A partial ordering on P is a relation < between elements of P which satisfies the following conditions:

374 1° 2°

Chapter 4 The Classical Banach Spaces reflexivity: x < x. antisymmetry: x < y andy < x implyx = y.

3° transitivity: x < y andy < zimplyx ~ z. ~< is called a linear, or total, ordering if it satisfies the additional condition :

4°

trichotomy: x, y € P impliesx ~ y or y < x.

If for every A C P, A # Q, there exists a € A such that a < x for every x € A then< is

called a well ordering. ‘We call (P, y for y ~ x and x > y for y « x.

Let (P, 0.

3°

subadditivity: p(u + v) < p(u) + p(v) foru,ve

V.

Remark 15.63. (a) Conditions 1° to 4° in Definition 15.1 for anorm ona linear space imply conditions 1° - 3° in Definition 15.62. Thus a norm on a real linear space is a sublinear functional on the real linear space. (b) If p is a sublinear functional on areal linear space (Vv, R) then forz, v € V anda e [0, 1]

p(au + (1 —a)v) < ap) + (1 —@)p). Thus a sublinear functional is a convex function.

Theorem 15.64, (Hahn-Banach Theorem for Real Linear Spaces) Let (V, R) be a real linear space and let p be a sublinear functional on V. Let M be a linear subspace of V and let f be a real-valued linear functional on M such that f(v) < p(v) for v € M. Then

there exists a real-valued linear functional g on V such that g(v) = f(v) for v € M and

8X) < p@) forveV.

§15 Normed Linear Spaces

375

Proof. 1. If@ = V, there is nothing to prove. Thus we assume M + V. Let ug € V and vo ¢ M. Let Vo be the linear subspace of V spanned by M and vp, that is,

(

Vo = {u + Avo: u € M,A eR}.

Let us show that there exists a real-valued linear functional fo on Vo such that fo(v) = f(v) for v € M and fo(v) < p(v) for v € Vo.

With an arbitrary c € R let us define a functional fp on Vo by setting

@

fol) = fou +Avo) = fu) +c

forv =u+Avg € Vo.

Let us note that since vp ¢ M the representation ofv € Vo as v = w+ Ac withu € M and 4 € R is unique and therefore the definition of fp by (2) is well defined. Now fo is a real-valued functional on Vo. Let us show that fo is a linear functional on Vo. Let v1, v2 € Vo and a, a2 € R. Now vy = uy + Arvo and v2 = uk + Aguo for some

uy, ua © M and Ay, Az € R. Thus we have

fo(avy + 202) = fo( (aims + aur) + (aA + @2A2)u0) = f (@yuy + ou2) + (aA + a2dg)e

= an { f(41) + Arc} + o2{ fr) + Arc} = a1 fo(v1) + 2 fo(v2).

This shows that fo is a linear functional on Vo. Clearly fo(v) = f(v) for v € M. Let us show that the constant c € R in the definition of fo by (2) can be so chosen that fo(v) < p(v) for v € Vo, that is,

3)

ft) +Ac < p(utAvo) forallac Manda eR.

Note that if 4 = 0 then (3) reduces to f(u)

< p(w) for all

wu € M which is valid by our

assumption on f. Thus dividing (3) by 4 0 and applying the linearity of f on M and condition 2° in Definition 15.62 we obtain a set of conditions on c:

F(%) +e < p(k +)

for all u € M and A > 0,

f(-%)-e are real-valued linear functionals on the real linear space V. Proof. 1. To prove (1) note that for every v € V we have

Q)

f@v) = AiG») + ifr)

and by the homogeneity of f we have

@)

FG») = if (ve) = HA) + iA} =ifi@) — AO).

Equating the real part of (2) to that of (3) we have the second equation in (1), and equating the imaginary part of (2) to that of (3) we have the first equation in (1).

2. To show that f; and /) satisfy the additivity condition on V, let u, v € V. Then

(4)

f@+v=fAtetr)

+ifut v).

On the other hand by the additivity condition on f we have

6)

fut) = f@M+/MO = [A+

AM} +{AW + AO}.

Equating the real part of (3) to that of (4) and equating the imaginary part of (3) to that of (4) we have

Atv)

= fi@)+ fic),

flutv) = fp) t+ flv).

§15 Normed Linear Spaces

379

This proves the additivity of f; and fo. 3. Let j = 1 or 2. Suppose f; is not identically vanishing on V. To show that f; does not satisfy the homogeneity condition we show that there exist vy € V and y € C such that

FjQvv) F fj). Now for any v € V andy € C, y = @ + if where a, f € R, we have

©

fyv) = filvv) + ifa(yy).

On the other hand by the homogeneity of f we have

)

f(y») = vf @) = {a + BHA) +i} = {afi(v) — BAW)} + i[BAO) +oA(v)}.

Equating the real part of (6) to that of (7) and equating the imaginary part of (6) to that of (7) we have

8)

| Avy) = afi) — Bfa(v) = vf) — IBA) — Bh), Aalyv) = BA) + af) = yfav) — iBfa(v) + BAC).

If f; is not identically vanishing on V we can select v € V such that f,(v) 4 0. If we select B # Othen fi(yv) ¥ yfi(v). If fo is not identically vanishing on V we can select v € V such that fo(v) 4 0. If we select8 4 0 then fo(yv) # yfo(v). 4, Let us regard V as a real linear space. Let us show that f; and f> are real-valued linear functionals on the real linear space V. We have verified the additivity of f, and fo in 2 above. To verify the homogeneity of f; and f) on the real linear space V, let v € V and a € R. Then we have

9)

f@v) = fi@v) +ifav)

and by the homogeneity of f on the complex linear space V and by the fact thata e RC C we have

(10)

f@v) = af(v) = afi) + iafr(v).

From (9) and (10) we have

A@v) = afi), flav) = afo(v).

This verifies the homogeneity of f; and f> on the real linear space V. Thus fi and f> are real-valued linear functionals on the real linear space V.

Lemma 15.67. Let (V, C) be acomplex linear space. Let f be a complex-valued functional on V satisfying the following conditions : (

f@t+v=fWO+f)

(2)

fv)

(3)

f@iv) =if(v)

=af(v)

forve

foruvev V,acR

forve V.

380

Chapter 4 The Classical Banach Spaces

Then f is a complex-valued linear functional on V.

Proof. By (1), f satisfies the additivity condition 2° of Definition 15.26. It remains to verify that f satisfies the homogeneity condition 1° of Definition 15.26, that is,

(4)

f(iyv)=vf)

forve V,y €C.

Let v € V and let y € C written as y = o + if where a, 8 € R. Then by (1) we have

F(yv) = fav + ify) = fav) + fGpv). By (2) we have f (av) = af (v). By (3) and (2) we have f(ifv) = if (Bv) = iBf(v). Thus we have f(yv) = af(v) = iff (v) = yf(v). This proves (4).

Definition 15.68. Let (V,C) be a complex linear space. A functional p on V is called a

sublinear functional on V if it satisfies the following conditions: 1° real-valued: p(v) € R for every v € V. 2° positive homogeneity: p(yv) = |y|p(v) for v € V andy €C. 3° subadditivity: p(u + v) < p(u) + p(v) foru,ve V.

Remark 15.69. (a) A norm on a complex linear space (Vv, Cc) satisfies conditions 1° - 3°

of Definition 15.68 and is therefore a sublinear functional on V.

(b) As we noted in Remark 15.3 a complex linear space (V, C) can be regarded as a

real linear space. Since conditions 1° - 3° in Definition 15.68 imply conditions 1° - 3° in Definition 15.62 a sublinear functional p on a complex linear space V is a sublinear functional on the real linear space V. The following extension of the Hahn-Banach Theorem for real linear spaces to complex linear spaces is due to H. F. Bohnenblust and A. Sobdezyk.

Theorem 15.70, (Hahn-Banach Theorem for Complex Linear Spaces) Let (V,C) be a complex linear space and let p be a sublinear functional on V. Let M be a linear subspace of V and let f be a complex-valued linear functional on M such that |f |< pon M. Then

there exists a complex-valued linear functional h on V such thath = f on M and |h| < p ony. Proof. Let us write f = fi +if2 where f; and f2 are the real and the imaginary parts of f. Let us regard the complex linear spaces V and M as real linear spaces. By (c) of Observation 15.66, f; and f are real-valued linear functionals on the real linear space M. Now

IA@)! = {LAMP + LAMP}

1/2

=If@)| < p@)

forve M.

The sublinear functional p on the complex linear space V is also a sublinear functional on the real linear space V by (b) of Remark 15.69. Thus by Theorem 15.64 (Hahn-Banach) there exists a real-valued linear functional g on the real linear space V such that g = f on

§15 Normed Linear Spaces

381

M andg < pon V. Thus we have

gutv)=g)+g(v)

@ (2)

g(av) =ag(v)

G3)

gv) = fi(v)

(4)

gv) 0. Then there

exists a K-valued bounded linear functional f on V such that f = 0 on M, f (vo) = d,

and|| fle = 1.

§15 Normed Linear Spaces

383

Proof. Let Vo be the linear subspace of V spanned by M and vp, that is,

aw

Vo = {ut+yu:ue M,y € KR}.

Let us define a K-valued functional fo on Vo by setting

(2)

Sov) = fou + yu) = yd

forue Mandy €K.

Sinced = inf,cm ||u—voll > 0, vo ¢ M so that the representation of vy € Vo by v = u+yuo with # € M and y € K is unique and therefore fo is well defined by (2). It is easily verified that fo is a linear functional on Vo. Also f = 0 on M. v EMC Vo we have v = uw + Ovg with u € M so that by (2) we have

Indeed for

fotv) = fou +09) =0-d=0 forve M.

(3)

Since vp € Vo and vg = 0 + vo, we have by (2)

(4)

folvo) = fo(O+ lu) =1-d =d. Let us show that fo is a bounded linear functional on Vo with | foll1 = 1. According to

Theorem 15.36 it suffices to show that if we let 5 = {v € Vo: ||v|| = 1}, then we have 6)

sup | fo(v)| = 1. veS

Now

s= {2lull :vevou do} = {Ht le +

cwemy eKutym 40.

yvoll

Thus forv € S we have

©

6

IFoo)! vj

=

u+yuo )

be (ie ya

—_—_————

|

lfou + yv0)|

a

=

lyld

=

_,

Iu+yv0ll — Iu + yvoll

Since vp 4 0 € Vo, we have Tel € S. By (4) we have

vo

—

fe (Ga)

j=

fo(vo)

Ivoll

ad

=—

[voll

>0.

Thus we have

M

sup|fo@)|= veS

sup —_| fotv)|.

veS, fo(v) 40

Now fo(v) # 0 for v € S implies y # 0 by (6). Then by (6) and (7) we have

ee

suy

lvyld ———=

d ———

ee go [+ 700] — veeyo I + vol

v)|=

sup

= sup

vem [vo — wl

=1.

sup d

itfwea vo —wl|

d

384

Chapter 4 The Classical Banach Spaces

Thus we have|| fol|« = 1. We have just shown that fp is a K-valued bounded linear functional on Vo such that fo = 0 on M, fo(vo) = a, and | follx = 1. Then by Theorem 15.65 ifK = R and by Theorem 15.71 if K = C, there exists a K-valued bounded linear functional f on V such thatf = fo on Vp and | f lx = Il folle = 1. SinceM C Vp and fo = 0 on M by (3), we have f = fo =00n M. Since up € Vo, we have f(vp) = fo(vo) = dby (4). & Corollary 15.73. Let (Vv, K, || - ) be a normed linear space over the field of scalars K whereK = Ror K =C. If vp € V and up £0 € Vthen there exists a K-valued bounded linear functional f on V such that f (vo) = ||voll and | f lle = 1.

Proof. Consider the linear subspace M of V consisting of 0 € V only, that is, M = {0}.

Since vp # 0 we have d = infycy ||# — voll = |lvol| > 0. Thus by Corollary 15.72,

there exists a K-valued bounded linear functional f on V such that f(vg) = d = || voll and

Ifle=1.

0

[X.3] Bounded Linear Mappings of a Dual Space into a Dual Space Proposition 15.74 . Let (Vv, K, || - ) be a normed linear space over the field of scalars K

where K = R orK = C. Let (V*,K, | - lle) be the dual space of (V,K, | - ||) and let S¥(0) = {f €V*: ||flle = 1}. Then for every v € V we have

qd)

loll=

sup

ESTO) FESTO)

|f()I.

Consequently for each fixed v € V the mapping of V* into K defined by f > f € V* is a K-valued bounded linear functional on V* with norm equal to ||v||. Proof. For every f € S{(0) we have | f(v)| < |lfll«llull = llvl|

for all v € V so that

sup |f(v)| < [lull foreach ve V.

@ Ifv = 0 €

f(v) for

SESTO)

V, then ||v|| = 0 and f(v) = 0 for every f € V™ so that (1) holds trivially. If

v € V and v £ 0 € V, then according to Corollary 15.73 there exists f € S{(O) such that Ff (v) = |lv||. This fact and (2) imply (1). Observation 15,75, Let A and B be two arbitrary indexing sets and let co,g € R fora €¢ A and § € B. Then

q)

sup { sup ca,p} =

acd * peB

SUP _ Ca,p = SUP | SUP Ca, f-

acA,peB

BeB

‘aca

Proof, For each a € A we have Co,p < 5UPgeR Ca,p- Thus

Q)

sup

acA,peB

Cag

sup

{ sup ca, } = sup SUP Ca,p f-

weA,peB * BeB

acdé * peB

§15 Normed Linear Spaces

385

On the other hand supga Ca, < SUPye4,peR Ca,p SO that (3)

SUP]

aecA *

SUPCy,pf


Ny.

Thus |f| < Ilfalloo + } ae.

on X form > Nx. This shows that f is essentially bounded on X and thus f € L©(X, QM, uw). From (4), we also have ||f — falloo < } forn > Nx. This shows

that lim If — floc =.

Theorem 16.44. Given a measure space (X, A, 2). Let (fy :n € N) be a sequence and f be an element in L(X, A, p). If Jim, llfn — f lloo = 0, then

1° Tim Il falloo = If lo:

2° 3°

lim | Sn = f uniformly on X \ E where E is a null set in (X, A, 2),

1

(fn in €N) converges to f on X in measure ju.

Proof. Since | - |loo is anorm we have \Ilfa lloo — If lloo|

Ifa —F lloo by Observation 15.6.

Then lim lfn — flloo = 0 implies lim || frlloo = II flloo- This proves 1°. 2° follows

from Theorem 16.43. To prove 3°, note that by 2° for every ¢ > 0 there exists N € N such that |f, — f| < eon X\ E forn > N. Then p{X : |f, — f| = e} =O forn > N. This shows that dim, wx = lf. — fl = €} = 0 for everye > 0, that is, (f, : n € N) converges to f on X in measure p.

©

Remark 16.45. Let (f, : n € N) be a sequence and f be an elementin f € L™(X, A, ). The condition dim, Sn = f ae. on X and the condition jim, I falloo = If lloo together do not imply

lim

n-00

| f, — flo = 0, that is, Theorem 16.28 does not hold for L°(X, 2, jz).

Example. In (R, 9, 4,), let Sn(x) = {

forn €

N and

let f

=

10,0).

0 nx

1

Then

forx € (—oo, 0), forxe [o, i ;

forxe Gg im

fa

=

ov),

foR,

|lfrllo

=

1 forn

EN,

If lloo = 1 so that im, Fnlloo = I flloo, but Ilfn — floc = 1 for everyn € N and thus

jim, I fn — filloo = 14 0.

414

Chapter 4 The Classical Banach Spaces

Definition 16.46. Let (X, 2, 2) be an arbitrary measure space. (a) Let (fa 1m € N) be a sequence and f be an element in L1(X, 2, 4). We say that the sequence (fy, :n € N) converges to f weakly in L'(X, 2, w) if we have

lim [J fuga = [/ fadu for every g ¢L™(X,%, u). jim, (b) Let (f, :m € N) be a sequence and f be an element in L®(X, A, 2). We say that the sequence (f, :n € N) converges to f weakly in L(X, A, 2) if we have

tim f[ fas du= f | fad foreveryg ¢LMX,%, u). jim, Theorem 16.47. Let (X, A, j4) be an arbitrary measure space. (a) Let (fy 1 n € N) be a sequence andf be an element in L'(X, A, ).

if jim, fn — fll1 = 0, then (fy :n € N) converges to f weakly in L'(X, A, p).

(b) Let (f, : n € N) be a sequence and f be an element in L®(X, Al, wu).

Ff lim

fa — flloo = 0, then (fn :n € N) converges to f weakly in L®(X, A, 1).

Proof. To prove (a) suppose lim

1

llfn — fila = 0. Let g € L(X, 2, 4). Now we have

| [mean fi teau|< fiche Nslan= We Deli < fi Fillo by Hélder’s Inequality (Theorem 16.40). Thus the assumption im, Ifn—F ll: = 0 implies

lim | n>00

x

Sng dp —

x

fg dp| =0, thatis, lim | frgdu= | fedu. nO

SX

xX

This proves (a). (b) is proved in the same way using Theorem 16.40 again. Remark 16.48. Theorem 16.33 which is valid for p € (1, 00) does not hold for p = 1, that is, if (f, :m € N) is a sequence and f is an element in L'(X, 2, #) such that im, fh=f ae. on X and || f,ll1 < B for all

€ N for some B > 0, the sequence (f, : n € N) need

not converge to f weakly in L'(X, &, 2). See the Example below.

Example. Consider £1((0, 1), 9t, M (0, 1), #,) and L°((, 1), Mt, NO, 1), 4,). Let

fx = nlosjn forn € Nand f = 00n (0,1). Now Ifalli = foo *losmde, = 1

for every n € N so that (f, : # © N) is a sequence in L!((0, 1), Mt, N (0, 1), #,) and

falls = 1 foreveryn ¢ N. Alsof ¢ L'((0, 1), Mt,

everywhere on (0, 1).

(0,1), H,) and limf, = f

Letg = 1 on (0, 1). Then g € L™((0, 1), Dt, NO, 1), w,). Now for everyn € N we have Sor Arg du, = Soy fn dp, = 1 while Soy fedu, = Jo Ode, = 0so that the equality lim fio1) fn8 dH, = Soo.1 fg du, does not hold. Thus (f, : 2 € N) does not n—>00

converge to f weakly in L'(X, A, y).

§16 The L? Spaces Theorem

16.49.

415 Let (X, 2, 4) be an arbitrary measure space.

a sequence and f be an element in L©(X, A, uw).

Let (fy,

If im, tn = fae

: n € N) be

on X

and if

\fnlloo < B for alln € N for some B > 0 then the sequence (f, :n € N) converges to f weakly in L(X, A, 2). Proof. Letg € L'(X, &, z). Then ik Bl\g|du < oo, that is, B|g| is jz-integrable on X. Now lim, ink = fg ae. on X and |f,g| < Blg| ae. on X forn € N. Thus by the Dominated Convergence Theorem (Theorem 9.20) we have

jim, [ teau= fi feau. This shows that the sequence (f, : m € N) converges to f weakly in L(X, A, 4). Given a measure space (X, 2, 4).

If 4(X) = co then for some f ¢ L(X, A, yn)

we may have || f||p = 00 for every p € (0, 00). An example of such a function is given by f = 1 on X. We show below that for an arbitrary measure space (X, A, yz), if f €

L™(X, 2, 4) and if || f lly < 00 for some po € (0, 00) then || f|lp < 00 for every p > po and jim, If llp = IfllooTheorem 16.50. Let (X, 2, 4) be an arbitrary measure space. be such that f € L?0(X, Q,

4) for some po € (0,00).

P = po and moreover jim, If llp = If lloo-

Let f € L©(X, A, 4)

Then f € L?(X, A, pz) for every

Proof. 1. Forp > po we have p — po > 0. Then |f|? = | f |?-?°| F179
2) is an increasing sequence in A with im

> 00

By =

geo Be = B. We have _

ot (Be) 0 be arbitrarily given. Then the convergence above implies that there exists ky ¢ N such that

@

Ira,

Podu

dp =fir

du fou

PO du —

Po

du po. Since X is the disjoint union of A, B*, B \ B*, and C,

@

f If! du = | fl? dwt f fl? dwt f xX

A

=f

Bt

AUB*

B\B*

fl? du + [ fl? dp Cc

irtaes fisrau. B\B*

For the first integral on the right side of (4) we have

6)

f‘AUB* [flP dp Sf llbou{A U BY).

For the second integral on the right side of (4), recall that | /| € (0, 1) on B \ B*. Thus for Dp > po we have | f|? < | f|?° on B \ B* so that by (3) we have

©

ffB\Bt redusf B\ Bt ifimau 0 in (7) , we have

Wftle={ fer du}” 1,p,q>7r,

be related by

1

8

1-38

r

Pp

q

Show that for every extended complex-valued %-measurable function f on X we have

Wflle = FID IF? Prob. 16.4. Given a measure space (X, A, 2). Let f be an extended real-valued Ameasurable function on X such that ik lf |? dw < oo for some p € (0, 00). Show that for every ¢ > O there exists M > O such that for the truncation of f at level M, that is,

fl! = (f AM) Vv (—M), we havefy |f — fl? du 0 and order P € (0, 00) on I if we have | f(x’) — f(x”)| < C|x! — x” |? for any x’, x” € I. Thus the Lipschitz Condition is a Hilder Condition of order 1. Prove the following theorem: Theorem. Let f be a real-valued function on an interval I C R. Suppose f satisfies a Holder Condition with coefficient C > 0 and order p € [1, 00) on I. Then we have:

(a) f satisfies a Lipschitz Condition with coefficient C > 0 on I. (b) f satisfies Condition (L) on I. (c) f is absolutely continuous on every [a, b] C I.

Prob. 16.8. Let f be a real-valued function on [a, 6] c R with £([a, b]) = 1. Assume that f satisfies a Lipschitz Condition with coefficient C > 0 on [a, 6]. Show that f satisfies a

Hélder Condition with coefficient C > 0 and order p € (0, 1] on [a, b]. Prob. 16.9. Let Sia.b) fl? du, (a) Show that f (b) Let F be an with coefficient

f be an extended real-valued 20, -measurable function on [a, b] such that < 00 for some p € (1, 00). Letg € (1, 00) be the conjugate of p. is jz, -integrable on [a, b]. indefinite integral of f on [a, b]. Show that F satisfies a Hilder Condition | f ||p and order i on [a, b], that is, we have

FQ) — F@")| S I fllp lx’ —x"1'4

for x’, x” € [a, 5).

(c) Show that for every x € [a, b], we have lim? & +

hiO

— F@)

nia

_

o.

Prob. 16.10. Let f be an extended real-valued 9)t, -measurable function on [0, 1] such that to 1] |fl? du, < 00 for some p € [1, 00). Let g € (1, co] be the conjugate of p, that is,

5+} =1. Leta € (0, 1]. Show that

lim

1 lim ——

[, ld,dp,

=0.

Prob. 16.11. Prove the following theorem: Theorem. (Uniform Absolute Continuity of Integral Relative to Measure for a Class

426

Chapter 4 The Classical Banach Spaces

with Bounded Norms in L? Space) Let (X, &A, 2) be a measure space and let po € (0, 00).

Let {fg 2 (a)

© A} C LPO(X, QU, x) such that supyc, || fallpy < 00. Letp € (O, po). Then

tim, supf J{X:| fal?>A} fal? du =0.

400 wed

(b)

For every « > 0 there exists 5 > 0 such that for every E € & with u(E) < 4 we have [ \fal? du 00 2° |lfallp < M foralln EN. Show that {a) lim, Sg fxg du = f, fg dy for everyA € A for every g € L7(X, A, pn). )

im, ta ind = f, f du for every Ac A.

(Hint: Egoroff’s Theorem) Prob. 16.14. Let (X, 2, 4) be a o-finite measure space and let p € [1, 00}. Let f be an element and (f, : n € N) be a sequence in L?(X, 2, 4) such that im, llfn — fllp = 0.

(a) Show that for every ¢ > 0 there exists 5 > 0 such that for every E € & with w(E) < 6 we have

[imran 0 there exists A € 2 with (A)

f

X\A

< co such that

\fal?du 0 there exists 8 > 0 such that for every E € M with w(E) < 6 we have f |fnl?du O there exists A € & with p(A) < 00 such that

[

X\A

\fal?du O there exists § > 0 such that | fx(x') — fa(x”)| < eforallac A when x', x! € I are such that |x’ — x"| < 8. Prove the following theorem: Theorem. Let { fy : a € A} be a collection of real-valued functions on [a, b] C R such that

Ja is differentiable and the derivative fy is bounded on (a, b] and supye, fa,sj | fiPdu, < M where M > 0. Then {fa : a € A} is equicontinuous on [a, b].

Prob. 16.17. Given a measure space (X, A, 4). Let D € A and let (f, : n € N) and f be

extended real-valued 2{-measurable and j:-integrable functions on D such that 1°

2°

lim

n> 00

f, = f ae. on D,

lim fy lfaldu = fry lfldu.

Show that for every E € 2 such that E C D we have

@

lim frlflde = fy lfldu,

b)

lim fa dus fy f du.

Prob. 16.18. If condition 2° in Prob. 16.17 is replaced by

3°

lim fp fndu = fy f du,

then (a) and (b) in Prob. 16.17 do not hold. Construct an example to show this.

Prob. 16.19.

Consider a Banach space L?(X, Mi, 4) where p € (0,00).

Let f be an

element and (f, : m € N) be a sequence in L?(X, A, 4) such that

1°

2°

«lim f, =f ae.onX,

n>oo

dim Ilfallp = If llp-

According to Theorem 16.28, assumption of 1° and 2° implies that im,

fn —f lp = 9.

Show that this is not true for a Banach space L(X, 2, 12), that is, jim, Sn = fae.onXx,

and lim | falloo = lIflloo do not imply lim Ifa — flloo = 0.

Prob. 16.20. Consider a real-valued function f (x) = e~*’? for x € R. Show that f € L?(R, 9, 4,) for every p € (0, co] and moreover lim PB

Ifllp = Il flloo-

428

Chapter 4 The Classical Banach Spaces

Prob. 16.21. Construct ameasure space (X, 2, 4) and areal-valued 2{-measurable function f on X such that f ¢ L?(X, A,

2) for everyp € (0, 00) but f ¢ Loo (X, A, 2).

Prob. 16.22. Let (X, &, 2) be a o-finite measure space with (X) = co. (a) Show that there exists a disjoint sequence (E, : n € N) in & such that L),ony En = X and p4(E,) € [1, 00) for everyn € N.

(b) Show that there exists a real-valued &%-measurable function f on X such that f ¢

L'(X, &, yw) andf € L?(X, A, 4) for all p € (1, co].

Prob. 16.23. Let (X, 2, 44) be an arbitrary measure space. Let f be an extended complexvalued &-measurable function on X such that |f| < co, y-ae. on X. Suppose that

fg €L\(X, &, u) for every g € L1(X, A, 4). Show that f € L°(X, A, 4).

Prob. 16.24. Let (X, A, 4) be ao-finite measure space with 4(X) = oo. Let p, g € [1, 00]

be such that 3 + i = 1. Let f be an extended complex-valued 2%{-measurable function on X such that |f|


0. To show that

€ [1, 00), note that fy lfP du, = Yen”? /e*. To

show that this series converges, note that by I’ H6pital’s rule

jim, n? fe” = 0 so that there

exists N € N such that n?/e” < 1 forn > N. Then ),,yn?/e* < sy This shows that fy | f|? du, < co and therefore f € L?(X, Dt, 1X, u4,). Remark 17.6.

For p € [1, 00], as we showed in Theorem

increases. | f | p however does not have this property.

1/e” < 00.

17.4, Mp(f) increases as p

Example. Consider ([0, 2), 99%, 1 [0, 2), #,). Let f = 1 on [0, 2). Then | fli = 2 and If ll2 = V2. Thus || fla > lf llo[11] Approximation by Continuous Functions Let us consider approximation of functions in L?(X, 21, 4) by continuous functions on X.

The notion of a continuous function on X is predicated on a topology on X, Here we consider the case X = Rwith its Euclidean metric topology and (X, A, 4) = (R, mM, H,)- In §19,

we consider a locally compact Hausdorff space X with a Radon measure yz on the Borel o-algebra By of subsets of X.

Definition 17.7. Let C,(R) be the collection of all complex-valued continuous functions vanishing outside a compact set in R which depends on the individual function. Since every compact set in R is contained in a finite closed interval and a finite closed interval is a compact set, a complex-valued continuous function on R is a member of C,(R) if and only if it vanishes outside a finite closed interval. Thus if f € C,(R), then there exists B > 0 such that f vanishes outside the interval [—B, B]. Then we have

SUpR | f| = max__az,5)|/| < 00. Thus f is bounded on R andhencef € L©(R, Dt, u,)Also for every p € [1, 00), we have fa |f|’ du, < (supg|fl)”2B < 00 so that f € L?(R, 9t,, #,). This shows C,(R) C L?(R, 2, #,) forp € [1, oo]. Lemma 17.8. Letf € L?(R, 9, 4.) withp € [1, 00]. For everyn €N, let a truncation

432

Chapter 4 The Classical Banach Spaces

of f be defined by fa) = |

F(x)

forx €R such that |x| 0. By 4° of Lemma

17.8, dma

l fn — fillp = 0 and thus there existsN ¢ N

such that

&

qd) By definition,

If — frllp [maxzar,..e ail]? = Ilxll2 and

thus ||x||p > [lxlloo. The inequality (1) is from (3) of Proposition 17.24.

Thus for every

P €[1, 00) we have |[x||p > [lxlloo and &-'/? [x||p < [|xlloo. This shows that | - ||, and

I + lloo are equivalent norms. This implies that any two norms in {|| - ||p : p € [1, 00]} are equivalent. &

Problems Prob. 17.1,

Consider the measure space ([0, 1), 99%, |10,1), #,)-

where Dg = [VE} 271,

SE, 27) fork e N.

Then [0,1) = gen Dr

Let f be a function on [0, 1) defined by setting f(t) = 2 for t € Dy fork € N.

Determine the values of p € (0, 00) for which we have f € L?((0, 1), 9, |[0,1), #1)Prob. 17.2. Given a measure space (X, &, 2). Let f € L?(X, A, uz) for some p € (0, 00)

andg € L(X, A, 4). Show that fg € L?(X, A, pw) and [I fgllp < If llpllglloo-

Prob. 17.3. Prove the following theorem: Theorem. (Bounded Convergence Theorem for Norms) Let (X, A, j2) be a finite measure

444

Chapter 4 The Classical Banach Spaces

space and let1 < p < po. Letf € L?(X, A, x) and

(f,:n EN) C L(X, A, pw) be

such that

1

2°

jim, tn=f, prae. on X,

I fallp

< C for all n € N for some C > 0.

Then we have lim n-

I fn — fllp = 0.

Prob. 17.4. Definition. Let (X, A, 2) be a measure space and let pi, pr € (0, oo]. We define L?!(X, A, w) + L?*2(X, A, u) to be a linear space consisting of all f, + fo where

fi © L?1(X, 2, w) and fo € L?(X, A, pw).

Prove the following theorem: Theorem. Let (X, &, 2) be a measure space and let0 < pi < p < po < 00. Then

LP(X, Mw) C [L?'(X, MW, w) + LK, W, w)], that is, every f € L?(X, &, 2) canbe represented as f = f|+ fy where fy € L?\(X, &, w) and fr € L?(X, A, u). Prob. 17.5.

Given a measure space (X, M, 2) with u(X)

€ (0,00).

Let f be an arbi-

trary extended complex-valued 2{-measurable function on X. By Definition 17.1, we have

M,(f) = Wf llp/#(X)? for p € (0,00) and Moo(f) = II flloo- Prove the following

statements:

(a) Forp € (0, 00), we have

lim

pon

(b) Forp ¢€ [1, 00), we have Mp(f)

Mp(f) = Moo(f). t Moo(f) as p to.

(c) Forp € (0, 00), we have im, IF lp = If lloo(d) Forp & [1, 00), we have ||flp t Il flleo as p t 00, provided that 4(X) € (0, 1]. Prob. 17.6. Given a measure space (X, U, 4) with 4(X) € (0, co). Let f ¢ L(X, M, yw) be such that || flloo € (0, 00) and let a, = te lf du = || ff forn € N. Show that we «Ont have lim —~ = |[f lloo. 500 On

Prob. 17.7. Let (X, 2, 4) be a measure space with j4(X) € (0, 00) and let 1 < p, < pz < oo. Then according to Theorem 17.4 we have L??(X, A, w) C L?1(X, A, wz) and thus the norm || - ||p, on Z?!(X, A, 2) serves as a norm on L??(X, A, y).

Show by constructing an example that L?2(X, 2, 2) need not be complete with respect to to the norm | - ||p, and thus L?2(X, M, 2) need not be a Banach space with respect to the

norm | - [Ip (Hint for an example: L?((0, 1], 99t, (0,11, #,) C £1((0, 11, Bt, o,1y He) )

Prob. 17.8. Let (X, 2, 2) be an arbitrary measure space. Let f be an extended realvalued 2-measurable function on X such that f € L1(X, A, w) N L(X, A, pw). Show

thatf € L?(X, A, w) for every p € [1, oo].

Prob. 17.9. Let (X, A, 2) be a measure space and let 0 < pi < pz < oo. Then whereas L?1(X, Ml, w) 1 L?2(X, A, w) # G, neither L71(X, A, wz) D L2(X, A, w) nor LP1(X, A, w) C L?2(X, A, w) holds in general.

theorem on L?1(X, M, x) 1 L?2(X, A, 2):

(See Prob. 17.11.)

Prove the following

§17 Relation among the L? Spaces

445

Theorem. Let (X, A, 4) be a measure space and let 0 < p, < p < pz < 00. Then we have

[L?P1(X, MW, 2) 1 LP2(X, A, w)] C LP(X, A, w).

Indeed for every f € L?\(X,A, w) NL” (X, A, pz) there exists A € (0, 1) such that

Ifllp SFI, IFllp,* < 00. Prob. 17.10. Consider L?(X, A, 44) for p € (0, oo]. By Remark 16.21, Theorem 16.23 and Theorem 16.43, || - ||p is a norm on L?(X, A, j2) if and only if p € [1, o]. Prove the

following theorem: Theorem.

Let (X, A, 4) be an arbitrary measure space and let 1 < p < q < 0.

For

brevity, let us write L?(X) for L?(X, M, 4) and L4(X) for L4(X, A, x). Then L?(X)N L4(X) is a linear space. Let us define a function | - ||, on L?(X) M L4(X) by setting

Iflle = fll + fle forf € LPO) N L4(X). Then | - ||« is @ norm on LP(X) 1 L4(X) and moreover L?(X) 1 L4(X) is a Banach space

with respect to the norm || - ||xProb. 17.11.

Given a measure space (R, Dt,, u,).

following statements by constructing counter examples:

(a) (b)

LetO

< p

< q < oo.

Prove the

L?(R, M,, u,) D LIBR, M,, w,). LR, Dt,, w,) Z LIBR, M,, u,).

Prob. 17.12. Given a measure space (X, 2, 42). (a) Suppose (X, 2, 2) has a 2-measurable set of arbitrarily small positive measure, that is, for every ¢ > O there exists E € A such that 4(E) € (0, ¢]. Show that there exists a disjoint sequence (E,, :n € N) in 2 with ~(£,) > 0 for every n € N such that im, uA(E,) = 0. (b) Suppose (X, 2, jz) has a 2{-measurable set of arbitrarily large finite measure, that is, for every M > O there exists EF € 2l such that w(Z) € [M, 00). Show that for every y > 0 there exists a disjoint sequence (E, : n € N) in & with u(E,) € [y, 00) for every n € N.

(This problem is a preparation for the next.)

Prob. 17.13. Prove the following theorem: Theorem. Let (X, 2, 2) be an arbitrary measure space and let 0 < p < q < 00. Then we have L?(X, UA, w) Z L4(X, A, wu) if and only if (X, A, 4) has a A-measurable set of

arbitrarily small positive measure.

Prob. 17.14. Prove the following theorem: Theorem. Let (X, A, 2) be an arbitrary measure space and let0 < p < q < 00. Then we have L?(X, A, uw) C L1(X, A, 2) if and only if there exists c > 0 such that every E € A with 2(E) # 0 has u(E) > c. Thus we have L?(X, U, w) C L4(X, A, p) if and only if

inf {u(E) : E € Wand w(E) ¢ O} > 0.

446

Chapter 4 The Classical Banach Spaces

Prob. 17.15. Prove the following theorem: Theorem. Let (X, U, jz) be an arbitrary measure space and let0 < p < q < oo. Then we have L?(X, A, 2) D LY(X, A, w) if and only if (X, A, 2) has a A-measurable set of

arbitrarily large finite measure.

Prob. 17.16. Prove the following theorem: Theorem. Let (X, 2, 2) be an arbitrary measure space and let0 < p 0 such that p(E) < B for every E € &. Thus we have L?(X, A, 2) D L4(X, A, uw) if and only if 4(X) < 00. Prob. 17.17. Let f € L?(R, Dt, 4.) where p € [1, 00). Withh e€ R, let us define a function f;, on R by setting f,(x) = f(x +h) forx € R. Show that f, ¢ L?(R, Mt, w,) for every h € R and jim If — fallp =0. =

Prob. 17.18. Let f € L?(R, Dt,, w,) andg € L7(R, Mt, w,) where p,q € (1, 00) and

are conjugates. Let us define a real-valued function F on R by setting

Fh) = [ f@+Ag@)u, (dx) R Show that F is a bounded continuous function on R.

forh eR.

Prob. 17.19. Consider (R, Mt,, #,). LetE € 9, with 4, (E) < oo. Forh € R, consider the translated set E — h. Let us define a real-valued function F on R by setting F(A) = u,((E —h)n E)

forh ER.

Show that F is a bounded continuous function on R.

Prob. 17.20. Letf ¢ L?(R, 9, w,) andg € L4(R, Mt,, w,) where p,g € (1, oo) are conjugates. Withh € R, let f;,(x) = f(x +h) and g;,(x) = g(x +A) forx € R. Show that

(a) jm0 fag — f gli =O. (b) jim Ifa 8h — Ff alls = 0. 0

Prob. 17.21. Let (a, : n € N) and (5, : n € N) be two sequences of real numbers.

P,@ € (1, 00) be such that 1/p+ 1/¢ = 1. Show that »

lanBul < {>

neN

lanl? }/?f >

neN

Let

[bn |? }/,

neN

Prob. 17.22. Let (a, : n € N) and (5, : n € N) be two sequences of real numbers.

p € (0, 1) andg € (—o0, 0) be such that 1/p + 1/¢ = 1. Show that

Let

YD lanbal = {So lanl}? alt

neN

neN

neN

Prob. 17.23. Let (€, : n € N) be a sequence of positive numbers. Let p € (0, 00). Prove

Yen < 00 => ) oe? < 00. neN

neN

§17 Relation among the L? Spaces

447

Prob. 17.24, According to Theorem 17.21, we have £? C £4 for0 < p < q < oo. Show

that 2? 4 £7,

Prob. 17.25. Let 0 < p < g < oo. Show that £? Z £7.

448

Chapter 4 The Classical Banach Spaces

§18

Bounded Linear Functionals on the L? Spaces

[I] Bounded Linear Functionals Arising from Integration Let (X, Ml, 44) be an arbitrary measure space. Consider the Banach space L?(X, X, 1) with p € [1, 00) consisting of the equivalence classes of extended complex-valued 2measurable functions f on X such that f, x |f |? dy < 00 with the norm on the space given

by Ifllp = Le lel? dul”.

For such a functionf we have |f| < oo ae. on X. For

p = 00, the Banach space L™(X, 2, jz) consists of the equivalence classes of essentially bounded extended complex-valued 2{-measurable functions f on X, the norm on the space If lloo being the essential supremum of f on X.

We shall show that for p € [1, co] if

g € L1(X, A, 4) where gq is the conjugate of p and if we define a complex-valued function

Lg on L?(X, &, jz) by setting Ly(f) = tx fgdp bounded linear functional on L?(X, QA, jz). For £ € C, let

for f

< L?(X, A, 4), then Ly isa

gig? fors £0,

sgng=

o

for

Then

=0.

1 forg 40,

Isgng]=

0 forg =0,

and

g-sgn¢

=f]

forevery¢ €C.

Theorem 18.1. For an arbitrary measure space (X, A, 2), consider L?(X, A, 2) where p €[1, oo]. Let g € L7(X, A, 2) where g is the conjugate of p, that is, 1/p +1/q =1. {a) A complex-valued function Ly on L?(X, A, j4) defined by

a)

L,(f)= [ fad for f € L?(X,2%, u) is a bounded linear functional on L?(X, U, j).

(b) if p € 1, oo] and q € [1, 00), then

Q)

ILelle = llgllgIf (X, A, 2) is o finite, then (2) holds for the case p = 1 andq = & also.

Proof. 1. By Hélder’s Inequality (Theorem 16.14 for the case p, g € (1, 00) and Theorem 16.40 for the cases p = 1 and g = 00), we have

Q)

wNi=|f

tedul < f ifeldu < Uf lplely for f < L?(X, Mi, u).

Since ||f|lp < 00 and ||gllg < 00 we have |Ly(f)| < oo, that is, Ly(f) € C. The linearity of L, follows from the linearity of the integral with respect to the integrand. The

§18 Bounded Linear Functionals on the L? Spaces

449

inequality (3) shows that ||g|lq is a bound for the linear functional L, and L, is a bounded linear functional on L?(X, M, 4) with ||Lgll+ < llgllq- To prove (2), it remains to show the

reverse inequality |[Lg le > Ilgllg-

2. Let S(O, 1) = {f € L?(X, A, w): lf llp = 1}. Then |[Lglle = suprese,y |Eg(PI

by Lemma 15.36.

We construct fo € S(O, 1) such that |Zg(fo)l =

llgllg-

With the

existence of such fo € S(O, 1), we have ||Lgll« = Sup ses¢o,1) [Eg(F)| = |Eg(fo)l = llelle-

The construction of fo is divided into three cases depending on the value of p € [1, oo]. Ifg = 0 € L4(X, A, wz), then Lg(f) = 0 for every f € L?(X, MA, wz) so that Ly is the identically vanishing linear functional on L?(X, 2, 2) with ||Lglle = 0 = [lglg and (2) holds trivially. In what follows we assume g # 0 € L?(X, &, jx) so that ||g||z¢ > 0. 2a. If p € (1, 00) so thatg € (1, 00) also, let

fo = Wally? lel? sen. Sincep = q/(q — 1) and p(q — 1) = q, we have

fol? = [ lg? |g/@-Y? [sgn gl? du

= leliz? [ Isl*Isangl? dp = Isllz? {f {X:g=0} odu+ fist’: ian {X:g40} = lll? [ Ig? dw = Nelly *Igllg = 1.

Thus || follp = 1 so that fo € S(O, 1) C L?(X, A, w). By Lemma 15.36, we have

IZglle =

sup » ILg(f)| = [Le Go)| FeSO,

-| [ fox dul = tell* [ let tsame edu

= Ist"? [ igi? |gldu = IIgit-# [ ial? dye = llelly “lglg = lglg. Thus we have ||Lg||* = |Zg(fo)| = llgll in this case. 2b. Ifp = o0 and g = 1, let fo = sgng.

L™(X, 2, 4). By Lemma 15.36, we have HZglle=

Then ||folloo = 1 so that fo € §(0,1)

sup [Zg(f)| > |Lg (fo) FESO,1)

=|f foedn|=|f x

=f leldu = Wells. x

x

sane del

c

450

Chapter 4 The Classical Banach Spaces

Thus we have ||Lg||x = |Zg(fo)| = llglli in this case.

2e. Consider the case p = 1 andg = ov. For an arbitrary ¢ > 0, let Ap = {x EX: lg(x)| > Ilglloo — €}. Assume that¢ > 0 is so small that ||g|loo — € > 0. Since ||g loo is the infimum of all essential bounds of g, ||g|loo — € is not an essential bound of g and therefore H(A,) > 0. If we assume that (X, A, 2) is v-finite, then there exists a 2-measurable subset A of A, such that (A) € (0, 00). Let 1 f= utay'4

Lyall

=—_ emgidu wa1 f,| SPB

S8n g.

an

= 1 aw1

ft

=1.

Thus fo € S(O, 1) c L'(X, A, w). Then by Lemma 15.36, we have

HZglle=

sup [Lg(f)| 2 [Lg Fo) FeS0,1)

=| ff fora] =

1

5 | f seme eau

=— [ glau ie => —_{lalloo aw1 fl gy1 lel ~— 80)e}u(A = [glleo—6. By the arbitrariness of¢ > 0, we have ||Lgllx = [Lg(fo)| = llglloo Let us note that when p = 1 and g = oo, the equality ||Zg||+ = [lglloo may not hold without the assumption that (X, 2, 2) is o-finite. Example. Let (X, 2, 4) be a measure space which has an atom with infinite measure, that is, there exists A € 2{ with j4(A) = oo such that for every 2&-measurable subset Ap of A we have either z(Ao) = 0 or (Ap) = (A). With such an atom A, let g = 1,4. Then g €

L©(X, 2, 2) with ||glloo = 1. Consider the bounded linear functional L, on L'(X, 2, 1)

defined by Le(f) = fy fg du for f € L'(X, U, 2). For an arbitrary f € L1(X, A, 4), we have fy |f| dy < 00. This implies that the set E = {x € X : | f(x)| 4 0} is a o-finite set according to (f) of Observation 9.2. Thus there exists a sequence (E, : 2 € N) in 2 such that ), O, then we have z{x € X : |fo(x)| = 1} > 0. Since | fo] < 1 on X, we have || folleo = 1 and

fo € L™(X, A, w). This shows thatg € L1(X, 2, z).

Problems Prob. 18.1. Let (X, UA, jz) be an arbitrary measure space and let p € (1,00). Consider (fn in €N) C LP(X,A, 2) andf € L?(X, A, wu). Show that the sequence (jf, :n € N)

converges weakly to f in L?(X, M, y) if and only if 1°

2

|lfallp < M for alln € N for some M

> 0,

lim fy fndu =f, f du for every E € & with u(E) < oo.

n->00

Prob. 18.2. Let (X, A, 2) be a c-finite measure space. Let p € [1, 00) and g € (1, 00] be

conjugates, that is, 1/p + 1/q = 1. Let g1, go € L7(X, A, 4). Assume that we have

[ teau= ff tean for every f € L?(X, A, 4). Show that g1 = gz, jt-a.e. on (X, 2, 2).

Prob. 18.3.

Let (X, 2, 4) be a a-finite measure space.

Let p € [l,oo].

Let fo €

L?(X, &, wz) be such that for every bounded linear functional L on L?(X, A, 2) we have L(fo) = 0. Show that fo = 0 € L?(X, A, p).

Prob. 18.4.

Let (X, 2, 2) be a o-finite measure space.

Let p,q

€ [1, 00] be such that

1/p+1/q = 1. Let f be an extended real-valued 2-measurable function on X such that lf | < 00, g-a.e. on X. Define a sequence of functions (jf, : n € N) by setting x)=

f@)

f@) | 0

if|f@1 V, forr,s € Q such thatr Vo and Vi > K.

Applying Proposition 19.6 to the compact set K contained in the open set V, we have an

open set Vo such that Vo is a compact set and K C Vp C Vo C V. Applying Proposition 19.6 again to the compact set K contained in the open set Vo, we have an open set V; such

that V; is a compact set and K CVC

Vi CMC VOC

V. Let {rj : j € Z} be an

arbitrary enumeration of the elements of Q with rp = 0 and r; = 1. Now suppose for some n > 1, we have selected open sets V;),..., Vr, such that V,,,..., V;, are compact sets and

Vi, C ¥,, for r; > r;. Consider rz41. We have0 = ro < rag < 71 = 1. Let 7; be the greatest among those of ro, ... , 7, that are less than 7,41 and let r; be the least among those of 79, ..., 7 that are greater than 7,41. By Proposition 19.6, there exists an open set V,,,,,

such that V,,,, is a compact set and V;, C Viyi; C Vryu: C Vy. By induction onn € Zy,

we have a collection of open sets {V, : r € Q} satisfying conditions (i), (ii), and (iii). With the collection of open sets {V; : r € Q}, let us define two collections of functions {f; :r © Q} and {g, : s € Q} on X by setting

fr@) = |

r

ifxeV,,

0

otherwise,

and g(x) = |

1

ifxe Vs,

s

otherwise.

Let f = sup,cg f, and g = infseg gs. The characteristic function of an open set is lower semicontinuous and that of a closed set is upper semicontinuous by Observation 15.81. Thus J; is lower semicontinuous and g, is upper semicontinuous. Then by Theorem 15.84, f is lower semicontinuous and g is upper semicontinuous. Clearly f(x) € [0, 1] forx € X, f(x) = 1forx € K, and supp{f} c V. Since f is lower semicontinuous and g is upper semicontinuous on X, to show that f

is continuous on X it suffices to show that f = g on X. Let x € X be arbitrarily fixed.

Suppose f,(x) > gs(x) for some r,s € Q. Then the definitions of f, and g, imply that r > s,x € V,andx ¢ Vy. Butr > s implies V, C V,, contradicting the statementsx € V, andx ¢ Vs. Thus f-(x) < gs(x) for all r, s € @. Now suppose f(x) < g(x). Then there exist r,s € Q such that f(x) 0,40 € Ox, O D Eand p(O\ E) 0,4K € Rx, K C Eand w(E \ K) 0,4K

€ &x, K C E and p(K) > M.

§19 Integration on Locally Compact Hausdorff Space Proof.

1. Let us prove (1). Let E € Sy

471

and u4(£) < oo. Suppose yz is outer regular for

E, that is, we have w(E) = inf {u(O) : O > E, O € Dx}. Then for every¢ > 0 there

exists O € Ox such that O > E and u(O) < u(E) +6. Then since (EF) < 00, we have

uO \ E) = 40) — w(E) 0 there exists O € Ox

such that O D E and

H(O \ E) < e¢. Then since p(E) < oo, we have 4(O \ £) = 2(O) — u(£) and thus B(O) — p(B), &. This shows that for every e > 0 there exists O € Oy such thatO > E

and (0) — u(E) < ¢, that is, w(E) = inf {u(O) : O D E,O € Ox}. 2. (2) is proved by the same argument as above. 3. Let us prove (3). IfE € Bx and w(£)

= ov, then for everyO € Ox such that O D

E we have u(O) > #(E) = oo. Then inf {u(0) :ODE,O€ This shows that jz is outer regular for E.

Dx}

= 0 = “L(E).

4, Let us prove (4). Let E € By and u(£) = oo. Suppose y is inner regular for E. Then we have sup {u(K) :KCE,Ke Kx} = (E) = oo. Then for everyM > 0 there existsK € Ax such that K C E and w(K) > M.

Conversely suppose for every M

> 0 there exists K € Rx such that K C E and

u(K) > M, Then we have sup {u(K) : K C E,K € &x} = co = p(E). This shows

that 4 is inner regular for E.

©

We show next that if a Hausdorff space X is such that every open set in X is ao-compact

set, that is, a countable union of compact sets, then every Borel measure yz on X is inner

regular for Ox. Note that R” is such a space by Observation 19.13.

Observation 19.17. Let X be a Hausdorff space such that every open set in X is aa -compact set. Then every Borel measure z on X is inner regular for O x. Proof.

Let O € Ox.

(Ky,: 2 € N) in Ry

Since O is a o-compact set, there exists an increasing sequence

such that

jim, Kn =

Unen Kn=

O and then jim, BAK)= BO).

Now if 4(O) < oo, then for every & > 0 there exists n ¢ N such that (Ky) > w(O)-8, that is, 4(O \ K,) < e, so that y is inner regular for O by (2) of Lemma 19.16. On the other hand if 4(O) = oo, then for any M > 0 there exists n € N such that u(K,) > M.

Thus yz is inner regular for O by (4) of Lemma 19.16.

Outer regularity for %y or inner regularity for $y can be used to prove the identity of two Borel measures that are equal on Ox or equal on Rx respectively. Proposition 19.18. Let 2; and jz2 be two Borel measures on a Hausdorff space X. (a) If u1 and 12 are outer regular for By and1 = 2 on Dy, then 1 = v2 on By. (b) If 41 and22 are inner regular for By and 1 = pz on Rx, then p, = bz on By. Proof. Suppose j1 and j22 are outer regular for By

and 41 = 42 on Oy.

Then for every

472

Chapter 4 The Classical Banach Spaces

E € By we have by (1) of Definition 19.15

m(E) = inf {41(0): OD E,O € Dx} = inf {y2(O)

: ODE,O

ex}

= H2(E). Thus jz; = j42 on 3x.

19.15.

©

This proves (a). Then (b) is proved likewise with (2) of Definition

Definition 19.19. A Radon measure is a Borel measure pp on a Hausdorff space X such that x, and finite on Rx, that is, # is outer regular for Bx, inner regular for 1° outer regularity for By: u(E) = inf {u(0) :ODE,O€

Dx} for every E € Bx,

2° inner regularityfor Ox: (0) = sup {u(K) :KC0O,Ke

Rx} for every O eDx,

3°

finiteness on Rx: j4(K) < 00 for every K € Ky.

A Radon measure jz on a Hausdorff space X is outer regular for 8 y butitis not required to be inner regular for 3x. We show next that if a Radon measure yz on a Hausdorff space X is a a-finite measure then yz is inner regular for %$y and hence regular for Bx. Proposition 19.20. (a) Every Radon measure yt on a Hausdorff space X is inner regular for every E € Sx with u(E) < oo. (b) Every o-finite Radon measure on a Hausdorff space X is inner regular for By and hence regular for Sx. In particular every finite Radon measure is regular for Bx. Proof. 1. Let us show first the z is inner regular for every E € By with u(E) < oo. By the outer regularity of js for E, for every ¢ > 0 there exists O € Oy such that O > FE and u(O \ E) < § by (1) of Lemma 19.16. Then by the inner regularity of 41 for our O € Ox with 4(O)

< oo, there exists C € Rx such thatC Cc O and p(C) > u(O) — 5 by (2) of

Lemma 19.16. Since yu is outer regular for O \ E € Bx and since u(O \ E) < oo, there existsV € Ox such that V > O \ E and p(V) < w(O \ E) + & < 5 by (1) of Lemma

19.16. Let K =C\V=CNV¢ € &y. Then K C O\(O\ E) =E and u(K) = w(C\ V) = w(C) — WC NV) > W(O) — 3 — WV) > WE) —e.

This shows that jz is inner regular for Z by (2) of Lemma 19.16. 2. Let E € Sy. If u is a o-finite measure on By, then every E € Bx

is a o-finite

set. If u(E) < oo, then p is inner regular for E by our result in 1. If 4(£) = oo then the o-finiteness of E implies that there exists an increasing sequence (E,, : nm € N) in Bx such that 4(E,) < 00 forn € N and cn En = E. Let M > 0 be arbitrarily given. Since

jim, #(En)

= 2(E) = oo, there exists N € N such that u(Ey)

> M+

1. Since

§19 Integration on Locally Compact Hausdorff Space

473

Ey € By and p(Ey) < ©, zis inner regular for Ey by our result in 1. Thus there exists K & &x such that K C Ey and p(Ew)

— 1 < w(K) by (2) of Lemma

19.16. Thus for an

arbitraryM > 0, we haveK € Rx such that K C Ey C E and w(K) > p(Ew)-1> M.

This shows that jz is inner regular for E by (4) of Lemma 19.16. This shows that jz is inner

regular for By.

wo

Corollary 19.21. Every Radon measure 1 on a compact Hausdorff space X is a finite Radon measure and hence regular for By. Proof. Since X is a compact Hausdorff space we have X € Rx. Then since yw is a Radon measure we have 4(K)

< oo for every K € &y and thus in particular we have 4(X) < oo.

Thus y is a finite Radon measure and then yz is regular for 93 by Proposition 19.20.

Proposition 19.22. Let X be a o-compact Hausdorff space. Then every Radon measure p on X is inner regular for By and hence regular for Bx.

Proof. o-compactness of X implies that there exist a sequence (K, : n € N) in Rx such that X = U,cy Kn. Since wz is a Radon measure on X, we have u(Kn)

< oo forn € N.

Then pz is a o-finite Radon measure on X and thus by Proposition 19.20, uz is inner regular for By.

As we showed above, a o-finite Radon measure jz on a Hausdorff space X is regular for 33x, that is, both outer and inner regular for 3x. This has the following consequence. Theorem 19.23. Let 4 be a o-finite Radon measure on a Hausdorff space X. We have: (a) For every E € Bx and e > 0, there exist an open set O and a closed set C such that

CCEC

Oandp(O\C) 0 there exists an open set O, > E, such that H(O, \ En) < yet by (1) of Lemma 19.16. If we let O = J, ey On, then O is an open set, O > E, and

0\E=0n8*=0n(Us) neN

=0n(()z%) neN

=Ulan(ne)|c LJ (0.7 £8) =U, \ En) neN

neN

474

Chapter 4 The Classical Banach Spaces

so that

w(O\ E) =) wOn\ En) < neN

&

neN

seer = 5

Thus we have shown that for every E € %3x and ¢ > 0, there exists an open set O such thatO D E and u(O \ E) < 5. Applying this to E° € %3x, we have an open set V such thatV > E* and w(V \ ES)


E, O € Ox} for E € &. w is inner regular for Ox that is, u(O) = sup {u(K) : K C O,K © Sx} for

Oe

Dy.

Then we have: 5° 6°

pis inner regular for A, that is, wCE) = sup {u(K) :KCE,Ke Sx} forE EA For every E € & and ¢ > O there exist an open set O anda closed set C such that

T°

For every E € & there exista Gs-set G anda F,-set F such that F C E C G and

CCEC Oandu(0\ C0) 0 there exist an open set O and a closed set C such that CCEC Oandu,(O\C) < «. Actually this is a particular case of Theorem 19.24. This can be shown as follows. To start with, R is a Hausdorff space and SR

C Mt,. The

measure 2, on Dt, is o-finite and finite on Ag. 2, is outer regular for Mt, according to Proposition 3.30. Every open set in I is a o-compact set by Observation 19.13. Then

#4, is inner regular for Og by Observation 19.17. This shows that (IR, 9t,, ,) satisfies

conditions 1° to 4° of Theorem 19.24. Thus by 5°, 6°, and 7° of Theorem 19.24, jz, is inner

regular for 99, , for every E € 2t, and e > O there exist an open set O and a closed set C

§19 Integration on Locally Compact Hausdorff Space

475

such that C C E Cc O and px, (O \ C) < , and for everyE € DN, there exist a Gs-set G and an F,-set F such that F C E Cc Gand 4, (G \ F) =0.

[III] Positive Linear Functionals on C,(X) Definition 19.26. Let X be a locally compact Hausdorff space and let C,(X) be the linear space of real-valued continuous functions with compact support on X. Let I be a realvalued linear functional on C,(X), that is, I is real-valued, I(f + g) = I1(f) + I(g) for f.8 © C,(X) and I(Af) =AI(f) for f € C-(X) andk ER We say that a linear functional I on C,(X) is positive if I(f) > 0 for every f € C-(X) such thatf > 0 on X.

We shall show that if 7 is a positive linear functional on C,(X), then there exists a unique Radon measure jz on %3x such that [(f) = tx F du for every f € C,(X). The

construction of such a measure jz is given in the next propositions.

Definition 19.27. Let X be a locally compact Hausdorff space and let C,(X) be the linear space of real-valued continuous functions with compact support on X. (a) For O € Oy and O # Gand for f € C.(x) such that0 < f < 10n X, we write

Ff < O ifsupp{f} Cc O.

(b) For K € Rx and f € C,(X) such that 0 < f < 1 on X, we writef > K iff > 1x, that is, f =1on K.

Definition 19.28. Let X be a locally compact Hausdorff space. Let I be a positive linear functional on C,(X). Let us define a set function y on Dy by setting y(®) = 0 and

(1)

y(O) = sup {1(f): f € CX), f < O} forOeDx,0 FH.

Let us define a set function p* on JB(X) by setting

(2)

w*(E) =inf {y(0): OD E,0 € Ox} for Ec P(X).

Proposition 19.29. Let X be a locally compact Hausdorff space. Let I be a positive linear junctional on C,(X). For the set functions y and js* in Definition 19.28 we have: (a) y is monotone and subadditive on DO x.

(b) p* = y on Dx.

(c) * is an outer measure on X.

Proof. 1. Let 01, 02 € Ox and O1 C O2. Letf € C,(X). Iff < O1, then f < O2 also. Thus {f € C.(X): f < O01} C {f € C.(X) : f < Oo} and hence y(O1) < y(O2). This shows that y is monotone on Ox.

Next let us show that y is subadditive on Dy, that is, for 01, O02 € Ox, we have

(1)

¥(Q1 U O2) < y(O1) + ¥(O2).

Let f be an arbitrary member of C,(X) such that f « O01 U Oo. Such function f exists by Theorem 19.11 (Urysohn’s Lemma). Then the compact set supp{ f} is contained in the

476

Chapter 4 The Classical Banach Spaces

open set 0; U 02 so that by Theorem 19.12 (Partition of Unity), there exist hi, ho € C.(X) such that hy < O1, hz < Op, and hy + ho = 1 on supp{f}. Then fhi, fho ¢ C,(X) and

fh, < O; and fh2 « Op. Furthermore fhi+ fho = f{h1 +h2} = f on supp{f}. Since Ff =0 outside supp{f}, we have fh, + fho = f on X. Thus we have

Q)

T(f) = 1(fhi + fha) = 1(fhi) + 1(fha) < y(O1) + y(O2),

where the inequality is by the definition of y by (1) in Definition 19.28. Then by (1) in Definition 19.28 and by (2), we have

y(O1 U Op) = sup {1(f): f € Co(X), f < O1 U Oo} < v(O1) + y (02). This proves (1).

2. The fact that z* = y on Dy is immediate from (2) of Definition 19.28 and the

monotonicity ofy on Oy.

Indeed if O, 0;

€ Ox

and O C O; then y(O)

that we have *(O) = inf {y(O1) : 01 D O, 0; € Ox} = y(O).

< y(O)

so

3. Let us show that j:* is an outer measure on X. Clearly 4*(£) € [0, co] for every E € $(X) and *(B) = 0. To verify the monotonicity of * on $8(X), let £1, Ey € P(X)

and £; C E>. If Ez c O for some O € Dx, then FE; C O also. It follows from (2) of

Definition 19.28 that z*(E1) < u*(E2).

Let us prove the countable subadditivity of 2* on 93(X), that is, for every sequence (E; : j € N) in $8(X), we have

@

w( U8) < wep. jen

jen

Let ¢ > 0 be arbitrarily given. By (2) of Definition 19.28, there exists O; €

x such that

Oj D Ej and y(O;) < u*(Ej) + 33. Consider the open set LU ey Oj. Letf € Co(X) be

such that f < Ujen O;. Since {O; : j € N} is an open cover of the compact set supp{ f},

there exists N € N such that supp{f} C Uj; Oj. Then f < UL, 0; so that by (1) of Definition 19.28 we have

a

N

N

j=l

j=l

1A 0, we have (3).

Proposition 19.30. Let X be a locally compact Hausdorff space. Let I be a positive linear functional on C,(X). For the outer measure * on X defined by Definition 19.28, we have By

C M(u*), the o-algebra of u*-measurable subsets of X.

§19 Integration on Locally Compact Hausdorff Space

4717

Proof. If we show that Ox C Itz"), then we have By = o (Ox) C Mt(pz"*). Let O € Dy. To show that O € Mt(u*), it suffices, according to Observation 2.3, to

show that for every A € 93(X) we have

(a)

H(A) > w*(ANO) + B*(AN OY).

Let A € $3(X). For an arbitrary « > 0, by the definition of j:* by (2) of Definition 19.28, there exists an open set G > A such that y(G) < n*(A}+¢. Nowif AN O = G, then (1)

reduces to .*(A) > 0+ 2*(A) which is trivially true. Thus we consider the case ANO # B. Then since G 5 A, wehave GMO # @. Take an arbitraryf € C,(X) such that f x GNO. Then since the compact set supp{ f} is contained in GN O C O, by Proposition 19.6 there exists an open set V such that V is a compact set and supp{f} C V C V C O. Nowif O > A, then (1) reduces to 4*(A)> w*(A) + 0 which is trivially true. Thus we consider the case O Z A. Then O ZG. This implies V Dp G so that G\ V # @. Take an arbitrary

g € C,(X) such that g < G\ V. Since supp{f} C V and supp{g} Cc G \ V, we have

O< f+g Ki D -++ > Ky. Since f(X) = [0, 1], wehave K; # Oforj =0,1,...,n.

The continuity of f on X implies that K; is a closed set. Being a closed set contained ina compact set Ko, Kj is a compact set for j = 1,...,7. Let us write

Ko = (Ko \ Ki) U (Ki \ Ko) U--- U (Kt \ Kn) U Ka. Note that f(x) € [45", non 4) forx € Kj-1\ Kj for j =1,...,mand f(x) = 1 forx € Ky. Forj = 1,...,, let us define a function f; on X by

(2)

- A} vojal Fi) =[{F@)

forxe x.

Then f; is continuous and0 < f; < 1/n on X and

Q)

Fj@Q=%

0

forx ¢ Kj-1,

f@)-—G-D/n

€ Kj-1\ Kj, forx

1/n

forx € Kj.

480

Chapter 4 The Classical Banach Spaces

Thus we have

(4)

supp{ fj} = Kj-1,

)

fj € CX),

(6)

Mx, < fj sx.

M

f= Lisi. jal

The last equality is verified as follows.

If x ¢ Ko, then f(x) = 0 and f;(x) = 0 for

j =1,...,n so that (7) holds at x. Supposex € Ko. Then x is in one of the n + 1 disjoint sets Kg \ K1, Ki \ Ko, ..., Kn_1\ Kn, Kn. Supposex € Kjo_1\ Kj for some jo = 1,..., 7. Then

I/n

for j =1,...,40—1,

0

for j = jotl,....n.

fi) =} F)—Go—Din for j= jo This implies

,

yAe="

*+7@)-2— = fe.

Ifx € Ky, then f(x) = 1 and f;(x) = 1/n forj = 1,...,n so that f(x) = Dj-l This proves (7). Now from (6) we have

1

ntKi) for j = 1,...,. integral, we have

Since nf;

1

< [ fjdus 2K)

Summing over j = 1,...,” and recalling (7) and the linearity of the

ix

re)

Fj).

nk Kj.

< I(fj). Therefore we

have 14(K;) < (fj) < 4a(Kj-1) forj = 1,...,n. Summing overj = 1,..., and recalling (7) and linearity of the functional 7, we have

@)

1X

a MK) j=l

i

< HP) n H(Ko). Since (Ko) < 00, letting n — 00 we have |I(f) — fy f du| = 0 and thus for our particular f € C,(X) we have I(f) = fy f du. Next consider f € C,(X) such that f > 0 on X. The equality 7(f) = ty fap holds trivially if f is vanishing identically on X. If f is not identically vanishing on X,

then@ := maxzex f(x) > 0. Let fo = anf, Then fo € C.(X),0 < fo < 1, and Maxxex fo(x) = 1. Thus by our result above, we have (fo) = tx fod. Multiplying

the equality by a , we have 1(f) = fy f du by the linearity of J and the integral. For an

arbitrary f € C,(X), let us write f = f+ — f-. Since f+, f- € C.(X) and f+ > Oand f7~ = 0on X, we have

IN =1)-109 = [ ftdu- [ frdp

= firt-rhan= ff rau. 2. Let us prove the uniqueness of a Radon measure v on 8x such that I(f) = ik fdv for every f

€

C,(X).

Let u be the Radon measure in 1, for which we have shown

I(f) = fy f dp for every f € C,(X). Let O be an arbitrary nonempty open set in X. Let f € C,(X) be such that f x O. Then since0 < f < 10n X and supp{f} C O, we have f 1x so that fy fdu =1(f) = fy fdv = fy 1x dv = v(K). For our Radon measure 4, we have according to Lemma 19.31

w(K) = inf {1(f): f € C,(X) andf > K} = inf { fy fdu: f € Co(X) and f > K}

> v(K).

482

Chapter 4 The Classical Banach Spaces

Since this holds for every compact set K, the inner regularity of jz and v on the open sets implies that 4(0) > v(Q) for every open set O. Then by the outer regularity of 2 and v on 38x, we have 4(E) EeSBy. a

> v(E) for every E € By.

This shows that ~(£) = v(£) for every

Applying the Riesz-Markoff Theorem we obtain the following regularity condition for a Borel measure. Lemma 19.35. Let X be a o-compact Hausdorff space and let 2 be a Radon measure on X. Let v be a Borel measure on X that is finite on Aix. If v = pon Dx, then v is a regular

Radon measure on X.

Proof. Since X is a o-compact Hausdorff space, a Radon measure yz on X is regular for 98x by Proposition 19.22.

Since v is a Borel measure on X that is finite on Ry, to show

that v is a regular Radon measure it remains to show that v is both outer and inner regular for By.

1. Let E € By. Since pz is a Radon measure on a o-compact Hausdorff space X, pw is ao-finite measure. Then by Theorem 19.23, for every ¢ > 0 there exist an open set O and a closed set C such that (lb

CcECO

and

p(O\C) M. Thus o(E) = Ky € Rx, Ky CC C E, and v(Ky) > M. This shows the inner regularity of vfor E by

(4) of Lemma 19.16.

3.2. Consider the case v(C) < oo. Then v(E) — v(C)= v(E\ C)< v(O\C) 0, are immediate. Thus / is a positive linear functional on C,(X). By Theorem 19.34 (Riesz-Markoff), there exists a unique Radon measure v on (X, 23x) such that

(2)

I(f) = [

trav

forf € C,(X).

Since X is a a-compact Hausdorff space, the Radon measure v on X is inner regular for %8x and hence regular for 3x by Proposition 19.22. Now by (1) and (2) we have

(3)

[ rau= [rar for f € CX).

Letus show that (3) implies 4 = von Dx. Let O € Oy. Since O is an open and o-compact set, there exists an increasing sequence (f, : 2 € N) in C,(X) such that

jim, fa =1o

by Proposition 19.14. Then by the Monotone Convergence Theorem (Theorem 8.5) and by (3), we have

wo) = f | todu = tim, iim, f fe fadu = tim, [ fed» = f n-+00

xX

xX

tod»

=v(O). This proves 4 = v on Oy. The fact that v is a Radon measure on a o-compact Hausdorff space X and yz is a Borel measure on X that is finite on Ry such that ~ = v on Ox implies that is a regular Radon measure on (X, 3x) by Lemma

19.35.

[IV] Approximation by Continuous Functions Theorem 19.37. (Lusin) Let X be a locally compact Hausdorff space and let jz be a Radon measure on the Borel o-algebra Bx of subsets of X. Let f be an arbitrary extended real-valued B x -measurable function on X which is finite a.e. on X and vanishes outside ofa set A € Bx with w(A) < oo. Then for every & > 0 there exists g € C,(X) such that p{x € X: f(x) # g(x)} < ©. Moreover g € C,(X) can be so chosen that we

have infyex f(y) < g(x) S supyex f(y) for x € X and in particular sup, O there exists g € C,(X) such that g(x) € [0, 1] forx ¢ X and p{X: f Ag}

0, there exists a compact set K,, such that K, C B, and u(B,\ Ky) < yaar By the outer regularity of 4 on 8 y, there exists an open set V,, such that w(V, \ Bn) < maT

Since B, C V, we can take V, C V by intersecting with V if necessary.

For the sets

Ky C Bn C Vn, we have U(Va \ Kn) = (Vn \ Bn) + H(Bn \ Kn) < ga + pa = 3

Since K, C V,, by Theorem 19.11 (Urysohn’s Lemma) there exists h, € C,(X) such that hy < V, and hy > K,. Let us define a function g by setting g = nen dha. Since hy(x) € [0, 1] forx € X foreveryn € N, we have g(x) € [0, 1] forx € X. By the uniform

convergence of the series in the definition of g, g is a continuous function on X.

Since

hy < Vn, we have supp{,} C Vn C V so that h, = 0 on V° for everyn € N. This implies

thatg = 0 on V° so that {X : g £0} C V and then supp{g} C V. Since V is a compact set, we have g € C,(X).

Itremains to show u{X : f # g} < e. Recall thatg = Dncy Hhn and f = Dycy Vn

Let us compare dhn and w,. By hy > Ky, we have h, = 1 on K, and then dhn K,. On the other hand we have 2", K,, and thus dihn

= & on

= 1, and B, > K,. This implies that ¥, = x on

= w, on K,. Next hy ~< V, implies hk; = 0 on Vf and then dihn =Oon

V,. On the other hand 2"y, = 15, implies ¥, = 0 on BF > V,. Thus we have ah, =n

on V{ also. Therefore we have ah, = vn on K,UVE. Then {X : ah, =n} > K,UVE and consequently

{X: Bhn # Va} C (Kn U VE)’ = KEN Vy = Va \ Ko. Since {X : g = f} > Qhen{X: ah, = Wn}, we have

(Xie # FHC [(]{X: Min = Yah] =Unen (Xda = va)” aeN

= U {X: neN

hn # Wn} C Unen Va \ Kn.

Then w{X:¢ 4 f} < Drew H(Vn \ Kn) < Den 8/2” =.

§19 Integration on Locally Compact Hausdorff Space

485

2. Next let f be a real-valued %3 y-measurable function on X such that f(x) € [0, 1) for x € X and f =0on A‘ where A € Sy and (A)

< oo. Then by (a) of Proposition 19.20,

#- is inner regular for A. Thus for an arbitrary ¢ > 0 there exists a compact set K C A

such that 2(A \ K) < 5. Let fo = f1x. Then fo(x) € [0, 1) forx € X and fo = Oon K°. Thus by our result in 1, there exists g € C,(X) such that g(x) € [0, 1] for x € X and

uAX : fo # 8} < 5. Since X = K U(A\ K) UA‘, we have

{X:fAg}c{K:fAg}UCA\K)U{A: f Za}. Since f = foon K, wehave {K: f A gu}={K: fo#s} c {X: fo #8}. Similarly, since f = fy =00n A we have {A°: f A g}={A°: fo#e} Cc {X: fo #8}. Thus

{xi f #a}c{X: foFs}UA\K)

and therefore u{X: f

#g} 0, there exists N € N such

that u(By) < §, that is, u{X : |f| > N} < §. Let fl = (f V-N) AN. Then fa) € (—N —1,N +1) forx € X and f™] =0 on A®. By our result in 4, there exists g € C,(X) such that g(x) € [—N —1,N +1] forx € X and y{X: f'%] 4 g} < §. Now

{(X:f As} ={X: Il N.f Fe} UlXlfl> NF Fah. On {X : |f| < N}, we have f = f'!, Thus

{Xi lS Nf Aah = {Xi lflsN}O(X: fF AB}

c{x:f=Ffn{x: Ff 48} c{x: fl ¢ gh.

486

Chapter 4 The Classical Banach Spaces

We also have {X : |f| > N,f £2} c {X: |f| > N}. Therefore we have

{X: f Ae} c{x: fF 4g} Ulx:lfl> N}, and then

WX: f Ag} sa{X:

Ze} + uf{XsIfl>N} 0, thereexists

g € C,(X) such that u{X : f # g} < € as we showed in 5 above. Let us define a function h on X by settingh = (g Vinfyf) A supy f. Thenhk € C,(X), infyf < h < supyf

and {X: f Ah} c{X: f Ag} sothatu[X: f Ah} 0 such that |f| < M on X. Thus we have

f x fl? du = [ supp{f} If! du < MPu(supptf}) O there exists g € C.(X) such that ||f — gllp < ¢. Let So(X, Bx, 44) be the collection of equivalence classes of complex-valued simple functions on (X, %$x) each vanishing outside a set in 8 y with finite measure. According to Theorem 18.3, So(X, Bx, 12) is a dense subset of L?(X, By, 4). Thus for an arbitrary f €L?(X, By, w) and ¢ > 0, there exists g € So(X, Bx, 4) such that ||f — gllp < 5.

Leta = sup,cy |v(x)|. Ifa = 0, theng = 0 so that¢ € C,(X) and we are done. Consider the case a > 0.

Now g = Ry +iSg

and Rg and S¢g are real-valued 93 y-measurable

§19 Integration on Locally Compact Hausdorff Space

487

functions on X vanishing outside a set of finite measure so that by Theorem 19.37 (Lusin) there exist real-valued g1, g2 € C,(X) such that eP

sup |gi| < sup |Re| 0 there

488

Chapter 4 The Classical Banach Spaces

exists O € Ox such that O D E and (411 + 42)(O (441 + 2)(E) < 00 implies p;(Z) < oo fori = Bx and E € Sy, there exists O; € Ox such that Lemma 19.16. Let O = 0, M Op. Then O € Dx,

\ E) < & by (1) of Lemma 19.16. Now 1 and 2. Since yz; is outer regular for O; D E and y2;(O; \ E) < 5 by (1) of O D E, and

(1 + 42)(O \ E) = 41(0 \ E) + w2(O \ BE) = w1(01 \ E) + p2(O2\ E) M by (4) of Lemma

19.16. Then (j41 + 442)(K) > M. This proves the inner regularity of 41 + 22 on O by (4) of Lemma 19.16. Consider the case (441 + 42)(Q) < 00. To show that 1 + j22 is inner regular for O, it suffices to show that for every ¢ > 0 there exists K € (1 + H2)(O \ K) < © by (2) of Lemma 19.16. Now (#1 #i(O) < co fori = 1 and 2. Since ji; is inner regular for Oy, that K; C O; and p;(O \ K;) < 5 by (2) of Lemma 19.16. If

K € &x, K C O,and

Ax such that K C O and + 42)(O) < 00 implies there exists K; € Ax such we let K = Ki U Ko, then

(1 + #a)(O \ K) = 41(0 \ K) + u2(0\ K) < w1(O1 \ Ki) + w2(O2 \ Ko) M+1.

Since (Ey) K € Sx H(K) >

< 00, w is inner regular for Ey by our result above.

such that K uw(Ey) —1

C Ey > M.

and w({Ey) —1 < Thus for every M

Thus there exists

p(X) by (2) of Lemma 19.16. Then > 0, there exist K € Ax such that

K C Ey C E and u(K) > M. This shows that yz is inner regular for E by (4) of Lemma 19.16. 2. Let us prove (b). Suppose j2 is a Radon measure. Then the o-finiteness of j2 implies

that 22 is regular, that is, both outer and inner regular, for 98 by Proposition 19.20. Let us show that jz is a regular Radon measure. Now since j42 is a Radon measure, {22 is finite on Ry. Then 4; < 2 implies that 21 is finite on Ay. It remains to show that 21 is both outer and inner regular for Bx. Let E € %y be arbitrarily chosen. If j4(Z) = ov, then y is regular for E by (a). Let us consider the case j2;(E) < oo. By the assumption in (b), we have jz2(Z) < oo. By the outer regularity of 42 for E, for every ¢ > 0 there exists O € Oy such that O > E and

H2(O \ E) < € by (1) of Lemma 19.16, Then 41(O \ £) < u2(O \ EZ) < &. This shows

that 421 is outer regular for E by (1) of Lemma 19.16. Similarly the inner regularity of p2 for E implies that for every ¢ > O there exists K € Rx such that K C E and p2(E\K) 0 there exists N € N such that (2)

lfm(x) — fa(x)| < § for allx ¢ X when m,n > N.

Letn > N and x © X. By (2) we have fn(x) € (fn(x) — §, fu(x) + §) for all m > N. Then f(x) = im, Sm(&) € (fa(x) — &, fax) + €). Thus we have

forallx € X whenn > N.

lf) — fa@)| 0 such that

IEA) S Mllfllu for every f € C(X). Let C(X)* be the dual space of C(X), thatis, the linear space of all bounded linear functionals

on C(X). According to Theorem 15.38, if we define a function | - ||, on C(X)* by setting for every L € C(X)*

[Zlle = inf {M > 0: |L(f)| < Milflla for f € CCX)}, then | - [lx is a norm on C(X)* and (C(X)*, | - llx) is a Banach space. Proposition 15.30 we have

IL) < [lL llell fle Let B(fo, r) = {f € C(X) : lf — follu 0. Then according to Lemma

[Zlle=

sup

feBO,1)

According to

[L(f)|=

15.35 and Theorem 15.36, we have

sup

feSO,1)

[L(f)I.

Consider C+(X), the subset of C(X) consisting of nonnegative continuous functions on X. As in Definition 19.26, a linear functional L on C(X) is called a positive linear functional if L(f) > 0 for every f € C+(X). We shall show that for every bounded linear

§19 Integration on Locally Compact Hausdorff Space

493

functional ZL on C(X) there exist two positive bounded linear functionals L+ and L~ on

C(X) such that L = L+— L-.

Observation 19.51. Let L be a positive bounded linear functional on the Banach space (cj,

I-ll «) of real-valued continuous functions on a compact Hausdorff space X. Then

(1)

Lf) SL(|fl)

(2)

Zll4 = £0).

for every f < C(X),

Proof. For every f €¢ C(X), we have —f, |f| € C(X) and f, —f < |f|sothat|f|-—f >0 and |f| + f > 0, that is, |f| — f, |fl + f € Ct(X). Then since L is a positive linear

functional on C(X), we have L(|f|— f) = O and L(|f|+ f) = 0, thatis, L([f|) = L(f)

and L(|f|) = —L(f). Therefore |L(f)| < L(|f|)- This proves (1). Let f € SO, 1) = {g € C(X):

1—|f|

llellu = 1}. Then || flu = 1 so that |f| < 1. Then

= 0so that 1 — | f| € C+(X). Then since L is a positive linear functional on C(X),

we have L(1 — |f|) > 0, that is, L(1) > L(|f|). Thus recalling (1), we have

Zlx=

sup |L(f)|< sup L(|fl) < £0). FeSO,1) feS(0,1)

On the other hand, since1 € S(O, 1), we have ||L ||. > |E(f)| = LQ). Thus ||Z||x = £(1). This proves (2).

Lemma 19.52. Consider the Banach space (C(X), | - llu) of the real-valued continuous

functions on a compact Hausdorff space X and its dual space (c (X)*, Il - Il u)- Let C+(X)

be the subset of C(X) consisting of nonnegative continuous functions on X. Corresponding toeach LE (c()*, l - lls) let us define a real-valued functional Az, on C*(X) by

A,(g) = sup {L(@) :h © Ct(X),0 sh 0. (©) additivity: A, (gi + g2) = A, (1) + A, (g2) for g1, g2 € CT(X). Proof. 1. (a) is immediate from the definition of A,.

2. Let g1, g2 € C+(X) and gi < go. Ifh € C*+(X) andO 0so that

Q)

L*Cf) =A, Cf + cy) — A, Cy)

=c{A,(f tv) -A,W)} =cL(f), where the second equality is by (d) of Lemma 19.52. Next for the case c = —1, we have

L*Cfy + L*(f) =L*Cf + f) = Lt @) = A, 19.52. Thus we have

(10)

= 0 by (6), (5) and (c) of Lemma

Lt(fy=-L*(f).

Finally for c < 0, we have

ql)

LY (cf) = L*(—|elf) = —L* (lelf) = -lelL*(f) = eL*(f),

where the second equality is by (10) and the third equality is by (9). This proves (7). For

g € Ct(X), we have L+(g) = A, (g) = 0 by (5) and (c) of Lemma 19.52. This proves (8). 3. Let us show that the linear function L+ on C(X) is a bounded linear functional. By

(4) and by the fact that A, is nonnegative valued, we have

—A, (If lle) SLC) S AL(f + Ifllu) According to (c) of Lemma

for f € CCX).

19.52 we have A, (g) < ||Lllallgllu for every g € C*(X). Thus

—[Lllellflla < £t(f) < [Lief + Uf lull, forf ¢Ccx). Now ||f + Iflle||, < Uf llu + Uf lla = 2Ilfllu. Thus we have [E*(f)| S2[Lllellflle forf € C(X).

§19 Integration on Locally Compact Hausdorff Space

497

This shows that L+ is a bounded linear functional on C(X) with ||L* ||. < 2||L|l«. 4, Let L~ = L*+ — L. Since L and L+ are bounded linear functionals on C(X), so is L-. For g € C+(X), we have L*(g) = A, (g) > L(g) by (5) and by (a) of Lemma 19.52. Thus L~(g) = L+(g) — L(g) = 0. This shows that L~ is a positive linear functional.

We have L = L*+ — L~. Then by Observation 19.51 we have

(12)

[Ll = LF — £7 lle SLTI + ET lls = E71) + LQ).

To prove the reverse inequality, leth € C+(X) be such that 0 < A < 1. Then [2h —1|

so that 2h — 1 € BQO, 1) ={f € C(X): If lla < 1}. Thus

(13)

IZlk=

E} for E € 98(X). Then the u*-measurability condition (1)

BX(B) =e (BOE)+u*(BNE*)

for every B € BX)

is equivalent to the condition (2)

B*(A) = w*(AN E)+ w(ANE*) for every A € A.

Proof. Since an algebra 2 of subsets of a set X is a covering class for X, this theorem is a particular case of Theorem 2.24.

[I] Extension of an Additive Set Function on an Algebra to a Measure Definition 20.4. Let 1 be a nonnegative extended real-valued additive set function on an algebra U of subsets of a set X with (0) = 0. We say that yt is extendible to a measure on Xif there exists a measure space (X, §, v) such that A C F and wp =v on A.

Theorem 20.5. (Hopf Extension Theorem) Let 1 be a nonnegative extended real-valued additive set function on an algebra A of subsets of a set X with w(@) = 0. Then p is extendible to a measure on X if and only if 2 is countably additive on A. Proof. 1. Suppose jz is extendible to a measure on X, that is, there exists a measure space (X, §, v) such that 2 c ¥ and ~ = v on A. To show that jz is countably additive on 2, let (A, : n € N) bea disjoint sequence in 2 such that Unen A, € QW. Since A Cc Fandp=v

on & and since v is countably additive on ¥, we have i (new An) = v (Une An) =

Den V(An) = Dew (An). This proves the countable additivity of 4 on 2. 2. Conversely suppose yz is countably additive on 2. Then according to Theorem 20.1, if we define an outer measure yz* on X by setting

w*(E) =int | Yen (An) : (An

€N) CM, Upen An > E]

for E € $(X), then (X, Mt(u*), w*) is a measure space such that A # = u* on A. Thus yz is extendible to a measure on X. wo

C

Mt(p*) and

[IIT] Regularity of an Outer Measure Derived from a Countably Additive Set Function on an Algebra Definition 20.6. Let 2 be an algebra of subsets of a set X. We define Aq as the collection of all countable unions of members of & and define Uys as the collection of all countable intersections of members of Aq.

§20 Extension of Additive Set Functions on an Algebra Observation 20.7.

507

For an algebra 2 of subsets of a set X,

the collections A, and A,5

are not closed under complementation in general and thus they are not algebras in general. However 2, is closed under countable unions since a countable union of countable unions of members of & is still a countable union of members of 2. Also A, is closed under intersections and hence closed under finite intersections. To prove this, let A, B € Ay.

ThenA = (Jnen An and B = (Jey Ba where (A,

: n € N) and (B, : n € N) are

sequences in &. Then AM B = (Umen Am) O (Unen Bu) = Unnen(Am M Bp). Since MW is an algebra, A,, M B, € Wand thus AN BE Ay.

Let yz be anonnegative extended real-valued countably additive set function on an algebra A of subsets of a set X with 4(@) = 0. Let z* be an outer measure on X based on p, that is,

pA(E) = ink { Sonew #(An) : (An: EN) CA, Unew An D E}

for E € $8(X). Consider the c-algebra Jt(*) consisting of those members E of $3(X) which satisfy the .*-measurability condition

p(B) = w*(BN E)+p*(BNE*)

for every B € $B(X).

By Theorem 20.1, 2 C 29(*) and thus we have the following chain of inclusions:

ACA, C Ass CoM C MUu*). Also z = u* on Qi according to Theorem 20.1. Let us consider approximation of a set E € $8(X) by members of A, and 2,5 in terms of pz*. Lemma 20.8. Let js* be an outer measure based on a nonnegative extended real-valued countably additive set function 4 on an algebra X of subsets of a set X with p(B) = 0. Then for every E € 98(X) and e > 0, there exists A € Uz such that E C A and

BSCE) < w*(A) < w*(E) +6.

Proof. Let E € 93(X). By the definition of ,z* as an infimum, for every e > 0 there exists a sequence (A, : n € N) in 2 such that ),.7 An > E and uw*(E) < Dew (An) < B*(E) + 6.

(Note that since z*(E) may be equal to 00, a strict inequality may not be

possible.) Let A = nen An € Ac. By the monotonicity and countable subadditivity of the outer measure jz* and by the fact that 4 = u* on & according to Theorem 20.1, we

have w*(E) < u*(A) < Den (An) = Den (An) 2, then (B, : n € N) is a disjoint sequence in ‘A with Gn) = SB) neN

neN

neN

=D") =2"(U Bs) = 2"), neN

neN

where the second equality is by the countable additivity of v on %, the third equality is from the fact that 2 = v on Ql, the fourth equality is from the fact that 4 = j2* on 2 and the fifth equality is from the countable additivity of * on Dt(uz*). This shows that v = z* on Ay. 2. Let us show that y = y* on Ags. Let A € Ags. Then A = (yey An where (A, : n € N) is a sequence in &,. The o-finiteness of jz on the algebra MX implies that there

exists a disjoint sequence (X;

: j €¢ N) in & such that Ujen Xs

= X and p(X;)

forj 0, there exists A € A

Proof. Let 21 be an arbitrary algebra of subsets of X such that o (2) = ¥ and let yz be the restriction of v to 2. Then jz is a nonnegative extended real-valued countably additive set function on & with (4) = 0. If we let 4* be the outer measure on X based on the set function yz on A, then A Cc Mt(uz*) and wp = u* on 2 by Theorem 20.1. Since Mt(u*) is a o-algebra containing 21, we have ¥ = o (2) C Mt(u*). Since v is a finite measure on %, its restriction yz to 2 is a finite set function on A. Thus by Theorem 20.12, we have v=p* ong. Let E € ¥ and € > 0. By the definition of j* as an infimum, there exists a sequence

(Ay : n € N) in & such that L,cy An D E and w*(E) < Dyey

(An) S w*(E) + §-

Since v = p* on ¥ and v = pw on A, the last inequalities are equivalent to

a)

v(E) =) v(An) $ VE) + 5. néeN

Now (Uj_; 4% : 2 € N) is an increasing sequence in A C F and Z_, Ak t Upen An 80 that v (Uf At) t » (pen An). Since v (nen An) < 00, there exists N € N such that

312

Chapter 5 Extension of Additive Set Functions to Measures

¥ (nen An) = 3 < ¥(Upen An)- Then Q)

#(L4e\ Ue) = (Uae) -o(U a) neN

neN

Let A = UM, An € 21. Since E C Ucn An, we have

B)

vE\ A) = v(U4n\ A) oo inf Ay) = 0.

Proof. Let E € o(&). For everyn € N, there exists A, € & such that z(HAA,)

< i by

Theorem 20.15. Now {X : |1a, — 1z| > 0} = EAA, so that for everyn € N, we have B{X : |1a, —1z| > 0} = #CEAA,) < 1. Thus for every ¢ > 0 we have lim {X

n—00

: |14, —1z] > e}
0}

n>00

= lim ,(EAA,)< lim 1 =0 nove0 noon that is, the sequence (14, : 2 © N) converges in measure to 1g. Therefore there exists a subsequence da, : n € N) which converges to 1g ac. on X. Let us write Ay for An,.

Then (Ax : & € N) is a sequence in 2 such that jim 1,4, = 1z ac. on X. This proves (1). OO.

Now we have (3)

lim

ko

14, =1z

on EG,

§20 Extension of Additive Set Functions on an Algebra where Eo is a null set in (X, o (20), 4). Supposex ¢ EM EG. By (3), we have jim N

é N such that 14,(%)

=

1 fork

> 00

>

14,@)

513

= 1g(x) = 1. Thus there exists

N, that is, x € A, fork

>

xe lim inf Ax. Thus EN £5 C lim inf Ax and consequently we have 00

N, in other words

> OO

4)

EM EG C (liming Az) N 5.

Conversely supposex € (lim inf Ax) N EG. By (3), we have iim, 1a, @) = 1z(x). Since xe liminf Ar, there exists N” &€ N such that x € Ax fork g > N. k>

Na so6 that jim n da)

(lim inf Ax)

= 1. This implies 1z(x)=

Then 14,(x) =

1 for

1 so thatx € E. Thus we have

Ee C E and consequently

©)

lim inf Ay) (liminf 42) N1 EjE§

c Cc

E12 EG.8

By (4) and (5), we have EN Ej = (liminf Az) 1 EG. Therefore EAliminf Ay Cc Eo so

+00

that u(EA liminf Ax) < #(Eo) = 0. This proves (2).

+00

> OO

Proposition 20.17. Let (X, 0 (Ql), 4) be a finite measure space where A is an algebra of subsets of X. Let 8 be the collection of all nonnegative simple functions on X based on A, that is, functions of the type y+ = Ye cla, where c; > 0, A; € A, fori =1,..., pand p €N. Then for every nonnegative extended real-valued o (21)-measurable function f on X, there exists a sequence (yr, : n € N) in 8 such that iim, Vn = f ae. on (X, o (QW), 4)

and im, ty ¥ndu = fy f dp. Proof. If f is a nonnegative extended real-valued o ()-measurable function on X, then there exists an increasing sequence of nonnegative simple functions (gy, : n € N) on (X, o Q, 2) such that g, + f on X by Lemma 8.6. Let g, be given by

(1)

Pr

Gn = >on fle,js j=l

where ¢y,j > O and E,,; € o(&) for j =1,..., px. The Proposition is trivially true iff is an identically vanishing function on X. Thus assume that f is not identically vanishing and assume that ¢, is not identically vanishing so that max g, > 0 forn € N. Now for an arbitrary ¢ > 0, corresponding to Ep, ; there exists An,; € 2% such that (2)

€ H(En,j44An,j) > en, j14,jj=l

314

Chapter 5 Extension of Additive Set Functions to Measures

Then {X : |gm — Yal > 0} C

4

UF; (En,jAAn,j) 80 that by (2) we have Pa

&

,

1

WX: Wen —Vol > 0} < Do (Ems) < ay min |

an,

This implies }°,, 0} < Donen gr = & < 00 80 that by Theorem 6.6 (Borel-Cantelli), we have

6)

(Him sup{X : In — ¥n| > O}) = 0.

According to Lemma 1.7, the statement (5) is equivalent to the statement (6) Since

HX : Gy, # Wp for infinitely many n € N} = 0. tim, Pn (x)

= f(x) for every x € X, (6) implies iim, Vn)

= im, Pn(x) = f(x)

for ae. x € X. By (4) we have

[foau- fi vndu| < f ten —valdn O}

w(Fa,j,

Ei).

dn=1

By (4), the disjointness of the collection @ C G and the countable additivity of jz on

SG

and by (5) we have

(6)

HE) =

x

neN

[Oo a. VED] = A.B). fx=l

neN

Then by (2) and by the definition of ji on a(G), we have m

O

ma=Suey=> [Sanz] => [aa, nz]; i=1

Since A, = A,

A = A,

i=l

neN

(UP, Ei) =

néN

i=l

Ui (An 2 E;), the additivity of 7% on «(G)

implies that )77_, #(An M E;) = (A,). Substituting this in the last equality, we have H(A) = Dex #(An). This proves the countable additivity of # ona(G). & Corollary 21.7.

Let p be a nonnegative extended real-valued finitely additive set function

on a semialgebra © of subsets of set X with ~(@) = 0 and let ji be the extension of pt to a(S) as defined in Lemma 21.5.

(a) & is monotone and finitely subadditive on a(©)

and js is monotone and finitely

subadditive on ©.

(b) If is countably additive on G, then ji is countably subadditive on a(©) and yu is countably subadditive on G. Proof.

1. By Proposition 21.6, jz is finitely additive on a(G).

Since a(G) is an algebra,

monotonicity and then finite subadditivity of f@ on «(G) follows from the additivity of on a(S) by (a) of Lemma 1.22. Since G C a(G) and yw = fi on G, the monotonicity and

322

Chapter 5 Extension of Additive Set Functions to Measures

finite subadditivity of ji on «(G) imply the monotonicity and finite subadditivity of 4 on S. 2. If ~ is countably additive on G, then jz is countably additive on a(G) by Proposition 21.6. Since a(@) is an algebra, the countable additivity of 7 on a(G) implies its countable subadditivity on a(G) by (b) of Lemma 1.22. Then since G C a(G) and p = 7 on G, pw

is countably subadditive on G. Theorem 21.8.

Let 2 be a nonnegative extended real-valued finitely additive set function

on a semialgebra © of subsets of set X with 4(B) = 0. Then us is countably additive on G

if and only if 1 is countably subadditive on ©. Proof.

If jz is countably additive on G, then jz is countably subadditive on G by (b)

of Corollary 21.7. Conversely suppose yz is countably subadditive on G. To show that u is countably additive on G, let (E, : n € N) be a disjoint sequence in G such that

Unen En € G. Consider the extension ji of js to a(G). For an arbitrary N € N, we have

Us En © Unen En. Since US; En € o(G) and since Ucn En € G C a(G), the

monotonicity of j¢ implies

(Us) neN

HL) >a) neN

N

N

n=l

n=1

= Vee) = Vee) Since this holds for every N € N, we have

(Unen En) = Donen (En). On the other

hand by the countable subadditivity of 4 on G, we have 4 (U,cy En) < Donen (En). Thus we have 4 (Un En) = “nen #(En)- This proves the countable additivity of 2 on cS.

a

[1] Outer Measures Based on Additive Set Functions on a Semialgebra Let be anonnegative extended real-valued countably additive set function on a semialgebra S of subsets of set X with 4(@) = 0. The extension ji of jz to the algebra #(G) generated by G is a nonnegative extended real-valued countably additive set function on a(G) by Proposition 21.6, Thus by the Hopf Extension Theorem (Theorem 20.5), 2 can be extended to a measure on o (@(G)) = o(G) and therefore our set function 44 on G has an extension to a measure on o(G). Let us address the uniqueness question of extensions of jz to a

measure. Now since a semialgebra of subsets of X is a covering class for X, the set function z* on $3(X) defined by

e*(E) = inf { Dnen #(En) :(En:n EN) CG, Unen En D E} for E € §8(X) is an outer measure on X according to Theorem 2.21. function ji* on 93(X) defined by

Similarly the set

fi*(E) = int { Dyen (An) : (An? €N) CaS), Unew An > E}

§21 Extension of Additive Set Functions on a Semialgebra is an outer measure on X.

(X, DU"), H*).

§23

Let us compare the two measure spaces (X, Mt(z*), u*) and

Lemma 21.9. Let y be a nonnegative extended real-valued countably additive set function on a semialgebra © of subsets of set X with (0)

= 0 and let ji be the extension of 1 to

the algebra a(@). For the two outer measures * and ji* on X based respectively on the set function jt on © and the set function ji on a(G), we have p* = ji* on YB(X) so that

Mw") = Mi") and (X, Mu"), w*) = (X, DIG"), 7").

Proof. Let E € $3(X). Since G C a(S), the collection of all sequences in G covering E is asubcollection of the collection of all sequences in a(G) covering E. Then since % = js on

©, we have ji*(E) < *(E). On the other hand, every member of «(G) is a finite union of members of G so that a covering of E by asequence in a(G) is acovering of E by asequence in G. Also for every A € a(G) there exists a finite disjoint collection {Z; : i = 1,...,m}

in © such that f(A) =

77, #(E;). Thus u*(E) < ji*(E). Therefore * = ji* on $8(X).

This implies that IN(u*) = Mt(H*) and (X, Mt(u*), u*) = (X, ME*), 2").

a

Theorem 21.10. (Uniqueness of Extension to Measure) Let jz be a nonnegative extended real-valued countably additive set function on a semialgebra © of subsets of set X with BQ) = 0 and let u* be the outer measure on X based on the set function p. (a) * is a regular outer measure on X.

(bh)

GS C tu") and

wp = p* on GS.

(c) Let (X, §, v) be a measure space such that G C ¥ C Mt(u*) andv = on G. If pis o-finite on G, then vy = u* on §. (d) In particular, if is o -finite on ©, then an extension of to a measure on a (©) is

unique.

Proof. 1. Let jz be the extension of jz to «(G) and let ji* be the outer measure on X based on jt. By Lemma 21.9, 4* = jz* on $$(X). Since jz* is a regular outer measure on X by Proposition 20.9, * is a regular outer measure. This proves (a). 2.

We have ©

c

a(G).

Now

since 7 is additive on the algebra a(G), we have

a(S) C Mii") by Theorem 20.1. Then since Mt(zZ*) = Mt(u*) by Lemma 21.9, we have © C Mt(u"*). Since ~ = fi on S by Lemma 21.5 and jz = jz* on «(G) by Theorem 20.1 and j* = u* on $3(X) by Lemma 21.9, we have 4 = y* on ©. This proves (b). 3. If u is o-finite on G, then 7 is o-finite on a(G). If SG C F C Mu"), then since Mt(u*) = MIG") by Lemma 21.9 we have a(G) C F¥ C Mi(fi*). If v = won G, then v = fi on a(S).

Then by Theorem 20.12 (Uniqueness of Extension), we have v = ji* on

¥. Since fi* = y* on $3(X) by Lemma 21.9, we have v = y* on ¥. This proves (c). 4. (d) is a particular case of (c) with § =o(G).

Theorem 21.11. Let G be a semialgebra of subsets of a set X. Let 1 and 2 be two measures on the o-algebra a (©). If 41 = U2 on GS and 1, and pz are o-finite on ©, then

Hi = M2 ono(G).

Proof. If 41 and 12 are o-finite on the semialgebra G, then they are o-finite on the algebra

524

Chapter 5 Extension of Additive Set Functions to Measures

a(G). Since every member of a(G) is a finite disjoint union of members of ©, if 41 = pa on &, then 41 = 42 on a(G). Then since o(G) = o(a(S)), 41 = 2 on a(S) by

Theorem 20.14.

The *-measurability condition for an outer measure j4* based on a countably additive set function yz on a semialgebra G of subsets of a set X can be reduced to testing by the members of G only. Theorem 21.12. Let jz be a nonnegative extended real-valued countably additive set function on a semialgebra G of subsets of set X with 4(B) = 0. Then for every E € 9B(X), the p*-measurability condition

qd)

B*(B) = (BNE) + u*(BN E*) for every B < B(X)

is equivalent to the condition

(2)

(FP) =e (FOE)

+e (F NE)

forevery Fe 6.

Proof. Since a semialgebra G of subsets of a set X is a covering class for X, this theorem is a particular case of Theorem 2.24.

§22 Lebesgue-Stieltjes Measure Spaces

§22

525

Lebesgue-Stieltjes Measure Spaces

[I] Lebesgue-Stieltjes Outer Measures Let g be an arbitrary real-valued increasing function on R. g(x—)

=

lime, g(€) and the right-limit g(x+)

—00 < g(x—) < g(x) < g(x+) < 00.

Let us write g(—oc)

=

8 (00) < 00.

lim

+00

g(x) and g(oo)

=

=

For x ¢€

R, the left-limit

limg,, g(€) exist and furthermore lim g{x).

x00

Then —co

< g(—0oo)


x for every

neéNandé, |x. Then lim g-(6:) = hima @) and Jim g() = lim 8). Then since &r = g except at countably many points in R we can select (&, : n € N) so that g,(&,) = g (Eq) for every n € N. Then Jim, &r(En) = im, gn) so that lime), gr(&) = limg x g(&).

4.

To show that g, is right-continuous at every x € R, note that by 3° and by the

definition of g, we have lim), g,(€) = limgyx g(&) = gr (x).

5. Let (& : n € N) be a sequence such that & | —oo, Then dim, gr Gn) = gr(—00)

526

Chapter 5 Extension of Additive Set Functions to Measures

and lim g(§,) = (—00). By 2° we can select (&, : n € N) so that g,(6;) = gn) for every n € N. Then

lim g,(é,) = lim g(&,). Thus g-(—oo) = g(—00). We show n>00 n>00

similarly that g,(00) = g(oo).

6. Since g, and g are real-valued increasing functions on R, g, is continuous at x € R if and only if limg4x gr (€) = lime, g, (€) and similarly g, is continuous at x if and only if

limgtx g(€) = lig g(€). Now by 3° we have limes, g/(€) = limgyx g,(&) if and only if

lime 4, 9(€) = limg, g(€). Thus g, is continuous at x if and only if g is continuous at x. If g is continuous at x, then g(x) = limgy, g(€) = g(x+) = g7(x).

7. Since g and g, are real-valued increasing functions on R there exist two null sets £1,

and E2 in the Lebesgue measure space (R, Mt,, #,) such that g is differentiable at every

x € Ef and g, is differentiable at everyx ¢ E53. Let E = E, U Ep. Then £ is a null set in

(R, MN, u,) and both g and g, are differentiable at every x € E*. Letus show that g’(x) = g/ (x) ateveryx € E°. Now differentiability implies continuity.

Thus both g and g, are continuous at x. Then by 6° we have g(x) = g,(x). Let (&, : 2 € N) be a decreasing sequence such that &, > x for every n € N and &, | x. Since g and g, are differentiable at x, we have

lim

noo

&(Gn) — 8) & —

= g’(x) and

x

lim

n->00

8r En) — Sx) _ Ex

-

According to 2°, g = g, except at countably many points in R.

g(x). Thus we can select

(& : n € N) so that g(&) = g,(é,) for every n ¢ N. As we noted above, g(x) = g,(x). Thus we have

lim 8En) — 8) _ lim &r (Gn) — Br(x) no & x n>0 «EX

Therefore we have g(x) = g(x)

&

Definition 22.3. Let g be a real-valued increasing function on R and let g, be its rightcontinuous modification. Let us define a set function £. on the semialgebra Joc of subsets ofR by setting £,(@) = 0 and for (a, b] € Joc

£,((a, bl) = 8b) — (a), with the understanding that (a, 0]

:= (a, 00) and @,(oo)

=

Jim, &r(x).

Let uy be the

outer measure on R based on the set function £2 on Joc, that is, for every E € 98(R), we define

uX(E) =int { Doren Len): Un 2 EN) C Boer Upen In D El.

We call £, the set function on Jo, based on g and 43 the outer measure on R based on £4. Theorem 22.4. Let g be a real-valued increasing function on R and let £4 be the set function on the semialgebra Jo, and wy be the outer measure on R based on £,.

(a) £, is a nonnegative extended real-valued countably additive set function on the semialgebra Joc with £,(8) = 0.

§22 Lebesgue-Stieltjes Measure Spaces

527

b) ue is a regular outer measure on R.

(©) Joc C Dts) (implying Br C Meus) ), and Ly = wy on Doc. Proof. 1. Let us verify the countable additivity of the set function £2 on the semialgebra

Joc. Thus let (J, : n € N) be a disjoint sequence in J, such that LJ,en In € Joc. Let Jy = (Gn, by] forn € N and let J = Uncen ia = G, 2). Since the right-continuous modification g, of g is a real-valued increasing function on

R, the ordering by < in the set {a, b} U {a,, by : 2 € N} is preserved by the mapping g,.

In particular we have g-(an) < gr(bp) for everyn € N. Let Jn = (g,(@n), 8r(bn)] in case 8, (an) < g,(bn) and let J, = {g,(bn)} in case gy (Gn) = 8, (bn) for each n € N. Let

J = (gr(@), gr(6)]. Let Sg be the interior of J, for n € N. Note that if J, = {g,(bn)}

then J? = @. Since g, preserves the ordering by {ar(&n) — Br(an)}, neN

that is,

£51) = Yn). neN

This proves the countable additivity of £, on Joc. 2. Since £, is a nonnegative extended real-valued countably additive set function on the semialgebra 7,,,, the outer measure ue is a regular outer measure, Jp, C Bus (Ta) and

f=

we on Jo, by Theorem 21.10. Since o(Jy,) = BR and since Mus) is a o-algebra

containing Jc, we have BR

C ON (u3).

1

Definition 22.5. Let g be a real-valued increasing function on R. We call 45 the LebesgueStieltjes outer measure on R determined by g. Let jt, be the restriction of 43 to the o-algebra

528 INH5)

Chapter 5 Extension of Additive Set Functions to Measures of all uy -measurable members of 98(R) and call it the Lebesgue-Stieltjes measure

determined by g. determined by g.

We call (R, Dus), Hs) the Lebesgue-Stieltjes measure space on R

Remark 22.6. By Observation 3.2, the Lebesgue outer measure jz} on R

is also given by

us(E) =inf { Doyen £0) : Un im €N) C Soc: Unen Ix D e} for every E € 93(R). The identity function : on R is a real-valued increasing function on R and since itis continuous on R, itis identical with its right-continuous modification :,. Then

£(G@, b]) = b — a = 16) —t(@) = & (6) — tr @) = 2,(G, B)) for (@, 6] € Foc. Thus the Lebesgue measure space (R, 9t, , 4,) is a particular case of the Lebesgue-Stieltjes measure

space (RR, Dt(u), wg) with g =2.

Proposition 22.7. In a Lebesgue-Stieltjes measure space {R, Mus), Lg) determined by a real-valued increasing function g on R, we have

(1) #e(G@, b]) = g4+) — ga), (2) wg ([a, b)) = g(b-) — e(@-), (3) g(a, 6)) = g(b-) — gat), (4) ug(Ia, bl) = (6+) - g(a-), (5) ug({e}) = g(e+) — ge—-) force R. Thus jig(fe}) = 0 ifand only ifc € R is a point of continuity of g. If g is discontinuous at

c ER, then 2,({c}) = g(c+) — g(e—) > 0.

Proof. Note that (a, 5], [a, b), (a, b), [a, b], and {c} are all members of 3R. By Theorem

22.4, BR C tug).

Thus these sets are all members of t(uZ).

To show (1), note that

by Theorem 22.4, £, = jg On Joe 80 that

Hg((a, b]) = £¢((@, b]) = 87) — g-(@ = ght) — g(at). To prove (3), note that (a, b — 1] + (a,b) asn —> oo. Thus

ie ((46)) = i, g(a.— 41) = Jim, fe (63) — ar} = lim {g (b— 3+) — g(a} = 8) — g@).

To prove (5), note that {c} = (a, c] \ (a,c). Then by (1) and (3), we have

Hg({e}) = He (@, cl) — ug((@, 09) = {e(c+) — g@4} - {e@—-) - s@p} = g(ct+) — g(c—).

§22 Lebesgue-Stieltjes Measure Spaces

529

The equalities (2) and (4) are proved likewise.

Theorem 22.8, Let (R, t(u2), ug) be a Lebesgue-Stieltjes measure space on R determined by a real-valued increasing function g on R.

(a) QR, Nus), 7) is a o-finite measure space.

(b) 42g (R) = g(00) — g(—00).

(©) (R, Mus), Hg) is a finite measure space if and only if g is a bounded function on R.

Proof. {(@ —1,n] : 2 € Z} is a countable disjoint collection in Jo, C Mt(uf) with Unez@ — 1,7] = R and pg ((n — 1,n]) = gr(n) — a(n — 1) < 00. This proves (a). Since (—n,n] t R, we have ,(R) = lim ig (1, n]) = dim, {gr@) — g(—n)} = 8 (00) — g(—00). This proves (b). By (b), 42(R) < 00 if and only if both g(0o) and g(—00)

are finite. Since g is an increasing function on R, this is equivalent to the existence of some M > Osuch that —M

< g(—o0) < g(x) < g(oo) < Mforeveryxe

R.

Theorem 22.9. Let g and h be two real-valued increasing functions on R. Suppose g =h

on a countable dense subset D of R. Then (R, Wt(us), ug) = (R, Wt(uz), wn).

Proof. Letx € R. Since D is dense in R, there exists a sequence (x, : n € N) in D such that Xn | x asin —> oo. Since g and h

and h(x+) =

jim

hn).

are increasing functions, we have g(x+) =

tim, gan)

Since g(x,) = h(x,) forn € N, we have g(x+) = A(x+) for

every x € R, that is, g, = h, on R. Thus by Definition 22.3, £, = £; on Joe and wE = BE

on $B (R). Then Mt(u%) = Dut) and ug = sh.

Note in particular that if g, is the right-continuous modification of a real-valued increasing function g on R, then g, = g on R except at countably many points so that

(R, t(u§,), Hg,) = (R, Wt(u3), 4g) by Theorem 22.9,

[1] Regularity of the Lebesgue-Stieltjes Outer Measures The Lebesgue-Stieltjes outer measure ue on R determined by a real-valued increasing function g on R is always a regular outer measure as we showed in Theorem 22.4. Beyond. this, 2¢ has the following regularity properties. Lemma 22.10, Let yf be the Lebesgue-Stieltjes outer measure on R determined by a realvalued increasing function g on R. Let E € $B(R).

(a) For every & > 0, there exists an open set O in R such that O > E and

ui(E) = u(0) < us(E) +6.

(b) There exists a Gs-set in R such that G D E and u3(G) = 45 (E). Proof. 1. Let E € 98(R). By the definition of 43 (Z) as an infimum, there exists a sequence Gy in € N) in Jo, such that L,ey dn D E and wg(E) < Ven lg(n) < uy(E) + 5. There is no generality lost if we assume that /, is a finite interval for every n € N. (If J, is an infinite interval, we decompose it into countable disjoint intervals in the class Jo,.) Let

530

Chapter 5 Extension of Additive Set Functions to Measures

In = Gn, Bal for n € N. Then £,(J,) = gr(Bn) — gr(an). By the right-continuity of g,,

there exists c, > b, such that g,(cp)—2;(Bn) < pat contains EF and

Then the open set O = |), en(Gn; Cn)

HEE) < 030) = 13( Gn end) < neN

neN

45 (Gn cn))

= Do {e@a-) - 8@rt)} = > {8r(en) — ar(an)} neN

neN

sv {8e(@n) — r(@n) + rez} = La +, &

E such

that u5(E) < #g(On) < ug(E) + 2. If we letG = (),cn On, then G is a Gs-set, G D E, and *(G) < ut(On) < ux(E) + 5 for everyn © N so that u*(G) = u(E). © Theorem 22.11, Let (R, t(u*), 4g) be a Lebesgue-Stieltjes measure space on R determined by a real-valued increasing function g onR. For E € $3(R), the following conditions

are all equivalent :

@ E € Mut).

(ii) For everye > 0, there exists an open set O D E with ug(O \ E) E with u(G \£)=0. (iv) For every € > 0, there exists a closed set C C E with us (E\C) 0 be arbitrarily fixed. For every n € Z, consider the restriction of f to

[n—1—6,n+8]. By Theorem 12.10 (Lebesgue), there exists a null set A, in (R, 9t,, 4,)

contained in [n — 1 — 5,n + 8] such that f’ exists and is nonnegative at every point in [n —1-—6,n+8]\

A,

and f’ is St,-measurable on [7 — 1 — 6,

+ 8] \ An.

Let

A = Unez 4n. Then A is a null set in (R, 7t,, #,) and f’ exists and is nonnegative

at every point in R \ A and f’ is 9t,-measurable on R \ A. Since (R, Mt,, u,) is the

completion of (R, 3p, /4,,) by Theorem 5.7, there exists a null set B in (IR, Bp, w,) such that B > A by Observation 5.5. Then f” exists and is nonnegative at every point in R \ B and f” is Jt, -measurable on R \ B. If we let ge = f’ on R\ B andg = 0 on B, then g is a nonnegative extended real-valued DY, -measurable function on R. By Proposition 5.9, there

exist a null set C in (R, Sp, j,) and an extended real-valued $q-measurable function g

on R such that g = yon R\ C. LetE = BUC. Then E£ is a null set in (R, Br, 2,) and g =

R\E.

a

= f' on R\ E. Since g is By-measurable onR \ E, f’ is BR-measurable on

[111.1] Absolute Continuity of Lebesgue-Stieltjes Measures

Consider two Lebesgue-Stieltjes measure spaces (R, 0t(%), ug) and (R, Dt(uf), un)

determined by two real-valued increasing functions g and h on R. Since Wt(ws) and Dt(u};) may not be identical, Definition 11.4 for the absolute continuity of a measure with respect to another on the same o-algebra does not apply. Since both Mt(wz) and Mt(u;) contain BR as a sub o-algebra, we define the absolute continuity of a Lebesgue-Stieltjes measure with respect to another as follows. Definition 22.16. Let zz and 4, be two Lebesgue-Stieltjes measures determined by two real-valued increasing functions g and h on R. We say that jz, is absolutely continuous with respect to jig on Bp and write up, KX wg on Bp if u,(E) = 0 for every E ¢ BR with u,(E) = 0.

Proposition 22.17. Let 4, and j1}, be two Lebesgue-Stieltjes measures determined by two real-valued increasing functions g and h on R. Suppose pu, & tg on BR.

(a) Every null set in (R, M(u3), wg) is a null set in (R, Mu), wn).

(b) Dt(uZ) C Mt(uz).

© Ln K pg on MU(y2), that is, if E € Dt(w§) and w,g(E) = 0 then pa (E) = 0. Proof. 1. Let E be anull setin (R, Mt(u2), He). Since the measure space (R, Mt(uz), Hg) is the completion of the measure space (R, Br, 7) by (b) of Theorem 22.12, E is a subset of a null set B in (R, Bp, 4g) by Observation 5.5. Since py K wy on Bp and since

}ig(B) = 0, we have 4,(B) = 0. Thus B is a null set in (R, 9t(4}), zn). Then E, being

§22 Lebesgue-Stieltjes Measure Spaces

533

a subset of a null set B in the complete measure space

Q, Dt(uj), un). This proves (a). 2.

QR, Mtu}), Ln), is a null set in

IfE © Dt(uz), thenE = AUC where A € Bp and C is a subset of a null

set B in (R, Br, fz) by (a) of Theorem 22.12. Since 4, «K uz on BR, B is a null set in (R, Bp, ,). Then we have E € Dt(uf) by (a) of Theorem 22.12. This shows that

OTs) C MU (wh). This proves (b). 3. (c) follows from (a) and (b).

Theorem 22.18, if(R, (3), ug), (R, Mt(uz), un), and (R, (ws ,)s feg+n) are three

Lebesgue-Stieltjes measure spaces determined by real-valued increasing functions g, h, and

g+honk, then we have (@)

wg K Myth and pn K tg+n on BR.

(b) WRF.) C WBZ) N Wt(uz).

©

Bein = Ht uy on PR).

Proof. 1. Consider the semialgebra 3,,. consisting of all intervals of the type (a, b] in R and §. For every I = (a, b] € Joc, we have by (1) of Proposition 22.7

Beta) = (8 +4) GH — @ +h)G@+) = [Gt — g@t} + {hGH — hep} = bg (1) + wa). Thus the two o-finite measures 424, and 4g + a on the o-algebra BR are equal on the semialgebra J,,. Then since BR = (Joc), we have gin = Mg + Ue OD BR by Theorem 21,11. Then feg(Z) < g+n(E) for every E ¢ Bp so that jtg44(£Z) = 0 implies

Hg(E) = 0. This shows that zg < j2g4, on BR. Similarly uw, K gy, on BR. Then by (b) of Proposition 22.17, we have Wt(u2,,) C Mt(uz) and Mus.) C Mi(uj). Therefore (2, ,) C Wes) M Mt(u;). This proves (a) and (b). 2. To prove (c), recall that by Definition 22.3 for every subset E of R, we have (EZ) =

inf Den £g (dn) where £_((a, b]) = gr(b) — gr(@), gr is the right-continuous modification

of g, and the infimum is on the collection of all sequences (7, : n € N) in Jy. such that

Unen Ie D E.

Similarly for uf(E) and Hey (E).

Note that (¢ +A),

= g, +h, and

£g4n((@, b]) = £¢((a, b]) + £4((@, b]). Now for an arbitrary sequence (J, : n € N) in Foc such that J, cy Jn D E, we have

Yee na) = Do Len) + fan)

neN

neN

neN

> inf D> gn) + int D> fan) neN

= wE(E) + uh(E).

neN

Thus we have

()

Bon (E) = inf D7 lg in(In) = wE(E) + HRC). neN

534

Chapter 5 Extension of Additive Set Functions to Measures

To prove the reverse inequality, let (J, : n € N) and (J, : n € N) be two arbitrary sequences in Joc such that J,ey Jn D E and (Jen Jn D E. Then

#5 4,(Z) < min { > 24nd,

tensa}

neN

neN

= min { Yell) + neN

> Cnn), ¥en) + Dan}. neN

neN

neN

Now for any a1, a2, bi, bz € [0, 00], we have min {a1 +42, bi +2}

< May +a_ +b, +53}.

Thus we have

ut (E) < a Yella) + Yo eala) + J b(n) + Dts}. neN

neN

neN

neN

Then by the arbitrariness of the sequences (J, : n € N) and (J, : n € N), we have

Q)

why CE) < Eat) + wh (E) + ut) + wf} = wt) + uh).

By (1) and (2), we have Mean (E) = uz(E) + #;,(E). This proves (c). Theorem 22.19, Let g be a real-valued increasing function on R and let c > 0. Then for

the Lebesgue-Stieltjes measure spaces (R, Mus), Hg) and QR, Mus,), leg) determined by the two real-valued increasing functions g and cg, we have

(8) Meg K fg and pug K cg on By.

(b) Rus.) = W(H4).

(© He, = cugz on F(R). Proof. 1. For J = (a, b] € 3c, we have by (1) of Proposition 22.7

Heg (I) = (cg)(b+) — (eg)(at) = e{gb+) — g(at)} = cg). Thus the two o-finite measures jicg and cu, on the o-algebra Sp are equal on the semialgebra J,,. Since BR = o (Fac), we have ficg = cig on BR by Theorem 21.11. Let E € Bp. If ug(Z) = 0, then eg(E) = cug(E) = 0. Thus ficg < 4g on BR. Conversely if jtcg(E) = 0, then c1g(EZ) = 0 and thus jzg(Z) = 0. This shows that jug < pcg on Br.

Now fcg & fg on Br implies M(wZ) C Mt(us,) by (b) of Proposition 22.17 and similarly wg K fcg On Bp implies that Mt(w7,) C Wt(us). Therefore Mt(us,) = Wt(us). 2.

Let E be an arbitrary subset of R.

that Unenw dn D E, we have

Hg,(E) = cu(E).

&

cn leg)

For every sequence (J, : n € N) in Joe such

= € Cnenle(h).

From this it follows that

In Definition 22.3, the Lebesgue-Stieltjes outer measure u3(E) of a subset E of R is based on sequences (7, : n € N) in the class 3,, such that Unen I, D E. We show next

that the value of 42 (Z) is unchanged if we restrict to coverings of E by disjoint sequences (Jn in €N) in the class Toe.

§22 Lebesgue-Stieltjes Measure Spaces

535

Lemma 22.20. For the Lebesgue-Stieltjes outer measure yy determined by a real-valued increasing function g on R, we have

up(E) = inf { Dyan len) | Un im EN) C Soc, disjoint, Unen Jn > E] for every E € $3(R). Proof. Since the collection of all disjoint sequences (J, : n € N) in 3, such that _), E, we have

BE)

< inf (Den len) : Gn it € N) C Joc, disjoint, U,cn Jn D E}. It remains to

prove the reverse inequality. Now

Joc is a semialgebra of subsets of R.

Thus a(J3,-), the algebra of subsets of

R generated by Jy, is the collection of all finite unions of members of 3,, by Theorem 21.4, Also a finite union of members of J, is always a disjoint finite union of members of 3 according to Lemma 21.3. Let (7, : n € N) be an arbitrary sequence in J,, such that J,oji > E. If we let Ay = f, and A, = I, \ Uz i, forn > 2, then (An : n €N) is a disjoint sequence in the algebra a(Jo-) and (),cn~ An = Unen Jn. Since

An € @(Joc), An is a disjoint finite union of members of 3o,. Let Ap = Urns Faken where {Jn,i, : kn = 1,..-, Pn} is a disjoint collection in 3,,. Let us rename the collection {Fits sees Faypis F215 «+s J2,pos S315 ++ +s JB,pys a8 (Ji, J2, J3,...}. Thus we have a

disjoint sequence (J, : 2 € N) in Jo, such that Lnew Jn = Unen 4n = Une i 2 ENow

Yi fen) = > wen) = He U Jn) = e(U An) neN

neN

neN

naeN

= Doe(An) S P elle) = Yo ben). neN

neN

neN

This shows that for every sequence (J, : n € N) in Jo, such that cn In D E, there exists a disjoint sequence (J, : nm € N) in 3,, such that J, cn Jn D E and acy lg(Jn) < Den £gUn). Therefore we have

inf {Dlnew £e(Jn) : Gn it EN) C Joc, disjoint, ney Jn > E}

E} = 2 f (2). This completes the proof of the lemma. Theorem 22.21. Let jz, be the Lebesgue-Stieltjes measure determined by a real-valued increasing function g on R. Then jt, is absolutely continuous with respect to the Lebesgue measure 1, on Bx if and only if g is absolutely continuous on R. Proof. 1. Suppose g is absolutely continuous on R. To show that 2, is absolutely continuous with respect to 4, on Sp, we show that if F ¢ Bp and z, (EZ) = 0, then u,g(Z) = 0, Let us show that for every N ¢ N, if we let Ey = EM (—N, N] then ug(Ey) = 0. Lete > 0.

The absolute continuity of g on [—N, N] implies according to Lemma 13.7 that there exists 6 > 0 such that for any sequence of non-overlapping closed intervals (Ian, bine N)

536

Chapter 5 Extension of Additive Set Functions to Measures

contained in [—N, N] with >, e—j(bn — Gn) < 8, we have ) ney {8(bn)—

8 (an)} < &. Now

#,(E) = 0 implies u, (Ey) = 0. By Lemma 22.20, valid for any Lebesgue-Stieltjes outer

measure and in particular for the Lebesgue outer measure, there exists a disjoint sequence in

Joes say (Gn, By] : 2 € N), such that Jey (Gn, Bn] D Ey and D,cq(bn — Gn) < 5. Then ([4n, bn] : n € N) is asequence of non-overlapping closed intervals contained in[—N, —N] with D,cu(bn — Gn) < 8 so that Y,cw {g(On) — ¢(@n)} < &. Since g is continuous, it is identical with its right-continuous modification g,. Thus we have u,(Ey)

= u3(En)


0, we

have j1g(Ey) = 0. Then (E) = g( Jim Ew) = lim p(Ew) = 9.

2. Conversely suppose ji, is absolutely continuous with respect to 4, on BR. To show that g is absolutely continuous on R, that is, g is absolutely continuous on every finite closed interval in R, it suffices to show that g is absolutely continuous on [—N, N] for an arbitrary N € N. Let Bewy.y) = BRN[—-N,N] C Bp. By the absolute continuity of jg with respect to 4, on Sp, we have ug(E) = 0 for every E € By_y,n) with 4, (E) = 0. Let us show that this implies that for every ¢ > 0 there exists 6 > 0 such that Bg(E) < & for every E € Bl_y,n) with u,(Z) < 8. Assume the contrary, Then there

exists £9 > 0 such that for every m € N there exists Z, € B[_y,n) with w, (Zn) < x and

Hg(En) = €0. Then nen Hz (En) < 00 so that jz, (lim sup E,) = 0 by the Borel-Cantelli n00

Lemma (Theorem 6.6). On the other hand, since z,([—N, N]) < 00 by (4) of Proposition 22.7, we have jtg(limsup E,) > lim sup jg(En) > 9 by Theorem 1.28. Thus for our

n00

n>00

lim sup Ey € %B(—y,w}, we have j2, (limsup E,,) = 0 and j9(limsup E,) > 0. This is a noo

R00

contradiction.

noo

Now for every ¢ > 0, there exists 5 > 0 such that z,(E) < ¢ for every EF € Bi_w,n] with .,(E) < 8. Let ([@n, b,] : n € N) be a sequence of non-overlapping closed intervals

contained in [—N, N] with }°,

(bn — an)

< 6.

If we let E = Unen(Gn. bn], then

E € S|-w,yy and 4, (E) = ),ew(bn — Qn) < 4 so that zg(E)

< ¢. Thus for the right-

continuous modification g, of g, we have ) nen {8r (bn) — 87 (@n)} = Donen He (Gn, bn) = Hg(E)


0 so that there exists

an interval [a, 6] C R such that jia([a, B]) > 0. Let us define two real-valued increasing functions g, and g, on [a, 8] by setting gg(a)} = g.(~) = 0 and

| Balt) = Ha((@,x])+9(a)

(2)

8s(x) = ws (Co, x])

forx € (a, A], for x € (a, Bl.

Let us assume that g is right-continuous on R. Then by (1) of Proposition 22.7 and by (1)

above we have g(x) — g(a) = g(a, x]) = Ha((@, x]) + 44s((@, x]) and then by (2) we

have

8%) = Ha ((@, x]) + g(@) + Hs ((@, x]) = gale) + gs(2) forx € (@, B]. Since g, ga, and g, are real-valued increasing functions on [@, 8], we have g’ > 0, g/, > 0, and gi = 0, (Br, 4,)-ae. on [a, B] by Theorem 22.15. We also have g’ = g/, + g/, (Br, #,)-a.c. on [a, B]. As we noted above, g’ = 0, (Br, 4,)-a.e. on [a,b]. This implies that g/ = 0, (Bp, u,)-a.e. on [@, B]. By (2) and by the fact that pe « p,, we have

3)

8a(%) — g(@) = Ha((e, x]) =

Then by Theorem 13.15, we have gi, = dpa/dy,,

dua/du,

= 0, (Br, u,)-ae.

on [a, 6].

d, Ba du,. Tax] EH, (Br. u,)-ae.

on [a, B].

Thus

On the other hand, since ya([a, B]) > 0,

(3) implies that du,/dy, > 00na BR-measurable subset of [or, 8] with positive Lebesgue measure. This is a contradiction. Therefore 42 = 0 so that zz 1 yu, under the as-

sumption that g is right-continuous on R. If g is not right-continuous on R, consider its tight-continuous modification g,. The fact that g is a singular function on R implies that g, is also a singular function on R by 7° of Observation 22.2. Then by our result above, we have jig, | j2,. But according to Theorem 22.9, we have jig, = tg. Therefore pag 1 pz,.

2. Conversely suppose , 1 y,. Let us show that g is a singular function on R. Now

since g is a real-valued increasing function on R, g = ga + gs where g, is an absolutely

§22 Lebesgue-Stieltjes Measure Spaces

539

continuous increasing function and g, is a singular increasing function on R by Theorem 22.14. Let fg, and tg, be the Lebesgue-Stieltjes measures determined by the two realvalued increasing functions gq and g,. Then wg = fg, + fg,- Since gz is an absolutely

continuous function, we have zz, < j, by Theorem 22.21. Now since pg 1 y2,, there exist C1, C2 € Sp such that C1 C2 = G, C, UC2 = R, pe, (C1) = O, and pxg(C2) = 0

by Definition 10.16. Since jeg = jig, + pig,, we have peg(C2) = pg, (C2) + bg, (C2). Then Hg(C2)

= 0 implies z,,(C2)

= 0. This shows that xz,

1 y,.

Therefore we have both

Hg, & pw, and jeg, | je, and consequently zz, = 0 by Observation 11.12. Now yz, = 0 implies that the increasing function g, is constant on R, say gg = c on R where c is a real number. Then g = ¢ + gy. Since c and g, are both singular functions on R, sois g. & Example. Let t be the Cantor-Lebesgue function on [0, 1] as in Theorem 4.34 and let g be areal-valued function on R defined by 0 e(x) = ¢ t(x)

1

forx € (—00, 0), forx e [0,1],

for x € (1, 00).

Then the Lebesgue-Stieltjes measure 2, determined by g and the Lebesgue measure 2, are mutually singular, that is, there exist Ci, C2 € Sp such that C; 1 Co. = 8, C1 N Co = R,

Hg (C2) = Oand 4, (Ci) = 0.

Proof. Let T be the Cantor ternary set contained in [0, 1] as in Theorem 4.33. Consider the open set G = [0, 1] \ T. Let {J, : n € N} be the countable collection of disjoint open

intervals contained in [0, 1] and constituting G. The Cantor-Lebesgue function + is constant on each of these open intervals and so is g. Thus j4g(Jn) = 0 by (3) of Proposition 22.7 for every n € N and thus z2(G) = 0. Consider the four disjoint sets (00, 0), G, T, and (1, 00) in 3g whose union is R. Since g is constant on (—oo, 0) and on (1, 00), we have

Hg((—c0, 0)) = 0 and jzp((1, 00)) = 0 also. Thus jtg((—00, 0) U G U (1, 00)) = 0. On

the other hand we have 4, (T) = 0 by Theorem 4.33. Let Cy = T and Cz = (—00, 0) U GU (1, 00). Then we have C1, C2 ¢ Sr, C1 NC2 = B, C1 NC2 = R, fog (C2) = 0 and

(Ci) =0. 0

[TV] Decomposition of an Increasing Function Our aim is to show that if f is a real-valued increasing function on R then f =h+o+y where h is a real-valued, increasing and continuous function, ¢ is a real-valued, increasing and right-continuous singular function, and yf is areal-valued, increasing and left-continuous singular function on R. For a real-valued increasing function f on R the only kind of discontinuity at any point x € Risa jump discontinuity, that is, f(x—)

< f(x+), with a jump f(x+) — f(x—)

> 0.

Since there can be at most countably many jump discontinuities for a real-valued increasing function, if we let {& : » € N} be an arbitrary enumeration of the points where a jump discontinuity occurs, then the sum 7 ,cxy {f Ga +) — f(En—)} exists in (0, oo]. If we are to

340

Chapter 5 Extension of Additive Set Functions to Measures

sum the jumps as we move in the positive direction of R recording the result as real-valued increasing function on R, then since {&, : 2 € N} is an infinite set in (—oo, 00) we start

summing from an arbitrarily fixed point in R entering a jump in the sum as a positive number as we move in the positive direction and entering a jump in the sum as a negative number as we move in the negative direction of R.

Let f be a real-valued increasing function on R. For x € R, let us call f(x) — f(x—), F@+)— f(x), and f(~+) — f (x) the left jump, the right jump, and the jump respectively

of f atx. Then the jump of f atx is the sum of the left and the right jump of f atx. Also (a) f is left-continuous at x if and only if the left jump of f at x is equal to 0. (b) f is right-continuous at x if and only if the right jump of f at x is equal to 0. (c) f is continuous at x if and only if the jump of f at x is equal to 0.

We show first that a real-valued, increasing and right-continuous singular function on R with preassigned left jumps can be constructed. Lemma 22.25.

LetD = {&, : n € N} be a countable set in R and let (a, : n € N) bea

sequence of positive numbers such that for every finite interval I in R, we have

a)

YS

mH 0, forx 0 there exists § > 0 such that for every x € (x9, Xo + 6) we have

>

On xo be arbitrarily fixed.

We have

hens elt nl} @,

€ [0,e) by (1).

This

implies that for every ¢ > 0 there exists N € N such that Lfnewss eto. ml} a, & [0,£).

Consequently if {n < N : & € (xo, x1]} =@, then for any x € (xo, x1] we have > {nen

=

:€¢0,x1}

>

a, € [0, €).

{n>w: gre(xo,x]}

If {n < N: & € (xo, x1]} ZG, let

8 =min {& — x0: 2 SN,& € (x0, x1]}. Then for x € (xo, Xo + 8), we have >

On =

{netv:é,€(20,21}

>

a, < 8.

{n>W:E,€(z0,x]}

This proves (4). By (3) and (4), we have the right-continuity of g at xo. 2. For arbitrary x < xo in R, we have y(xo) — g(x) = L fnew: gretez0l} a, by (3). By

the same argument as in deriving (4), we have

(6)

Jim

x

li

{neN: §,€(x,x0)}

on == 0,

and then

(6)

lim

xta

a=}

a@, if xp € Dand xp = &,

” 0

{nen:4,¢¢,x0]}

ifxo¢ D.

"

Thus we have 7

lim

{¢(xo)

—

e(x)}

Jim (C20) — 0}

=

a,

| 0

ifxo € Dand x= &,

ifs ¢D.

This shows that ¢ is discontinuous at xo if and only if x9 € D. If x9 = &, then by (5) of Proposition 22.7, we have jzg({xo}) = ¢(xo) — ¢(o—) = en by (7). This proves (b). 3. To show that g is a singular function on R, it suffices to show that zy | pw, on (R, BR) according to Theorem 22.24. Let us show that for every M € N, we have

(8)

Hy((—M, M]\ D) =0.

342

Chapter 5 Extension of Additive Set Functions to Measures

Now by (1) of Proposition 22.7 and by the right-continuity of g and then by (2) and (1), we have

Q)

Hy((—M, M])

o(M + 0) — o(—M + 0) = g(M) — o(—-M)

YY

=

{nen:é, 0,

- Drece,0) {g@) - gtt—)}

forx 0. Thus g(&,) — g(,—) > Oforn EN. Leta = g(En) — 8(En—)

forn € N. Since g(x) — g(x—) = O forx ¢ D, the definition (1) of g can be transcribed as

(2)

9%) =

Lfnen: &r 0,

—Lfrenrtaee.n} 2

forx ) are two measurable spaces then the collection x Ay of subsets of X; x Xz need not be closed under complementations since the

complement of a product set need not be a product set. Thus &%; x 2 need not be an algebra of subsets of X; x X2. We show next that a finite cartesian product of semialgebras is a semialgebra. Lemma 23.2. Let G; be a semialgebra of subsets of a set X; fori ©, x --- x G, is a semialgebra of subsets of X1 x --- x Xp.

=

1,...,n.

Then

Proof. We have 8 = 6x ---x Be G, x--- x G, and X, x ---x X, € G@, x---x Gy. If Ey x ---x E, € @, x--x G, and Fy x --- x F, € G1 x -+- X Gy, then we have (A, x---x Ex) N(x +++ x Fy) = (E10 Fi) x --+ x (Ep NF) € G1 x ++ x Gy. Thus ©; x --- x G, is closed under intersection. Let us show that if Ei x --x E, € G, x--x G,, then (EZ; x --- x E,)° is a finite disjoint union of members of G x --- x G,,. Now since E; € G; and G; is a semialgebra

of subsets of X;, Ef is a finite disjoint union of members of G;. Thus there exists a disjoint

collection {Ej,x, : ki =0,..., pi} in @; such that Uo Ejx, = X; and Ej, = E;. Then

Xx

Pl

Pa

Pl

Pn

ki=0

kn=0

ki=0

ky=0

xk =(U Ein) xx (U Enix) = UU Gg ++ By):

Now {£,,x, X ++ X En, 2h =0,..., prj -..3 kn =O9,..., Pn} is a disjoint collection in G1 x --- xX Gp. Since Ey x --- x Ey = E1,9 x --- X En,o is a member of this collection and the union of this collection is the set X1 x --- x X,, (E1 x --- x E,)° is a finite disjoint union of members of G; x --- x G,. This completes the proof that G, x --- x G, isa semialgebra of subsets of X; x --- x Xn.

§23 Product Measure Spaces

549

Definition 23.3. Let €; be a collection of subsets of a nonempty set X; fori = 1,...,n. Let us write €; @ --- @ €, for o(€1 x --- x €,), that is, the o-algebra of subsets of X1 x--++ x Xp generated by €) x +++ x Ey.

Lemma 23.4. Let €; be a collection of subsets of a set S; fori = 1,2. Then we have

(a) o(€1) x Sy = o (€; x Sy),

(b) 8, x o(€2) = a(S x Ey). Proof. To prove (a), let 1 be the projection of S; x Sz onto §;, that is, 71 (x1, x2) = x1 for

(x1, x2) € Sy x Sy. Then x; 1(o(€1)

= (€) x Sy. On the other hand by Theorem 1.14,

we have x; !(a(€1)) = o(m,1(€1)) = o(€) x $2). Thus o(€1) x S = o(€1 x S). This proves (a). Similarly for (b).

Proposition 23.5. Let €; be a collection of subsets of a set X; fori € N. Then for every n > 2, we have

(1)

(«++ ((E1 @ €) @ €2) @--- @Ep_1) OE, = E1@---BEn,

and

(2)

C1 ® (C2 @---@ (Cn-2 @ (En-1 @ Ex) ---) = C1 W--- OEp.

Proof. We prove (1) by induction on n. For n = 2, (1) is obviously true. Now suppose (1)

holds for n = k for some k > 2, that is,

(3)

(--- (Er @ Ez) @ Ez) @--- BE 1)

OBO =]ElW--- OK.

Let us show that under the assumption of (3), the equality (1) holds form = k +1. For brevity, let us write € for the left side of (3). To show that (1) holds form = & +1, we show

that

@

CO Ce

= C1 WOE @ Ses.

Now by (3), we have € = €1 @---@ Ey = (Cy Ex Ce41 D1 x x Ee x Czy and then

()

x --- x Ey)

D Ey x --- x Ey. Thus

o(€ Xx Cy41) Da(Ey xX - + x Ce x Cea).

On the other hand, with an arbitrary Fy41 € €¢41, we have

€x Fey

=o(E1 x --x Cy) x Expy = (Er X--x Ky Xx Ex) Co(€y

x--+ x & x x41),

where the first equality is by (3) and the second equality is by (a) of Lemma 23.4. Since this holds for every Ey41 € €z41, we have € x €x41 C o(€y x «++ x Cy x Ej 41). Thus

(©)

o(€ x Exy1) CoE

x--- x Ee x Egy).

550

Chapter 5 Extension of Additive Set Functions to Measures

By (5) and (6), we have a (€ x €j41) = o(€1 X---K Ex K Ce41) = C1 @- - - DE, OERThis proves (4). Then by induction on n, we have (1). The equality (2) is proved by similar argument. Lemma 23.6. Given two measure spaces (X, A, 2) and (Y, %B, v). Define a set function

A on the semialgebra A x B of subsets of X x Y by setting A(E) E=AxB

€ &U x B with the convention thatoo-0

=

0-co

= (A)v(B) for

=

0.

Theniisa

nonnegative extended real-valued countably additive set function on the semialgebra A x 3B with 1(B) = 0. Moreover if (X, MX, 4) and (Y, %, v) are o-finite measure spaces, then d is o-finite on B x B. Proof. 1. Let us prove the countable additivity of 2 on 2 x 3. Thus let (Z, : n € N) be a disjoint sequence in 2 x % such that J,cy En € A x B. Let us write Ey = Ay x By

forn ¢ Nand L, u(An)v(Bn). neN

For every x € X andy € Y, we have

qd)

La@)12 (9) = Laxe@, ¥) = 10) oy nx Ba

Y)

=o 1sxa,6, 9) = D> {14,@)12,0)}, neN

neN

where the third equality is by the disjointness of the sequence (A, x B, : n € N). With y € ¥ fixed, integrating 1,4(-)13(y) as given by (1) on X with respect to yw and applying the Monotone Convergence Theorem (Theorem 8.5) to the sequence of partial sums of the series, we have

@)

wate) =| f 14) wax)| In)= f 1AGaG)u(as) =f Xttaerta,or}utds) = f 1,012, 0)mds) xX

xX

x neN

neN

x

=> [ [ ta, (a)n(da)| 12,09) =) w(An)1z, 09). neN

neN

Integrating with respect to v and applying the Monotone Convergence Theorem, we have

wayw(B) = w6A) f taorvtay) = f wcarteorvcay) = [ LmAnrta,orv(ay =X ff wands, odv(ay) Y

neN

Y

neN

=) un) [ 1p, (9) v(dy) = > w(An)v(Bp), acN

ncN

§23 Product Measure Spaces

551

where the third equality is by (2). This proves the countable additivity of 4 on & x B. 2. Suppose (X, 2, 4) and (Y, 3, v) are o-finite measure spaces. Then there exists a disjoint sequence (A, : n € N) in 2 such that ney An = X and (Ay) < 00 for every

n € N and there exists a disjoint sequence (B, : n € N) in % such that _),¢y Bn = Y and v(B;,) < 00 for everyn ¢ N. Then {Ay, x B, : m € Nandn

€ N} is a countable disjoint

collection in & x %B with U,,e~ Unen(Am X Bn) = X x Y and we have A(A, x By) = H(An)v(Bn) < 00 form € N and n € N. This shows that is o-finite on A x B. of

Theorem 23.7. For arbitrary n measure spaces (X1, M1, (21), ».-, (Xn, Wns Un), @ product measure space (1 XX

Xn, o (My x ++ K Ay), pf KK

Hn) exists. Moreover if the

n measure spaces are all o finite, then the product measure space is unique. Proof. 1. Consider (X1, 21, 21) and (X2, M2, tionA on the semialgebra 2; x Az of subsets of for E = A; x Az € My x Ay is a nonnegative function with 4(@) = 0. If we let A* be the outer

42). According to Lemma 23.6, a set funcX1 x Xz defined by A(EZ) = w1(A1)42(A2) extended real-valued countably additive set measure on X, x X2 based on A, then accord-

ing to Theorem 21.10, the complete measure space (X1 x X2, Mt(A*), 4*) has the properties

that A, xy

C MUA*) anda = A* on Ay x Ay. Then o (Ay x Ay) C M(A*). If we restrict

1 too (Qi x Az), then the measure space (X1 x X2, o (Qi x 22), 4*) is a product measure space of (X1, 21, 41) and (X2, Wo, 42) since A*(A1 x Ag) = ACA1 x Aa) = 1 (A1)#2(A2) for Ai x Az € Mi x Mp.

2, Suppose a product measure space (X1 x --- x Xx, o(Q1 x +++ x Mz), oa

+--+ X we)

of thek measure spaces (Xj, 21, 441), ..., (Xx, Mx, 44x) exists for some k < n. For brevity let us write (Y, %, v) for this product measure space. Consider the two measure spaces (¥Y, B, v) and (Xg+1, Mle41, We+1). If we define a set function 4 on the semialgebra 3 x

x41 of subsets ofY x Xg41 by ACE) = v(B) +1 (Agsi) for E = Bx Apis € Bx Agi, then by Lemma 23.6, d is a nonnegative extended real-valued countably additive set function with 1(Z) = 0. If we let 4* be the outer measure on Y x X%+1 based on A, then the complete

measure space (Y x Xx11, 1t(A*), A*) has the properties that 3 x M41 C Mt(A*) and A= A* on B x Mz41 by Theorem 21.10. If we restrict A* too (B

x Az +1) C MIi(A*), then

the measure space (v x Xp41, 0(28 x Mk), a*) is a product measure space of (Y, 35, v)

and (X41, Mr41, He+1) since A*(B x Agyi) = (BX Bx

Ags

€ B Xx Aes.

Agyi) = v(B)wi41(Ag41) for

Consider a product measure space (v x Xp41, 0 (2B xX Mzi1)}, vx E41) whose existence we have just established. Since Y = X, x --- x Xz, B = a(A, x --- x Ay), and

v =p

69)

X--- X pg, We have a measure space

(Xi x +

Kapa, 0 (Ar x +++ x Me) x Mess),oa

+ % Mes).

Now by (1) of Proposition 23.5 we have

oy

X-+- X My) = WA @--- OMA = (--- (Ai @ Az) @ Ms) @--- @Az_1) @ Az.

552

Chapter 5 Extension of Additive Set Functions to Measures

Then we have

a (oy

x ++» x My) x Mz)

=((--- (Bh @ Ar) @ My) @--- @Ay_1) @ Ar) @ Aes =

@--- @ MA, @ Mi y1 =o (My x +++ x M41),

where the second equality is by (1) of Proposition 23.5. Thus the measure space (1) is iden-

tical with the measure space (X1 x-- +x Xg41, 0 (Qa X--- x Mey), M1 X ++ pee44). This shows the existence ofa product measure space of (X1, li, 41), ..., (Xx+1, Me+1, Me+1)Thus by induction a product measure space of (X1, 21, 41), «.., (Xn, Wn, Hn) exists.

3. Suppose the measure spaces (X1, 2h1, 41), -.., (Xn, Mn, Mn) are all o-finite. Then the set function41 X--- X fy, on the semialgebraAM, x -- - x QW, of subsets of Xj x--- x Xp is o-finite. Thus by (d) of Theorem 21.10, the extension of jz; x --- X jn to. a Measure on

o(&, x --- x M,) is unique. Thus our measure p11 x --- X pi, On O(Ay X --- K Ay) is unique.

[II] Integration on Product Measure Space Definition 23.8.

Given two sets X and Y.

Let E C X x Y and let f be an extended

real-valued function on E. (a) Forx € X we call the set E(x, -) := {y € ¥ : (x, y) € E} C Y the x-section of E. For

y €Y we call the set E(.,y) := {x € X : (x, y) € E} C X the y-section of E.

(b) Forx

€ X we call the function f (x, -) on E(x, -} the x-section off. Fory € Y we call

the function f (-, y) on E(., y) the y-section of f.

Proposition 23.9. Consider the product measurable space (X x Y,a (A x 98)) of two measurable spaces (X, 2) and (Y, 3B). (a) If E € o (Ax B), then E(x, -) € B for everyx € X and E(-, y) € A for every y € Y. (b) If f is an extended real-valued o (A x %)-measurable function on E € o (A x B),

then f (x, -) is a B3-measurable function on E(x, -) € % for every x € X and f(., y)is a A-measurable function on E(-, y) € A for every y ¢ Y.

Proof. 1. Let ¥ be the collection of all subsets E of X x Y such that E(x, -) € B for every x € X and E(., y) € MW for every y € Y. To prove (a), we show that o (Ql x 93) Cc F. Let us show that # is a o-algebra of subsets of X x Y. Clearly X x Y € ¥. To show that F is closed under complementations, let E € ¥. Then for every x € X, we have

Ex, )={ye¥:@,yeE}=¥\{ye¥:@,y eE}=Y\ EG, EB, since E(x, -) € 3. Similarly E°(.,y) € M&M for every y € Y. Thus E° € ¥. To show that ¥ is closed under countable unions, let (EZ, : n € N) be a sequence in ¥. Then for every

§23 Product Measure Spaces

553

x € X, we have

(Um)@=[yer:@eU neN

neN

=Uvver:@nem}=U neN

neN

Bef

EG) €B,

since E,(x, -) € % for every n € N. Similarly (U,en En) (-, y) € Ol for every y € Y.

Thus # is closed under countable unions. Then ¥ is a o-algebra of subsets of X x Y. Let us show that 2 x Bc ¥F. Let E € A x B. Then E = A x B whereA € A and B € . Then for every x € X, E(x,-) = Bifx € Aand E(x,-) = Gifx € A*. Inany

case, E(x, -) € % for everyx € X. Similarly E(-,y) € 2 for every y € Y. Thus £ € F. This shows that & x % c ¥. Then since F is a o-algebra containing A x %, it contains the smallest o-algebra containing A x %, namely o (A x B). Thus o(A x B) c F. 2. Let f be an extended real-valued o (2 x 2%3)-measurable function on E € o (21x 3B).

Then for every a € R, we have F := {(x,y) © E: f(x,y) < a} € (A x B). This implies that F(x,-) € % for every x € X by (a). Then for every x € X, we have

{y € E@,-): f(x,y) x}

P

Yo r@)=L yeep.

{i:Al>x}

i=1

Thus v(E (zx, -)) is a 21-measurable function of x € X and

P

[ v(EG, )) ux) = [ Yo 14,@)»B)udz) i=l

P

P

-> [ f« «ucas| vB) = 7 a(Av(B) =i P

=u

LX

i=l

x v)(Al x BY) = (ux v)(E).

i=1

Similarly ty u(EC, 1.3. sequence

2(E(-, y)) y)})v(dy) = Let us show in ¥ and let

is a S-measurable function of y € Y and we have (u x v)(E). This shows that E € §. Therefore a (Ql x that ¥ is a monotone class. Let (Z, : n € N) be E = tim, E,. We ate to show that E ¢ ¥. Now E

the equality 3) c ¥. a monotone = Uncen En

or E = (\,en En according as (E, : n € N) is an increasing sequence or a decreasing sequence. Since E, € ¥ C o(&A x B), we have E € o (A x B). For every x € X, we have E,(x, -) ¢ E(x, -) or En(x, -) | E(x, -) according as E, t or E,, |. In any case, since v is a finite measure, we have

jim, v(En(x, )) = v( lim En (x, )) = v(E(, -)Since E, € ¥, v(En(x, -)) is a M-measurable function of x € X. Thus the limit function v(E(x, -)) is a M-measurable function of x € X. The sequence of 2-measurable functions of x € X, (v(En(x,-))

: » € N) is uniformly bounded on X by a

constant since

v(En(x, -)) < v(Y) < 00 for all x € X andn € N. Since u(X) < co, the Bounded

Convergence Theorem (Theorem 7.16) is applicable and we have

[ v(EG@, ))u(dx) = lim, [ v(En(x, -)) e(dx) x

x

= lim(u x v)(En) = (a x v)( lim En) = (u x v)(E), where the second equality is by the fact that Z,, € ¥ and the third equality is by the fact that

(E, : n € N) is a monotone sequence and yz x v is a finite measure. Similarly 4(E(., y)) is a 3-measurable function ofy € Y and f, u(EC., y))v(dy) = (u x v)(E). This shows

that ¥ is a monotone class. 1.4. We have shown above that the algebra a(&A x 3B) is contained in F and F is

a monotone class. According to Proposition 23.14, the smallest o algebra containing an algebra a (21 x 9B) is equal to the smallest monotone class containing the algebra a (A x B)

558

Chapter 5 Extension of Additive Set Functions to Measures

and is hence contained in the monotone class ¥ containing a(@& x 8).

Thus we have

a(a(A x B)) c F. But c(a(A x B)) = ao (A x B). Thus we have a(A x B) Cc F. This completes the proof of the Proposition for the case that (X, 2, 2) and (Y, %, v) are

finite measure spaces.

2. Let (X, &, 2) and (Y, %, v) be two o-finite disjoint sequence (A, : n € N) in 2 such that Unen and there exists a disjoint sequence (B, : n € N) v(B,) < oo forn € N. Then {A; x B; : i € N andj

measure spaces. Then there exists a A, = X and u(A,) < 00 forn EN in 8 such that Lew Be = Y and € N} isa disjoint collection in A x B

and Usen Ujen(Ai x Bj) = X x Y. Let 2; = 212 A; and Bj = BN B;. Consider the

product measure space (A; x By, o(Q; x Bj), wx v) of the two finite measure spaces (Aj, 2l;, 4) and (B;, 33;, v). Now QW; xB; = (ANA;) x GBNB;) = (Ax B)N(A; x Bj)

and

o (i; x Bj) = a((A x B) (A; x B,)) = 0 (A x B)N (A; x By)

by Theorem 1.15. Thus we have

(A; x Bj, 0 (Ql; x Bj), wx v) = (A; x By, 0 (A x B) (A; x By), wx v). Let E € o (Ax B) and let E;,; = EN(A; x By) € o (A x B)N(A; x Bj) = o (A; x Bj).

Then by our result in 1, v(E;,;(x, -)) is a 2;-measurable function on A;, 4(E;,;(-, y)) is a 38 ;-measurable function on B;, and

x \(E,) = [ (Eis, Yule) = [ Ess Nod. Letx € X. Then x € Aj for some i € N.

Thus v(E(x,:)) = Dyewv(Eis@, ))-

Since v(E;, ;(x, -)) is 2;-measurable and hence %-measurable, the sum v(E(x, -)) is 2measurable. Also

“x)=

YVexdea=vo>y [ v(Ei 56, )) eax) ieN jeN

ieN jeN

=

v(E,, 76, )) ex) =

= f (ee,

near.

v(E(x, -))w(dx)

Similarly 4(E(., y)) is %3-measurable and (1 x v)(E) = fy u(E(, y))v(dy). Theorem 23.17. (Tonelli’s Theorem) Let (X x Y, o (Ax), wx -v) be the product measure space of two a-finite measure spaces (X, UM, jt) and (Y,B,v).

Let f be a nonnegative

extended real-valued a (A x 9%8)-measurable function on X x Y. Then (a) F}()

=

ss f(x, -} dv is a A-measurable function ofx € X.

(b) F2(y) := Sx fC, y) dus is a B-measurable function ofy € Y.

© Sexy faux v) = fy Fldp = f, F? dv, that is,

[faux fi [ [re ar] was)= f [ [1 yan] v(dy).

§23 Product Measure Spaces

559

Proof. Note that by Proposition 23.9, the o (@& x %3)-measurability of the function f on X x ¥ implies the %3-measurability of the function f(x, -) on Y for every x € X and the -measurability of the function f(-, y} on X for every y € Y. 1, Consider first the case f = 1z where E € o (21 x 93). In this case, f(x, -) = lew,

and F\(x) = f, 1z¢,)dv = v(E(2, -)), which is a &-measurable function ofx € X by Proposition 23.16. Similarly F2(y) is a 93-measurable function of y € Y. This verifies (a) and (b). Now

a

[

XxY

fduxv)= [

XxY

Led(u x v) = (ux v\E).

On the other hand, we have

@

f [ [1 av was)= ff [ [ tad] was)= ff »(BC2, uaa)

and

8)

[ [ f fl, yan| v(dy) = f [ f deena

v(dy) = f u(EG, »))vdy).

According to Proposition 23.16, the right sides of (1), (2), and (3) are all equal.

Thus the

left sides of (1), (2), and (3) are all equal. This verifies (c) for this case. 2. If f is a nonnegative simple function on X x Y, then f = )“f_, a¢1x, where {E, : k = 1,...,n} is a disjoint collection in o (Ql x %) and a, > 0 fork = 1,...,n. Then (a), (b), and (c) hold for f by our result in 1 and by the linearity of the integrals.

3. If f is a nonnegative extended real-valued o (& x %3)-measurable function on X x Y, then there exists an increasing sequence (/, : n € N) of nonnegative simple functions on (X x Y, o(A x B)) such that f, ¢ f on X x Y by Lemma 8.6. By our results in 2, (a), (b), and (c) hold for f, for each n € N so that Fix) = tr Sr(x, -) dv is a A-measurable function ofx € X, F(y) = ik Jn(, y) dp is a B-measurable function ofy € Y, and

®

i

Sad

xv) = [ Lf in@)av

u(dx) = [

fe Ay) au] v(dy).

For each x € X, (fn(x,-) : m € N) is an increasing sequence of nonnegative simple functions with f,(x,-) * f(x,-} on Y. Thus by Theorem 8.5 (Monotone Convergence Theorem), we have

(5)

Fla) = [ F(x, dv = lim f ful, av.

Since fy fr(x, -) dv is a 2-measurable function ofx € X for every n € N by our result in

2 and since (fy fn(x,-)dv : n € N) is an increasing sequence, F(x) is a A-measurable function of x € X by Theorem 4.21. This proves (a). Similarly for (b). To prove (c), note that by Theorem 8.5, we have

©

tim f

n00

hdtux=f

faux).

560

Chapter 5 Extension of Additive Set Functions to Measures

Since (fy f,(x,-)dv :n €N) is an increasing sequence of nonnegative extended realvalued @&{-measurable functions of x ¢ X, applying Theorem 8.5 we have

Osim,

| [ fal, ao pldx) = [ Jim, [ [ fal, ao] pdx)

=f [ [re av u(ax) by (5). Applying (6) and (7) to (4), we have fy,» f d(u x v) = fy [fy F(x, -) dv] w(x).

Similarly fyyy f dtu x v) = fy Ly FC. y) du] (dy). Theorem 23.18.

(Fubini’s Theorem) Let (X x Y,o(& x 9), u x v) be the product

measure space of two o-finite measure spaces (X, A, ) and (Y, 3, v). Let f be a jp x v-

integrable extended real-valued o (21 x 9%8)-measurable function on X x Y. Then (a) The B-measurable function f (x, -) is v-integrable on Y for -a.e. &M-measurable function f (-, y) is -integrable on X for v-ae. y €Y.

x € X and the

(b) The function F'(x) := Sy £@, -) dv is defined for p-a.e. x € X, A-measurable and f-integrable on X. The function F2(y) := ik F¢, y) dy is defined for v-a.e. y € Y, B-measurable and

v-integrable on Y.

(c) We have the equalities: fy y f du xv) = fy F'du = f, F* dv, thatis,

[faux f [ ic ao] was)= f [ [1 yan] vidy). Proof. 1. Let f = f+ — f-. Then f* and f— are nonnegative extended real-valued

o(& x %3)-measurable functions. The jz x v-integrability of f on X x Y implies that f+ and f~ are yt x v-integrable on X x Y. Applying (c) of Theorem 23.17 to f+ and f— , we have

@

[ (fda x y= [ [ [ F090] u(dx)

= [f F*¢.9)du | vidy) 0. Since f = 0 on supp{ f}° and since supp{f}° > [—a,a]°, on [—a, a]*. Now [—a — 1,a + [-a@ — 1, a + 1] implies uniform continuity of f on [—a,a]* and

f = 0 on [—a,a]°. Thus f is uniformly continuous 1] is a compact set in R. Then the continuity of f on continuity of f on [—a — 1,@ + 1]. Then the uniform the uniform continuity of f on [—a — 1,a + 1] imply

§23 Product Measure Spaces

579

the uniform continuity of f on [—a — 1,@ + 1] U [—a, a]° = R. Now that f uniformly continuous on R, for every ¢ > 0 there exists 5 > 0 such that

x,y € Rand |y| |f@@+y)— f@)| 0 there exists 5 > 0 such that fory € R where |y| < 6 we have

IT,f— filp= Q@+ D)Pe.

This proves lim ||Tyf— fllp =0.

70 2. Now consider the general case f € L?(R, 9, w,) where p € [1, 00). According

to Theorem 17.10, C,(R) is dense in L?(R, 9, #,,) when p € [1,00). Thus for any f € LP(R, M,, w,) ande > O there exists g € C,(R) such that ||f — gllp < e. Now If -— Tyfllp

=

If - 8llp +llg- Tygllp + IIZyg — Ty f llp-

By Lemma 23.40 we have ||Tyg — Ty f llp = ITy(g — Alp = lle — fllp. Thus we have

If— Tyfllp < 21F — allp + llg— Tyallp < 2€+ lle— TyallpFor g € C,(R) we have lim |g — Tygllp = 0 by our result in 1. Thus we have y-'

lim sup || f — Ty fllp < 2e + lim |lg — Tygllp = 2¢. y0

yo

By the arbitrariness of ¢ > 0 we have lim sup lf - Tyfllp = 0.

lim in If — T,fllp = 0 and then lim If —"filly 1 .(dy)} O,

show that 4, u,(D) > 0. (f * g)(x) — then at least

(D) = 0 where Now D is the g(x) > 0} and one of 4, (D1)

and 1, (D2) is positive. Consider the case 4,(D)) > 0. Since (R, M,, u,) is a o-finite measure space, there exists a Jt, -measurable subset EZ of D, such that j2, (E) € (0, 00).

Let ho = 1g. We have ||h||2 = fy (lel? du = f, 1dy < 00 so thatho € L7(R, Mt, u,). Now by the definition of L 4, by (4), we have

L gug(ho) — f hoo di, = [mtcree -g}dy, = [ {(f*8)—9} du, >0 since (f « g) —g > Oon E C D, and p,(E) > 0. This contradicts (7). Similarly the assumption that 41, (D2) > 0 leads to a contradiction of (7). This shows that x, (D) = 0.

Therefore f *g=9, W,-a.c.0onR.

&

582

Chapter 5 Extension of Additive Set Functions to Measures

Lemma 23.43. Let p,q,r € (1, 00) be such that 5 +1 —1+}. Lets € (1, 00) be the conjugate of r, that is, 1 + 1 = 1. Then there exist &,n, € € (1, 00) such that 1

1

1

1

7Tr-F=7

@)

py tah

@

e+ a

P

144 1y1lt

3

.@4

Ey =$24l51 tate

Proof. Equations (1), (2), and (3) constitute a system of linear equations in three unknowns

Pe i, and re Solving this system we have

1

1

§

q

~=1-—e€(0,1),

tLe Hn

or

0

on,

tii -le@p. g

Bp

Thus &, 7,¢ € (1, 00). Also

111 1 1 1 1) 1 ptot+pal--45¢1-—=1-fi4+—}+ ota fon

¢

q

f°

P

r)or

so that (4) is satisfied. Theorem 23.44, (Young’s Convolution Theorem) Let p, g, r € (1, 00) be such that 1 1 1 —+—=1+-. P @ r

(1) Let

fe LPR, mt,

H) andg é L(R, m,, l,): Let us define a function f * g on R by

setting

@)

(f*—@) = [ f@— ye) u(y)

forx € Rfor which the integral exists in R. Then (f * g)(x) is defined and real-valued for ae. x € Rand f * g is M,-measurable. For every x € R for which the integral in (2) does not exist, we set (f * g)(x) =0.

3)

Then fxg

€ LR, Mt,

If «ally < If llpllalla

Hz) and

§23 Product Measure Spaces

583

Proof. 1. If f € LP(R, M,,

and g € L4(R, 9,, ,), then f and g are real-valued

#,-ae. on R. On the null set in

(R, Dt, 7) on which f is not real-valued, redefine f to

be equal to 0 so that f is real-valued and 9)t, -measurable on R. Redefine g similarly. Then

by Lemma 23.32, f(x — y)g(y) for (x, y) € R x R is real-valued and 971? measurable on RxR.

Lets € (1,00) be the conjugate of r, that is, +1

= 1. Leth © L3(R, 9%,, u,)

be such that A(x) € (0, 00) for every x € R. Then the function h(x) f(x — y)g(y) for (x, y) € R x Ris real-valued and ot? -measurable on R x R. Let us show that

4

[ z Ih) F@ — WeO) Hz x H,)(d@,y)) S If llpllgllallls.

Now p,q,r € (1,00), +3 = 1+} and? +1} = 1. Then by Lemma 23.43, there exist € (1, 00) such that &,,&

Djtli

Eon

pon

jit 6

ayaa

@

&

gs’

ayia, ”

Let a1, a2, bi, bz, c1, cz > 0 be defined by s

)

4=57, é

s

@=5, g

Pp

Dp

h=>, g

b=,

q

a=-, n

q

a=. c

Then we have 11

ata={p+,|s=1

1

bith=[+

1

}p=1.

1

1

ata={>+;]o=1

Thus the integrand in (4) can be written as

Ih) F@ — el

= (h@) IF

— WP HIF] — WI leO}{le@) 7 1A@)I” }-

Now &, 9, € € (1, oc) and i + i + : = 1. Then applying Theorem 16.56, an extension of Hélder’s inequality, we have

©

[ eore — e001, 0 there exists

M

—-C,o

R)fllp

=0.

> 0 such that

ITyx oR) F — Go R)fllp 0 such that

IF * a)@ +y) — CF #8)x)1 S ellgilg for ly] < 4. This proves the uniform continuity of f *« g on R for the case p € [1, 00).

Now consider the case p = oo. Inthiscase we haveq = 1 and thus g € L}(R, Dt, 42,).

Thus by our result above g « f is uniformly continuous on R. As we showed above we have

(f *g)(x) = fe F(®—y)8(y) w, (dy) € R for every x € R. Then (g + f)(x) = (f #g)(@)

for every x € R by Proposition 23.31. Thus the uniform continuity of g + f on R implies the uniform continuity of f * g on R. 3. Letus show that if p € (1, co) then f*g € Co(R), thatis, ei

Fe)

= 0. Now

forp € (1, 00) we haveq € (1, 00). Then f € L?(R, Mt,, 4,) andg € L7(R, Mt,, u,)

imply

®)

firvrau, 0 such that (9)

[

[-4,MF

\fl? du, 2M

c [—M, MI°.

Substituting (11) in (10), we have for x € R

Vi

wreoels{f [-M,MF ironman} tell

+ Fal fp Ort metann

y

0 there exists M > 0 such that

IF *8)@)1 < {IIfllp+ Iigllg}e for |x| > 2M. This shows that

lim

lelo00

(f * g)(x) =0.

4. To prove (d), note that for every x € R we have by (2)

If ea@l< [ |(r. © RYFO)lgO)1 u, (dy) a} > pw, {x eR: h(x) =c} =2>0. This shows that any a € [0,c) is not an essential bound of A by Definition 16.35 and therefore ||/||o0, the infimum of all essential bounds of h, is equal to c. This proves (2). For p € [1, 00) we have

all2 = [ HP dp, = 2c” [ R

= 2cP{ [

(0,1)

[0,00)

{xO A 1}? yw, (dx)

ty, (dx) + f

= 2c? {1 +{(+8)p—

[1,o0)

1p tp

x OF Oreny

y zy} |

= 2? {1+ {+8 —1}"}. Taking the p-th root we have (3). By (2) and (3) we have ||f ||» < 00 for all p € [1, oo].

Thus h € LP(R, 9,,, 2,) for allp € [1, 00]. This proves (4).

§23 Product Measure Spaces

591

2. To prove (5), note that the condition |g| < 4 on R implies ||g||p < ||Allp < 00 forall

P €[1, co] so thatg € L?(R, M,, w,) forall pe [1,00].

=

[IV.3] Approximate Identity in Convolution Product

Let p € [1,00]. Let f € L?(R,Mt,,4,) and g € L1(R, Mt, u,). We showed in Theorem 23.42 that the convolution product f « g is defined and f *g € L?(R, 0, u,).

‘We show next that the identity of multiplication in convolution product does not exist.

Proposition 23.51.

Letp

[1, co]. Letf € L?(R, Mt,, u,) andg < L1(R, Mt, u,)

and consider the convolution product f * g. Identity of multiplication for the convolution product does not exist, that is, there does not exist g < L1(R, 9t,, w,) such that

qd)

f*g=f

foreveryf « L?(R, M,, u,).

Proof. Suppose there exists g € L1(R, 9t,, 4,) such that (1) holds. Let us show that this leads to a contradiction. Now since g € L}(R, M,, u,) is u,-integrable on R, by Theorem 9.26 there exists § > 0 such that

@2

FsagBOD = x51

Let f = 1,-2,5). Then f € L?(R, 9, u,). Thus by (1) we have for p,-ae. x ER

3)

fR=(F #@) =

=f =

1-88]

* NG) = [ie

— y)FO) 4. @y)

se-vu@n=f 1-86] so-oucay

[x—8,x+8]

8) 4, @y),

where the second equality is by Proposition 23.31, the fifth equality is by Theorem 9,32 (Linear transformation) and the last equality is by Theorem 9.31 (Translation Invariance).

Since the equality (3) holds for jz, -a.e. x € Rand since 42, ([—6, 5]) = 28 > 0, there exists

xo € [—6, 6] for which (3) holds. For x9 € [—8, 6], we have f (xp) = 1. Thus

«

1= f(%0) = f [xo—8,x0

, gQ) #4, dy).

Then by (4) and by (2) we have

1=|f = fay

somes f BOM 0, the e-dilation of 9 is a function on R defined by Pe(x) = *9(2)

forx eR.

Observation 23.54, Let y, be the e-dilation of an extended real-valued function g on R. Then we have

@

1

sup lve] = — sup lel.

Unless ¢ is the identically vanishing function on R, we have

(2)

lim sup |g-| = 00

@)

supp{ye} = ¢ supp{y}.

20 Regarding the supports of the functions ¢ and y,, we have Thus as € — 0 the set supp{y,} becomes smaller and moves toward 0 € R. If supp{¢} is a compact set then so is supp{g-} and moreover we have

4) Proof.

tim w, (supp{ye}) = 0. 1. The equality (1) follows immediately from the definition of yg, in Definition

23.53. If g is not identically vanishing on R then sup |y| > 0. Then (2) follows from (1). 2. Let us prove (3). Now by definition we have supp{g} = {x € R: g(x) # 0} and supp{gs} = {x € R: g(x) ¥ 0}. Observe that

{x ER: ge(x) £0} = {xeR: *o(2 ) #0} =|zeR: o(= ) #0} = {ex ER: g(x) £0} =e{x € R: g(x) £0}.

§23 Product Measure Spaces

593

Then

supp{ys} = {x € R : g(x) # 0} = e{x ER: g(x) 4 0}

= e{xERo(x) :£0} =e supply}. 3. If supp{y} is a compact set, that is, a bounded closed set in R, then so is supp{y.} by

(3). Moreover if supp{} is a compact set then 2, (supp{y}) < co. Then

tim m, (suppige}) = lim y, (e supp{g}) = lim 2, (supp{y}) = 0, where the second equality is by Theorem 3.18 (Positive Homogeneity). This proves (4).

Proposition 23.55. Let p € L1(R, 0t,, 44,). For ¢ > 0 consider the &-dilation g, of 9. Then for every D € Dt, we have

a

f Gel) py, (dx) = [e@mtan,

and in particular we have

(2)

[ eC) by, (dx) = [ 96%) py, (dx),

Moreover for every M > 0 we have

3)

lim

£90 J iceR:|x|>M}

Ge(x) Hy, (ax) = 0.

Proof. 1, Let D € D0, and ¢ > 0. Then by (1) of Theorem 9.32 (Linear Transformation) we have 1

[ Gol) um, (dx) = [ 9(=) w(ax) =I, g(x) 1, (dx).

This proves (1). When D = R we have 1R = R. Thus (2) is a particular case of (1). 2. Let us prove (3). Let M

1p =

@)

> 0 and let D

{x €R: |x| > “}. By (1) we have

Lf {xeR:|x|>M} el) a, (ax)| = |

= {x € R:

{xeR:|x|>“}

|x| > M}. Then we have

(x) 4,(dz)

< f {xeR:|x|>4} |o)| wp (dx). Now since fy |p(x)| 44,,(dx) < 00, we have jim, JSecerixl>By |e@)| 4, (dx) = 0. With M > 0 fixed, if we let e > 0 then ¥ —

6)

lim

oo. Thus we have

£90 JiceRi[x|> 4}

lp@)| #z @x) = 0.

594

Chapter 5 Extension of Additive Set Functions to Measures

Applying (5) to (4) we have Ton, | Feeer:|e}>-ay Pe *) # (dx)| = O which implies (3). Proposition 23.56. Let y be an extended real-valued IN, -measurable function on R. For & > O let o, be the e-dilation of p. Then we have 1

pol 1

wo

Iivellp = (=) ” lllp forp € 1,00),

Q)

IPelloo = = IPlloo-

E

1

Thus, for p < (1, oo], if € LP(R, Mt, u,) then g. € LP(R, MT,,, u,). Proof. 1. Letp € [1, 00). Then we have

wool= f wel? au, = f ()"0(Z)” wstas)

= (1)"e [oc man = ( 0 on R and |lg||1 = 1. f *G. € L?(R, Mt, ,) and furthermore

qd)

Then for every e-dilation g, of @ we have

dim |lf *e — fllp = 9.

Proof. 1. The nonnegativity of ¢ implies the nonnegativity of g, for every ¢ > 0. Then by (2) of Proposition 23.55, we have

2)

f iweiau, = [i odu, = fodu, R

R

R

-f

R

leldu, = llglh =1.

This shows that g, € L!(R, Mt, 4,) and ||gel|1 = 1. Then by Theorem 23.42, we have

f *_ € L?(R, Dt, w,) and thus f * gy, — f € L?(R, Mt, u,).

2. Let us prove (1). Let g € [1, 00] be the conjugate of ourp € [1, co]. Let us define a linear functional L, on g € L2(R, Mt, 4,) by setting

8)

Le(g) = [ (F #96 — fed,

According to Theorem 18.1, L, is a bounded linear functional on L7(R, 0t,, 4,) with norm given by

@

Lele= If *%e — fllp-

§23 Product Measure Spaces

595

Thus, to prove (1) we show that lim et

||Zell+ = 0. To show this convergence we show that

for every 7 > 0 there exists 6 > 0 such that

)

Lele 0, by (3) of Proposition 23.55 there exists 6 > 0 such that

(11)

ite f

ur e(¥) Hy, Gy) < ; fore € (0, 4).

596

Chapter 5 Extension of Additive Set Functions to Measures

Then by (8) we have for e € (0, 5)

(12)

ILe(a)l < | [ [-M,M] ve) |T-yf — f],u.@y)} ele

+L

ane ROT rf — fly od}

For the first integral on the right side of (12), we have

af

13)

»

n 4 yf (dy) M}

IFC? u, (dx) M}

FP du, 0 on R and ||g||1 = 1, we have Jag) ,(dy)

#,), then (1) holds = 1. Thus for every

x ER, we have f(x) = fy f(x)g(y) u(y). Now (f *2)(x) = fe fe —y)g(y) uz (dy)

for 4,-a.c. x € R according to Theorem 23.42. Subtracting the first equality above from the second, we have (1). This proves (a) as well as (b).

2. 1g € Myrett,coi LP (R, Mt, ,), then (f #g)(x) = fo f(x — y)e(y) My, for every

x € R by Theorem 23.48. Then by (b), (1) holds foreveryx eR.

a

Theorem 23.61. Let f < L°(R, Mt, u,). Let p € L1(R, Mt, u,) be such thaty > 0 on Rand |g|l1 = 1. Fore > O let y, be the e-dilation of p. Iff is continuous at xo € R,

then lim (f *Qe)(Xo) = f (x0). Thus if f is continuous on R, then lim (f * Pe) (x) = f(x) ee

for everyx ER.

ee

Proof. 1. We have fy - du, = fade, = llgll1 = 1 by Proposition 23.55. Note also that since g > Oon R, wehaveyg, > OonR. Iff ¢ L°(R, Mt, 4,), then foreveryx € R

600

Chapter 5 Extension of Additive Set Functions to Measures

we have

|[

-reeor usta] < f [Fe -v]eornan < flo fe) mea) = Ilflloo < 00.

Then by (b) of Lemma 23.60 we have for every x € R

qd)

Ff * g)(@) — F@) = [ire —¥) — F®)} ve) oH, (@y). 2. If f is continuous at x € R, then for every 7 > 0 there exists M > 0 such that

@)

|f@o— ¥) — f@o)| 0 on R and ||¢||1 = 1. Assume further that ¢ is bounded and has a compact

support in R. For e > 0 let y, be the e-dilation of yp. If f is continuous at xo € R, then we have lim (f *@2)(40)

foreveryx eR.

= f (xo). Thus if f is continuous on R, then lim (f * G2) (2)

= f(x)

§23 Product Measure Spaces

Proof.

601

1. If y is bounded and has a compact support in R then for every € > 0, g, is

bounded and has a compact support in R by Observation 23.54, Then g € L}(R, 9, ,)

and y, € L2°(R, Mt,, y,) also. This implies that gs € (yett,oxj LP (R, Dt, w,) by Proposition 23.49. Then by (c) of Lemma 23.60, we have for every x € R

qd)

Cf * e)(x) — fF) = [ {£@ — y) — £(«)} Ge) m, (dy). 2. Iff is continuous at x9 € R then for every 7 > 0 there exists M > 0 such that

(2)

|f@o—y)— f@o)| 0 on R. Thus we have, applying (2),

(6))

|Cf * ge)() — F@o)| < [ |f@o—y) — Fo)| ge) uy) M}

|f@o—y) — fF Go)| Gey) u, (dy).

For the last integral above, we have

« =

fi {yeR:|y|>M} [feo —y) — F0)| ge) wy (4y) &€

Prexiyee

[Fo

-y)»

~)y wi

Feo) o(Z) Hut »

= fae [70 29) — F040)| 9) ns dy) {yeR:lyl>o)

where the last equality is by Theorem 9.32 (Linear Transformation). Substituting (4) in (3),

(5)

IF ge)(20) — Fo) < 9+ f {yeR:|y|> , | fo

— ey) — fGo)| GO) 4, @y).

Now since supp{¢} is a compact set in R, there exists B > 0 such that supp{y} c [—B, B]. For our fixed M > 0, there exists 3 > 0 such that u > B fore € (0,8). Then we have

[- ¥%, “] > [-8, B] > supp{y} and then {y € R: |y| > ¥} =[- %, ¥]° c supp{e}*

for ¢ € (0, 8). Thus for e € (0, 5), we have

©

Foo{yeR:lyl> a,>} [FC80 = 29) = Fa) 00) 2)

0 on R and ||g||, = 1. Suppose satisfies the condition that for some c > 0 and & > 0 we have

forx eR.

g(x) < ef[xI-O) Al}

a)

For ¢ > 0, let y, be the e-dilation of p. Then we have

2)

ge

()

LP(R, Mt,, u,).

pell,00]

Moreover for every p € [1, 00) and M > 0 we have

@)

lim

£0

Go(x)? uw, (dx) = 0.

J&eR:[x|>M}

Proof. 1. Condition (1) implies that ¢ € (1 pet1,00) L7 (R, Mt, #,) by (b) of Proposition 23.50. This then implies that 95 € (),c11,00] Er (R, Dt, #,) by Proposition 23.56. 2. To prove (3), let p € [1, 00) and M > 0. We have

@4

|

Ie

(x)?

(dx)

ly\P x\P = = (dx) = IF cesusan (5)’o(=)”

=(2)1\p-1 Fen” —

{a

‘a. Pp

tas)

1, (dx)

:= I@), where the second equality is by Theorem 9.32 (Linear Transformation). Let us estimate Z(¢). Let ¢ > 0 be so small that u > 1. Then by (1) we have

6)

1) M4} row? wan}? f {yeRlyl>M} ew, (dy)}

USL fag OM MeO) 2

Vy

\/q

where the first inequality is by Theorem 16.14 (Hilder) and the equality is by Theorem 9.31 and Theorem 9.32. For the case p = 00, we have

6)

[

{yeR:|y|[>M}

If @o — yl Ge) Hz (y)

Ife f

{yeR:|y|>M}

e(y) Ht, dy).

604

Chapter 5 Extension of Additive Set Functions to Measures

Substituting (5) in (4), we have for the case p € (1, 00),

6)

|CF * exo) — F(%o)| S19 + 1F Go)

{yeR:|y|>M}

¢2(y) uw, (dy)

+1ftlof f {yeR:|y|>M} 0)" (ay)

1/q

Similarly substituting (6) in (4), we have for the case p = 00,

| CF * Ge)(x0) — Fo) | < 1 + {IFO + II flloo}

tyeR:|y|>M}

s(y) u, (dy).

By Proposition 23.55 we have Tim Syerelyi> My Gey) #,(dy) = 0 and by Lemma 23.63 we have im D Siyeriyl>aqy ge(y)? 4, (dy) = 0. Thus for p € (1, oo], letting ¢ >

0 in (7)

and (8) we have iimn sup Ic Ff * (x0)— f' (x0)| 0 on R with ||g||1 = 1 and satisfies the condition that with some c > 0 we have

Q)

(x) 0, let gp, be the e-dilation of ~. Then we have

3)

gee

()

pé[1,00]

LP(R, 9, u,)-

Moreover for every M > Owe have

@

lim |

sup

50h emlsl>My

¥@)] =0,

and for every p € (1,00) and M > 0 we have

5)

lim

£90

JER: [x|>M}

Ge(x)? pw, (dx) =

Proof. 1. To prove (3), note that since ¢ is bounded onR wehavey € g € L®(R, 9, 14,).

Thus we have g € L}(R, 9t,, 2) M L2°(R, M,, ,) and then by Proposition 23.49 we

have 9 € Mpeti,oo LP(R, 9, u,) . Then we have g, € NM pett,co} LP(R, M1,, u,) by Proposition 23.56.

§23 Product Measure Spaces

605

2. To prove (4), let M > 0 be arbitrarily given. Since I tim v(x) = 0, forevery » > 0 x|+00 there exists B > 0 such that y(x)

< 7 for |x| > B. Lete

> 0 be so small that “

Then {x ER: |x|> 4) c{eeR: |x| > B)} and hence sup

v)
™}

sup

> B.

w) 0 be arbitrarily given. (x)

é1

(2)

Since I tim v(x) x|+00

yp

(dx)

(a) = 0.

= 0, there exists B > 0 such that

< 7 for |x| > B. Lete > O beso small that ¥ > B, Then

f

eke)

148)) (148 yds)< (1\8 aya 2) f feeRnto () Me

My

1\é6 =of2(;)1 ws f gy

148) +812048

p,Be (dx)

yg PAE -04) STP tals)

_ oft) sf]© _ 5 tg ty 8 =2(2)n

[l=

5M".

606

Chapter 5 Extension of Additive Set Functions to Measures

Since this holds for sufficiently small ¢ > 0, we have

13

timsup(*)'& f {xeRi|x|> M4} Ce) xO

p, (dx) < 2 ot,

Then since this holds for every 7 > 0, we have

1\é

tim sup (—) [ 20

SE%

SfreRi|x|>My

$x)

)x-O

pdx) = 0.

This implies that the limit inferior is also equal to 0 and then the limit exists and is equal to 0. This proves (7).

Theorem 23.66. Letp € [1, 00). Let f € L?(R, M,, u,). Lety € L1(R, Mt,, ,) be such that p > 0 on R and ||g||1 = 1. Assume further that 9 satisfies the following condition

a)

(x) Oand pf is a positive valued IN, -measurable function on Rand I him For ¢ > 0, let gp, be the e-dilation of . If f is continuous at xo € R then

2)

x] +00

v@)=0.

dim(S * G6) (0) = Fo).

Thus if f is continuous on R then iim(f * Qs)(x) = f(x) foreveryx eR. Proof. We have g; € LF (R, OM,., f,) for all p’ € [1, oo] by Lemma 23.65. Then by (c) of Lemma 23.60 we have for every x € R

8)

CF * G)@) — f@) = [ire —y¥) — F@)} p60) #, @y).

If f is continuous at x9 € R then for every 7 > 0 there exists M > 0 such that

(4)

If @o — y) — f@o)|

i}

Ge(y) bh, dy)

=n + H(e) + lf @o)lb().

§23 Product Measure Spaces

607

By Proposition 23.55 we have

©)

dim bh)= lim 80

e(¥) Hy) = 0

J{yeRily|>M}

Let us show that 7

im h@=

me

If Go — yee)

{yeR:[y|>M}

#4, @y) = 0.

Now f € L?(R, 1, u,) where p € [1, 00). We argue for (7) separately for the case p = 1 and for the casep € (1, 00). For the case p = 1 we have

@)

h@ = [ {yER:|y|>4} Lf (20 — »)lge(9) He, (dy)

M} sup ec} [ {yeR:lyl>M} LF(e0 — »)1 a, (@y) M}

efit.

By (1) we have

mo = 0(5) = celu()lz tf aev(2)br". Then we have

sup

(yeR:[yl>M} 1

M)

sup

fyeRyl>ay

y

1

¥(2)ir7 SE

¥(-)= >

sup

sp

veofifh.

( )

M eerixi>¥}

ve).

Substituting this in (8) we have

hO4}

Then by (5) of Lemma 23.65, we have im ,(€)= 0. This proves (7) for the case p = 1.

Por the case p € (1, 00), letg € (1, ~) be the conjugate of p. Then

@ ne-=f {yeR:|y|>af} IF Go — y) Ivey) 1, (€y)vy -~y)P Ud 0, Aj € &, and B; € B for j = 1,...,k. Then for every nonnegative extended

real-valued o (21 x 9%8)-measurable function f on X x Y, there exists an increasing sequence (yin EN) in P such that y, + f ae. on Xx Y.

Proof. By Lemma 8.6, there exists an increasing sequence (y, : n € N) of nonnegative

simple functions on (X x Y, 0 (& x %3)) such that g, + f on X x Y. Let g, be given by

ran) where cn,; > O and En;

Pa

n=) ¢njle,, jal

€ o (2 x 93). Let o(2 x 9) be the algebra generated by the

semialgebra& x %3. Then o(a(Q& x %3)) = o (A x 9) and E,,; € o(a(A x B)) so

§23 Product Measure Spaces

613

that by Theorem 20.15 for an arbitrary ¢ > 0 there exists F,,; € (2 (ux v)(En,j AF) < Pp Let

x 93) such that

Pn

(2)

Yn = Yen le,jj=l

Let us show that

lim ¥, = lim ¢, = f ae. on X x Y. Now we have a0 noo Pa

Pr

{Xx ¥ lon — Val > 0} cL {Xx Y= |e, —1,,1> 0} =) Bnj4Fajdel

j=l

This implies

Pa

&

(ux vY{X x ¥: ln — val > 0} S D0 x v)(EnjAFa,s) < 55° j=l

Then

DG

e

x VX XY slam — al > 0} < D5; < 00

neN

neN

so that by Theorem 6.6 (Borel-Cantelli), we have

(uw x v)(lim sup{X x Y : |g — Yn] > Of) =0. 100

By Lemma 1.7, this is equivalent to

(uw x v){X x Y¥ : G_ A Wn for infinitely many n € N} = 0. Thus

lim ¥, =

lim g, ae. on X x ¥ and then

lim

y, = fae.onX

x Y.

noo noo n>00 To complete the proof, it remains to show that w, € P forn € N. By Theorem 21.4 and Lemma 21.3, every set F € a(2 x %) is a finite disjoint union of members of the semialgebra 2 x 33, thatis, F = (Ai x B1)U---U(Ax x Bg) where {Ai x By, ..., Ag x Bg}

is a disjoint collection in & x B. Then Kk

k

j=l

j=l

Ir, y) = Yo 1ajxay@, 9) = This shows that y, given by (2) isa member of P.

14, @)12,0)-

&

Problems Prob. 23.1. Consider the product measure space (R x R, o(%p x 5x), , X u,). Let D={(@, y) €RxR:x= y}. Show that D € o(S3R x BR) and (w, x w,)(D) = 0. Prob. 23.2. Given the product measure space (R x R, o(Br x Bp), #, X 4). Let f be areal-valued continuous function on R. Consider the graph of f which is a subset of R x R

614

defined by

Chapter 5 Extension of Additive Set Functions to Measures

G={@,y)eRxR: y= f(x) forx € R}.

Let us write (R x R, 2, 4, x #,) for the completion of (R x R, o (Br x Br), f, x H,). Show that G is 2-measurable and (4, x “,)(G) = 0.

Prob. 23.3. Given the product measure space (R x R, o (BR x BR), Uy, X Hy). Let f be a real-valued function of bounded variation on [a, b]. Consider the graph of f defined by

G={(@,y)€RxR: y= f(x) forx € [a, b]}. Show that G € o (Bp x Bp) and (u, x 4,)(G) = 0.

Prob. 23.4. Let (X, A, 4) be a finite measure space and (Y, $3, v) be anono-finite measure space given by X=Y=(0,

1,

A = B = Bio), the o-algebra of the Borel sets in [0, 1], = “, and v is the counting measure.

Consider the product measurable space (x x Y,o(A x %)) and a subset in it defined by

E={(@,y)eXxYix=yh}. {a) Show that E € o (2 x B). (b) Show that

Sa Ufy le dv} du # fy {fx le du} dv.

(This shows that the o-finiteness condition on (X, 2, jz) and (Y, %, v) in Theorem 23.17 (Tonelli’s Theorem) cannot be dropped.) Prob. 23.5. Given two o-finite measure spaces (X, 2, 4) and (Y, %, v) where

X=Y=%, A= B = Y(Z,), the o-algebra of all subsets of Z, # and v are the counting measures. Consider the product measure space (X x Y, o (2 x %), 4 x v). Define a function f on XxY by 14+2* whenx=y, f@,y)=¢ -1-2% whenx=y+l1, 0 otherwise. Show that

@) firth favldu # fly f duav, (b) fy yy |Fla(u x v) = 00,

© Sexy f d(u x v) does not exist. Prob. 23.6. Consider the product measure space (X x Y, o (A x 93),

x v) of two o-finite

measure spaces (X, 2, 42) and (Y, %, v). Let f be an extended real-valued 2f-measurable and j-integrable function on X and g be an extended real-valued %3-measurable and v-

integrable function on Y. Let AG y=

f@g(y)

0

for (x, y) € X x Y for which the product is defined,

otherwise.

§23 Product Measure Spaces

615

(a) Show that h is o (Ql x %3)-measurable on X x Y.

(b) Show that A is wz x v integrable on X x Y and moreover we have the equality

Fxxy hae x v) = {fx f du} {fy gar}. Prob. 23.7. Consider the product measure space (R, x Ry, (3p x Br, 4, X ,).

Let gi and go be real-valued continuous functions on [a@, b] C R, such that g1(x) < go(x) for x € [a, b] and let

Dy, = {(x,y) € Ry x Ry : y € [g1(x), g2(x)] for x € [a, b]}. Let hy and hz be real-valued continuous functions on [c, d] C Ry such that h1(y) < ho(y) for y € [c, d] and let

Dy = {(x,y) € Ry x Ry : x € [h1(), h2(y)] fory € [c, d]}. Let D = D, N Dy and let f be a real-valued continuous function on D. (a) Show that D € o(%3R x BR) and moreover D is a compact set in R, x Ry.

(b) Show that f is 4, x j4,-integrable on D and furthermore we have the equalities:

frewxun=f Lf FC, y) oy (dy) oC) D [a,b] * J [gi (x), g2(x)] = [, [ IF smon Fy) oy 4x) ] 14, @). Prob. 23.8. Consider the following integrals:

@)

Sp fp? sin xy dy dx

©

hh Sy cos x? dx dy

b)

@ ©

©

Se

SP ay ax

So fy 6? dx dy

to 8 Sx2 yal dy dx

So fo

3 aydx.

Determine the domain of integration as a subset of the xy-plane and then evaluate the integral. Prob. 23.9, Consider the product measure space (Rx x Ry, oD, f be a real-valued function on R, x R, defined by setting

f(x,y) =e"@49)

x M,), py, X ,). Let

for (x, y) € Ry x Ry.

Show that f is 2, x 4,-integrable on R, x Ry and Jaexey fd(u,

X u,) =z.

(Hint: Apply the improper Riemann integral f°°, eM dx = Jn) Prob. 23.10. Consider the product measure space (R, x Ry, (MN, x Mt,), u, x Hz). Let f be a real-valued o (Jt, x t,)-measurable function on R, x R,. Suppose further

616

Chapter 5 Extension of Additive Set Functions to Measures

that 1°

f is 4, X w,-integrable on R, x Ry, that is, Ja,xr, Fd(u,

2°

for every x € R,, the improper Riemann integral g(x) = Leo FG, y) dy exists in R and g is 99t, -measurable on R,.

x 4)

€R;

3°

Leo g(x) dx exists in R and Leo g(x) dx = Jr, &(x) 4, x).

Show that we have

Sraxp, fA, * Hy) = fron [foo FV dy} dx.

(Thus Jr, xR, F d(u, X j,) can be evaluated as an iterated improper Riemann integral.)

Prob. 23.11. From the improper Riemann integral hon oe

(1)

dx = /1, we derive

foo) €** bax) = },/% fora € 0, 00).

Show that for every n < N we have

@)

S0,00) "=" tt, (dx) < 00 fora € (0, 00),

and indeed with an arbitrary c € (0, «) we have

@)

Foca) 2

oa ds) < BE aRe o

Prob, 23.12, From the improper Riemann integral hon e'dx = /m, we derive

qa)

fa) €** br, dx) = py/% fora € (0, 00).

Show that for every a € (0, 00) and n € N we have

(2)

Fio,c0) e082 yy (dx) =2- OTD 1.3.5... (On - )./mo73-,

and in particular

(3)

Ffo,c0) ex

pdx) = 2-4) 1.3.5.--(2n — IJ.

(Hint: Apply Proposition 23.37.)

Prob. 23.13. Let f € L}(R, IM, w,) and let

v= fe {ia FO — ») FO) H,@y)} ux).

Show that y > 0 and moreover we have te fdp,=+/y.

Prob. 23.14. Let f < L1(R, Mt,, w,). Leta, B € R \ {0} and let

¥:= fe {fa fx — y) f By) m, Gy)} u, Gx). Show that y > 0 and moreover we have Ie fdp,

=+./laBly.

Prob. 23.15. Let f € L1(R, 0t,, 4,). With h > 0 fixed, let us define a function g, on R

by setting

1

Gna) = oh

[x-h,x+h]

f®u,@s)

forxeR.

(a) Show that ¢;, is a real-valued uniformly continuous function on R.

(b) Show that g, € Z1(R, 9t,, 4,) and Ilgalli < If ll.

Prob. 23.16. Let (X, 2, 42) be a measure space. Let D € & and let f be a nonnegative

§23 Product Measure Spaces

617

real-valued {-measurable function on D. Consider the measure space (R, Bp, u,) and

the product measure space (X x R,o(Q x Bp), u x ,). Let E be a subset of X x R such that

E(x,) =8 { E(x, -) = [0, f(x))

forx € D*, forx e D.

Show that E € o(& x BR) and moreover (4 x 2, )(E) = to F(x) w(dx).

This page intentionally lett blank

Chapter 6 Measure and Integration on the Euclidean Space §24

Lebesgue Measure Space on the Euclidean Space

[I] Lebesgue Outer Measure on the Euclidean Space Let R” be the Euclidean n-space.

The topology on R” is the metric topology derived

from the Euclidean metric d(x, y) = {Sha

ly - yj2y? for x = (x1,...,%n) and

y = (y1,..-, Yn) in R". To fix the notations, let us repeat part of Definition 3.1.

Definition 24.1. Let 3, be the collection of all open intervals and © in R. Similarly let Je, Joc, and Ico be the collections of all closed intervals, all intervals of the type (a, bl,

and all intervals of the type [a, b) respectively, and 8, where —oo < a < b < 00 with (a, 00] := (a, 00) and [—o00, b] := (—00, b]. Let I =I, UT, U Joc U Fea, the collection

of all intervals and 9.

Notations. (a) Given n sets E1,..., E,, we write Xia Bi for the Cartesian product E, x ---x En. We write X mE for the n-fold Cartesian product ofa set Z. (b) Forn classes of sets €1,..., €,, we write Xie for

Ep xx We write Xj

Gy = {Ei x--- x Ey: Bye Gi

=1,..., a}.

€ for the n-fold Cartesian product of a class € of sets, thatis, €x---x € =

{E, x---x By: Ey € €,i=1,...,n}

Definition 24.2. We define classes of subsets of R” by letting

Toe = XierFoer Wo = KiarTeor W= Kio, Y= BUR 619

UBUT.

WH KiaTn

620

Chapter 6 Measure and Integration on the Euclidean Space

Proposition 24.3. 35, and 37, are semialgebras of subsets of R". Proof. It is obvious that 3%, satisfies conditions 1° and 2° of Definition 21.1. Let us verify

3° for 3G,. Take an arbitrary member of 3}, given by R=

X a1), bj]. We let Ij, =

(-o0, aj], Jj,1 = (aj, bj], and Ij,2 = (6;, 00] with (—00, —o0] := @ and (00, 00] :=

for j = 1,...,n, Then [i,j

+++

nig! fis ++» dn = 0, 1, 2} is a disjoint collection

of members of 3%, whose union is equal to R”. Since R = li,1 x --- X In,1, R° is a finite

disjoint union of members of 3%,,. This shows that 37, is a semialgebra of subsets of R”.

By similar argument, 37, is a semialgebra of subsets of R”.

Note that 33 and 37 fail to satisfy condition 3° of Definition 21.1. Proposition 24.4. o (37,,) = Bre, that is, the smallest o-algebra of subsets of R" containing J), is equal to the smallest o-algebra of subsets of IR" containing the open sets in R”. Similarly o (32) = BR, (95) = Be, o (7) = BR and o(I") = Bre.

Proof. Let us prove o(3,) = 93x. To show thato (34) C

pe, take an arbitrary member

R= XJjai(aj, bj] of M,. Fork € N, let Oy = X5_1 (aj,b; +p). Then (Ox: k € Nis

a decreasing sequence of open sets in R" and (1), Ef. Let us call (u)* the Lebesgue outer measure on R". (The fact that (u2)" is indeed an outer measure on RR" is by Theorem 2.21.)

§24 Lebesgue Measure Space on the Euclidean Space

621

We refer to a member E of 3” = 37, U 32, UT U 3 as a box in R” and v(E) as the volume of E. We show next that 37, in the definition of (”)" may be replaced by 3%, or Jj, or 32.

Proposition 24.6. For every E € 98(R”), we have

Q) (#f)*@) = inf (Leen vLy) | (Lek EN) C32, Usen Le > E}; (2) (@)*@ = inf [Leen (O00 : (Ox 1k EN) C 3%, nen Ok D E},

GB) (ut)"(E) = inf {yen (Ce) : (Cesk EN) CT, pen Ce D E}Proof. To prove (1), let us write 4(£) for the right-hand side of (1). Let ¢ > 0 be arbitrarily given. Let (R, : k € N) be an arbitrary sequence in 3, such that (J,.y Ry > E. Let

(Le : k EN) be an arbitrary sequence in 3”, such that Ly > Ry and v(Lx) < v(Re) +e/2* fork € N. Then ) yen v(Le) < Vyen v(Re) + &. Thus by the definition of A(E) as the infimum in (1), wehave A(E) < } pen v(Re) +e. Since this holds for an arbitrary sequence

(Re : k € N) in 3%, such that Len Re D E, we have A(E) < (u*)"(E) +e. By the

arbitrariness of ¢ > 0, we have A(E) < (*)*(£). Interchanging the roles of 3%, and 3”, in the argument above, we have (")"(E) < 4(E). Thus 4(E) = (u”)"(E). This proves (1). The equalities (2) and (3) are proved likewise.

Let us observe that the expression (2) has the advantage that the covering |_), E. Then

E = EN (Up Or) = Ua(E 0 Oy) and v(E) < OL, WEN Og) < DHL vCOr)-

Thus by the arbitrariness of the sequence (O; : k € N) in 33, we have

v(E) 0. Now for each k € N, Ry can be decomposed into a finite collection {Rx,1,..., Rx,p,} in 35, in such a way that 1°

{Rx1,..., Re, p,} is a disjoint collection and Ry = Ry,1 U-++U Reps

2°

the lengths of the edges of Rx,1, ..., Rg,p, ate all bounded above by 8/ (2,/n).

Note that 1° implies that v(Ry) = v(Ry,1) +--+: + v(Re, p,). Let the countable collection R=

(Rij,

t je = 1,..., Des k © N} be arbitrarily enumerated as 9t = {F,

We have Ucn Fin D £1 U Ep and

: m € N}.

Pk

(3)

Y oF) = OY oR) = Yo v(Re).

meN

keN fp=1

keN

Let us show that no member of Ht = {F,, : m € N} can intersect both Z, and E2. Assume the contrary, that is, forsomem € N the set F,, intersects both EZ, and £3. Then F,,£; 4 6 and Fy, 1 Ey AO. Letx, € Fy, Ey and x2 € Fy, 1 Ep. Thend = d(£1, Ex) < d(x, x2).

Now since x1, x2 € Fy and every edge of Fy, has length < 8/ (2,/n), we have d(x1, x2) < [n(8/ (2/2) yy" 2

3. Thus we have 6 < d(x1, x2) < 8/2, a contradiction. This shows

that no member of 9 = {F,, : m € N} can intersect both E; and £2. Let us classify {Fm : m € N} into three subclasses: {F), : m € N} consisting of those F,, which intersect

E}, {F” : m € N} consisting of those F, which intersect E, and {F’ : m € N} consisting of those F, which are disjoint from both E, and E2. Then we have £1 C U,,cn Fm and

E2 C Umen Fm and therefore

(4)

Yi od) = Oo) + YO oe + 2 ee

meN

meN

> oe) meN

meN

+o ory

meN

meN

> (u")"(E1) + (u2)" (2). By (3) and (4), we have (2). Now since (2) holds for every sequence (Ry: ke N) in 3,

such that J,ey Re D E1 U Eo, we have (iy (EU Ep) > (u?)* (ED) + (uz)" (E)). This proves (1) and completes the proof that (u)* is a metric outer measure. Then by Theorem 2.19, we have Bg C mM((u")") =v. a The following decomposition of a nonempty open set in R” into countably many disjoint binary cubes is often useful. Proposition 24.10. For k € N, let Q be the subcollection of V7, of n-dimensional binary cubes of the form (1, ap] xx (2, %ent ] where z1,...,%n € Z. The classes Qu, k EN, of members of 35, have the following ‘properties: | (a) For eachk € N, Deisis a countable disjoint collection of members of 3,,, and the union of all members of Qi; is equal to R".

(b) If Ry € Dy and Ro € Qe where k,

£e Nandk

< £, then either Ri N Ro =@or

624

Chapter 6 Measure and Integration on the Euclidean Space

Ry D Ry. Also every member of Qy is the union of (28-*)" members of Qu. (c) Every nonempty open set in R" is a countable disjoint union of members of 3’, in Usew 2% where N is an arbitrarily chosen positive integer. Proof.

(a) and (b) are immediate from the definition of O; for k € N.

To prove (c), let

N €N be arbitrarily chosen and let V be an arbitrary nonempty open set in R". Then for every x € V, there exists an open ball B containing x and contained in V. For sufficiently

largek > N, there exists R € OQ; such that x € R c B. Thus V is a union of members of

Upen Qe. Since U,>7 Qx is a countable collection of members of 3},, V is a countable

union of members of 3?,. In the collection of those cubes in );..y 4g whose union is V, select those which belong to Q.y and drop those which are in _),.y1 {2x and are contained

in any of the cubes that have been selected. In the collection of the remaining cubes, select

those which belong to 441 and drop those which are in L),..y42 2% and are contained

in any of the cubes which have been selected. Thus proceeding we show that V is a disjoint

union of members of J,

Qk.

[I] Regularity Properties of Lebesgue Measure Space on R” Lemma 24.11. (Borel Regularity of the Lebesgue Outer Measure on R”) The Lebesgue outer measure (u)* on R" has the following properties.

{a) For every E € $8(R") and ¢ > 0, there exists an open set O in R" such that O D> E and

(H2)"@ s (uC) s HY

+e.

(b) For every E € $8(R"), there exists a G3-set G in R" such that G D E and

(41)"@ = (ui). (c) (ut)* is a Borel regular outer measure. Proof, 1. Let E ¢ $8(R"). By Proposition 24.6 we have

(ut)E) = int [Deeg v(Ox) : (On 1k EN) CF, per Ox D E}Thus for every ¢ > 0 there exists (O, : k € N) C 3% such that

()

, E. Since G C Oy for every k € N, we have (u")"(E) < (u")"(@) < (u")"(On) < (u")*(E) + } by the monotonicity of the outer measure (")". Since this holds for every k EN, we have (u")"(E) < (u”)"(G) < (u%)*(E) and therefore (u*)*(E) = (u2)*(G). 3. By Theorem 24.9 and Definition 24.8, we have Bg.

c Mt! = mt((u")*).

Thus

(ur)" is a Borel outer measure on R”. Since a Gz-set is a member of Sg, we have G © Spe. Then by (b), (")* is a Borel regular outer measure. Theorem 24.12. For E € 9(R"), the following conditions are all equivalent:

@(ii) ForLe every M((ut)") & > 0, there exists an open set O

>E

with (ut)"(o \E)« 0 there exists an open set O such

that O > E and u"(0) < (u4)") +e by Lemma 24.11. This implies that inf {u"(O) : EC O, O isopen} < (ut)*) +e.

By the arbitrariness of ¢ > 0, we have inf [y(O) : E C O, O is open } < (u?)"(E). [1M] Approximation by Continuous Functions Proposition 24.18. Let E € 2)t” with 4"(E) < 00. For every é > O, there exists a disjoint finite collection {O1, ..., Op} in TG such that

us(eal

0;)

j=l

O there exists a sequence (Rx : k EN) in 3}, such that L),.y Ri D E and yey HE (RE) — 5S BR(E) S Ven BT (Re)Since p7(E) < 00, we have Deen Hi (Rk) < 00 and thus there exists N ¢ N such that Dien we (Rx) -— 5 < Dien wt (Rx), that is, yy HE(R:) < 3. Consider the finite

collection {Ri,..., Rw} in 33. Since 33, is a semialgebra of subsets of R”, there exists a disjoint finite collection {S;,..., Sp} in 3%, such that UN, Re = Uh S; by Lemma 21.3. Now PB

PB

Pp

j=l

j=l

j=l

ut(eas)) =a2(2\ Us) +#2(U 5) \ 2).

Since E\ U?_, Sj C Usen Re \ Utes Be C Usow Re We have P

un(E\(JS)) = j=l

k>N

wRe 0, there exists a real-valued continuous function g on R" vanishing outside a bounded set in R" such that fa, |f — g| duty 0 be arbitrarily given.

By Theorem 24.19, there exists y = ey

alo,

where a; € R, O; € 3}, with u"(O;) < cofori = 1,...,N,suchthat fp, |f—w|dut E}

= (u)"®.

where the third equality is from the fact that 7, is a one-to-one mapping of 3%, onto 3},. Theorem 24.27. (Translation Invariance of the Lebesgue Measure Space on R”) The Lebesgue measure space (R”, WV? , 7) is translation invariant, that is, for every E € 30¢ and x € R", we have E +x € 0) and w?(E + x) = wh (E).

Proof. The o-algebra 9t" = Mt((u)") consists of all subsets E of R” satisfying the Carathéodory condition (u")"(4) = (u2)"(A NE)+ (ut)"(A 1 E°) for every subset A of

R". (See Definition 2.2.) Suppose E € 90t((u")*) and x € IR”. Then for E + x we have

(#2) (ANE +2) + (EAN E+)) =(u2)* {AN E+ x)} —x) + DT ({A0 ( +2)} -2)

=(u1)5 (4-2) 0B) + (WI) (4-9 +) =(u2)(A — x) = (42) (4),

where the first equality is by Lemma 24.26, the third equality is by the fact that E satisfies the Carathéodory condition with A — x as a testing set, and the last equality is by Lemma

24.26. This shows that if E < 9t(()”), then E +x ¢ 9t((u”)*), thatis, if E ¢ Mt",

then E +x € 9t?. Then p?(E +x) =

3 (E) by Lemma 24.26.

Theorem 24.28. (Translation Invariance of the Lebesgue Integral on R”) Consider the

Lebesgue measure space (R", nt, ur). Let f be an extended real-valued function on a set D & 0}. Leth & R" and let g be a function on D — h defined by g(x) = f (x +h) for xeEeD—h.

(a) If f is 9t! measurable on D, then g is 2t!-measurable on D — h. (b) If f is tf -measurable on D, then

a

I FG) wi(dx) = f _ fem unas),

in the sense that if one of the two integrals exists then so does the other and the two are equal, In particular if f is nonnegative, then (1) always holds.

634

Chapter 6 Measure and Integration on the Euclidean Space

(©) If f is defined and I} -measurable on R”, then

Q)

[ _f@) ux) = [ SG +h) ui(dz),

in the same sense as above. Proof. This theorem is proved by using Theorem 24.27 in the same way Theorem 9.31 was proved using Theorem 3.16. [VI] Linear Transformation of the Lebesgue Integral on R” Let T be a linear transformation of R” into R’, that is, T is a mapping of R” into R” satisfying the conditions: 1°

T(ex) = aT (x) forx € R" anda € R,

2° Taty)=T@)+T() forx,y eR". It is a well-known fact in linear algebra that if J is one-to-one, then T maps R” onto R” so that its inverse transformation 7 —" is defined on R" and is a linear transformation of R” onto R”. Let us call such linear transformation non-singular. Let {e1, ..., én} be the standard basis for R” and let My be the matrix of T with respect to the standard basis. Thus Mr is an n X n matrix whose columns are the column vectors T(e1),..., T (én). T is non-singular

if and only if det Mr 4 0.

Observation 24.29. It is well known in linear algebra that every non-singular linear transformation T of R” into R” is a finite product of non-singular linear transformations of the following three types: 1° 2°

TG... Hj eee Xka ) = Ge Mee + ey, °++) for some j,k T(x1, X2,.--, Xn) = (@x1, X2,.-.,%) for some a € R,a £0.

=1,...,0,f 0 there exists

(Py ik €N) C T(3%,) such that yen Py D T(E) and (u*)"(T(E)) +2 = Deen v(P)Let Ry = T~1(Py) fork € N. Then (Ry: k € N) C 3%, and pen Re D E. Now by (b),

o(Pe) = 1 (Px) = 2 (T (Re) = [det Mra” (Re). Then we have

(uf)"(TCE)) + ¢ = Yo o(P,) = [det Mol Y> u2 (Re) = | det Mr|(u2)" (2). keN

keN

Since this holds for every e > 0, we have

7)

(42)"(T(E) = | det Mr|(u2)").

Now (7) holds for every non-singular linear transformation T of R” into R” and every E € 98(R"). Applying (7) to the non-singular linear transformation T—! and the set T(E) € PR"), we have

(u2)*(7-(P)) > | det Mp-al(u2)" (TE).

638

Chapter 6 Measure and Integration on the Euclidean Space

Now T-!(T(E)) = E and det Mp-1 = (det Mr)—!. Thus the last inequality reduces to

(8)

|det Mr|(u")"(E) = (u")*(T(E)).

By (7) and (8), we have (c). Theorem

24.31.

(Linear Transformation of the Lebesgue Measure on R") Consider

the n-dimensional Lebesgue measure space (R", U", 22). Let T be a non-singular linear transformation of R” into R". Then (a)

T(E) € Dt" for every E € Dt.

(b) u2(T(E)) = | det Mr|y2 (EB) for every E € Mt". Proof. By Theorem 24.15, 20t? is the completion of SR. with respect to 47. Thus if E € Mt", then E = AUC where A € Bxe and C is a subset of a set B € Bp with u2(B) = 0. Then T(E) = T(A) U T(C) and T(C) C T(B). Since A € Bp, we

have T(A) € Bre and w2(T(A)) = |det Mr|z%(A) by Proposition 24.30. Similarly T(B) € By and u"(T(B)) = | det Mr|w"(B) = 0. Then since T(E) = T(A) UT(C) where T(A) € Bg» and T(C) is a subset of T(B) € Bye with u” (T(B)) = 0, we have T(Be€ ot.

This proves (a). (b) follows from (c) of Proposition 24.30.

Theorem 24.32. (Linear Transformation of the Lebesgue Integral on IR") Consider the Lebesgue measure space (R", 99t), 4"). Let f be an extended real-valued function on a set D € 0}. Let T be a non-singular linear transformation of R” into R". (a) If f is 200? measurable on D, then f 0 T is DV;-measurable on T—(D) and

q)

f F(x) uj (dx) = | det orl f Cf oT) wi dx), D T-1(D)

in the sense that if one of the two integrals exists then so does the other and the two are equal. In particular if f is nonnegative, then (1) always holds. (b) if f is defined and

@

Ui -measurable on R", then

ff feompasy = laeemrl [oF ory uptan,

in the same sense as above. Proof. If T is a non-singular linear transformation of R” into R”, then the inverse transfor-

mation of T exists and is a non-singular linear transformation of R” into R”. Let us write $ for the inverse transformation of T. Applying (a) of Theorem 24.31 to the non-singular linear transformation S, we have S(E) € 90 for every E € Mt?. Thus T(E) = S(E) € ot for every E € Mt. This shows that T is a 9)t7 /9)t?-measurable mapping of R” into R”. If f is an extended real-valued 99t7-measurable function on a set D € Dt}, then for the function f o T defined on the set TD)

€

tt we have for every c € R,

(f oT) '([-00, el) = T"(F"([-00, ¢])) € OT

§24 Lebesgue Measure Space on the Euclidean Space

639

by the fact f-1([—oo, c]) € 99%" and the fact that T is a Mt" /2t"-measurable mapping. This shows that foT isaOt} -measurable function on T—!(D). To prove (1), let us note first

that if D, E ¢ St, then since the inverse transformation S of T is a one-to-one mapping,

we have S(D N E) = S(D)M S(E) = T—!(D) NT~!(E), Then by (b) of Theorem 24.31, we have "(DM E) = nu" ((T 0 S)(DNE)) = | det Mr|u" (S(DNE)). Thus for f = 12, we have

[ f(x) u2 dx) = [ 1g (x) ue (dx) = wy (DN E) = |det Mrlu*(T-1(D) NT-"(E))

= Idee Met ff “tgp OO MEA) = |det Mr| f

T-\(D)

1r(T@))

(dx)

=laetmel f T-\(D) (fot) unas), proving (1) for this case. Then by the linearity of the integrals with respect to their integrands, (1) holds for nonnegative simple function f on D. Applying Lemma 8.6 and Theorem 8.5 (Monotone Convergence Theorem), we have (1) for every nonnegative extended real-valued

20t" -measurable function f on D. If f is an extended real-valued Dt? -measurable function

on D, then we decompose f = f+ — f— and apply the result above to f+ and f-.

‘We call a mapping T of R" into R” of the form T(x) = ax forx € R" with fixeda ¢ R a dilation with factor w. For a dilation T with factor a € R, we have

T (yx) =a(yx) = y(@x) =yT(x)

forx eR" andy ER,

Tat+y)=e(+y) =ex+ey=Tx)+T(y)

forx,yeR’.

Thus a dilation of R” is a linear transformation of R” into R". Corollary 24.33. (Positive Homogeneity of the Lebesgue Measure Space on R") Let T be a dilation on R® given by T(x) = ax forx € R" where a € Randa #0. Then for everyE € SU!, we have T(E) = aE € 0? and

@

BEE) = |o|" 4" (E).

If f is a S00} -measurable extended real-valued function on a set D € DN}, then f oT is 200" -measurable on 1p and

@)

[teouan=iar f fax urcas)

in the sense that the existence of one of the two integrals implies that of the other and the equality of the two.

640

Chapter 6 Measure and Integration on the Euclidean Space

Proof. A dilation T(x) = ax for x € R" witha € R, a ¥ 0, is a linear transformation of R” into R” whose matrix My is equal to aJ where I is the n by n identity matrix. Since det Mr = a” # 0, T is non-singular. Thus (1) follows from Theorem 24.31 and (2) follows from Theorem 24.32. § Proposition 24.34. Forx € R” andr > 0, let B(x,r) ={y € R®: |y —x| 0, we have pit (BG, ar)) =anyt (BG, r)).

Proof. 1, Let us prove (a). To prove B(x’, r) = BG, r) + (x' — x), we show that for any

y € R", y € B(x’, r) if and only if y € B(x, r) + (x’ — x). Now we have

y € B(x,r) + (x —x) & y—(@’ —x) € BG, 1) ely -G@’ -x)-x|

2-2-8. on R”.

Proof. 1. Note that we have X7_,[—e,¢] C B(O,1) C Xj_y[-1, 1] for sufficiently small

> 0. Then we have 7 (X ieal—€, el) < u"(BQ, 1) |y —xo| — [xo —x| > ro +6 —5 = rq so thaty ¢ B(x, ro). Thus 1a¢,r)(y) = 0 for every x € B(xo, 6). Then Jim, 1z@,n)0) = 9 = 1s¢o,n))This proves (d). 3. To prove (e), first let y € B(xo, ro) be arbitrarily fixed. Ify = xo, then y € B(xo, r) for every r € (0, 00) so that 1p¢,),7)(y) = 1 for every r € (0, 00) and this implies that

Jim, 13@,7)(¥) = 1 = 1n¢0,-)(9). Ify € BCxo, 70) \ {xo}, let8 = ro — ly — yol € (0, 70).

Then y € B(xo, 7) for every r € (ro — 8, 00) so that jim, 13 (9,70) = 1 = 1a@,m)()Next, let y € B(xo, ro)° be arbitrarily fixed. Let 6 = |y — xol — ro € (0, 00) so that y ¢ B&xo, r) for everyr € (0, ro + 8). Thus 1g¢9,7)(y) = 0 for every r € (0, 79 + 6) and

then lim 13(2,)(¥) = 0 = La¢ao,r0)(y). This proves (e).

646

Chapter 6 Measure and Integration on the Euclidean Space

Lemma 25.6. For every f € &},,(IR", 9%", 1"), the function (A, f)(x) forx € R" and r & (0, co) has the following continuity properties: (a) For every r € (0, 00), (A, f)(x) is a continuous function of x € R*. (b) For every x € R", (A; f)(x) is a continuous function of r € (0, 00).

Proof. of x

1. Let ro € (0, 00) be fixed.

To show that (A, f)(x) is a continuous function

€ R", we show that for every x9

Since u"(B(x,70))

lim Sater f tet x—>x0

|xo| + 6+

79.

= 42 (B(xo,70)) =

Svea ry f aH

Then for every x

€

€ R” we have

im

=

(An f) (0).

> 0 for every x € R’, it suffices to show that Let 5 >

B(xo,5),

0 be arbitrarily fixed and let R

we have

fo € S},,(R", 9", 4"), f is w2-integrable on B(O, R). ve

a

nf)

B(x,7ro)

C

B(0, R).

>

Since

For every x € B(xo, 8), we

foB(x,79) sawn fiB(O,R) Amand aut.

Now |1 56,70) f1 < [1a@,2) | which is 27 -integrable on B(0, R) and by (d) of Observation

25.5 we have jim 156.70) = Lpo,r), #2-2-€. on R”. Thus by the Dominated Convergence Theorem (Theorem 9.20), we have

@

lim

4>%0

= i

JB(z,70)

BO,R)

f dy} = lim

47% J BOR)

L3G.)f dey = f

BGo,70)

lag) f du,

fdph.

2. Let x9 € R” be fixed. To show that (A, f)(xo) is a continuous function of r € (0, 00), we show that for every rp € (0, 00), we have jim (Ar Fo) = (Ay, f) (xo). Since

lim r"wi (BG, 1) = ro" 22 (BG, 1)) = #7 (B(xo, r0)) > 0, it lim 3 (BGo, r)) = ror ror

suffices to show that Jin, S360.) fap

= I B000,70) f dp®. Let 8 € (, ro) be arbitrarily

fixed. LetR > r9+5+|xo|. Then for every r € (ro—8, 79 +8), we have B(xo, r) C B(O, R)

so that

@)

f

B(xo,r)

fduy= f

BOR)

Ingo,f uy.

Then since we have |13(x9,r) f| < |12¢0,n)f| which is 7 -integrable on B(O, R) and since

lim 1aga,r) = Lagp,ro) #7 -2-€. on R” by (€) of Observation 25.5, we have

rT

tim fo rauz= fiB(x0,70) faut

710 J B(xo,r)

by the Dominated Convergence Theorem. We introduce the Hardy-Littlewood maximal function which will serve in estimating

the family of averaging functions {A, f : r € (0, c0)}.

§25 Differentiation on the Euclidean Space

647

Definition 25.7. For f < £1,,(R", 9", u"), the Hardy-Littlewood maximal function Mf of f is a nonnegative extended real-valued function on R" defined by setting for x € R”

(Mf\x)=

sup

re(0,00)

(4-|fl)%)=

sup

Ria1 AY

re(0,00) HP (8@, r))

BQ)

\fldup. *

Observation 25.8. Note that if f < £},,(R", 9%", 2"), then |f| € £},(R", Dt, uw")

also, and Mf is defined in terms of | f| rather than f itself. Thus Mf = M|f|. By (a) of Lemma 25.6, for every r € (0, 00), A,| | is a real-valued continuous function on R”. As

the supremum of a collection of continuous functions on R", Mf is a lower semicontinuous function on R” by (a) of Theorem 15.84. Then for every a € R, {R” : Mf > a} is an open set in R® by (d) of Observation 15.81 and so that Mf is a S3p»-measurable function on R”.

Let us note also that according to (b) of Lemma 25.6, for each fixed x € R”, (Ar|f|)(x)

is a continuous function of r € (0, oo). Thus for an arbitrary countable dense subset D of (0, 00), we have

(Mf)(x)=

sup

7é(0,00)

(Ar|f1)(&) = sup (4r|f1)@). reD

Then Mf, as the supremum of countably many continuous and hence %3g.-measurable functions on R", is a S$Rr-measurable functions on R" by Theorem 4.22. Let f be an extended real-valued 2-measurable function on a measure space (X, 2, 1). Fora € (0, 00) consider the set {x tlfl> a} € &. The set decreases as @ increases. Thus

u{X ; |f| > a} decreases as o increases. Let us consider the behavior of eu{X : | f| > a} as a: increases. We show that if f in jz-integrable on X then im op {X [fl >a} =0. Observation 25.9, (a) Let (X, 2, 2) be a measure space and let f ¢ £1(X, 2, 4). Then qd)

an{x:ifl>a}s f ifldu a} =0.

(b) Iff € £1(R”, Mt", 2”), then there exists a constant C > 0 such that (3)

ap" {R’:|f|>a} a} < C fora € (0, 00) does not imply that

loc

Proof. 1. If f is a js-integrable extended real-valued 2{-measurable function on a measure space (X, 2, jz), then for every a € (0, 00) we have

[iriaue D snoo flaw > au{X:|f| >a}.

648

Chapter 6 Measure and Integration on the Euclidean Space

This proves (1). To prove (2), let F = {X : |f| < oo}. Then F° = {X : | f| = co}. The p-integrability of f on X implies that f is finite z-a.e. on X by (f) of Observation 9.2. Thus

we have (F°) = 0. Fora € (0, 00), let Dy = {X : |f| > a}. Then DS = {X: |f| az(D,) = 0, (5) implies jim, a4(De) = 0. This proves (2).

2. Iff € £'(R", MY, 2"), then by (1) we have ap"{R":|f|>a} 0. This proves (3).

Let us construct a function f on R” such that f ¢ C},(R", 9", 4") and f ¢ £1(R", 9t", 2”) and (3) does not hold. Let y > 1 and let f be a constant function f =y onR". Then f,, |fldu" = yu"(R") = 00 so thatf ¢ £'(R", Ot", 2"). On the other hand since f is bounded on R” so that f is locally }-integrable and hence

fe SLR’, Mt", uw"). Now consider« = 3 € (0, 00). We have

an {R" : |f| > a} = Lun{R" sy > 2} = tur") = 00. Thus (3) does not hold.

3. To prove (c), we construct a function f ¢ L},,(IR", Mt", 4") for which (3) holds.

We do this for the case n = 1, Let f bea real-valued 2)t, -measurable function on R defined

by

1

for x € (0, 1),

f@) -| 0 forx € (0, 1)°.

§25 Differentiation on the Euclidean Space

We have fig 1) lfldu, = So.

649

1p, (ax) = oo. Thus f is not yz, -integrable on the bounded

set (0, 1) and hence f ¢ £},,(R", Mt", 2”). Let us show that f satisfies (3) with C = 1.

Now for a € (0, 1}, we have {R: | f| > &} C (0, 1) so that we have

ap, {R:|f| > a} a} = {x € ©, 1): x < 1}=(0,2) so that we have

apy, {R: |f| >a} =an,(0, 1) =a4=1.

Thus we have shown that a,

(R:|f| >a} e. Since gee B > K, there exists a finite subcollection {B1,..., By} of € such that Us, B; D> K. Let By = B@;,1r;) fori = 1,...,N. Renumber By,..., By if necessary so that r; > --- > ry. From the finite sequence (B; :i = 1,..., N), we selecta disjoint subsequence as follows. Let i; = 1. Discard all B; that intersect B;, and let B;, be the first of the remaining ones if there are any. Discard all B; that intersect B;,, and let B;,

be the first of the remaining ones if there are any, and so on. The process stops after a finite number of steps, say p steps. Consider the finite disjoint collection {B;,,..., B;,}. Every discarded B; intersects some B;, with 7;, > rj so that B; = B(x;,7;) C B(xi,, 37;,). Thus

we have UN, By C UPL; BGtig, 37%). Then

e a} r0

c {R": M(f —8)+lg—f| >a}. Now

{R" : M(f — 8)

}. Thus we have

(3)

ui" (Eq) < un{R" : Mf — 8) +1 — fl >a}

Sui {R": MCF — 2) > $}+ #2 (R" le — fl > Zh By Theorem 25.11, we have 2

-— 9) > §}< anf un{R" : M(f

2

If - gldut < ae

Also

Se sle—sl>s}= € = {E,(x) : r € (0, 00)} where

E,(x) is defined by letting E,(x) = E} (x) if r is a positive rational number and E,(x) = E/(x) if r is a positive irrational number. Then E,(x) C B(x, r) for r € (0, co) and if

we let « = min {a’, a} then u”(E,(x)) > ay"(B(x,r)) for r € (0,00). Thus € shrink

nicely to x. Let (qx, : k € N) be a sequence of positive rational numbers such that g; | 0 and let (pz : k € N) be a sequence of irrational numbers such that p; | 0. Then we have

(Ea) _ |, "Fa®) _ pt (Ei, @)) =y

Broo pl (Eq(x)) and.

0 Thus (Dev)(x)

=

lim

F>00

VEn@) 5 =

(Ep, (@))

»o(z,@))

790 ut (E,@))

V(En@) on =y".

2 uh (EY, G))

does not exist.

This contradicts the assumption that

(Degv)(x) exists for every collection € that shrinks nicely tox.

©

Definition 25.21. Let v be an extended real-valued set function defined on (Byr)y. Let x € R*. If for every collection & in BRa that shrinks nicely to x the derivative (Dgv)(x)

of v with respect to the Lebesgue measure

7, at x along E exists, then we say that v is

differentiable with respect to the Lebesgue measure uw

at x. We call (Dev)(x), which does

not depend on & according to Theorem 25.20, the derivative of v with respect to 4? at x and we write (Dv)(x) for it. Example. As an example of the derivative of a set function with respect to the Lebesgue 1 measure, let f € £},,(R”, 9t7, 4”) and consider a set function v on (pe), be defined loc by setting

v(E) = [ fdu"

for E ¢ (Bpe)y.

Letx € R” and let € = {E,(x) : r € (0, 00)} an arbitrary collection in Sg» that shrinks nicely to x. Then in particular for every x € A(f), we have by (2) of Theorem 25.17

D

(ev)

= lim

v(E,@))

10 w(E-(x))

_

4 1 = lim —___

” (dy)

=

190 w8(E-(x)) [ (a) FO) MEN = FO)

f(x).

§25 Differentiation on the Euclidean Space

Thus we have

657

forxe ACf). §

(Dv)(x) = f(x)

As further examples of differentiation of set functions, we consider below differentia-

tion of indefinite integrals of locally integrable functions, density of measurable sets, and differentiation of Borel measures with respect to the Lebesgue measure.

[MM] Differentiation of the Indefinite Integral Let f be a ,-integrable extended real-valued 9Jt,-measurable function on [a, b]. By Definition 13.12, an indefinite integral of f is a real-valued function F on [a, b] defined by

F(x)= Stax F dj, + for x € [a, x] where c is an arbitrary real number. To extend the notion of indefinite integral to functions defined on R’, let us replace the function F defined on [a, b] by a set function ®(E) = f, f du, for E € Be NM [a, 5).

Definition 25.22. Let f

L},,(R”, 97, 4"). Let us call the real-valued set function ®

defined on (Bp), by P(E)= ff fg f duz for E © Bp«)s the indefinite integral of f.

Theorem 25.23. Let ® be the indefinite integral ofa function f

£},,(R", 9, u).

Then for every point x in the Lebesgue set AC) of f and for every collection —& = {E,(x) :

r € (0, co)} in Bye that shrinks nicely to x, we have

@

OE)

(De)(@) = lim 3b (E-@) un = fG@).

In particular for the symmetric derivative Ds® of ®, we have (2)

(Ds®)(x) =

lim

PUG: r))

1 0 us ([email protected]))

= f@).

Proof. Let x € A(f). Then by (2) of Theorem 25.17 (Lebesgue Differentiation Theorem), we have

@(E,(x))_

790 wa (E,(x)) 70 pt Ee)

lL. FO) uMdy)= FG).

This proves (1). Since $ = { B(x, r) : r € (, 00)} is a collection in 3g to x, (1) implies (2). =

shrinking nicely

Let f be an extended real-valued 9)t, -measurable and j1, -integrable function on [a, 5]. Consider the function F defined by F(x) = Ja.b) Ff du, for x € [a,b]. According to the

Lebesgue Differentiation Theorem for the indefinite integral (Theorem 13.15), the derivative F'(x) exists and is equal to f (x) for a.e. x € [a, b]. Let us show that this result is contained

in our Theorem 25.23 as a particular case.

658

Chapter 6 Measure and Integration on the Euclidean Space

Corollary 25.24. Let f be an extended real-valued IN, -measurable and 2, -integrable function on [a, b). Let a function F on [a, b] be defined by

FQ) = [ FO) mea) forx « (a,b) Then the derivative F'(x) exists and is equal to f (x) for a.e. x € [a, b]. Proof. To show that the derivative F’(x) exists and is equal to f(x) for a.e. x € [a, b], we

show that the right-hand derivative and the left-hand derivative of F exist and are equal to F(x) for ae. x.

Let us extend the definition of f to R by setting f = 0 on R \ [a, b]. Then we have

f € £},,(R, Mt, u,). Asin Definition 25.22, let us define a set function © on (S3p)s by

setting

Oo) = [70

u,(dy)

forE € (Br)p.

To find the right-hand derivative of F at x € R, let E,(x)

= (x,x +1) forr

€ (0, 00).

Then E,(x) C B(x,r) and 4, (Er(x)) =r = 4u,(BG,1)) for r € (0,00) so that & = [E,(z) : r € (0, 00)} is a collection in Bx that shrinks nicely to x. Letx € AC). Then we have

.

1

tim

dim 7{F@ +r)—F@)}= lim - ean fQ) 4, @y)

(EG)

= f@)

~ 190 u,(E-@)) by Theorem 25.23.

This shows that the right-hand derivative of F exists and is equal to

f(x) atx € ACf).

To find the left-hand derivative of F at x € R, let E/(x) = (x —1,x) for rr € (0, 00).

Then & = {E/(x) : r € (0, 00)} is a collection in Bg that shrinks nicely to x. Now

lim lim

.

{F(x —r)-F@)}= iim “{F@) - F@-n}

1

= oe lim — Iecnx)

P(E)

a fO) 4, @y) = lim 4, ” (E/@) @)) = Ff)

by Theorem 25.23. This shows that the left-hand derivative of F exists and is equal to f (x) at x € A(f). Therefore the derivative of F exists and is equal to f(x) at every x in the

Lebesgue set A(f) of f. Since u, (A°(f)) = 0, the derivative of F exists and is equal to F(x) for ae. x € [a,b].

©

[IV] Density of Lebesgue Measurable Sets Relative to the Lebesgue Mea-

sure

Definition 25.25. For E € 907, consider a set function v on (Br), defined by setting

v(B) = 2" (ENB) forB € (Bpz)p. Forx € R" andacollection& = {E,(x) : r € (O, 00)}

§25 Differentiation on the Euclidean Space

659

in Bae that shrinks nicely to x, let us define &g(E, x) = (Dgv){x), that is,

_

v(E-(x))

_ BEN E,()) 4S Nim 10 u%(E,G))

lim “= (D be(E, e(E, x) x) == (Dev) (x) = bu (E@))

provided the limit exists. We call 8¢(E, x) the density of E. at x with respect to the Lebesgue measure ue along &. In particular for the collection 8 = {B (x, 7): r € (0, 00) } the limit

Box,

MEN BG,

fa(E,2) = Darya) = ro tim MPO) BE.) wt (B(x,r)) 20tpg MEO wR (B(x,r))

if it exists, is called the symmetric density, or the Lebesgue density, of E. at x with respect to the Lebesgue measure 1". If s(E, x) = 1, we call x a density point of E. If §s(E, x) = 0, we call x a dispersion point of E.

Observation 25.26. Let E € 9”. Letx € R" and let € = {E,(x) : r € (0, 00)} bea collection in 3g

that shrinks nicely to x. If 8g(Z, x) exists, then d¢(E, x) € [0, 1].

Proof. Since EN E, (x) C E,(x), wehave u"(ENE,(x)) < “"(E,(x)) forallr € (0, 00).

By Definition 25.16, there exists a > 0 such that 4"(E;(x)) > @u(B(x,r)) for all r € (0, 00) so that 42” (Z,(x)) > 0 for all r € (0, 00). Thus we have

HR(EN E-@)) € [0,1] forallr € (0, 00). u2(E;(x))

Then if the limit of the quotient exists as r > 0 then the limit is in the range [0, 1], that is,

if 5g(E, x) exists, then dg(Z, x) € [0, 1].

Remark 25.27. Let E € St}. Letx € R” and let € and & be two collections in Sge that shrink nicely to x. (a) Existence of 5g (E, x) does not imply that of 5g/(Z, x). (See Example 1 below.) (b) Existence of both 3g(Z, x) and dg/(F, x) does not imply 5g(Z, x) = 5g/(E, x). (See

Example 2 below.)

Example 1, In this example, 5g(E, x) exists for some collection € in BR that shrinks nicely

tox but 8g(E, x) does not exist. In R let fk = [satsr. tx] with w, (lk) = ster fork € Zy, that is,

with

b=([z- x) 4=[3- 3) 2 = (Uo) =

1

yu

4)

=

1

Bp

(2)

xb b=[p wh 1

= re

1

B, (3) = a?

see

Let E = Ukez, fe 1. If we let

= {(—r, 0) : r € (0, 00)}, then € is a collection in 3g that shrinks nicely

to 0. Now for € we have

_ H(ENCrO) ___—

8e(E, 0) = lim —+

CE

EN

(nO)

aD)

= lim —*—_-.

90 (7,0)

=

0.

660

Chapter 6 Measure and Integration on the Euclidean Space 2.

Next let us show that 5s(Z, 0) does not exist.

EN BOQ, ge) =U tess U ya U--- 80 that 1

1

Now for every k € Z,, we have

1

#,(E0 BQ, ge) = spect + pass + pees t= On the other hand EN B(0, wut)

#,(E

BQ, gist) —t_

Thus we have

= iyi U

1

goa

eye U yg U--- so that

1

1

= pers + ps + pee t= =—=z

4 1

=

=

4

1

es

a

(BOBO, p)) 4 1 2 1

(BQ, ze)

and

32RE2 3

# (ENB, pier) 4

1 2

1

(BO. gia) 329 2-6

Therefore the quotient 2, (EM B(O, r))/1,(B(, r)) assumes the value 4 for r = = and

the value § forr = 5171. This shows that 5s(E, 0) = Tim {u, (EN BO, 7))/u, (BO, 7))} does not exist.

©

Example 2. Let E = (0, 00) C R. Consider two collections € and &’ in Sg that shrink

nicely to 0 € R given by € = {(—r, 0) : r € (0, 00)} and &’ = {(0,r) :r € (O, 00)}. Then

_ Bi (EN(r,0))

5e(E, 0) = lim —*+—__—+

e(E,

0) = lim

#((7,0)

=

(9)

—-—

70 4, ((—r, 0))

= 0,

Ser(E,0)= tim MEMO") _ pg Hel") _ 79

(0,7)

79, (7)

Theorem 25.28. Let E € St? and x € IR". Suppose for every collection & in Syn that shrinks nicely to x the density 5¢(E, x) of E with respect to x} at x along E exists. Then 5¢(E, x) does not depend on E, that is, if & and &" are two collections in Syn that shrink nicely to x then 5g/(E, x) = Sgv(E, x). Thus, in this case, for every collection & in By that shrinks nicely to x, we have 5g(E, x) = 6s(E, x). Proof. By Definition 25.25, 5g(E, x} = (Dev)(x) where v is the set function on (BR), defined by v(B) = wi (EN B) for B € (Brey. Thus our theorem is a particular case of

Theorem 25.20.

For every E € 29%, we have lr € £},,(IR", 9", 2").

Consider the Lebesgue set

A(1g) of the function 17. We show below that if x € A(1z) then for every collection € in Sp» that shrinks nicely to x the density 5g(Z, x) exists.

§25 Differentiation on the Euclidean Space

661

Proposition 25.29. Let E € I}. Then (a) E° U (E°° c A(1z). (b) 8E 5 A(z).

(©) IfE is a null set in (R", 99%", 2"), then E Cc A°(1z).

Proof. By Observation 25.14, if f < £},,(R", 9%*, 4") is continuous at x € IR” then

xeéA(f). Let Ee mt. Ifx € E°, then 1g is continuous atx so that x € A(1z). Thus E° c A(1ig). Similarly ifx € (£°)°, then 1g is continuous at x so that x € A(1z) and

thus (E°)° C A(1z). Therefore E° U (E°)° C A(1g). Since 2E = (E° U (E*)°)", we have 9E D A°(1z).

Suppose E

IfE £G,letx

is a null set in (R”, 0t", uw"). IfE = @, then E C A°(1z) holds trivially. € E. Then 1g =0 ae. on R” and 1¢(x) = 1 so that

=

1 lin ———— jim (BG) 1 lim ———

(a ste | 1 £(y) —-112(@)| u(y) Lu"

(dy)

=140.

7b ue (BG,P) Iocan MeO? = 1 #

This shows thatx € A°(1g). Therefore E C A°(1z).

#

Theorem 25.30. Let E € mr and x € A(1p), the Lebesgue set of the function 1g. Then

for every collection & = {E,(x) zr € (0, 00)} in Spe that shrinks nicely to x, the density d¢(E, x) exists. Moreover we have

i

BE(ENE,(@)) _

be(B, x) = Te

ur(E,@))

1 ifxe ENA(s),

|

0 ifx e E°N A(z).

Proof. If E € 9", then ly € £},,(R", 9, 2”). By (2) of Theorem 25.17, for every

point x in the Lebesgue set A(1z) and every collection {Z,(x) : r € (0, 00)} in Bye that

shrinks nicely to x, we have

B(ENE,(x))

730 wi(E-@)) im At

= Li

1

70 ut(E-@)) Je MR

1g) wy @y) = 1g).

Thus 5¢(E, x) exists and 5¢(E, x) = 1g(x). Therefore dg(EZ, x) = 1 forx € EN Ag) and 6g(Z,x) =Oforx ¢ E°N A(z). ©

Corollary 25.31. LetE € 990? andx € R". Then for every collection {E,(x) : r € (0, 00)} in Bae that shrinks nicely to x, we have

_,,

e(Bo 2) =

MA(ENE(e))

ur(E-@))

|

[1 fxek’,

0 x

(Ey.

662 Proof.

Chapter 6 Measure and Integration on the Euclidean Space By Proposition 25.29, ifx € E° then x € EM A(1g) and similarly ifx € (E°)°

then x € E°M A(ig). Thus the Corollary follows from Theorem 25.30. Corollary 25.32.

(Lebesgue Density Theorem) Let E € 20].

Then for the symmetric

density 55(E, -) ofE, we have

1

8s(E, x) =

@

3B,

x)

1

forx€

EN A(z),

| 0

forxe

E°N A(1g),

and therefore @)

1

2

forae. xe E,

63(E,x) = s(E. x) | 0 foraex€

E*.

Proof. Since {B(x,r) : r € (0, 00)} is a particular case of a collection in Sy» that shrinks

nicely to x, (1) is a particular case of Theorem 25.30. Then since u(A°(1z)) = 0 by Theorem 25.15, (2) follows from (1). In fact if we compare E with its subset EM A(1z),

then E\ (EN ‘o3}} =EN(En wate = EN(E°UA*(1g)) = EN A“(1z) so that we (E\ (EN Adz))) Remark 25.33.

= 0. Similarly BE(ES\ (E°N Adg))) =0.

Let E ¢ St? and x € A(1z).

According to Theorem 25.30, for an

arbitrary collection € = {E,(x) : r € (0, 00)} in Sg» that shrinks nicely to x, 8¢(E, x) always exists and is equal to 1 or0 according as x € E orx € E°. Ifx ¢ A(1z), d¢(E, x)

may still exist for some particular collection & = {E,(x) : r € (0, 00)} in Sg» that shrinks nicely to x.

Example 1. (In this example 5¢(£, x) exists for some x ¢ A(1z).) In R, let E = [0, 00). Then R \ {0} = E° U (E°)° c A(1z) by Proposition 25.29. On the other hand we have

lim ——__ is (BO)

bin oy1 [t-te

= lim —

non | 1z() E() ——120)| 12(0)|

ud#,(@y)

ane)

1-1

= lim_ 21 [2 - [ way} =5 1 #0. Thus 0 ¢ A(1z) according to Definition 25.13.

Nevertheless for the collection in By,

§ = [B(0,r):r € (0, 0o)}, that shrinks nicely to 0, we have

53(Z, 0) = lim

(EN BO,r))

,(BQ,r))

im’

Example 2. Leta € (0, 1) and let 3 € (0, 27) be such that zx

the plane R? sustaining an angle # at 0 ¢ R?. Then we have 2(EN BO.

5(E,0) = tim Mi r>0

=!

ror 2" =a.

BO.7)) _ lim” =a.

we (BO, r))

13027

Let E bea sector of

§25 Differentiation on the Euclidean Space

663

Since a is neither 1 nor 0, we have 0 ¢ A(1z) by (1) of Corollary 25.32. Proposition 25.34. Let E be the Cantor ternary set in R. Then we have A(z) 5g(E, x) exists and 53(E, x) = O for every x € R.

= E° and

Proof. By Theorem 4.31, £ is a closed set in R and is also a null set in (R, Sp, 4,). Thus E* is an open set and therefore E° = (£°)° C A(1z) by (a) of Proposition 25.29.

Since

E is a null set in (R, 91, u,), we have E C A°(1z) by (c) of Proposition 25.29. This

implies E° > A(1g) and therefore A(1z) = E°.

Since E is a null set in (R, 9,, 4,), EO BG, r) is anull set in (R, Mt, , w,) for every

x € Randr > 0. Thus u,(EN B(x,r))/m, Pe r)) = 0 and consequently we have 43(E, x) = Tim (4, (EN B(x, r))/4,(B@,r))} =0. ©

Theorem 25.35, Let E € 200". Let D(5s(E, -)) be the domain of definition of 5s(E, -).

Then D(8s(E, -)) € Mt", 5s(E, -) is Wt" -measurable on D(8s(E, -)) and f 6s(E, y) ui (dy) = wp (E).

Proof. 1. We have D(5s(E, -)) = {x € R" ; 6s(E, x) exists}. By Theorem 25.30, 3(E, x) exists for every x € A(1xz). Thus we have

a)

D(8s(E,)) > Adz).

Let (2)

A= D(855(E, )) N A(z).

Then A(1gz) NA = @ and

(3)

D(6s(E, )) = AUz) UA.

By Theorem 25.15, A(1z) € ot

and A°(1,) is a null set in the complete measure space

(R", Mt", w7). Thus the subset A C A°(1z) is a 90t"-measurable set. Then by (3), D(6s(E, -)) is a 90t"-measurable set. 2. Let us show that 5g(E, -) is 90t"-measurable on D(8s(E, -)). By (3), D(6s(E, -)) is the union of two disjoint Ot} -measurable sets A(1z) and A.

Thus it suffices to show

that D(8s(E, -)) is t?-measurable on each of these two sets. Now A(1z) is the union

of two disjoint 9t?-measurable sets EM A(1z) and E°M A(1z) and 4s(£, -) assumes the constant value 1 on the first of these two 9)t?-measurable sets and the constant value 0 on the second according to Theorem 25.30. Therefore 5g(Z, -) is

t"-measurable on A(1z).

Since A is a null set in the complete measure space (R", mt,

ur), every subset of A is a

20t} -measurable set and then every extended real-valued function on A is 29t/-measurable on A. In particular és(£, -) isOt} -measurable on A.

664

Chapter 6 Measure and Integration on the Euclidean Space

3. We have R? = {EM A(1z)} U {E°N A(1z)} U A°(1z), a union of three disjoint 99t" measurable sets. Now A°(1g) is a null set in (R”, 99t", ”). Thus we have

© — fweran=f Re = [

ENA(iz) ENA(z)

erate fase. aut ENA(z) lap? + [

EnAGz)

Ody"

= un(EN A(z). Now EN A(ig) = E\ A°(1z) = E\ (EM A(1g)) and since EN A°(1z) is a null set in (R", 91", 4") we have w"(EM A(z) = w"(E) — u®(EN A°(1g)) = u2(E). Substituting this in (4), we have

fg, 6s(E, y) u? (dy) = wi (E).

&

[V] Signed Borel Measures on R” Definition 25.36. (a) We call a measure on the a-algebra SiR of subsets of BR" a Borel measure on R*. (b) We say that a Borel measure v on R" is outer regular if for every E € By» we have

v(E) = inf {v(O) : O > E and O is an open set}.

(c) We say that a Borel measure v on R° is inner regular if for every E € Bx we have

v(E) = sup {v(K) : K C E and K is a compact set } .

(d) We say that a Borel measure v on R" is regular if it is both outer and inner regular. (e) By a signed Borel measure on R" we mean a signed measure on Brn.

(f) We say that a signed Borel measure v on R” is outer regular, inner regular, or regular if both its positive part v* and its negative part v~ are outer regular, inner regular, or regular respectively.

If we restrict Lebesgue measure we on mM

to Br, then we have a Borel measure on

R® which is outer regular by Proposition 24.17 and inner regular by Proposition 24,16, Note also that «7 is finite on the compact sets in IR" since a compact set in R" is a bounded set in R”. Observation 25.37. Let v be a signed Borel measure on R” which is finite on the compact sets in R”,. Then we have: {a) v is o-finite signed measure on BR». (b) Ifv< we on (R’, Spr), then every version dun of the Radon-Nikodym derivative of

v with respect to 2” is in £},(R", Sm, uw"). Proof.

1.

For every k € N, the closed ball B(0, k) is a compact set in R” and thus

v(B(O, k)) € R. Then since R* = (ey BO, k), v is a o-finite signed measure.

2. Suppose v < yz? on (R", Bp). Since py? is a o-finite positive measure, v < pu? implies the existence of the Radon-Nikodym derivative of v with respect to 4? by Theorem 11.16. Take an arbitrary version of the Radon-Nikodym derivative a . Then for every L

§25 Differentiation on the Euclidean Space

665

k EN, we have S50, sk) By oe due du" = v(B(O,k)) € R since B(O, ~ k) is a compact set in R”.

Since every bounded Borel set £ in R” is contained in some B(0, k) with sufficiently large

k and since v (B(0, &)) € R, wehave f,, #% du = v(E) € R by (a) of Lemma 10.5. This a e £1 (R", Br, un). shows that ae

a

L

Definition 25.38. Let v be a signed Borel measure on R" which is finite on the compact sets in R". We define the maximal function Mv of v by setting for x € R"

My)(x)

Ie] (BG, 7) au

=

HOO) = ee) uh (BCx, 7)

where | v| is the total variation of v, that is, |v| = vt + v~ and v* and v~ are the positive and negative parts of v. Proposition 25.39. The maximal function Mv of a signed Borel measure v on R” which is finite on the compact sets in R" is a lower semicontinuous function on R" and in particular it is a Bpx-measurable function on R". Proof. Since Mv is a nonnegative extended real-valued function on R", to show that it is lower semicontinuous on R’ it suffices to show that the set Dy = {R” : Mv

> a} is an open

set in IR" for every a > 0. To show that Dy is an open set, we show that for every x € Dy

there exists 6 > 0 such that B(x, 5) C Dy. Nowifx

€ Dg, then (Mv)(x) > a so that there

exists p > 0 such that |v|(B(x, »)) = Bu2(B(, p)) for some B > a. Let 8 > 0 be so small that (1 + 3) < a)", or equivalently, (p + 6)" < pre, Consider B(x, 6). To show B(x, 8) C Da, fet y € B(x, 8) be arbitrarily chosen. Then B(y, p +4)

by the translation invariance of 47 and by Proposition 24.34 we have

> B(x, p) so that

lvl (BO, 0 +8) 21 v1 (BG,0) = Bui (BG. 0))

= Ba) MBO. p+) > aut (Bly, o +8), and therefore

(Mv)O) = r€(0,00) sup

| (Bo )) v 7 un(By, r))

a.

This shows thaty € Dy. Thus B(x, 8) C Da. Theorem

25.40. (Maximal Theorem

for a Signed Borel Measure)

For the maximal

function Mv of a signed Borel measure v on R" which is finite on the compact sets in R", we have

ut {R": Mv > a} < = oi(e")

fora € (0, 00).

666

Chapter 6 Measure and Integration on the Euclidean Space

Proof. For a € (0, 00), let Dy = {IR" : Mv > a}. Then for everyx € Dy, we have (Mv)(x) =

sup PI@G”)

>

re(0,oo) #7 (BC, r))

so that there exists r, € (0, 00) such that |p|(B(x, r,)) > «47 (B(, rz). Consider the

collection of open balls € = {B(x,r,) : x € Dg}. Since Usep, B(x, rz) is an open set, it is a Sya-measurable set. By Lemma 25.10, for every real number c such that

c < 2"(U,en, BC, rx)), there exists a finite disjoint subcollection (By, ..., Bp} of € such that

< Dd» uty

and therefore

(6))

Dgv =

dv dyn

ae. on (R", Bre, 2”).

Proof. If v is a signed Borel measure on R” which is finite on the compact sets in R* and ifv < wz on aR’ BR), then by Observation 25.37, v is a o-finite signed measure on

(R", Bp), and4di

€ £},.(R°, Br, 4”). This implies according to Theorem 25.17 that that shrinks

nicely to x, we have dv

dun

.

1

(x)= lim ——__

70 w(E,(x))

=

Jz, ann

(y) wy) =

HL) =

This proves (1).

tim UE)

wh(E,(x))

With 8 = {B(x,r): 7 € (0,00)} asacollection in Br thatsb have (2) from (1). By Theorem 25.15, A°(#; dur

nicely to x, we , ut). By Lemma

24,11, there exists a Gp-set G D A°( such that 8(G) = 0. Now Ge cAte =). Thus by (2) we have (Dsv)(x)= # ¥-(x) for x € G°. Since G € BR, (3) holds.

a

Theorem 25.42. Let v be an outer regular signed Borel measure on R" which is finite on the compact sets in R". If v 1 ® on (IR", Bm), then for a.e. x in (R", MY, w") and for every collection & = {Er (x): r € (0, co)} in Bye that shrinks nicely to x, we have @

im

2Er®)

(Dev)() = am 750 un (E(x)

In particular for a.e. x in (R", I", u"), we have @

sv) (x)=

him “BO: r))

am 750 ws (BG 7)

and therefore QB)

Dgv =0

a.e. on (R’, By, ur).

Proof. Let |v| be the total variation of v, that is, |v] = vt + v~ where v+ and v7 are the positive and negative parts of v. Now v 1 x” on (R", Spe) implies |v] | 4% on

668

Chapter 6 Measure and Integration on the Euclidean Space

(R", Bx) by Proposition 10.27. Thus there exists a set A € Se such that |v](A) = #7 (A‘) = 0. Since v is an outer regular signed Borel measure on Br, vt and v~ are outer regular positive Borel measures on 93 and then so is |v]. For brevity, let us write

_ [email protected])

(2;lvI)@) = “u2(BO.D)

so that

B(x, (Dstot) 09 = ig EEC) — tn. (Q, v1) 00. 7 >

To prove (1), let us show first that for a.c. x in (R”, 99", z”), we have

4)

Ds(Iv1)@) = lim (0, |v])@) = 0.

To prove (4), it suffices to show that for a.e. x in (R”, 97%", 12"), we have

(6)

lim sup (Qr[1)(@) = lim,

sup

(,|v])@) =0.

Note that supo0

the union |_),.1y Fx, to prove (6) it suffices to show that ue (F) = 0 for every k ¢ N. Now

since |v](A) = 0 and since |v] is an outer regular Borel measure on Sq", there exists an open set O, > A such that |v](O,) < ¢. For each x € Fy C AC Ov, there exists an open ball B, with center at x such that B, C O, and |v] (Bx) > pun (B,) by the definition of the set F,.

Consider the collection € = {B, : x € Fj}.

Since Uren B, is an open set,

Uxer, Br € Bre. Ife € Randc < yu" (Uzcn, Bx), then by Lemma 25.10, there exists a

disjoint subcollection {B,,,..., By} such that Pp

p

j=l

j=l

c< 3" > uM(By,) < 3k >

B

LIB) =3"elvl(U Bs,)

< 3k v1( U Br) 0, we have 4” (Un, Bx) = 0. Since Fy C Uxen, Bx, and since

§25 Differentiation on the Euclidean Space

669

(R", M9", 7) is a complete measure space, Fy € Mt" with 4"(F,) = 0. This proves (6)

and completes the proof of (4).

By (4), there exists a null set

in (R", 997, 4”) such that lim (Q,|v1)@) = 0 for r>

x € E*, Let € = {E,(x) : r € ©, 00)} be acollection in Sg» that shrinks nicely to x.

By the fact that E,(x) C B(x, r) and the fact that there exists a constant a > 0 such that

ap" (B(x, r)) < u"(E,(a)) < u2(B(x, r)) forr € (0, 00), we have

WIG) _ BG.) - LIB.) _ © (Orble. un(E,(@)) ~ wR (E,(x)) ~ o (B(x, r)) Thus forx € E°, we have

tim PIE) _ 9 70 u"(E,(x))

(7)

Since |v(Z,(x))|_ O such that f is

Hy -integrable on B(x, rz).

Prob. 25.2. (a) Show that if f is a real-valued 9)t/-measurable function on R” and f is

bounded on every bounded subset of R”, then f € zhae (R”,U7, 4"). (b) Show that if f is a continuous real-valuedfunction on R”, then " € G(R", Dt, wu).

Prob. 25.3. Definition. Let {h, : r € (0, 00)} be a collection of real-valued functions on R*. Consider {h,(x) : r € (0, 00)} forx € R”.

(a) We say that lim h, (x) = & if there exists & € R such that for every rt

> 0 there exists

5 > O such that |h,(x) — &| < e forr e€ (0, 4).

(b) We say that lim hr (x) = 00 iffor every M > 0 there exists 8 > 0 such that h,(x) > M forr € (0,6).

r>'

(c) We say that lim hr (x) = —0o iffor every M > 0 there exists 5 > 0 such that h,(x) < r—>

—M forr é€ (0,6). (d) We define lim sup A,

oe

I

7) = Hg { sep

Observation. Observe that always

¥

exists in R. Similark

h,

Go}

and

lim inf h

r(x)= fim fin {, int

sup h,(x) decreases as 8 —> 0 so that Jim {

re(0,8)

y

int, yf (x) increases as 8 —> 0 so that Tim {

A-()}. sup h,(x)}

re(0,4)

inf_h,(x)}

re(0,8)

always exists in R. Therefore we ‘have ——oo < liminf,_,9 A, (x) < limsup,_,9 h; (x) < 00.

Theorem. Let {h, : r € (0, 00)} be a collection of real-valued functions on R". Consider {h, (x): r € (O, co)} for x € BR’. (a) iim hy (x)= & € R ifand only if lim inf h, (x) = Jim sup h(x) = &

(b) iim h, (x)= 00 if and only if liming

(x) = limsuphy(x) =

(c) lim hy (x) = —00 ifand only i Timing, (s) ==fimsup

(x) =

Prove the | theorem above.

Prob. 25.4. Let f be a real-valued function on R defined by f(x) = 0 if x € R is rational and f(x) = 1 ifx € Ris irrational.

(a) Show that f € £1,.(R, Dt, u;). (b) Find Af.

(©) Show that Af = f ae.on (R, Mt, u,). Prob. 25.5. (a) Show that if f ¢ £'(R", 990”, uv”), then for each r € (0, 00) the function A, f is uniformly continuous on R”.

(b) Show that if f 0, & > 0} C R?, where (1, £2) isan orthogonal coordinate system in R? . Show that A(z) = (@E)°. Prob. 25.10. Prolog. Consider (R", 907, 47). a collection in 3»

Forx € R” we write & = {E,(x) : r € (0, 00)} for

that shrinks nicely to x. Note that 5 = {B(x,r):

particular case of €. Let E € 9)t'. By Definition 25.25 we have

r € (0, co)}isa

5e(E, x) = 70 Yim iC) and in particular &3(E, x) = 70 tim MiCE™#@9) wt (z-G) ut (BG,7)) provided that the limit exist. By Theorem 25.28, if 5¢(E, x) exists for every € then 5g(E, x) is independent of €. By Theorem 25.30, if x € A(1z) then 6g(E, x) exists for every € and hence 8g(E£, x) is independent of €. In particular, if x € A(1z) then 53(£, x) exists.

Problem. Consider (R?, 917, 12). Let E = {(1,&) € BR? : & > 0,8 > 0} Cc R’,

where (£1, &) is an orthogonal coordinate system in R?. By Prob. 25.9, 3 is the union of the positive £-axis and the positive &-axis in the plane R? and moreover A(1z) = (8E)°. (a) Let x1 = (1,0) ¢ A(z).

(a.1) Show that 8s(E, x1) = 4. (Thus 8 is an example of € such that 5g(E, x1) = 3.) (a.2) Construct a collection € such that 5g(E, x1) = 1. (a.3) Construct a collection € such that dg (EZ, x1) = 0. (a.4) Construct a collection € such that 5g (EZ, x1) does not exist.

(b) Let xo = (0, 0) ¢ A(1z). Show that 4s(Z, xo) = i.

Prob. 25.11. Leta € (0, 1) and let ¥ € (0, 277) be such that z

=a.

Let E be a

sector of

the plane R? sustaining an angle @ at 0 € R?. Find A(1g). Determine 4s(E, -) and show

672

Chapter 6 Measure and Integration on the Euclidean Space

that 5s(E, x) exists for every x € R2. Prob. 25.12. a>

Consider (R”, D7, wt). Let E = B(0,a) = {x € R® : |x| < a} where

0. Find A(1z). Determine 5s(£, -) and show that 5s(£, x) exists for every x € R”.

Prob. 25.13. Let E be the set of all rational numbers in R. (a) Show that A(1z) = E°. (b) Show that 5s(Z, x) exists for every x ¢ R and és(F, x) = 0.

Prob. 25.14, Let E be the set of all irrational numbers in R. (a) Show that A(1z) = E. (b) Show that 5s(Z, x) exists for every x € R and és(F, x) = 1.

Prob. 25.15. According to Proposition 25.29, for every E € 207 we have 9E D A°(1z). Construct an example such that EF \ A°(1z) 4 9.

§26 Change of Variable of Integration

§26

673

Change of Variable of Integration on the Euclidean Space

[1] Change of Variable of Integration by Differentiable Transformations Consider a mapping of R” into R” given by y = T(x) € R" forx € R". IfT is atranslation,

that is, T(x) = x +h for x € R” where h € R” is fixed, then by Theorem 24.28, for every

nonnegative extended real-valued 20t?-measurable function f on a set D € Dt? we have fof) #2 @y) = Sram o T)(x) #2 (dx). Now since not only T\(D) € ON? but also T(D) € 207 for every D € Mi by Theorem 24.27, we have

[ pf OmaD = [ (F oT)y) way) for every E € DU. If T is a non-singular linear transformation of R” into R”, then T(D) € D0? for every D € Mt? by Theorem 24.31 so that Theorem 24.32 can be restated as

f

T(E)

F@) up (dx) = [det Met f (F oT) wydy) E

for every E € IN". These are examples of change of variable of integration. We consider next a change of variable by a nonlinear but differentiable transformation. Let Q be an open set in R” and let T be a one-to-one mapping of @ into R” whose component functions have continuous first order partial derivatives on &. We shall show that for an extended real-valued 2T?measurable function f on &, we have

f

TQ)

f@) ui (dx) = f (Ff 0 T)(y)| det JrQ)| wi dy) Q

where Jr is the Jacobian matrix of the transformation T. In preparation for this change of variable theorem, let us review some facts in the calculus of several variables.

Differential of a Mapping from R” into R”. Let T be amapping of an open set Q in R” into R™. We say that T is differentiable at a point p € Q if there exists a linear transformation L of R’ into R” such that if we let

@

R(p; h) = T(p +h) — T(p) — L@)

forh € R",h # 0, such that p +h € Q, then

2)

R(p;h) _ im a =0eR”.

We call the linear transformation L the differential of T at p and write dT(p; -) for it. If dT (p;

QB)

-) exists, then for every u € R" with |u| = 1, we have

dT (p, 4) = him

T(p + Au) — T(p)

2

674

Chapter 6 Measure and Integration on Euclidean Space

We call dT (p; u) the directional derivative of T at p in the direction w. If T is differentiable

at p, then T is continuous at p.

Uniqueness of the Differential. If the linear transformation dT (p; -) satisfying conditions (1) and (2) exists for some p € Q, then it is unique.

In particular, if T itself is a linear

transformation of R” into R”, then for every p € R" the linear transformation dT (p; -) exists and dT(p; -) = T. Existence of the Differential. For R” = R, x--- x Ry, let 7; be the projection of R™ onto R;, that is, 7;(q) = q; forg = (q1,..-,4m) € R”. For a mapping T of an open set Q in R* into R™, let (gi, ..., gm) be its component functions, that is, g; is a real-valued function on Q defined by g; = x; 0 T on Q. If all the partial derivatives 5£ for i =1,...,m and j=1,...,

@)

exist at p € Q, then we call the m x n matrix

Be) Jr@=|: Hep)

EO) : e(p)

the Jacobian matrix of the mapping T at p. We say that T is a mapping of class C1(@) if all the partial derivatives se

fori =

1,...,m andj = 1,...,# exist and are continuous on &.

If T is a mapping of class C!(), then T is differentiable at every p € Q and furthermore the Jacobian matrix Jr(p) is the matrix of the linear transformation dT (p;

-).

Chain Rule for Differentials. Let T be a mapping whose domain of definition D(T) is an open set in R” and whose range $4(T) is a subset of R”. Let S be a mapping whose domain of definition D(S) is an open set in R” containing X(T) and whose range 94(S)

is a subset of R&. Consider the mapping S o T of D(T) into R¢. If T is differentiable at some p € D(T) and S is differentiable at T(p) € D(S), then S o T is differentiable at p

and its differential at p is given by

6)

d(S oT)(p;-) = dS(T(p); -) 0 dT (p; -) = dS(T(p); aT (p; -)),

and, assuming the existence of the first order partial derivatives of the component functions of T at p and those of S at T(p), the Jacobian matrix ofS o T at p is a product of two matrices given by (6)

Isor(p) = Is(T(p)) Ir(p).

Proofs of these statements can be found for instance in R. C. Buck [5]. To show that if T is a mapping of an open set Q in R” into R” of class C1(Q), then T(E) is a null set in (R", Dt", 2”) for every null set E in (R", Mt", 2”) contained in Q, we prepare a covering theorem for a null set in (R", nt, un) in the following Lemma. Lemma 26.1, Let E be a null set in (R", 9907, 2"). Then for every ¢ > 0 andn > 0, there exists a countable collection of open balls {B@;, rpijeé€ N} such that x; € E,rj 0, O is acountable disjoint union of n-dimensional cubes with lengths of the edges less

than n—'/2n, Drop from this collection of cubes those which are disjoint from £. Let the

remaining disjoint countable collection be denoted by {Q; : j € N}. We have jen Qj 2 E. Let a; be the length of the edges of Q; for j ¢ N. We have aj < n—'/2n, The length of the diagonal of Q; is equal to n/a; 0. By (3) of Proposition 26.10 and Theorem 23.17 (Tonelli), we have

L en atei +33} we (dG, x2)) =

{0,00)x(—m,7)

en, 2 (d¢, #))

=[ [ n,(4)][ [ vag

bak a).

686

Chapter 6 Measure and Integration on Euclidean Space

The integrand in the second factor is nonnegative. Thus computing the Lebesgue integral as an improper Riemann integral, we have [.

re

ar?

CO

u, (ar) -[

re

[Pepe dr=bga jim,[e Fag le = on

Substituting this in the previous equation, we have

@)

[‘R2 oobi} 2 (d(x, m2) = =.a

Now by Theorem 23.17 (Tonelli), we have

—afxf+x3} ,,2 [ ott a (da, 22)

=

[ po

ax?

xd (@x)][ oe

(dx2)].

The last two factors are identical. Taking the square root and recalling (1), we have

@)

[ ey, (ds) = i

Then

@B)

n [ ete? undx) =T][ [ eeu, dx] = (")" i=l

by Theorem 23.17 (Tonelli) and (2). In R? the orthogonal coordinates (x, y) and the polar coordinates (r, #) are related by x =rcos?, | y=rsind,

for (r, 8)

€ (0,00) x (—x, 7) and in R® the orthogonal coordinates (x, y, z) and the

spherical coordinates (r, y, #) are related by Z=rcos@, x =rsingcosd,

y=rsingsind, for (7, g, ®) € (0, 00) x (0,7) x (—7, 7).

follows.

We generalize these to R" for all mn > 2 as

Proposition 26.11. (Spherical Coordinates in R") For n > 2, let Q, be an open set in R” defined by Q2 = (0, 00) x (—7, 2),

()

Ms = (0, co) x (0, x) x (—z, 7),

Qq = (0,00) x 0, 2) x «++ (0, 2) x (Em, #).

§26 Change of Variable of Integration

687

Note that the first factor in &,, is (0, oo), the last factor is (—z, 2), and every factor in between is (0, 2). For the coordinates of a point in Q,,, we write (7, 1, «++ 5 Pn—2; On—1) € (0, 00) x (0, 7) x --- x (0,2) k (HH, 7) = Qn. For n > 2, we define a mapping T, of Q, into R” inductively as follows. For n = 2, let

(2)

Th(r, g1) = (cos gi, 7 sin gi),

for (7, g1) € Q2. Forn > 3, let

(3)

Ta, Or «++ Gn—1) = (r cos G1, Tri sin g1, G2, ---, Gn—1)),

for (7, 91, .-., @n—1) € Qy. We write (x1, ..., %n) for the component functions of T,. Thus by (3) and (2), we have

4)

T3(r, g1, 92) = (r-cosg1, Tar sin gi, g2)) =(r cos 91, r siny1 COs @, 7 sin g sin g2)

so that

x1 =rcos gi, x2 =r sin1 cos 92, x3 =r sing) sin g.

Similarly by (3) and (4), we have Ta(r, 1, G2, 93) = (r cos yr, T3(r sin g1, go, Y3)) =(Fr cos 91, r sin @1 COS gy, r sin ¢] sin ¢2 COs Y3, 7 Sin G1 Sin 2 sin 3) so that

x1 =Prcosgi, x2 =r sing cos g, x3 =T sing sin 2 cos ¢3, x4 =r sing Sin g

sin 3.

In general for T,,, we have x1

=Trcos 1,

x2 =r sing] cos $2,

(5)

3

COS 3 #1 sin y cos

x3 =P sing sing)

¢3,

Xn-1 = 7 SiN Q] - - SiN G__2 COS Gy_1,

Xn = 7 sinpy +++ SiN Gp—z SiN @p-1,

688

Chapter 6 Measure and Integration on Euclidean Space

that is, with the understanding []}_, sing; = 1, we have i-l x =rcosg; | [ sing;

(6)

j=l

From (5), adding x2 + xy

fori =1,...,2-—1,

and

n-1 wn = rT] sing;.

j=l

teeet x? we have

n

1)

vie ear’. i=l

Let us define

®)

An = {x € R” : x, = Oand x,-1 € (00, O]},

An = R"\ An = {x € R" : x #0 or xn_1 € (, 00)},

that is, A, is that half of the n — 1 dimensional hyperplane in R” determined by equation Xy, = 0 which contains the negative x,,_1-axis but not the positive x,_1-axis. Let us show that 7, maps (,, one-to-one and onto Aj. It is easily verified from (5) that ify = (7, G1, ..-, On—1) € Qu, then x = T,(y) € R". To show that x ¢ A,, let us note first that for sing; and cos yj forj = 1,...,2 — 1 in S), we have

sing; € (0, 1) sin @—1 € [—1, 1],

@)

cosy; € (—1,1) cos Y,-1 € (—1, 1].

for j =1,...,n —2, for j =1,...,n—2,

If x € Ag, then x, = 0 which implies sing,_1 = 0 by (5) and (9) and then g,_1 = 0 so that cos g,-1 = 1 and thus x,_1 > 0 by (5) and (9), a contradiction. Thus x ¢ A,. This shows that 7, (Qn) C An.

It remains to show that for every x € A,, there exists a unique y € @, such that

T,(y)

= x.

Let us prove this by induction on x.

Now for n = 2, it is obvious that for

every x € Ag, there exists a unique y € 2 such that 72(x) = y. Suppose the assertion

is true form — 1 where n > 3. Letx = (x1,...,x,) € Ay, be arbitrarily chosen. Let r =|x| > 0. Then a € (—1, 1) since x, 4 0. Thus there exists a unique 3; € (0, 7) such thatr cos 3}; = x1. Now since (x2, ..., x.) € A,—1, by our induction hypothesis there exists a unique point (p, #2, ..., Pn—1) € Qn_1 such that T,_1(p, 82, ..., On—1) = (x2, ..., Xn)»

Then we havep = {x3 +--- + x2}? = {r?- xy? = {r’ — 7’ cos” a}? =rsind,.

This shows that 7,1 maps the point ( sin 1, 32, ..., 0,1) to the point (x2, ..., Xn). Thus according to the definition of T,, by (3), J, maps the point (7, 1, ..., In—1) € Qn to {x1,..+)4n) € An. From (5), it is obvious that all the first order partial derivatives of x1, ..., x, with respect to the variables r, 1, ..., @»—-1 are continuous on the open set 2,. Thus 7, is a class c 1(Qn) mapping of , into R". As we showed above 7,, maps &,, one-to-one and onto Ap, = R* \ Ap. Note that Ap, being a subset of an m — 1 dimensional subspace of R”, has

“3 (An) = 0. Expanding the Jacobian determinant det Jz, along the first row, we obtain

§26 Change of Variable of Integration

689

det Jr, (7, 91, -.-, @n—1) = rsin"-? gi Jp,_,(7, G2, ..-» Pn—1). Using this recursion formula and recalling that det Jz, (7, 91) = r, we have

an

(10)

det Jr, (7, G1, «++» Gn—1) =r" 1 TT] sin" J 9. j=l

Let us note that det Jz, > 0 on Qa. Observation 26.12. The volume of a ball in R” can be expressed in terms of the gamma function. The gamma function is a real-valued function on (0, 00) defined by

0)

T@) = f

(0,00)

x"1e* 4, (dx)

foru € (0, 00).

It has the properties that (2)

Tw+1)=al(%)

(3)

rd) =1,

4

r=,

foru € (0, 00),

and in particular

6)

Tan+1D=m!

formeN.

Proof. 1. Let us show that '(z) € (0, co) for wu € (0, 00). For fixed uw € (0, 00), the integrand in (1) is a strictly positive continuous function of x € (0, 00) so that '(u) > 0. To show that '(%) < 00, let us note that

|

Since Then [

*

.1]

xtle

Be, (dx) 0.

[1,00)

x le* pdx)

= f

[1,co)

xttle-*y—2 y (dx) < Bf

[1,00)

x? p, (dx) < 00.

Therefore "'(u) < 00.

2. To prove (2), note that since the integrand in (1) is nonnegative, the Lebesgue integral can be evaluated as an improper Riemann integral. Then by integration by parts, we have

wa=

Tet

=f

‘OO

= uf

x“e* dx oO

0

[H(-

=|x(-e*

xt let dy = ul (u).

I]. + oO

+f

oO

1

ux"—*e™ dx

690

Chapter 6 Measure and Integration on Euclidean Space 3. M1) ={~ e*dx =1. 4. With a change of variable of integration x = y? for y € (0, 00), we have

rays

Lr 0

x We-* dy -[° ye" 2y dy = af” oe” dy = Jn. 0

0

5. (5) follows by iterated application of (2).

Fora > Oletus write B? for {x € RR” : |x| < a}, an open ball with center at 0 € R” and

radius a. Let BY be the closure of B%, that is, B? = {x € R” : |x| 3.

For an arbitrary n > 3, let us write R" = R? x R’~?. For an arbitrary point (x1, x2) € R?, consider the R?-section of B? at (x1, x2), that is,

(4)

BR(1, x2,-) = {3,-0.4 tn) ER"? : Gr, an) © BEY = {G3,...,4n) € RY? rxf +--+ +32 2, let us define a measure pu

a

on (@, 00), BQ,00)) anda

measure oft on (3, Bn-1) by setting

a)

pe) = f rly, (dr) forE € Be);

2)

of *F) =n(uz o@")(O, 1] x F) for F € By. The set function p*—! on 38 (0,00) defined above is a measure by (c) of Proposition 8.11.

The fact that the set function ot! on B st defined above is a measure is verified directly.

In particular the countable additivity of of =! on 3 sit follows from the countable additivity of 47 on BR\\0}Theorem 26.18.

Forn > 2, consider the measure space (R” \ {0}, Brey oy, ut) and the

measurable space ((0, 00) x set 0 (Be) x B sp-))- Let v be the image measure of pu? on the o-algebra o (280,00) X B set) by the mapping © of R” \ {0} into (0, 00) x St}, that is, v(G) = (2 0 &-) (G) for G € o (Be) x B sot). Then v =p"! xo? on

6 (Be) x By).

Proof. The set function v on o (380,00) x B 9-1) is a measure by Theorem 1.44 (Image Measure). Let us show that v is identical with the product measure pt!~x 2(Be,v) x Bg).

ont

on

694

Chapter 6 Measure and Integration on Euclidean Space Let 3(0,00) be the collection of @ and all intervals in (0, 00) of the type (a, A]. Itis easily

verified that 3(0,00) is a semialgebra of subsets of (0, 00) and ¢(3(0,00)) = Bo,00)Lemma 23.2, 30,00) x 3 st is a semialgebra of subsets of (0, 00) x st

BY

Furthermore

&(J(0,00) x By-1) = 6 (2B (0,00) x Bsn-1) by Lemma 24.23. Therefore to show that v = pe?! xo?! on the o-algebra o (3 (0,00) x B sot), it sufficesto show that v = p?"1 x07! on the semialgebra 39,00) With Fe

Bor

according to Theorem 21.11.

DB sp} and a > 0, consider the two subsets of R? \ {0}:

©-1((0, 1] x F) = {x € R” \ {0} : |x] € (, Land [x|-'x € F}, ©-1((0, a] x F) = {x € R* \ {0} : [x] € (0, a] and |x|-1x € F}. For anyx € R” \ {0} with |x] € (0, 1] we have |x| € (0, a] and for any y € R” \ {0} with

ly| € (0, a] we have dy

€ (0, 1]. Thus &—!((0, a] x F) = w@—'((0, 1] x F) and then

un” (®-1(@, a] x F)) = au" (-1((, 1] x F)) by Corollary 24.33. Thus for0 < a < B, we have

v((a, 61x F) = u2(&—"((@, 61 x F))

=H (© (©, 61 x F)) — ai (@ "(a1 x F))

=(6" — 2")ui (®"(O, 11x F)) = 6" —a")ot *)

= [ eat

mean ot) = 2 (a, Alot)

=(e""! x of !)(@, 61 x F). This shows that vp = p?—! x of Theorem 26.19.

on the semialgebra (9,00) x B set.

For n > 2, let F be an extended real-valued 2(Be,00)

x Boo-1)-

measurable function on a set D € 0 (B(0,00) x B st): Let ® be the mapping of R” \ {0} into (0, 00) x set as in Definition 26.15. Then

Ls Fom@utan =f Fea (e" xot (aca). in the sense that the existence of one side implies that of the other and the equality of the two. In particular if F is nonnegative, then the equality holds.

Proof. The mapping ® of R” \ {0} into (0, 00) x st

is Bre \(0)/0 (Beo,00) x ®B -1)-

measurable by Lemma 26.16. Consider the image measure v of 2" on a (93(0,00) x 3se) by the mapping ©, that is, v(G) = (u" 0 @-!)(G) for G € o(Be,00) x By-1). Theorem 9.34 (Integration by Image Measure), we have (dy

KG

a ®)(x) wy (dx) = [re

By

§26 Change of Variable of Integration

695

According to Theorem 26.18, » =”! x a?! on 6 (830,00) x Byy-1). Thus

@)

[ Fdv= [ F(r,2) (0! x 0? )d¢r, 0).

With (1) and (2), we have the proof.

#

A function of x € R? is called a radial function if it depends only on |x|. In other words, a function g on R" is a radial function if there exists a function h on [0, co) such that g(x) = h(|x|) for x € R". For integration of radial functions, we have the following

specialization of Theorem 26.19:

Theorem 26.20. For n > 2, let f be an extended real-valued %B ¢,.0)-measurable function on (0, 00). Then

Lg Steven =)

[seo mao)

in the sense that the existence of one side implies that of the other and the equality of the two,

Proof. For D = (0,00) x sr} we have ®—1(D) = R* \ {0}. Given an extended real-valued $8 9,o0)-measurable function f on (0, 00), consider an extended real-valued

6 (Be) x B s-1)-measurable function on D defined by F(r, z) = f(r) for (r, z) € D. Since (x) = (|x|,x*) € D for x € R” \ {0}, we have (F 0 ®)(x) = F(®(x)) = F (|x|, x*) = f (|x|). Thus by Theorem 26.19, we have

[ ayo, SRD HEED) = Po coese FO)(0 nm

_—

for

= Lf,

n—1

x o? (de,2)

wtar][ fi, of *az)] "

}, 00)

yf

=o

n-1

w, (dr). FQ"!

We apply our integration formula for radial functions to compute the volume of an open sphere with radius a > 0 in terms of the surface area of a unit sphere in R”. Since the volume of a sphere is already known, we can find the surface area of the unit sphere from this relation. Theorem 26.21. For n > 2, the volume

(Bf) of the open ball B” with center 0 and

radius a > 0 and the surface area of~'(S"~') of the sphere St—| with center 0 and radius 1 are related by

w

ut (Bt) = Lato (st), 1

696

Chapter 6 Measure and Integration on Euclidean Space

and thus

@

of

nf

(St!) =m (BD)= Fea+1)" nr

Proof. By Theorem 26.20, we have

#5 (Ba) = wy (@2\000) = ff‘R"\{0} tages upcax) =f R"\ {0} dootted unas) = af 1(S) | (0,00) Ayoar(ryr** (ar) =of “lst- yf

rod =

da"ot” 1(sn- 1).

This proves (1). Then since .” (B%) = a"x"/2/T

from (1).

(§ + 1) by Theorem 26.13, (2) follows

With the measure of ? on the spherical hypersurface st”—1 with center 0 and radius 1 in R", we can define a measure on the spherical hypersurface S37—1 with center 0 and radius a > 0 in R" by projection as follows. Lemma 26.22. Forn > 2, let P be a mapping of S{~ 1 into S®—| defined by P(x)= forxe ST” 1 (a) P maps sr one-to-one and onto S®~' with inverse mapping P-\(y) = a~'y for

ye sro}

(b) P is B st {3B 's-1-measurable. Proof.

(a) is obvious.

To prove (b), let E

€ B got.

Since Boot

= Ben

gaol by

Observation 26.14, we have E = FN seo} for some F € Sgr. Then

P(E) =a! (FN St) =a Fast Since a1F

€ Spe, we have a FO set

€ Ben

=a set

Fast. = B oot.

Thus we have

P(E) e %B got. This shows the 5 gx-1/98 s.-1-measurability of P. Definition 26.23. For n > 2, consider the measure space (st 1 3 gs of) measurable space (sz, 8 at). With the mapping P(x) = ax forx € st

and the

let us define

a measure of on (Sz~', Byr-1) by setting of) = a®\(o2-! o Pu), that is, a8} is a" times the image measure of oT | on (St-1, B ont). Observation 26.24, Needless to say, ot oP

isameasureon (S2-!, 8 set) by Theorem

1.44 (Image Measure). Then 2-1 = a"—!(af—! o P-") is a measure on (S271, B got).

§26 Change of Variable of Integration

697

For the o—! value of S?—!, we have

of MSI) = atop o PY (SI°1) = atof-'(st-1)

and thus by (2) of Proposition 26.21 and (2) of Proposition 26.13, we have

“ten

nn? of 1(s2-1) =rg ”

1 onhi Ba)

forn EN.

In particular we have

oa(S) = Ha,

o2(S2)=a,

oF(88)= Ba’, ol(S)=a’,

..,

o2(S2)= "Fa", of(S2)= 75 a*, of(S8) = Tksa%, of (St)= Proposition 26.25. Forn > 2, the volume (BQ) of the open ball BY with radius a > 0 in IR" and the surface area ot} (sz-') of the spherical hypersurface Sao} with radius a in R" are related by

a)

d

atte Ba) = 04 *(S5~'),

and

(2)

dr = un (B2). [ af1(s"-!)

Proof. Since 42? (B?) = n/a" /T (3 +1) and of !(s2-1) =nn"a"1/T (3 + 1), the equalities (1) and (2) are obvious.

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Chapter 7 Hausdorff Measures on the Euclidean

Space §27

Hausdorff Measures

[I] Hausdorff Measures on R” Definition 27.1. Consider 9$3(R"), the collection of all subsets of R". The diameter |E| of E € PR"), E FX G, is defined by |E| = sup {|x — y|: x, y € E}. We define |O| = 0. Note that |Z| € [0, 00] for E € $$(R") and |E| < ov if and only if E is a bounded set

in R". IfE # @, then || = 0 if and only if £ is a singleton. If E consists of two pointsx and y only, then |Z] = |x — y| > 0.

Observation 27.2. (a) For every E € 93(R"), we have |E| = |E|.

(b) For every E € 93(R") and y > 0, there exists an open set G such that G > E and

IG| 0.

If x, y € E then there exist xp, yo € E such that |x — xo| < ¢ and |y — yo|

< ©. Then

|x — yl < |x — xo] + [xo — yol + lo — yl < |Z] + 2e. Thus |E| < |E| + 2e. Since this holds for every¢ > 0, we have |E| < |E|. Therefore || = |EI.

2. Let > 0. LetG = U, ez B(x, $). Then G is an open set andG > E. Let

x,y € G. Then x € B(xo, 3) and y € B(yo, £) for some xp, yo € E. Now we have

lx — yl < |x — xl + [x0 — yol + lyo — yl S El +9. Then |G] °;c% |Vil* an estimate of H3(E) by the 8-cover (V, : i € N) of E. Note in particular that a oo-cover of E is just a sequence (V; : i € N) in $8(R") such

that jen Vi D E without any restriction on the diameters. Observation 27.4.

Let (V;

: i € N) be a 5-cover of a set

FE € $(R”).

Suppose we

replace a member V; in the é-cover by two disjoint subsets V’ and V” of V; such that ENY¥; = EN(V'U VY”), that is, V’U V” still covers that part of E which is covered by V;.

Let (W;

: i € N) be the resulting sequence.

Then (W;

: i € N) is a 5-cover of

E with Usen Wi C Usen Vi. but the estimate of 1((E) by the 6-cover (W; : i € N) of E may not be smaller than that by the 6-cover (V; : i € N), and in fact we may have Dien (Wik > Liew |Vil*. This is shown by the following examples.

Example 1. Let E = [0, 4] U [2 1]. Consider a 1-cover (V; : i € N) of E consisting of Vj = [0, 1] only. We have Dien Vili =

|Vil’ = 1° = 1. If we replace V, with two

disjoint subsets W, = [0, }] and Wo = [2, 1], then (W; : i ¢ N) = (Wi, Wa) isa 1-cover . ys) (1s 1\F o < /1\s 1) of E with Den Wil’ = 3) + (3) = 2(3)’. Since lim (3) =1, we have 2(3)’ > 1 for sufficiently small s > 0 and for such s > 0 we have Dien Wil’ > Dien Vil. Example 2. Consider a subset of R” defined by

FE =(2(0,3)\

2,3} U{80,0\ 86,3}.

For a 2-cover (V; : i € N) of E consisting of one member only given by V; = B (0, 1), we have }°;en IVilf = [Vil’ = 2°. Fora 2-cover (W; : i € N) of E consisting of two members

W and W, only, given by Wi = {B (0, 4) \ BO, })} and W. = {B (0, 1) \ B (0, 2)}, we

have Den

|Wil® = [Wil® + |Wol’ = 15 +25. Thus > en Wil? > Dew

Vil.

Proposition 27.5. For every s € [0, 00) and & € (0, oo], the set function Hj on 98(R") is an outer measure on R", that is, Hs satisfies the following conditions:

(1) (2) GB)

3(E) € [0, co] forE € $(R"). 3) =0. IE) < IGF) FEC F.

4)

HS (Uiew Fi) < Lien GED.

Proof.

(1) is immediate from Definition 27.3. To prove (2), note that by using a 5-cover

of @ consisting of only one member @, we have 1(;(@) < |@|' = 0 for every s € [0, 00)

§27 Hausdorff Measures

701

including the case s = 0 for which we have \o\° =0 by our convention. Thus 7(5 (0) = 0.

(When s # 0, by using a 5-cover of @ consisting of only one member B(x, r) with r < 3, we have H3(@)

< (2r)*.

Since Tim (2r)*

= 0 when s # 0, we have H{(;(@) = 0. When

s = 0, no matter what nonempty set V € $$(R”) we may use as a 5-cover of 9, we have |V|° = 1. Without using @ as a 6-cover of with the convention |9|° = 0, we would have

58@) =1)

(3) follows from the fact that if E C F, then every 5-cover of F is also a 6-cover of E, that is, the collection of all 6-covers of F is a subcollection of the collection of all 6-covers of E.

To prove (4), note that if }*,

i€N jeN

IVs = {GED + £} = Do IGE) +6. ieN

ieN

Since this holds for every ¢ > 0, we have (4).

Theorem 27.6. For fixed s € [0, 00) andE € §B8(R"), Hi(E) t asd | 0. Thus HS(E) := him H4(E) exists in [0, 00]. Furthermore KH is an outer measure on R". Proof. Let 0 < 6; < 62 < oo. Then every ,-cover of £ is also a 52-cover of E. Thus the collection of all 6;-covers of E is contained in the collection of all 52-covers of E. This

implies that 15 (E) < H5, (EZ). Thus HG(E) t as 5 | 0. The fact that F(* is an outer measure on R” follows from the fact that H(; is an outer measure on R” for every 6 € (0, 00) and F§ t Ft* as 6 | 0. In particular the countable subadditivity of F(* can be verified as follows: By (4) of Proposition 27.5 and by H t 3°

as 8 | 0, we have HS (Ujew Zi) < Dien HG(E:) < Vien HC (Ei). Then letting é | 0, we have J¢* (sex Ei) = Dien CE).

0

Definition 27.7. For s € [0, 00), we call the set function H® on J8(R") defined by H'(E) = lim H3(E) = sup H5(E) 540 5>0 for E € $3(R") the s-dimensional Hausdorff measure on R". Let us note that the s-dimensional Hausdorff measure }(* defined above is only an outer measure by Theorem 27.6. Since it is an outer measure, the collection Nt(H*) of all H*measurable subsets of R” is a o-algebra of subsets of R® by Theorem 2.8 and if we restrict

HS to this o-algebra, then (R", INCH), I*) is a complete measure space by Theorem 2.9. Remark 27.8. The s-dimensional Hausdorff measure H*(£) of a subset EZ of a Euclidean

space does not depend on the Euclidean space containing E on which F(* is defined. Suppose

702

Chapter 7 Hausdorff Measures on the Euclidean Space

we regard E € 93(R") as a subset of R"+* for some k € N. If we let a = ‘H*(E) defined for E as a subset of R® and 8 = ‘H°(E) defined for E as a subset of R*+*, then a = £. Proof. For6 € (0, 00], lets = H5(E) defined for E as a subset of R” and let 8s = 3(3(E) defined for E as a subset of R°+*. Then a = limas

and B = im Bs. Now every 5-cover

of E by subsets of R” is also a 6-cover of E by subsets of R"+*. This implies £3 < as. On the other hand, if (V; : i € N) is a 5-cover of E by subsets of RM, then since [Vi NR*| < |V;| < d fori € N, (Vi OR" : i € N) is a d-cover of E by subsets of R”.

Now Dien Vi NR"

< Yen IV.

Thus corresponding to every 5-cover (V; : i € N)

of E by subsets of R***, there exists a 5-cover (W;

:i € N)of E by subsets of R” with

Dien Wik < Liew |Vil’. This implies ws < fs. Thus os = fs for every 6 € (0, 00). Letting 5 | 0, wehavea = 6. 44; (E) is a nonnegative extended real-valued function defined for a triple of variables (s,6, Z)

€ [0,00) x (0, 00] x $8(R").

Let us fix one or two of the three variables s,

6, and E and study the behavior of J(;(£) as a function of the remaining one or two

variables. We have shown above that for fixed s and 5, 1(;(£) is an outer measure on R”, We have shown also that for fixed s and E, (;(E) as a function of 6 increases as § | 0 and F°(E) := lim 5G) is an outer measure on R”, We show next that for fixed s and E, H5(E) as a function of 6 € (0, oo] is either identically vanishing on (0, oo] or strictly positive on (0, oo]. Observation 27.9. Let E € 93(R") and 6 € (0, oo].

(a) If H(E) = 0 for some 5 € (0, oo], then E = 8.

(b) 16 30(B) = 0, then E = . Proof.

1. Suppose 3Q(E)

= 0. Then for an arbitrary « € (0, 1), there exists a 5-cover

(Wj: i €N) of E such that 1. |Vil° < &. Now if E # @, then V; # G so that |V;|° = 1

for at least one i ¢ N and consequently ~~

2. Suppose F(Z) = 0. Since (E)

we have FQ(E)

= 0. Then by (a), wehave

|V;l° > 1 > ¢, acontradiction. Thus E = 9.

> 5(9(E) for any 6 € (0, co] by Theorem 27.6, FE = 9%.

Lemma 27.10. Let E € 9$(R") ands € [0, 00) be fixed. Consider }(;(E) as a function of

6 € (0, oo]. {a) If FG, (E) > 0 for some 8p € (0, oo], then F(3(E) > 0 for every 8 € [89, oo]. (b) If H"(E) > 0, then H5(E) > 0 for every 5 € (0, oo], or equivalently, if HG, (E) = 0 for some 8 € (0, 00], then H°(E) = 0.

(©) H5(E) as a function of § € (O, 00] is either identically vanishing on (0, 00] or strictly Positive on (0, 00). Indeed if H5(E) = 0 then 3(5(E) is identically vanishing on (0, 00]

and if H5(E) > 0 then H;(E) is strictly positive on (0, 00].

Proof. 1. To prove (a), suppose H{;,(Z) > 0 for some dp € (0, co]. Since H3(E) + as 6 | 0, to show that 3(;(E) > 0 for every 6 € [89, 00] it suffices to show that J(3,(E) > 0.

To show H5,(E) > 0, assume the contrary, that is, HS,(Z) = 0.

§27 Hausdorff Measures

703

Consider the case s € (0, 00). Now H$,(£) = 0 implies that for an arbitrary ¢ > 0, there exists a oo-cover (V; : i € N) of E such that D)-y|Vil* < e%. Then |Vjl° < &°, that is, |V;| < ¢, for every i € N so that (V; : i € N) is actually an e-cover of E. Then

HE(E) < Viex |Vil* < £*. This implies 1°(E) = lim 5€5(E) < lime? =Osinces > 0.

Since Hi, (E) < H'(£), we have He, (£) = 0, contradicting the assumption HG, (E) > 0. Now if s = 0, then52, (E) = 0 implies that E = @ by (a) of Observation 27.9. Then

Fy (E) = Hey (@) = 0, acontradiction.

2. To prove (b), suppose H*(E) > 0. Since F(j(E) + H*(E) as 6 | 0, there exists 69 € (0, 00] such that H3(Z) > 0 for every & € (0, do]. By (a), 45, (EZ) > 0 implies that

H5(E) > 0 for every 6 € [80, 00]. Thus Hj (EZ) > 0 for every 6 € (0, oo]. 3. To prove (c), note that since H3(E) < H%(E) for every 6 € (0, oo], if H'(Z) = 0 then 3(5(E) = 0 for every 5 € (0, 00]. On the other hand, if (*(Z) > 0, then H(E) > 0 for every 5 € (0, 00] by (b). Remark 27.11. (a) If £ is a bounded set in R” then for every s € [0, 00) and 6 € (0, co] we have Hj(E) < oo, Indeed if E is a bounded set then E is contained in a cube Q = x hal—r, r] with sufficiently large r > 0. Now for any 8 € (0, 00) there exist finitely

many cubes, say Ci,..., Cy, with |C;| < 5 fori = 1,...,N such that UNG

Then H{(E) < YN, IC;

< ON, 8 < 00.

> @.

(b) For a bounded set E in R", H'(E) = kim 54 (E) may not be finite. In Proposition 27,32

below, we show that ‘1° is a counting measure on the -algebra ‘f$(R"). This implies that if E has infinitely many elements then J(°(E) = oo whether E is a bounded set or not. In Theorem 27.33 we show that if H{**(Z) € (0, 00) for some s* € [0, 00) then H'(E) = co for all s € [0, s*) whether E is a bounded set or not. Theorem 27.12. For every s € [0, 00), the s-dimensional Hausdorff measure }(° is ametric

outer measure on R". In particular we have By is }{*-measurable. Proof.

C M(IL*), that is, every Borel set in R"

To show that FC* is a metric outer measure, we show that if E, F € $8(R") and

a(E, F) > 0 then HS(E U F) = 3(E) + H8(F). Let 8 € (0, d(E, F)) be arbitrarily fixed. Let (W; : i € N) be an arbitrary 6-cover of EU F. Now ifx

€ E andy € F, then

|x — y| = d(E, F) > 4. Then for each i € N, since |W;| < 5, W; cannot contain both x and y and thus W; cannot intersect both E and F. Let {U; : i € N} be the subcollection of {(W; : i € N} consisting of those W; which intersect E and let {V; : i € N} be the subcollection of {W, : i € N} consisting of those W; which intersect F. Then (U; : i € N) is a 6-cover of E and (V; : i € N) is a d-cover of F. Thus

Yow = Swit + OM ieN

ieN

ieN

Since this holds for an arbitrary 5-cover (W;

= IG) + 9G 0F).

: i ¢ N) of E, we have (EU

F)

>

3(E) + H5(F). On the other hand since F(§ is an outer measure, it is subadditive and thus the reverse inequality holds. Therefore F({(E U F) = H4(E) + H4CF). Since this

704

Chapter 7 Hausdorff Measures on the Euclidean Space

equality holds for every 8 € (0,d(E, F)), letting 6 | 0 in the last equality we have JS(E U F) = IE(E U F) + F(E U F). This shows that 3(* is a metric outer measure. Then BR C MICH") by Theorem 2.19. m [11] Equivalent Definitions of Hausdorff Measure [11.1] Covering by Closed Sets and by Open Sets The s-dimensional Hausdorff measure H(* has equivalent definitions using 5-covers by closed sets only or using 5-covers by open sets only. We show that if we use 5-covers by closed sets only, then 3(; remains unchanged for every 5 € (0, oo] and thus H(* is unchanged;

if however we use 6-covers by open sets only, then #(; changes but H(° is unchanged. Theorem 27.13. With s € [0, 00) and

€ (0, oo], let us define set functions F; and S§ on

PR") by w

F3(E) = inf [Dien |Fil’ : (Hs i € N) is a 8-cover of E by closed sets} Si(E) = inf [yen |Gil : (Gi si € N) isa b-cover of E by open sets}

forE € $(R"). Then

@)

FRE) t

and GRE) +

asé 0.

Let us define set functions F* and $* on $8(R") by setting

Q@)

FE) = lim 33)

and 9°(E) = lim $5(2).

Then F5, 5, F and S are all outer measures on R". Furthermore for0 < 5 < 8', we have

4

$3 (E) 5 F5(E) = H5(E) = 95(E),

and 6)

$'(E) = F*(E) = HE(E).

Proof. The fact that F; and 9§ are outer measures on R’, F3 ¢ and S; t as 6 | 0, and F* and S* are outer measures on R” can be proved by the same argument as in Proposition 27.5 and Theorem 27.6 for Ft; and 3°. Let us prove (4). Since the collection of all 5-covers of F by open sets is a subcollection of all 6-covers of E, we have H3(E)

< $3(E). This proves the second inequality in (4).

By the same reason as above, we have H3(E) < 33(£). Let us show that the reverse inequality holds in this case. Now if H(3(E) = oo, then F3(E) < Hj(£) holds trivially. Suppose H(3(Z) < oo. Then for an arbitrary 7 > 0, there exists a 6-cover (V; : i € N) of

E such that H(E) +n > Dien Vil". Now V; is a closed set and |¥j| = |Vj| fori € N.

Then

(V; : i € N) is a 6-cover of E by closed sets and thus we have ) jen |Vi|’ > F$(E).

§27 Hausdorff Measures

705

Thus H5(E) +9 > Diew Vill = Dien lvl? > F5(E). By the arbitrariness ofn > 0, we have H3(E) > F3(E). Therefore $3(E) = H3(E). This proves the equality in (4). Let us prove the first inequality in (4). If 53(E) = oo, then 93,(Z) < 5(E) holds trivially. So let us assume F3(E) < oo. Then for an arbitrary 7 > 0, there exists a 6-cover (F, : i € N) of E by closed sets such that

©

SF ieN

< FE) +2.

Since0 < 6 < 8’, for eachi € N there exists e; > 0 such that

“

{Alt 26} 0, we have Nien |Gil? < 45(E). This proves the first inequality in (4).

Finally letting 6 | 0 in (4), we have $%,(E) < 3*(E) = ‘H*(E) < 9°(E). Then letting

5’ | 0, we have (5).

If we use 5-covers by balls only then the resulting outer measure is no longer equivalent to H(5 but still comparable. Proposition 27.14, With s < [0, 00) and

be defined by

(n

€ (0, oo], let set functions 8; and Bs on $B(R")

S8§(E) = inf [Dew |S’ : (S; : i € N) is a 8-cover of E by closed balls} Bi(E) = inf {Dey Bil’ : (Bj : i € N) is a 8-cover of E by open balls}

for E € $$(R"). Then 8} and Bs are outer measures on R" and 8;(E) + and Bi(E) t as 5 | 0. If we define set functions 8* and BS on $B3(R") by setting S(E) = lim $B),

@)

BF (E) = lim BEE) ayo”

then 8° and B* are outer measures on R". Furthermore for 8 € (0, 00), we have

@)

FG(E) < S§(E)

and 85;(E) < 2°HG(E),

706

Chapter 7 Hausdorff Measures on the Euclidean Space

(4) (5)

H(E) < 8°(E) < 290 (E), 83(E) < Bj(E)

and By3(E) < a°83(E) for everya € (1, 00), BS(E) = 8(E).

©

Proof. $5, Bs, 8°, and B* are outer measures on R" by the same argument as in Proposition 27.5 and Theorem 27.6. Let 6 € (0, co). The first inequality in (3) follows from the fact that the collection of all

5-covers of E by closed balls is a subcollection of all 6-covers of E. To prove the second

inequality in (3), let (V; : i € N) be an arbitrary 5-cover of E. For eachi H5(E). prove the reverse inequality. For an arbitrary 5-cover (V; : i € N) of E, consider

(A(V) si € N) where h(¥;) is the convex hull of V;. By Proposition 27.17, #(V;) is aclosed

convex set. By Proposition 27.23, |A(V;)| = |A(V;)| = [Vil < 8. Thus (A(V;) : i © N)

is a 6-cover of E by closed convex sets. This shows that corresponding to every 5-cover (V, : i € N) of E there exists a 5-cover (h(V;) : i € N) of E by closed convex sets such

that Dien GA) = Diew Vil’. This implies that Kj(E) < H5(£). Therefore we have K;(E) = 13 (E). Then letting 5 | 0, we have K°(E) = H°(E). w [I] Regularity of Hausdorff Measure

Lemma 27.26. Lets € [0, 00) and E € $3(R”).

{a) For every 6 € (0, 00] and n > 0, there exists a 5-cover (F; : i € N) of E by closed sets

such that Dye [Fil < HE(E) +n.

(b) For every 5, 8’ € (0, 00] and n > 0, there exists a 8 + 5'-cover (G; : i € N) of E by open sets such that Ye |Gil’ < H3(E) +n. Proof. 1. By the definition of $3 (£) in Theorem 27.13, there exists a 5-cover (F; :i € N) of E by closed sets such that >, oy [Fi < 3(E) + 9. By Theorem 27.13, 15(E) = F3(E).

Thus Dien Fl’ < GCE) +9.

2, For an arbitrary 6 € (0, oo] and 9 > 0, by Definition 27.3 for 7(;() there exists a

6-cover (V; : i € N) of E such that V°jcy |Vil¥ < H§(E)+ 3. Letd’ € (0, 00] be arbitrarily given. For each i € N, by (b) of Observation 27.2 there exists an open set G; > V; such that

IGi] < [Vil +6’ H*(E) = 00. Thus R” is an open set containing EZ with H*(R”) = 5° (E) and we are done. Consider the case H(* (EZ) < oo. By (b) of Lemma 27.26, for every i € N, there exists a

2 cover of E by open sets (O;,; : j € N) such that Dijen 10,31 < Hy) + i If we let G=lie Ujen O;,;, then G is a Gs-set containing E. For eachi € N, (O;,; : j € N)isa

}-cover ofG so that H§,,(G) < Djen 01,5" < Hf,,(E) + }. Since HG t Has 5 > 0,

we have H'(G) =

lim 13(G) = lim 53,,(G) < lim {3, + +} = H°(E). On 5-0 00 itoo the other hand, since G D> E, we have H'(G) > H*(£) by the monotonicity of H*. Thus we have H°(G) = H5(E). Since G € Bx C MUCH") by Theorem 27.12, H° is a Borel regular outer measure. 2. Let E € S(H"*) and H'(£) < oo. Let us show that for every ¢ > 0, there exists an F,-set F, C E such that

(1)

HA (F,) = HO(E) —e.

Now by (a), there exists a Gs-set G > E with H*(G) = H°(E). Let G = 1); H8(E) —6.

By (4) and (2), we have HS (F, \ E) < 7

(Niex 0; \ E) = 0. By (a), there exists a G3-set

G > Fy \ E such that 1°(G) = 30 (F, \ E) =0. Leth = &\G= ENG. SinceG

is a Gs-set, G° is an F,-set. Since F, defined by (4) is a closed set, F, is an F,-set. Now

Fe = Fa VG® C Fa \ (Fe \ BE) = Fe (Fe NEY =F,A(FEUE)=FNECE and H(F.) = HS(Fx \ G) > FOC) — H(G) = HF) completes the proof of (1).

> HCE) — & by (5). This

Now for every & € N, there exists an F,-set F, C E such that H' (Fy) > H*(£)— } by

(1). Let F = pen Fe. Then F is an F,-set and 3° (F) > FO (Fy) > HCE) — } for every k € Nsothat H’(F) > H5(£). On the other hand since F C E, wehave H'(F) < H*(E). Therefore H*(F) = H°(E). This completes the proof of (b). 3. To prove (c), let E € Di(H*) and H*(£) < oo. By (b), there exists an F,-set FC E with H*(F) = H*(E). Since F is an F,-set, we have F = U,en Ce where (Cy : & € N) is an increasing sequence of closed sets. Since H* is a measure on the o-algebra NU(H*) containing Bre and since Cy € Ba and Cy ¢ F ask + oo, we

have H(F) = J¢( Jim Cr) = jim HE(Cy). Since H'(F) = HS(E) < oo, for every e

00

>

0, there exists ko

00

€ N such that H°(C;,)

FO CE \ Cyy) = HE(E) — HO (Cay) = HF)

>

H°(F)

— e.

Thus we have finally

— H(Cyy) H§(E) for

every8 € (0, 00].

© If H'(E) < 0, then H'(E) = 0. (d) fH! (E) > 0, then H3(E) = 00. (©) There exists at most one point s* & [0, 00) such that H*" (E) € (0, 00).

Proof. 1. Recall that we have H(§(E) = inf [Den |Vil* : (Vi: i € N) is a d-cover of E} for 6 € (0, co] and H'(E)

= lim 5G).

Consider 6 € (0, 1) and let (V; : i € N) be an

arbitrary 5-cover of E. Then since |V;| < 6 < 1 and0 [V;|' so

that Vien Vill > Dien |Vil’. Therefore H§(E) > H(E) for8 € (0, 1) and this implies kim Hy(E) = kim 45 (E), that is, H*(E) > H'(E). This proves (a). 2. To prove (b), let & € (0, oo] and let (V; : i € N) be an arbitrary 6-cover of £.

O 0} = sup {s € [0, 00) : H*(E) = oo}. (@) dim, E = inf {s € [0, 00) : H*(E) = 0} = inf {s € [0, 00) : H*(EZ) < oo}. (e) HE) € (0, 00), then s = dim, E. (f) The converse of (e) is false, that is, s = dim, E does not imply H'(E) € (0, 00). For instance dim, @ = 0 but FO@) = 0 and dim, R" = n but H"(R") = 0. Proof. (a) and (b) are from Lemma 27.30. (c), (d), and (e) are from Theorem 27.33.

&

Proposition 27.36. Regarding the Hausdorff dimension, we have the following: (a) FE, F € $(R") andE Cc F, then dim, E < dim, F.

(b) For (E; :i € N) C PR"), we have dim, Use E; = sup { dim, E; : i € N}.

In particular, dim, \(J"_, Ei = max { dim, E1,..., dim, Ey}.

Proof.

1.

Suppose E C F.

measure F{(* we have H5(E)

For every s € [0,00), by the monotonicity of the outer
0. Then for every i e N we have t +

> dim, E; so that H'+*(E;) = 0 by (d) of Theorem 27.35.

Then we have H't*( jen Hi) < Dien H'**(E;) = 0 by the countable subadditivity of

4,

Then by (d) of Theorem 27.35, we have dim, ); 0, we have dim, J;.y Ei 0 with coefficient c > 0 if it satisfies the condition

IT@)-TO)| Selx — yl? forx,y €E. (b) A Holder mapping of exponent 1 is called a Lipschitz mapping. Let us note that a Hilder mapping of a set E € 93(R”) is uniformly continuous on

E. Indeed for an arbitrary ¢ > 0, if we let 6 = (2), then for any x, y € E such that |x — yl < 8, we have |T(x) — T(y)| < clx —yl* < ¢(£) =e. Theorem 28.4. Let T be a Holder mapping of exponent a > 0 with coefficient c > 0 of a set E € $3(R") into R”.

Then

(a) 5/4 (T(E)) < c5/*H5 (EB), that is, HS (T(E)) < c*H*(E), for every s € [0, 00). (b) dim,, T(E) < 1 dim, E.

In particular if T is a Lipschitz mapping with coefficient c > 0, then

(c) HS (T(E)) < e*H*(E) for every s € [0, 00). (d) dim,, T(E) < dim, E. Proof.

1. For 5 € (0, 00), let (V; : i € N) be a d-cover of EZ. Now forx,y

€ ENV,

we have |T(x) — T(y)| < clx — yl* < cl¥j|* so that |T(EN V;)| < clVil* < c8%. Since EC Ulen Vi. we have E C Ujen(E 9 Vj). This implies T(E) C Ujen T(E 9 Vi). Thus (T(EN Vj) : i EN) is acd%-cover of T(E). Therefore for every s € [0, 00), we have

HE (T(E)
(clit)? = ieN

ieN

IVE.

By the arbitrariness of the 5-cover (V; : i € N) of E, we have sls (T(E)) < c/*H5(E). Then letting 5 | 0, we have H(*/*(T(E)) < c*/*H5(E). This proves (a). 2. By (a), we have H°/*(T(E)) < c*/*H*(E). Since c'/* > 0, the inequality implies that {s € [0, 00) : H*(EZ) = 0} c {s © [0, 00) : H*/*(T(Z)) = O}. Then by (d) of Theorem 27.35, we have

dim, T(E) = inf {s € [0, 00) : H*(T(E)) = 0}

= inf {£ € [0, 00): H*/"(7(E)) = 0}

= 1 inf {s € [0, 00) : H°/#(T(E)) =0}

< Linf {s € [0, 00) : (2) = 0}

= dim, E. 1

The result above for Hélder mappings has the following extension.

720

Chapter 7 Hausdorff Measures on the Euclidean Space

Theorem 28.5. Let T and S be two mappings of IX" into R™ satisfying the condition that there exist c > 0 and a > 0 such that

IT@) — TY)| < elS@) — SQ)|*

forx,y eR".

Then for every E € J$(R"), we have

(a) H*/*(T(E)) < c'/*HS (S(E)) for every s € [0, 00).

(b) dim,, T(E) < } dim, S(E). Proof. Lets € [0, 00). For arbitrary ¢ > 0 and 5 € (0, oo], let (V; : i € N) be a d-cover

of S(E) C R™ such that Den |Vil? < H§(S(E)) +

< HS (S(E)) +6. Let (W;

7 EN)

be a sequence of sets in R” defined by W; = T (S~1(V;)) fori € N. Let us show that the sequence is a c5*-cover of T(E). Now since S(E) C en

Vi. we have

Ecs‘\(s@)cst (u x) =Us'm. ieN

ieN

Then T(E) C T(Ujen S71(0¥)) = User T(S1(V)) = Ujery Wi. To estimate |W; |, let

w’,w" € W;. Then there exist z’,z” € S—(V;) such that w’ = T(z’) and w” = T(z"). We have |w’ — w”| = |T(z’) — T(z”)| < elS@’) — S(z")|*. Since 2,2” € S1(Y),

we have S(z’}, S(z”) € V; so that |S(z’) — S(z")| < 6. Thus |w’ — w”| < cd%. This implies |W;|

< cd*.

Therefore (W;

: i € N) is a cé*-cover of T(E).

|w! — w"| < ¢|S@2’) — SQ") < c]V;|* and thus |W;| < c|V;|*. Therefore

HS (T(E) < Sowa = Doe icN

icN

We also have

1ViI" < of! (5¢*(S(E)) +e}.

From the arbitrariness of e > 0, we have sole (T(B)) < c/"5 (S(E)). Letting 8 | 0, we have H°/*(T(E)) < c°/*H5(S(E)). This proves (a). Then (b) follows by the same argument as in the proof of (b) of Theorem 28.4.

&

As an example of Lipschitz mapping we have the following: Proposition 28.6. Let T be a mapping of class C1(Q) of an open set Q in R® into R, that is, the first order partial derivatives of the component functions g1,..., 8m of T are

all continuous on Q.. Let E be a convex set contained in Q. If all the first order partial derivatives of 21, ..., 8m are bounded on E, then T is a Lipschitz mapping of E. Proof. By our assumption on the first order partial derivatives of g1,..., 8m, there exists a

constantB > 0 such that

iL ()

< B foreveryp € E,i=1,...,m,andj =1,...,n.

Letp € E and let h = (41,..., 4x) € R” be such that p +A

€ E. Foreachi =1,...,m,

by the Mean Value Theorem for functions of n variables applied to g;, there exists a point py} on the line segment joining p to p +h such that g;(p + A) — g;(p) = Yja1 sf (Ph; Thus we have

OP 88 we - 0A} Wo? < BAP. 24 12 lee +A) — ait)? 2 < [Beep] j=l axj

j=l

§28 Transformations of Hausdorff Measures

721

Therefore we have

[Tip +h) -T@)/ = > |gi(p +4) — gi(P)? < mnBh/, i=l

Then |T(p +h) — T(p)| < ./mnB\h|. Thus we have |T(x) — T(y)| < ./mnB\|x — y| for x,yeE.

Definition 28.7. (a) We say thata mapping T of a set E € 38(R") into R" is a bi-Lipschitz mapping with coefficients c, and c, where 0 < cy < €2 if T satisfies the condition

cilx — yl 0. If T is not identically vanishing on R", then ||7|| = SUP|x[ 0 and IT (x) — T(y)| < IIT|| |x — y| for x, y € R”. This shows that T is a Lipschitz mapping.

Suppose T is non-singular on R". Then T maps R” one-to-one and onto the ndimensional linear subspace T (R") of R”, The inverse mapping T~" is defined on T(R”)

722

Chapter 7 Hausdorff Measures on the Euclidean Space

and is a linear transformation of T (R") to R”. Let x, y € R". Then T(x), T(y) € T(R")

and |7—' T(x) —TT(y)| < ITT @)—TO)|, thatis, |z—y| < TI |T@)—TO)|-

Thus we have ||7~1||-!|x — y| < |T@) — TO) < IIT I |x — y|. This shows that T is a

bi-Lipschitz mapping. Theorem 28.10.

©

Let T be a bi-Lipschitz mapping with coefficients c, and cy of a set

E € $(R") into R”. Then

(a) c{H8(E) < H(T(E)) < G50 (E) for everys € [0, 00).

(b) dim, T(E) = dim, E.

In particular if T is an isomery, then

(c) H (T(E) = H°(E) for every s € [0, 00).

Proof. If T be a bi-Lipschitz mapping with coefficients c, and c2, then T is a Lipschitz mapping of E with coefficient cz so that by (c) of Theorem 28.4 we have the estimate HS(T(E))

< cj3*(E)

for every s € [0,00).

mapping of T(Z) with coefficient c!

By Observation 28.8, T—

is a Lipschitz

so that by (c} of Theorem 28.4 again we have

a(T—1(T(E))) < c[ °F (T(E)), that is, {3° (E) < I (T(E), for every s € [0, 00). This proves (a). Also by (d) of Theorem 28.4, we have dim, T(E)

< dim, E as well as

dim, T—!(T(E)) < dim, T(E), thatis, dim, E < dim, T(E). This proves (b). Since an isometry is a bi-Lipschitz mapping with coefficients c1 = cz = 1, (c) follows from (a).

©

Theorem 28.11. (Invariance of S)t(H*) under Isometry) Let E € 93(R") and let T be an isometry on R". Then T(E) € UH) if and only ifE € DH’). Proof. By definition (KH) consists of all E € $3(R") such that for every A € 9(R") we have H*(A) = H'(AN E)+ HS(AN E*). Suppose FE € M(H‘). Let us show that

T(E) € DUH).

Now if T is an isometry on R", then T is a one-to-one mapping of R"

onto R” so that the inverse mapping T—' exists and furthermore T~' is an isometry. For an arbitrary A € 93(R”), (c) of Theorem 28.10 implies

FE(ANT(E)) = HE (TT (A) N T(E))

= (T(T-1(A) NE) = 50 (TA) NE), and similarly

HE (ANT(E)) = HE (TT

(A) N T(E)

=HS(T(T—1(A) N E*)) = H(T-1(A) N E*). Then by the fact that

ZH € S)t(H*) and by (c) of Theorem 28.10 we have

HE(ANT(E)) + (ANT (EY) = H8(T—1(A) NE) + HE (TA) NE’) = H8(T~!(A)) = H(A). This shows that T(E) € 98(R*) if E € M(H").

Conversely if we assume that T(E) €

M(H), then since E = T—!(T(E)) and since T— is an isometry, we have E € tH")

by our result above.

§28 Transformations of Hausdorff Measures

723

[II] 1-dimensional Hausdorff Measure We show that for every E € 98(R), we have H!(E)

= ui (E).

The difference between

the 1-dimensional Hausdorff (outer) measure H(! and the 1-dimensional Lebesgue outer measure ur is that while us is an outer measure on R, 4! is an outer measure on R? for

anyn € N.

Theorem 28.12. For every E € (R), H}(E) is independent of8 € (0, 00] and in fact HEE) = w*(E) for every 8 € (0, oo]. Thus H'(E) = u*(E). Proof. A subset of R is a convex set if and only if it is an interval. Thus the collection of closed convex sets in R is the collection of closed intervals 3,. Then by Theorem 27.25, for every EZ € 93(R) and 6 € (0, oo], we have

+(E) = inf { Zien lil : Jy € Fe, Fil S 8,7 € Nand Use Ji D Eh. Recall that according to Observation 3.2, we have

px (E) = inf { Xiew il : dp € Te, i € Nand Usen di D EI. The collection on which we are taking infimum for a (E) is contained in the collection on

which we are taking infimum for #*(E). Thus closed interval J in R is the union of at most with disjoint interiors such that |J,| < 4 for p sequence (J; : i € N) in 3, such that |),-y J;

wf (EZ) < ah (£). On the other hand countably many closed interval Jp, p € N and |J| = Dpen |Jp|. Thus for D E, there exists a sequence (J! : i €

every € N, every N) in

3 such that |J/| < 6 fori € N, Unen Jj > E, and Den 1J/] = Cen lil. This implies HCE) < (EB). Therefore 4} (E) = 45 (E). Then H!(E) = lim Hg (E) = pi(E). 0

Remark 28.13. We showed in Theorem 28.12 that a} (£) = we (€) for every 5 € (0, co]

and in particular 3¢1,(£) = 4; (E) as long as E € $B(R). For E € 98(R”) where n > 1, wy (E) is undefined.

Example 1. Consider two line segments in R? given by E, = {(x,0) < R?: x € (0, 1]} and Ey = {(0, y) € R?: y € (0, 1]}. LetE = Ey U Ep. If we regard E; as a subset of R, then by Theorem 28.12 we have SO(E)

= wE(E1) = 1. By the invariance of $(! under

an isometry ((c) of Theorem 28,10), we have J(!(E2) = 3{!(E1) = 1. Then since Jt! is a measure on 3g

and since Ej N Er = G, we have H!(E) = H'(E) + H' (Eo) = 2.

On the other hand the triangle in R? defined by V=

contains E so that HL(E)

< |V| = /2.

{@, ye

Thus H1,(£)

R?: x, y2Ox+y< 1}

< H'(E£).

5{1(E) = 2 € (0,00), we have dim, E = 1 by (e) of Theorem 27.35.

Note that since

Note also that

we (E) = 0 while 2, (E) is undefined since £ is not a subset of R.

Example 2. If £ is a finite set in R" with & elements, then °(E)

= & by Proposition

27.31, On the other hand for any set V € 93(R") such that V > E we have |V|° = 1. Thus

HS(E) 0 such that | f(t’) — f(t’)| < elt! — t”| for 2’, t” € [a, b]. LetP € 98,4 be given by P : a = fp < --- < fm = b. Then

Vif?) =

k=1

if) — FGI on [a, 5].

t’ < ¢” then g(t’) A g(t”). Then for at least one of the component Say 97, we have g;(t’) # g;(t”). This implies that for the total variation have ¢;(t’) < ;(t”). Then by (2) in the Proof of Lemma 28.19, we 9; (¢”) — 9; (t’) > 0. This shows that ¢ is a strictly increasing function

§28 Transformations of Hausdorff Measures

g(b)

727

Now that 9 is a strictly increasing continuous function on [a, b] with g(a) = 0 and = £(G), its inverse function g@ is a strictly increasing continuous function on

[0, £(G)] mapping [0, £(G)] onto [a, b]. Let us define a mapping y of [0, 4(G)] into R”

by

3)

v6) =(gog")G)

fors €[0, £(G)].

Then y is a one-to-one continuous mapping of [0, £(G)] into R” and in fact we have

y (10, £G)}) = (g 0 @*)((0, £G)]) = a([a, B]) =G. Thus y : [0, £(G)] —

R” is a parametric representation of G.

Let us call the variable

s = p(t) fort € [a, 5] the arclength of G and call y(s) for s € [0, £(G)] the parametric

representation of G by the arclength. For s’, s” € [0, £(G)] and s’ < »”, let t’ and t” be the unique points in [a, b] such that s’ = p(t’) and s” = g(t”). Then y(s’) = g(t’) and y(s”) = g(t”). Thus by (2) we have

(4)

s"—s' =(t")— g(') = £ (Gee,20) =£(Gy6,y@9) = Ivo) — vol.

This shows that y is a Lipschitz mapping with coefficient 1. Theorem 28.23. Let G be a Jordan curve in R". Then H(G) Proof. 1. Let g : [a, b] >

= £(G).

R” bea parametric representation of a Jordan curve G in R". Let

,t" € [a, bland?’ < 2” and consider Gg, segment in R” joining g(t’) to g(t”) and let T containing [g(t’), g(¢”)]. Then for any x, y T is a Lipschitz mapping with coefficient 1.

gy) = all, 2”). Let [g(t’), g(¢”)] be the line be the orthogonal projection of R” onto the line € R” we have |T (x) — T(y)| < |x — y| so that Note also that T (Gen, ee) > Le’), ge").

Thus by (c) of Theorem 28.4, we have

@

HO (Geen) =H" (T (Geer,20)) = H (le),

= u,([a@), 8@"0)) = le) — 81

1)

where the first equality is by the fact that [g(’), g(t”)] C R and H'(E) = u* (E) for any E € $(R) according to Theorem 28.12. ForP : a = to < +--+ < tm = b, we have

m

VE(g, P) = > lg(te) — 8D k=1

m

< 0H! (Gen .em) = HO), k=l

where the inequality is by (1) and the last equality is from the fact that Ggiy_1),e@) is a compact set in R” by Observation 28.15 so that Germ ),e(4) € Ber C MH") and by the fact that |J7_1 Ggc_1),e(%) = G and moreover the intersection of any two of the sets Gg4_1),g(4)) * = 1,..., m, is either an empty set or a singleton for which the 5! measures

are equal to 0. Then we have

)

LG) = sup Vole, P) G such that both 1"(O) and

(u")"(O) are finite.

Consider first the case (u")"(G) < oo. By Proposition 24.6, there exists a sequence (Oj : j € N) in J} such that joy OF D G and Y joy (u2)"(0;) < 00. If we let O = Ujen Oj, then O is an open set containing G and (u2)*(0) G such that both H"(O) and (u2)"(0)

are finite. Since G is a G;-set, there exists a decreasing sequence (G; : j € N) of open sets containing G such thatG = lim G;. With the open sets O D G, we have joo

§29 Measure of Integral and Fractional Dimensions

G=GNO= and since

lim (G;

0). Since H” and (u2)" are measures on the o-algebra By

jroo

K(G;N 0) E such that

H"(G’) = 1("(E). By Lemma 24.11, there exists a Gs-set G” D E such that (1")"(G") = (u8)*(é).

Then since E C G'NG" C G’, we have H"(E) = H"(G’N G"). Similarly by

EC G'NG" CG", we have (u*)"(Z) = (u")"(G'N G"). Since G’ and G” are Gs-sets,

so is G’ G”. Then by our result in 3, we have H"(G' NG") = kn(u")*(G! NG"). Thus H"(E) = x, (u2)*(E).

Corollary 29.3. For every n € N, we have DUH") = Mt((u")"\E Nt"), that is, the o-algebra of H"-measurable subsets of R" is equal to the o-algebra of (ut)*-measurable subsets of R®.

Proof. Let E ¢ 93(R"). Then E € S0t(H(") if and onlyif for every A € 98(R") we have @

H(A) = HUAN E) + HAN E),

and E € 99t((u”)") if and only if for every A € 93(R”) we have

@

(4f)°A) = 2)" NB) + (HD)*(AN EY).

According to Theorem 29.2, H"(F) = ky(u")"(F) for every F € $8(R"). Thus if E satisfies (1), then it satisfies (2) so that 99%(3(") c 9((u")”) and similarly if E satisfies

(2), then it satisfies (1) so that 9t((u")") c MH"). Thus MH") = Me((u")").

a

[1] Calculation of the n-dimensional Hausdorff Measure of a Unit Cube in R” Let Q be a unit cube and S be a unit ball in R”. We show that x, = H{”(Q) and yy, = y22(S) are related by kp Yn, = 1. Definition 29.4. Forx = (11,..., %,) € R" andi =1,...,n, let x" be the point whose i-th coordinate is the negative of that of x. We say that a set E in R" is symmetric with

respect to the hyperplane H; = {@a, weep Xn) € Ri xy = 0} if for every x € E we have [i] xe E,

732

Chapter 7 Hausdorff Measures on the Euclidean Space

Theorem 29.5. (Steiner Symmetrization) Let E be a bounded set in Syn. i=1,...,n, there exists a set S;E in R" with the following properties:

For every

(a) SE € Bp.

b) u2(S/E) = uh(E).

(© |SjE| < |El. (@ S;E is symmetric with respect to the hyperplane H; = {(x1, ...,%n) € R” : x; =O}.

(e) If E is symmetric with respect to a hyperplane H; for some j = 1,...,n, then S;E is symmetric with respect to both H; and H;. Proof. For simplicity in notation, we prove the theorem for i = 1. Let 7 be the projection

of R* = R xR"! onto", that is, (x1, ...,%n) = (a, ....4n) for (x1, ...,%n) © B®. Thus for every E € §3(R") we have

x(E) = {y € R" : there existss € R such that (s, y) € E}. If £ is a bounded set in BR, then z(£) is a bounded set in R*-1, For every y € R"-! let Et, y) ={s €R: (s, y) € E} and call this subset ofR the section ofE at y. Note that if

y ¢2(E), then E(., y) = 8. Now 8m = 0 (Xj_13x) = 0 (Br x Byr-1) by Theorem

24.24 and Proposition 23.5. Thus E(-, y) € Sx forevery y € R"~! by Proposition 23.9 and

4, (EC, y)) is a real-valued 93.g,-1-measurable function of y € R"—!. For every y € R"}, let us define a set in R by

(1)

ly = (-34, (EC), 542(EC »)))

with the understanding that J, = @ if u, (EC. y)) = 0. Note that for y € R"—! such that #,(EC., y)) > 0, J, is an open interval of length 4, (E(-, y)) with center at the origin of R. Let us define a set in R” by setting

Q)

ME=

LJ 4x).

yeR1

The section of this set at y € R"—1 is equal to I,, that is,

@Q)

GiE)C, y) = Jy. Consider the hyperplane H, = {(x1,...,%n) € R" : x1 = 0}. Let us show that S1Z

is symmetric with respect to Hi. Let x = (41,...,%n) € SiZ. By (2), x1 € J, where y= (%,..., 4 € R®—1 is such that Ty #@. Now x1 € J, implies —x, € I, by (1). Thus (—21, x2,.--,%n) € SiE by (2). This shows that 5,Z is symmetric with respect to Hj. To show that S,£ ¢€ Sx, consider the nonnegative real-valued %3g,-1-measurable

function }4,(E(-, y)) of y ¢ R"-1. By Lemma 8.6, there exists an increasing sequence (g; : 7 € N) of nonnegative simple functions on (R"—!, 93pn-1) such that we have g;(y) t 4p, (EG, y)) for every y ¢ R"!. Let B; = {(s,y) € R” : |s] < 9,(y)] forj € N. Since

(X41, ...,%n) F [x1] is a real-valued 93g7-measurable function on R” and (s, y) +> g;(y) is also a real-valued %$R»-measurable function on R", we have B; € Bye by (3) of Theorem

4.16. By the fact that gj(y) t du, (EC, y)) for every y € R"! and by (1) and (2), we

have S,E = Ujen B;. Thus S,E € BR.

§29 Measure of Integral and Fractional Dimensions

733

The fact that 27 (51 £) = 47 (E) follows from Proposition 23.16 for a measurable set in a product measure space. Indeed we have

WESLE) = (1, x WE) SE) = fm, (SEDC) wey) = [

Hey) wy) = [

be (EC) HE ey)

= (4, x ut!) ©) = #7). where the second equality is by Proposition 23.16, the third equality is by (3), the fourth equality is by (1) and the fifth equality is by Proposition 23.16. Let us show that |S, Z| < |£|. For every y € REC,

us define a finite closed interval by setting

(4)

y) is a bounded set in R. Let

J, = [inf EC, y), sup EC, y)] CR.

Since Jy D> E(-, y), we have by (1)

6)

#, (Jy) = sup EC, y) — inf EC, y) = a, (EC, y)) = 4, Gy).

Let x’ and x” be two arbitrary points in $;E. By (2), we have x’ = (s’, y’) where s’ € Iy and y’ € 2(E) and similarly x” = (s”, y”) where s” € Iyv and y” € (E). Let c’ and c” be respectively the midpoints of the closed intervals J and J, defined by (4) with our y’ and y”. Then by (1) and (5) we have

©

Is’ — 8" LEI.

Suppose there exists x € SE

with |x| > 4|E|. Now x € SE implies that —x € SE also. Thus |x — (—x)| < |SE|. But

|x — (—x}| = |x| 4+|—x] = 2|x| > |Z| > [SE], a contradiction. This shows that SE is contained in a closed ball with center at 0 € R” and radius AE. The volume of such a ball

is given by 2/2 (AJEZ|)” /T (% +1) = yalEl" by Theorem 26.13.

Proposition 24.10 is a covering theorem for open sets in R” by binary cubes in R". The next Proposition is a covering theorem for open sets in R” by balls in R”. Proposition 29.7. Let G be a bounded open set in IR". The for every © > 0 there exists a disjoint sequence (Sy : k € N) of closed bails with diameters not exceeding € contained in

G such that w" (G \ Open Sz) = 0-

Proof. 1. Let ¢ > 0 be arbitrarily given. Let (S; : k € N) be a disjoint sequence of closed balls contained in G with diameters not exceeding ¢ and satisfying the condition that if we let G; be the collection of all closed balls with diameters not exceeding ¢ and contained in the open set G \, Uh S; and if we let d, be the supremum of the diameters of the members

of G, then |Se41| > Ade.

Let us show that a sequence (S; : k € N) of the description above exists by mathematical

induction. Let S, be an arbitrary closed ball with |51| < ¢ and contained in G. Then G \ Sy is a non-empty open set so that G, # 9. Thus we can select $2 € © with |S)| > hd. Suppose for some k € N we have selected k disjoint closed balls 51, --- , 5, contained in G with diameters not exceeding ¢ and |S;| > dit fori = 2,...,k. Then since UE, Sjisa

§29 Measure of Integral and Fractional Dimensions

735

closed set contained in the open set G, G \, UE, SS; is a non-empty open set. Then G; 4 G and we can select Sg41 € Gg with |Sk41| > 3g. Thus by induction (5;: k € N) exists.

2. Let us show that 2” (G\ Uzen Se) = 0. Suppose 2? (G\ Uzer Se) > 0. Let yn

be the volume of a ball with diameter 1 in R*. Then 47 (Sk)= |Sk|"yn- By the disjointness of the sequence (S; : k € N) and by the fact that G is a bounded open set in R”, we have

Yn D> ISel” = Yu (Se) = w2(L) Se) = uB(G) < ov. keN

keN

keN

Thus > yen | Sel" < 00, which implies jm, |Sq|” = 0 and then im

|Sz| = 0. Let 5; be

the closed ball concentric with 5; and having [Sel= 41S¢l.- Then pue” (5) = 4%" (Sz) so that Den a?(Sk) = 4 ren #7 (Sk) < 00 and therefore

mn, Dien

2 (54) ="0. Now

since 2” (G \ Uzen Se) > 0, for sufficiently large N € Nw we © have

ut( UJ &) 2 H(i) < ut(G\ (J &). k>N

keN

Thus there exists x9 € R” such that N

()

x €G\|(J&CE\LJ& keN

k=1

and x0 ¢ J &. k>N

Since Urea Sx is a closed set contained in the open set G, there exists a closed ball S contained in G with center at xp and |S| < ¢ such that S CG \ Us

S; and then

N

2)

soUs=s

Now if SA UE, S; = @ for some k € N, then S Cc G\Uh, [S| < dy < 2|Sg41|.

Since jim

> 00

S; so thatS € ©, and then

|Sz| = 0 and |S| > 0, this cannot hold for everyk € N.

Therefore S intersects S; for some k € N. Let 5S;, be the first in the sequence (S; : k € N) to intersect $. Then

Q)

SOS

AB

and

By (2), we have ko

> N.

Then by

SN(S,U---US,-1) =B (1), we have xo ¢ Sia:

Since SA Sz

4 @ and

xo ¢ Sh, we have 41S] + 41Skol > 4 (41S:ql) that is, [S| > 3]Szo]. On the other hand, since SNS,

U---U Sp.-1) = GB by (3), we have S € G,,_1 so that |S| < djy-1 < 2|5})|. Thus

3[Skq| < |$| < 2|Skq|, a contradiction. Therefore 12"(G \ Ugen Sz) = 0.

Theorem 29.8. Let k, = 3("(Q) and yn = 7 (S) where Q is a unit cube and S is a unit bail in R", that is, a ball with diameter 1. (a) ky, and yp are related by the equality knyn = 1. (b) The n-dimensional Lebesgue outer measure (ut)* and the n-dimensional Hausdorff

736

Chapter 7 Hausdorff Measures on the Euclidean Space

measure ‘{" are related by the equality y,-! (28)"(E) = H"(E) for every E € BR"). {c) H"(S) = 1 fora unit ball S in R". Fora ball B in R", we have H("(B) = |B/". Proof. 1. Consider Q = (0,1) x --- x (0, 1), an open unit cube in R". By the equivalent definition of 3(" by 5-covers of closed sets (Theorem 27.13), H}(Q)

=

F3(Q)

where

F3(Q) = inf { Vien Fil" : (Fi tt € N) is a 6-cover of Q by closed sets} forevery8 > 0. Thus for an arbitrary ¢ > 0 and 5 > 0, there exists a 6-cover (F; : i € N) of Q by closed

sets such that

(1)

YAM < HQ) +e =HGQ)+e 0, we have H"(Q) = wi(Q)/yn. Since H"(Q) = ky and w7(Q) = 1, we have knyn < 1. This

completes the proof of the equality k,y, = 1.

2, By Theorem 29.2, H"(E) = kn (22)"(2) for every E € $$(R"). According to (a), ky = - Thus we have (”)"(E) = yn3t"(E). This proves (b). 3. For every E € $8(R"), we have H"(E) = «,{(u")"(E) by Theorem 29.2. If § is a unit ball in R”, then (2) = Yq by (2) of Theorem 29.6. Thus we have H"(S) = kn(u2)"(S) = kan. Since knYn = 1 by (a), we have H"(S) = 1. Let B = B(x, 3), an open ball with center at x and diameter 4. Now B (x, 4) = AB (A~!x, 4) by Proposition

24.34. Then H"(B) = H" (AB (a-!x, 3)) = ad" (B(aq+x, })) = a7 = [BI by Theorem 28.1 and by the fact that B (41x, 5) isa unitball in R” for which the n-dimensional Hausdorff measure is equal to 1 as we showed above.

[11] Transformation of Hausdorff Measure of Integral Dimension Let T be a non-singular linear transformation of R" into R". By Theorem 29.2 and Theorem 24.31, we have

H" (T(E) = «x (uf)* (TED) = Kal det Mr|(u7)*(E) = | det Mr|9t"(E) for every E € 93(R"). Let us consider a non-singular linear transformation of R* into R". Observation 29.9.

Let T be a linear transformation of R* into R”.

It is a well-known

fact in linear algebra that there exists a linear transformation 7* of R* into RX, called the adjoint of T, such that (Tx, y} = (x, T*y) for everyx € R* andy € R® where (-, -) is the inner product. The matrix of T, My, is an n x k matrix and the matrix of T*, My«, isa

& x n matrix and the two matrices are related by Mr« = Mj. The composition T*T is a linear transformation of R* into R¥ and its matrix Mrr+ = Mr+Mr is ak x k matrix. The transformation 7*T is positive semidefinite, that is, (7*Tx, x) > 0 for every x € R*, and

det Mr7+ > 0.

Now let T be a non-singular linear transformation of R* into R" where k O satisfying8 < (cp — 1)89 and8 < (1—c')éo.

Then select L; € Lo so that ||L, — dT (x, -)|| < 6. Then for every z € R*, we have by the

triangle inequality of the norm in R” and the definitions of 5 and 69

[Lr(z)| < laT (x, 2) + [Lr(@) — aT (x, 2)| < |dT (x, 21 + lz < |dT (x, 2) + (Co — 1)S0lz| < |€T(, z)| + (Co — DIT G, 2)| = coldT (x, z)ISimilarly we have

[L,()| = |€T (x, 2) — [Lr (@) — aT (x, 2)| = |€T (x, 2)1 - lz > [dT (x, 2) — (1-5 )éolzl = dT, 2) — (1—c9')ldT, 2)| =e9 dT (x, 2)1.Thus we have cot |Lr(2)| < |dT (x, z)| < co|Z,(z)|, that is, x satisfies (4). Since L, is a non-singular linear transformation, we have n = inf|z)~1 |LZ,(z)| > 0. The differentiability

of T at x implies that there exists p € N such that

(8)

IT) —T@) -dT@, y — x)| < enly —x| < elL,-(y —x)] forx € Q such that |y — x| < i.

Then we have

IT(y) — TO) Ss ITO) — T@) - aT, y — x)| + |dT@, y — x)

SelL-(y — x) + colLr(y —x)| op |Le(y — x)| —elLe(y — x)| =e 1L-Gy —x)]. This shows that x satisfies (5). Thus x € E;,p. Therefore 2 = ew Upen Erp For our E;,p € Bp, let {Ey : g € N} be an arbitrary countable collection in Bp such that the diameter |E;,p,g| < 5 for q € N and Ujen Erpq = Erp. Now

§29 Measure of Integral and Fractional Dimensions

2 = Uren pen Ugen Erp.q-

741

Let (Ay : j € N) be an arbitrary numbering of the

countable collection {Ey,p,g:q €N, p €¢ N,r € N}. If Aj = £,,p,q in this renumbering,

let Lj = Ly. Hwe let By = Ay and B; = A, \ UZ)B; forj > 2, then (Bj : j e N)isa

disjoint sequence in Sp and the sequence (L; : j € N) of the corresponding non-singular linear transformations of R* into R* satisfy conditions (1) and (2).

Theorem 29.14. Let 1 < k &}

are two open sets in R*. Clearly they are disjoint. Since T (x1) = 0 < &, we have x1 € O1, and since T (x2) > & we have x2 € O2. For anyx € E, sinceé ¢ T(E) we have T(x) # & so thatx € O01 U O2. This shows that E C O; U O2. Thus we have shown that for any pair of distinct points x, and x2 in £, there exists a pair of disjoint open sets O, and O2 in X such that x1 € 01, x2 € O2, and E C 01 U O2. This shows that E is a totally disconnected set.

For examples of sets with Hausdorff dimensions in the range (0, 1), we define Cantor

sets C;, with ratio A € (0, }) next. The Cantor ternary set defined in §4 is a particular case

of C, with A = }.

Definition 29.17. Let 4 be a positive number in the range (0, 4). We define a sequence (Be: k € Zy) of classes of closed intervals in R inductively as follows. Let 3o = {Jo,1} where Jo, = [0, 1]. Let 31 = {Ji,1, J1,2} where J,,1 and J,,2 are the two closed intervals

of length 0 obtained by removing an open interval of length 1 — 2A from the center of Jo,1. Suppose for some k © N, the collection 3x_1 = {Jg-1,i 23 = 1,..., aly

of

4 closed

intervals of length )*—| has been defined. We define 3, = (Jk; : i = 1,...,2*} as the

collection of 2* closed intervals of length 1 obtained by removing an open interval of length (1 — 24)a*-! from the center of each of the members of 3-1. We continue the process

indefinitely. Let 3 = yer, Se For each k © Zy, let Ey = U2, Jes. We define the Cantor set C), with ratio 4. € (0, 4) by setting C, = Mex, Ek

Note that Ey € Bp for every k € Z so that Cy € Br. Also 4, (Ex) = 2*a* = (2a)*

and 4, (C,) = jim

> CO

i, (Ey) = O since 2A < 1.

Proposition 29.18. The Cantor set C;, has the following properties: (a) C, is a null set in (RB, Sr, Hy): (b) C), is an uncountable set.

(©) C, is a compact set in R. (d) C, is a perfect set inR.

(©) Cy is nowhere dense in R, that is, (Ci)° = @. Proof. These statements can be proved in the same way as Theorem 4.33 for the Cantor ternary set. ©

Lemma 29.19. Regarding the classes of closed intervals 3; fork € Z, and 3 = (rez, Sk in Definition 29.17, we have the following: (a) Ifx is an endpoint ofa member of J, then x € Cy.

(b) Ifx € C, and x is not the left endpoint of any member of 3, then there exists a sequence (mim

€N) such that xp, is the left endpoint of a member of J, Xm < X, and Xm t x.

Similarly ifx ¢ C, and x is not the right endpoint of any member of J, then there exists

744

Chapter 7 Hausdorff Measures on the Euclidean Space @ sequence (Xm :m €N) such that x» is the right endpoint of a member of J, Xm > x, and Xm | x.

(c) Ifx ¢ C,, and there exists a point € € C, such thatx < &, then there exists a € Cy such thatx < a, ot is the left endpoint of a member of J and [x, a) C), = @. Similarly ify ¢ C) and there exists a point n € C, such that n < y, then there exists B € C) such that B < y, B is the right endpoint of a member of3 and (B, y) MC, = @. Proof. 1. If x is a left endpoint of a member of J, then x is the left endpoint of a member of 3; for some k € Z,. Then x is the left endpoint of a member of J,

for every m > k so

thatx ¢ E,, for every m > k. Thenx € Cy = (ez, Ee. Similarly for the case that x is a right endpoint of a member of J. 2. Suppose x € C, and x is not the left endpoint of any member of J. Now since x € Cy = Mrez, Ek, we have x € Ex for every k € Z 4. For everyn € N, there exists k € Z i=1,...,2*.

such that raed i,

Since x € Ex = U2, Jxi, we have x € Jz; for some

Since eal = Mog

i the distance between the left endpoint &, of Jj,; andx

is less than i Thus & < x andx —&

< i Then

tim, fn =x. Let x», = max{&,...,&m}

form ¢ N. Then (xp, : m € N) is a sequence of left endpoints of members of J, x», < x,

and xm + x.

Similarly for the case that x € C, and x is not the right endpoint of any

member of J. 3. Supposex ¢ C,, and there exists a point

€ C, such thatx < &. Then [x, 00) NC,

is nonempty and is a compact set. Let a be the nearest point in the compact set [x, 00) NC), from x. Then [x,@) 1 C, = @ so that a must be a left endpoint of a member of J by (b).

Similarly for the case that y ¢ C, and there exists a point 7 € C, such thaty < y.

©

With the notations introduced in Definition 29.17, let C’ = C,NJi,, and C” = C,NJi,2. Then C’N.C” = §, C'UC” = Cy. Moreover it is evident from the construction of C,, that C’ = AC,, a dilation of C, by A, and C” = C’ + (1 — A), a translate of C’ by 1-4 ER, as can be seen by the construction of C,. Since C, € BR C M(H") for an arbitrary s € [0, 00), we have C’ € Nt(H*) by Theorem 28.2 and, since a translation is an isometry,

C” € MUCH") by Theorem 28,11. Then since 1H

is a measure on the o-algebra M(H),

we have H*(C,) = #5(C’) + 3°(C"). Now H°(C’) = A*H(*(C,) by Theorem 28.1 and H(C"”) = H*(C’) by (c) of Theorem 28.10 so that we have qd)

HE (Cy) = 20. F5(Cy)

for every s € [0, 00).

Now consider the nonnegative extended real-valued function H{5(C,) of s € [0, 00). The fact that C, is an infinite set implies that (Cy) = oo by Lemma 27.31 and the fact that C, Cc R implies that H*(C,) = 0 for s > 1 by (c) of Lemma 27.30. Since C, # 9,

3{*(C,) has exactly one point of discontinuity and the discontinuity occurs at a point in [0, 1] according to (d) of Theorem 27.33. Now oo, a finite positive number, and 0 are the three possible values of F(°(C;) at its point of discontinuity s < [0, 1]. If we assume that

HS(C,) € (0, 00), then dividing (1) by this positive number H{*(C,) we have 1 = 2a5. Thus log2 + slogA = 0 and therefore (2)

s=

log2

logd

log2

log”

§29 Measure of Integral and Fractional Dimensions

745

The calculation above for the point of discontinuity s = log 2/ log 4—! of the function H*(C,) was based on the assumption that the value of F{*(C,) at the point of discontinuity

is a finite positive number.

In what follows we define s* = log2/log’—! and show that

vie (C,) is indeed a finite positive number.

dim, Cy = s* = log2/log a}.

By (e) of Theorem 27.35, this implies that

Lemma 29.20. For4 € (0, 4), let s* = log2/loga~! = — log2/loga € (0, 1). (a) For every x € R, we have 2* = aoe

(b) If I € Big for some ky € Zy, then |J" = Ty, cz le al” for any k > ko. (©) For every k € Z1, we have a

[ei

= 1.

(d) For an arbitrary open interval in Randk < Z,, we have Y"y,.cr |Sei Proof.

ko, the interval

J contains 2'—" members of 3x = {Jes si = 1,...,2"}. Since |Jgi] = A* for every i=1,...,2%, wehave Dy, 07 Jail” = 2k a". Now 2*-40 = 40D" py (a). Thus

ST ial” = a8

Fict

This proves (b). 3. (c) is a particular case of (b).

Se =

= glo Sp”,

Indeed Jo, = [0, 1] € Jo contains all members of

[Jessi =1,...,2*} for an arbitraryk € Z, so that by (b) we have

1=\oil*=

ak

So Weil =o dei”.

FiiCJo,1

i=1

4. Now suppose that an open interval J in R contains no members of the class J, =

{Jz,::i=1,...,2*} for some k € Z,. Then the inequality in (d) holds trivially. Suppose

I contains some member of Jz. Let ko be the smallest nonnegative integer such that J contains some member of Jz. Then ko < k. Let Jig, j,,--++ Jka,j, be all the members of Ji that intersect J. Then p < 4 for otherwise I would contain some member of J4,-1.

Now since p < 4 and Jy.,j, C I form = 1,..., p, we have

4

P

= onl n=1

where the equality is by (b).

P

=>

SO

n= JkiCJig,in

eal = YO eal, Ieacl

746

Chapter 7 Hausdorff Measures on the Euclidean Space

Lemma 29.21. Let p € [0, 1] andA

ad

€ [0, 1]. Then for any a), a2 > 0, we have

{aa + (1—A)aa}? > aa? + (1 -Aak,

(2)

al +a? > {a +a2}”.

Moreover if a, > Oforn € N, then

Q)

al +-+-+ah > {aj+---+ay}? foreveryN EN,

@)

ez (Ya)’.

neN

neN

Proof. Consider the function f(x) = —x? forx € (0, 00). Since f'(x) = —pxP-) is an increasing function of x € (0, 00), f is a convex function on (0, 00) by (b) of Theorem

14,10. The convexity of f implies f (Aa; +(1—A)az)

< Af (@1)+(1—A) f(@). Multiplying

both sides by —1 we obtain (1).

To prove (2), we may assume without loss of generality that a, < a2. Assume further

that a, < a2. Consider 0 < ap < a1 < az < a; + a2. Applying Proposition 14.4, we have

F@) — f@) . f@) — F@o) . fla) — f(a) Ma-G

~

{ay +a}? — af ~ a ,

ay > {a1 + a2}? —a a

aq

that is, a? + af > {a1 + a2}?. For the case a, = ap, by letting a2 | a1, we have af + af > {a: +41}?. This proves (2). By repeated application of (2), we obtain (3). Letting N — oo in (3), we obtain (4) by the continuity of the power function.

&

Theorem 29.22. For the Cantor set C,, with 4 € (0, ), ifwe let s* = log2/logA—", then

1 } and HS"(Cy) = tim HE (Cy) > }. Thus } < H*"(C,) < 1. Then dim,, C, = s* by () of Theorem 27.35. Corollary 29.23. For everys € (0, 1) there existsa subset E ofR withdim, E = s. Indeed

for the Cantor set Cy, with 4 = exp { — 1 log2} we have dim, (C,) = s. Proof.

According to Theorem 29.22, for a Cantor set C, with A €

dim,(E) = log2/logA—!.

(0, 4), we have

Now the function (A) = log2/loga-! for A € (0,4) is

a strictly increasing continuous function mapping (0, 4) one-to-one and onto (0, 1). Its inverse function g~1 is then a strictly increasing continuous function mapping (0, 1) one-

to-one and onto (0, 4). Solving s = log2/logA—! for A, we obtain the inverse function 1=l(s) = exp{- i log 2} fors € (0, 1). Then fors € (0, 1), for the Cantor set C,-1(5) we have dim, Cy-1¢) = ¢(y"1(s)) = 5 by Theorem 29.22. & We show next that actually a" (C,) > 1. This is a shaper estimate than the estimate

HS" (C,) > } in Theorem 29.22. However the simple argument in obtaining (°"(C,) > } can be generalized in other situations where precise value of the Hausdorff measure is not necessary.

Theorem 29.24, For the Cantor set C, with } € (0, 4), we have 3O"(C,) = 1 where s* = log2/loga—'.

Proof. 1. For an arbitrary 6 > 0, let J = (a, 6) be an open interval such that a, b € Cf and

IMC, # G. Then there exists § € C,, such thata < &. Thus by (c) of Lemma 29.19, there

exists a € C, such that a < a, a is the left endpoint of a member ofJ = Urez,. J, and

(a, a)

Cy = G. Similarly there exists 8 € C, such that 6 < b, f is the right endpoint of

a member of 3, and (8, b) 1 C, = @. Thus we have I = (a, b) = (a, a) U [a, 8B] U CB, db)

and (a, a), (8, b) C Cf. Now if a point is the left (resp. right) endpoint of a member of J, then it is the left (resp. right) endpoint of a member of 3; for some k € Z, and then itis the left (resp. right) endpoint of a member of J, for every £ > k. Therefore there exists k € Z, such that a is the left endpoint of a member of 3; and £ is the right endpoint of a member of the same Ji. Let (Jez, + p = 1,...,¢} be the collection of all members of J; contained in [«, 6]

and enumerated from left to right. We have

q)

INC, C Ji

U-U Siig:

§29 Measure of Integral and Fractional Dimensions

749

Let us show that

(2) Now ifq =

ITI > Jeg” Ho + Jia 1, then since J >

J;,;,, we have \\s*

>

eal”

so that (2) is valid in this

case. Suppose q > 2. Consider the collection © of q — 1 open intervals, each between two consecutive members of {Trip :p=1,...,q}.

ftwo members of G have the same length

then there must be a member of & with a greater length. Since the member of © are all contained in [a, 8], there exists a unique member with a maximal length. Let G = (@’, 6’) be the member of @ with the maximal length. Let Fy = [a, «’] and F) = [’, 8] so that {F1, G, Fo} is a disjoint collection and [a, 8] = F, UGU F). Now G is contained in aclosed

interval J in the class J¢ for some £ < Z, with |G|/|J| = 1—2a. Let J’ and J” be the two closed intervals in the class 32,1 which constitute J \ G. Then |J’|/|J| = |J”|/|J| = 4. The fact that G is the unique open interval with maximal length in the collection G implies

that F, C J’ and Fy C J”. Thus |F,|/|J| < |J’|/|J| =. Consequently

Vl _WF/JI IG|

so that [Fi]
Ifa, Bll = Fil + 1G] + |Fal

= (775) a

+ a |4l+x ln| tea Il see

= (75) (ut

2

ae

IFal}

shin +1Fl}

1

= yy ilFAl + |Fol}.

Since s* € (0, 1), we have (}{a; +a5})"> haf" + }a$" for a1, a2 € (0, 00) by (1) of Lemma 29.21. Thus

Al+12l* A ql" > {! ail + | alt > Al

wre

=e

LAP [Fal

7 Oe

.

By (a) of Lemma 29.20, we have 2a" = 1. Thus

Ih > WAY + LI. Applying this process of reduction to F,; and F, in the place of [a, 8] and repeating, we finally have (2).

2. Let ,..., Zy be open intervals with endpoints in Cf such that C, C Ura I;. By dropping those intervals in the collection that are disjoint from C, we do not increase the sum )> jai Wy }*". Thus let us assume that none of the open intervals is disjoint from C,. By our result in 1, for each j = 1,..., N there exists k; € Z4 such that for the collection

750

Chapter 7 Hausdorff Measures on the Euclidean Space

{Je;.ip | P= 1,---,4,} of all members of the class 3, contained in Ij we have according

to (2)

. aq . i” = Sean lp=

Letk = max;—,...w kj. By (b) of Lemma 29.20, foreach j = 1,..., Nandp=1,...,4;

we have Wijipl

= LV tiCejip CSE; \Jeal- Thus for each j = 1,..., N, we have ay

BI 2 >>

YO

vet= YO el”.

YO

wel = Yoel

P=l SkiCIiyip

Tea

Then we have N

QB)

“

igre jal

a

“

Fes Ca

iol

”

=1

where the equality is by (c) of Lemma 29.20. 3. Let > 0. Let (K; : j € N) be a 8-cover of C, by closed convex sets in R. Recall that a closed convex set in R is a closed interval. Let ¢ > 0 be arbitrarily given. Now since C, has no interior points, we can select an open interval I; such that 7; > K;, the endpoints of I; are in Cf, and

4

Yui

jeN


1. Then

= inf {Xjen |\Kj ar (Kj : j €N) isa 8-cover of C, by closed convex sets} >1.

Since 33"(C,) = K3"(C,) by Theorem 27.25, we have H"(C,) > 1. Thus we have 50°(Cy) = Him H65'(Ca) > 1. Then since 4(*"(C,) < 1 by Theorem 29.22, we have 30°(C,) = 1. 0

Appendix A

Digital Expansions of Real Numbers A decimal number 0.i,i2i3---i,---,

where i,

€

{0,1,...9}

for every n

€

N, is not

so much a real number as it is a representation of a convergent series of real numbers Dnen 197: Moreover this way of representing a real number is not unique. For instance we have 0.03 = 0.02999. - Let us reflect that within the set N of all positive integers there is nothing special about the positive integer 10. We show below that for every p € N such that p > 2, every x & [0, 1) can be expressed as rey

where i, € {0,1,...,

p — 1} foreveryn ¢ N

nen?”

and thenx has the representation x =0.i,i2i3

---

wherei, € {0,1,..., p — 1} foreveryn

€ N.

In particular for p = 2, we have the binary numbers x =0.i,igi3---

wherei, € {0, 1} foreveryn € N.

For p = 3, we have the ternary numbers x =0.i)i2i3

---

wherei, € {0, 1, 2} for everyn € N.

For p = 10, we have the decimal numbers x = 0.i,igi3 --- where i, € {0,1,...,9} for every € N.

[I] Existence of p-digital Expansion Observation A.1. Let p € N and p > 2. Then we have

(1)

@)

y=

—

nen pe

ray 751

752

Appendix

In particular for p = 2, p = 3, p = 4, and p = 10, we have

1 —=1

1

neN a

Proof.

1

i

neN ”

1 1 — ==

2

neN ”

1

1

Lin —_=-,=3

3

neN

For r ¢ R, the geometric series with ratio r, >,

yr”, converges if and only if

|r| < 1 and when this is the case we have

ares

neN

|

A

a}

= -3

ts)

te neN ep

“Ele=

Now ) nen7pr is a geometric series with ratio r = ; and |r| = ? < 1. Thus we have

This proves (1). Then applying (1), we have

_

=(p-1) yt

neN

=1.

neN »

This proves (2).

Proposition A.2. Let p € N and p > 2. Considera series of real numbers defined by yy

where i, €

{0,1,..., p—1}foreveryneN.

neN

The series always converges and the sum s € [0, 1].

Proof. To show that the series Donen

*, converges, we show that the sequence of partial

sums (s,: n € N) converges. Now s;, is defined by n

Sn =

y

*

q

=

iiP

forn € N.

Thus (s, : 2 € N) is an increasing sequence of real numbers and then to show that (s, : 2 € N) converges it suffices to show that (s, : n € N) is bounded above. Now

wees

SL

neN

ep

by (2) of Observation A.1. Therefore we have s, < 1 for every n € N and then (s, :n € N)

is an increasing sequence of real numbers that is bounded above by 1 and hence the series Donen P converges and the sums € [0,1]. &

A. Digital Expansions of Real Numbers

753

Definition A.3. (p-digital Expansion and p-digital Representation) Let p € Nand p > 2. Let . x=)

where

i, € {0,1,..., p — 1} for everyn EN.

neN

We call the expression

to represent om

be

We call it a p-digital representation of the real number x.

Observation A.4. Let p € N and p > 2. Let us decompose the interval J = [0, 1) into p disjoint left-closed and right-open subintervals of length ; Ji,

and let

where

i, € {0,1,..., p— 1}

31 = (4, 2H € (0,1,..., p>— 1}.

Then let us decompose each J;, € J, into p disjoint left-closed and right-open subintervals of length 2 Fiz where i2 € {0,1,..., p—1} and let n=

{ Firin :h,i2 € {0,1,...,p— 1}}.

Then let us decompose each J;,;, € Jz into p disjoint left-closed and right-open subintervals of length 4, P Jini,

and let

where

i3 € {0,1,...,p—1}

Ta = [Inns : ti, fa, is € {0, 1,...,2 — Uh}.

We iterate this process of division indefinitely. Then for every n € N, we have

q@)

Bn = (Fair. tH, Ha, ---y in €{0,1,-.., Pp — TH}

where each Jj,i,,.;, of the p” disjoint intervals of length » in J, is given by

@

i

kg

hh

Titania = [D+ a tote

to

kh

a ttt

in

1

et »

In the notation J;,;,__;,, the letter n indicates that the interval J;,;,__;, is a member of the collection 3, and the indices iji2 ...i, indicate the location of the interval Jj,;,._;, inside the interval [0, 1) as shown by (2). Theorem A.5, (Existence of p-digital Expansion) Let p ¢ N and p > 2. x € [0, 1) has at least one p-digital expansion, that is, we can write x=) aeN

a

where i, € {0,1,..., p— 1} foreveryneN

Then every

754

Appendix

in at least one way. Proof. Let x € [0, 1). For every n € N, consider £,, as defined by (1) in Observation A.4, that is,

Tn = [Firion

tis ta, «stn € (0, 1,..., p— I}.

Now the union of all members of J, is equal to [0, 1). Thus there exists a member of J, that contains x and thus by (2) of Observation A.4 we have

i *€ Iino = [2+

ok

tt in ty+

ig

in tet

1 a):

Then for every n € N we have

i in 1 yep meee ctii ey pet

@

Pp

Pp

P

PP

pp

Now according to Proposition A.2, nen pe converges and Dye i € [0,1]. Letting n —

oo in (1), we have i

i

tim {4 tee + p nrolp*

i

.

1

{+2p 4.. +2prlnL *+ im Po"\ tee

.

pal

2

mm .

oo) .

The right endpoint of the interval in (5) does not exceed the left endpoint of the interval in (6). Thus the two intervals are disjoint. Then the point x is contained in each one of the two disjoint intervals. This is an impossibility. Therefore x cannot have two distinct non-terminating p-digital representations.

[11] Cardinality of the Cantor Ternary Set Observation A.11. Let p € Nand p > 2. According to Theorem A.10, everyx € (0, 1) has

a unique non-terminating p-digital representation. On the other hand, by (2) of Observation A.1, we have

0.09 — DeP-DE-D--=

p-l1 neN

P

"

=1.

Thus every x € (0, 1] has a unique non-terminating p-digital representation. (As we noted in Observation A.6, the real number 0 has a unique p-digital representation but it is a terminating representation.) Definition A.12, When p = 2, 3, or 10, we call a p-digital expansion ofx € [0, 1]a binary, ternary, or decimal expansion of x and we call a p-digital representation of x € [0, 1] a binary, ternary, or decimal representation of x.

Observation A.13. Let E be the collection of all binary representations of real numbers in [0, 1]. Then &z is the collection of all 0.i1 2 i3 --- i, --- where i, € {0, 1} for every n € N.

Then we have

qd)

card(Z2) = 2%0,

Let &} be the subset of S2 consisting of all terminating binary representations and let £5 be the subset of £2 consisting of all non-terminating binary representations. Then we have

(2)

By = 5,U8)

and

8,985 =9.

758

Appendix

Now 3} consists of 0.000--0.4,000---

where i; = 1,

0.41 i2000---

where i; € {0, 1}, #2 =1,

0.4, i2i3000---

0.4, ---i,000---

where i}, i2 € {0, 1}, i3 = 1,

whereij,...,i,-1 € {0, 1},%, =1,

Then we have

@)

card()) = 14242? 4-2" 1 4...=No.

Then (1), (2), and (3) imply

(4)

cand(B3) = 2%,

Let 6 = 0.111 ---, the non-terminating binary representation of the real number 1. Then

6)

card(S2 \ {b}) = card(@2) = 280.

According to Theorem A.10, every x € (0, 1) has a unique non-terminating binary representation. This establishes a one-to-one correspondence between (0, 1) and © \ {b}. Then we have

©

card ((0, 1)) = card(B3 \ {6}) = 2".

Let f be a real-valued function defined on (0, 1) by setting

f@=x-4

forx€

1.

Then f maps (0, 1) one-to-one and onto (— i, 4) and thus establishes a one-to-one corre-

spondence between (0, 1) and (— i, 4). Thus we have

@

card ((— }, 3) = card ((0, 1)) = 2%.

Next let g be a real-valued function defined on ( _ 1, 4) by

g(x) =tanzx

forx € (—}, 3).

Then g maps ( - i, 4) one-to-one and onto R and thus establishes a one-to-one correspondence between (— i, 4) and R. Thus

(8)

card(R) = card ((— 3, 5)) = 2%.

A. Digital Expansions of Real Numbers

759

Definition A.14, (Cantor Ternary Set) Let T) = [0, 1]. Let us remove the open middle third of the interval To. Let T; be the resulting set. T, consists of 2 disjoint closed intervals of length }.

Let us remove the open middle third of each of the 2 closed intervals constituting T,. Let Ty

be the resulting set. To consists of 2? disjoint closed intervals of length (uy. Let us remove the open middle third of each of the 2 closed intervals constituting T,. Let Ts be the resulting set. Ts consists of 2° disjoint closed intervals of length (yy.

Let us iterate this process of removal indefinitely. Then for every n € N, T,, consists of 2" disjoint closed intervals of length ()*. Now (T, : n € N) is a decreasing sequence of subsets of To. Let

T=()h. neN

We cali T the Cantor Ternary Set. Theorem A.15. For the Cantor Ternary Set T, we have catd(T) = 28°, Thus T and R have the same cardinality.

Proof. Consider the ternary representation of x € Tp given by 0.i1 i2i3 --+ i, -+- where iz € {0, 1, 2} for every k e N.

Then x € 7; is represented by 0.i, ing «++ ig -+-

where iy € 0,2 and iz € (0, 1, 2} for every k > 2.

Then x € 7) is represented by 0.i, i2i3 --+ ig --- where i1, iz € 0, 2 and i; € {0, 1, 2} for every k > 3.

Then x € 73 is represented by 0.4 fig +++ ig ++

Where ii, iz, 73 € 0, 2 and iy € {0, 1, 2} for every k > 4,

and so on. Forn € N, x € T, is represented by O.i,igi3 ---+ ig +++

whereij,...i, € 0, 2 and iz € {0, 1, 2} for every k >n +1.

Thus x € T = (),enTn is represented by 0.i1 i2i3 «+--+ ig +++ where iz € (0, 2} for every k € N.

The cardinality of the collection of these representations is 280.

This page intentionally lett blank

Appendix B

Measurability of Limits and Derivatives [1] Borel Measurability of Limits of a Function Let f be an arbitrary extended real-valued function defined on the real line R. Let A be an extended real-valued function defined by setting

AG) = lim FQ) for x € R for which jim

F(y) exists in the extended real number system R. Let D(A) be

the domain of definition of A, that is, the set of points x € IR such that

lim f(y) exist in

yor

R. The main objective here is to show that D(A) is a Borel set and A is a Borel measurable

function on D(A).

Let us fix the notations and definitions. We write $m for the Borel o-algebra of subsets of R, that is, the smallest o-algebra of subsets of R containing all open sets in R. For x € R

andr > 0, we write B(x,r) = {y € R: |y — x] 0 there exists 8 > 0 such that IfG) - &| < & fory € Bo(x, 5). We say that iim

£0)

= o if for everyM

> 0 there

exists } > O such that f(y) > M fory € Bo(x, 4). Similarly we say that jim FO) =—0O if for every M > 0 there exists 6 > 0 such that f(y) < —M fory € Bo(x, 4). Let us define

liminf you fQ) =lim 840 and

Now the set Bo(x,5)

inf | fO)

yeBo(x,8)

limsup f(y) =lim sup f(y). yx 540 yeBo(x,8) | as 8 | Oand thus

inf

yeBo(x,3)

f(y) tand

sup

yeBo(x,d)

f(y) | as

6 | 0. Then lim inf f(y) and lim sup f(y) always exist in R and — oo < liminf f(y) < yor yon yor lim sup f(y) < 00. Moreover lim f(y) exists in Rif and only if lim inf f(y) = lim sup f(y) you yor you yo 761

762

Appendix

and when this occurs, then we have

jim FO) = lim inf f() = Him sup £(). Theorem B.1. Let f be an extended real-valued function defined on the real line R. Let g(x) = limint f(y) and (x) = lim sup ) forx € R. Then y and w are Brmeasurable extended real-valued functions onR. Proof. Let us prove the 3x-measurability of g. By definition we have

@1

= liminf f) =lim inf. f) feforx € eG) =liminf

R.

Let N be the set of all positive integers and let Z be the set of all integers. For each n € N, let us decompose R into intervals J,,, = ((k — 1)2-", k2-] for k € Z. Writing 1, for the characteristic function of a set A, let us define an extended real-valued function g, on R by

@

n(x) = > inff FO) -Una(e) forx eR. kez”

Since J,,, € Bp, 1);,, is a SpR-measurable function on R and then g, is a 53R-measurable

extended real-valued function on R. For every x € R, (v(x): n € N) is an increasing sequence in R and thus im, ¢n(x) exists in R. Then im, Qn is an extended real-valued function on R and the Be. “measurability of g, on R for eevery n € N implies the Bpmeasurability of lim | , on R. 1 To show that ¢ is a Sy-measurable function on R, we show that 9 = lim Mt

@n except

on a countable subset of R. Let £ be the collection of the end points of the intervals J,,4

fork € Zandn

€ N. Then £ is a countable subset of R. Let us call x € R a local strict

minimal point of f if there exists 7 > 0 such that f(x) < f(y) for all y e¢ (* — 4, x+4).

Let S be the collection of all local strict minimal points of f. Since any collection of disjoint open intervals in R is a countable collection, § is a countable subset of R. Let us show that for the countable subset

FE U S of IR, we have

forx € (EUS).

lim gn) =e)

For each n € N, we have Ubez Ink = R. Let J,,z be the interior of the interval J, ;, that

is, Ine = (& — 102, k2-). Let x € (EUS). Then since x is not an end point of I,,x

for any k € Zandn € N, for each n € N there exists k(n) € Z such that x is in the interior In,kin) Of In,eqn). Let 8, > O be so small that B(x, 8n) C Jn,k~n). Then we have for every

neN

@

ole) = =

ato yelver, a) inf

yEIn kn)

FOE

fO)=

inf, FOdE inf, £0)

ye Bo(x,8n)

inf

yElnin)

FO) =¢nG).

B. Measurability of Limits and Derivatives

763

By (1), for every e > 0 there exists 7 > 0 such that

4

e)—e
N. Then we have forn > N

(5)

inf

YEIn, k(n)

fO)>=

inf

~ yeBCr.n)

fO)=

inf

fo)

yeBo(x,n)

where the last equality is by the fact that x ¢ S. With (3), (2), (5) and (4) we have e®)>=n%)= > g(x)—e

inf

Elan)

fO)>

inf

fo)

yeBo(x,n)

forx € (EUS).

Thus we have shown that im, n(x) = g(x) for x € (E U S)°. This completes the proof that ¢ is S$-measurable on R.

The %$,-measurability of y on R follows from the fact that

¥@) = lim sup f() = —liminf(—f)()

forx eR

and the S%p-measurability of jim inf (—f)(y) on R as we proved above. Theorem B.2. Let f be an arbitrary extended real-valued function defined on R. Let us define an extended real-valued function A by setting

AG) = lim FO) forx € R for which jim f(y) exists in, Let D(A) be the domain of definition of A. Then D(A) € Be and A is BR-measurable on D(A). Proof.

Consider the two extended real-valued functions g and yon R in Theorem B.1. Now

for x € R, jim

F(y) exists in R if and only if g(x) = y(x). Then the BR-measurability

of g and ¥ on R implies

D(A) = {x eR: 4 lim f() eR} = {x ER: g(x) = v(z)} e Br. Forx € D(A), we have A(x) = iim FO)

= g(x) = +(x). Then the Bp-measurability

of g on D(A) € Bp implies the BpR-measurability of Aon D(A).

©

For another case of Borel measurability of a function implied by the existence of limits, we have the following theorem.

764

Appendix

Theorem B.3. Let f be an arbitrary real-valued function on R. If the left-hand limit lim f(y) exists in the extended real number system R for every x € R, then f is a Br-

measurable function on R. Similarly if the right-hand limit kim F (y) exists in R for every x ER, then f is a BR-measurable function on R.

Proof.

1.

Suppose lim fO) exists in R for every x € R.

Let us define an extended

real-valued function A_ on R by setting

A(x) =lim fQ) ytx

forx ER,

Let us show that A_ is a S$,-measurable function on R.

positive integers and let Z be the collection of all integers.

Let N be the collection of all

For each n € N, consider

the decomposition of R into intervals J,,, = ((k — 1)2-", k2"] for k € Z, and define a

real-valued function f, on R by

noas(S

)

forx € I, fork € Z.

Let x € R be fixed. For each n € N, there exists a unique k(n) € Z such that x € I,.x(n)-

Since

lim 2~" = 0, we have (k(n) — 1)2~ + x asm — oo and thus 100

:

.

k@)—-1

sim, fa) = Jim, (“,

—) = A-@)-

Since f, is a BR-measurable function on R for every n € N, AL

=

im, Sn is a Br-

measurable function on R. 2. Now that A_ is a 9$R-measurable function on R, if we show that f = A_ except on a countable subset of R, then f too is a %3q-measurable function on R. For this purpose,

let us define

E = {x eR: A_(x) 4 f(@)}

and let

I(x)= |A_@) — F@)| = |}im FO) — F@)| € ©, 00] For everyn € N, let

forx € E.

E, = {x¢ E: I(x) >n7}}.

Then we have E = |_), 0 such that (x — 6, x) NA = @. Since any collection

of disjoint open intervals in R is a countable collection, any set A in R can have at most countably many left-isolated points. Thus, if E,, is not a countable set for some n € N, then the subset E* of E, consisting of points of EZ, which are not left-isolated points of E,, is an uncountable set. Let x € Ef. Since x is not a left-isolated point of Z,,, there exists a strictly increasing sequence of points (y; : k € N) in E, such that y, t x. Since

B. Measurability of Limits and Derivatives ye

€ Ey, we have | iim

ITE

£0)

-— fox

765 > n,

Let {za : & © N} be a sequence in R

such that yyp-1 < ze < yx and [f(ze) — f(ye)| > (2n)~!. Then we have a sequence {Z1, 1, 225 2, 23, ¥3, °°} in R which is strictly increasing and converges to x, but the

sequence {f'(z1), £01), f@2), f(y2), f @s), f(s), -- +} does not tend to any number in R since | f (zz) — f (ye)| > Qn)7! for every k € N. This contradicts the assumption that

lim FQ) exists. Therefore £,, is a countable set for every n € N and so is E. This shows yt

that f = A_ except on a countable subset of R and then the 93y-measurability of A_ on R implies the %$n-measurability of fon R. a

[11] Borel Measurability of the Derivative of a Function Let f be an arbitrary real-valued function defined on R. The derivative (Df)(x) of f at x € Ris defined by

qa)

(DAG) = im

£0) - f@) you

provided that the limit exists in the extended real number system R. Let us write D(Df) for the set of points x € R at which the derivative (Df)(x) exists. We address the question

of Borel measurability of the set D(Df) and Borel measurability of the function Df on D(Df). For this purpose we consider the lower derivate and the upper derivate of f. With

x € Randr

> 0, letus write B(x, r) = {y eR:

The lower derivate of f at x € IR is defined by @)

|y—x|
0 on [a, b]. Then by Theorem C.2, f is absolutely continuous on [a, 5]. Then by Theorem 13.16, f’ exists

a.e. on [a, b]. Let D be the set of points at which f’ exists. Then for every x € D we have

f@=

tim £@ +9 — FQ) 10

t

Since f satisfies a Lipschitz condition with coefficient M > 0, we have | f(x +1) — f(x)| < M|t|. Then we have

LF @1 = lig

IF@+D- FOV

2 yy

'

Remark C.4. According to Theorem C.2, if a real-valued function f satisfies a Lipschitz condition on [a, 5] then f is absolutely continuous on [a, b]. However absolute continuity

of f on [a, b] does not imply that f satisfies a Lipschitz condition on [a@, b]. The reason for this is that though the derivative of an absolutely continuous function can be unbounded, a Lipschitz condition implies that the derivative is bounded according to Theorem C.3.

Appendix D Uniform Integrability [I] Uniform Integrability Definition D.1. Let (X, , 4) be a measure space. Let p € (0, 00]. We write L? (X, A, u) for the particular case of L?(X, &, 2) in which the elements are extended real-valued, not

just extended complex-valued on X.

We call every f € LP(X, A, ) where p € (0, 00) a p-th order yt-integrable function on Xx Observation D.2. Consider the function || - | p on LP(X, &, w) where p € (0, 00) defined

by

1

Wtle={ fied}? torF © LPC, Mw).

According to Theorem 16.23, | - ||p is a norm on the linear space LP(X, A, w) when Pp €[1, 00). For p € (0, 1) we showed in the preamble to Lemma 16.51 that | - ||p does not

satisfy the triangle inequality and thus | -||p is not a norm. According to Theorem 16.52, the function p, defined by

Pplf, 8) = [ lf—gldu=Ilf—gll>

for fig € L7(X, , uw)

is a metric and in particular it satisfies the triangle inequality of a metric. Proposition D.3. Let (X, 2, 2) be an arbitrary measure space and let f € L?(X, A, 4) where p € (0, 00). Then for C > 0, we have

lim

C00 Six: f/P>C}

fl? du =0.

Proof. Iff ¢ L?(X, A, 1) then fy |f|? du < co and this implies that | f|? < coae. onX and thus we have 4{X : | f|? = 00} = 0. Foreveryn € Z,, let A, ={X:|f|? > nye A Then (A, : m € Z) is a decreasing sequence of sets in 2{ and thus

(1)

lim noo

An= () An ={X: |f|? = 00}. nek,

77

772

Appendix

Now 1- 4(A1) < Ie |f|? du < 00 and this implies that 42(A1) > oo. This and (1) imply

(2)

jim, w(An) = we lim An) = w{X : [FIP = co} = 0.

Now (1) also implies lim 14, = 1(x.|f,7=00}-

noo Since 14, -|f|? 0, let [C] be the greatest nonnegative integer not exceeding C. Then [C] € Z+

and [C] < C so that

Also

{Xi 1fP > Ch {Xi fl? > (Ch.

lim implies lim . Then we have C00 [C]+00

lim C00

Nx: FP>C}

This completes the proof.

fda
ich

#

Remark D.4. Let (X, 2, 4) be a finite measure space. Let p € (0,00). extended real-valued 2{-measurable function on X. Then we have

ql)

[ fim

0 S{X: F/P>C}

Let f be an

lf? du =0] = [f € LEC, 2, w)].

Thus the converse of Proposition D.3 is valid, provided that (X, 2, jz) is a finite measure

space.

Proof. Assumption of the left side of (1) implies that for every ¢ > 0 there exists C > 0 such that Sexisieocy |f |? du < e. With such C > 0 we have

fisedu =f {X:|f|?sC} rans f {X:]f P>C} fl dw x 0 we have lim

sup f

C00 eA J{X:| fal >C}

fal? dp =0.

D. Uniform Integrability

773

Proposition D.6. (Discrete Formulation of Uniform Integrability) Let (X, 2, 2) be an arbitrary measure space and let ¥ = {fy : « € A} C LP(X, A, 2) where p € (0, 00). Then $ is a p-th order uniformly integrable collection, that is, ¥ satisfies the condition that for C > Owe have

qd)

lim

C0

sup [

gcd

1X: fal?>C}

lfal? du = 0,

if and only if for£ € Z4. we have

(2)

Jim sup f £0 ved

(X:LfalP>4}

fal? du = 0.

Proof. Clearly (1) implies (2). It remains to show that (2) implies (1). Let us assume (2).

For C > 0, let [C] be the greatest nonnegative integer not exceeding C. Then [C] € Z, and C > [C] := 2 so that

{Xi | fal? > C}C {X: |fal? > €}. Then we have

so that

f (XL falP>C} [fal du s f {il falP >} fal? du, sup f

a@eA J{X:| falP>C}

Thus

[fal dy < up [ acd

J{X:|fa|P>é}

fal? du.

tim, sup f [fol du < £500 jim aed supffl? du = 0. aed J{X:| falP>C} JX: fal?>t}

C+

This completes the proof.

um

Theorem D.7. (A Criterion for Uniform Integrability) Let (X, 2, 4) be an arbitrary measure space and let ¥ = {fy : a € A} C LP(X, A, 4) wherep € (0, 00). Then F isa p-th order uniformly integrable collection if and only if for every such that

> 0 there exists £9 € Z.

sup i ful? dp < s. a€A SIX: falP>£0}

Proof. For £ € Z,, consider a sequence (yg : £ € Z:) where 7; is defined by setting

ve= sup f

acAJ{X:| fal? >t}

fel? du.

According to (2) of Proposition D.6, ¥ is p-th order uniformly integrable if and only if the sequence (yg : £ € Z) converges to 0. Now (yz : £ € Z,) is a decreasing sequence of

TTA

Appendix

nonnegative real numbers. Thus (yz : £ € Z) converges to 0 if and only if for every ¢ > 0 there exists 2p) € Z1 such that y, < «, that is,

sup [ J(X:|falP>£o} fal? du 2}

Lil? du = ka.

Thus we have

sup

[

LEX, Ww) YEAS IPE}

= 00, \f|? du > supka keN

and then lim

sup

[

$7 PKA, u) AEAF PDE}

lf |? du = 00 £0.

This shows that L? (X, &, 4) is not a p-th order uniformly integrable collection. Theorem where p uniformly uniformly

#

D.10. Let (X, 2, 12) be an arbitrary measure space and consider LP (X, &, 1) € (0,00). {fa :a@¢ A}, {fp: B € BY C LP(X, A, ws) are both p-th order integrable then {fy :y € AU B}={fa:a@ € A}U{fp : B € B} is p-th order integrable.

Proof. If {f, : « € A} is p-th order uniformly integrable then by Theorem D.7 for every € > 0 there exists 2; € Z+ such that

sup [ fal? du < 5. aed J{X:| falP>ei} &

D. Uniform Integrability

715

Similarly if {fg : 8 € B} is p-th order uniformly integrable then by Theorem D,7 for every € > 0 there exists £2 € Z; such that

sup f

BeB

&

J{X:|fal?>£2}

lfsl’ du < 2"

Let £9 = max{é1, £2}. Then since Soxipte>g | fal? dw decreases as £ € Z, increases, we

have

€ sup f fel? du lo}

and similarly sup f

Bes J{X:| fal? >Lo}

é \fpl? du < 2

Then we have

sup

yEAUB

[

V{X:|fy|P>£o}

lf? dus wup f

aeA J{X:| fo|P>£o}

Lal? du-+sup f

BEB J{X:| fp|?>£o}

lfpl? du Jim sup

C00 wed

HX:|I fall’

i"

>C}

|lfal|? dz =0.

On the right side of (2) we have | | fel | which is actually equal to | f,,|. Thus the left side of

(2) is identical with the left side of (1). This shows that { f, : @ € A} is p-th order uniformly integrable if and only if {| f.| : # € A} is p-th order uniformly integrable.

Theorem D,12. Let (X, 2, 1) be an arbitrary measure space and consider L? (X, A, 1) where p € (0, 00). Then {f,:a € A} C L?(X, SA, p) isa p-th order uniformly integrable

collection if and only if {+ f. : a € A} is a p-th order uniformly integrable collection.

776

Appendix

Proof. By Theorem D.11, we have {+ fq :a@ € A} is p-th order uniformly integrable 0 we have

lim

C00 J{X:| fxlP>C}

fel? du = 0.

This implies that for 2 € Z we have

(a)

lim

£00 JAX: fal? >e}

fel? du = 0.

Now we have

@

wf

k=1,..4.N

J{X1felP>e}

fil? dp< > “,(x: in>ey| fl du,

and then

tim, swf

£00 pa,....N I {Xs/ fal? >e}

-y a

{X:| fel? >£}

furan If < £400 im

| {il fel?>€} Lfil?du by 2)

\fl? du =S30=0 by (1). k=1

This shows that {f; :k =1,..., N}isap-th order uniformly integrable collection according to Proposition D.6. Theorem D.14. Let (X, &, 4) be an arbitrary measure space and let (f, :n € N) Cc LP(X, &, w) wherep € (0, 00). If Tim | ik |fal? dus = 0 then (f, : n € N) is p-th order

n

uniformly integrable.

Proof. The assumption that lim, f 'y |fal? dys = Oimplies that for every ¢ > 0 there exists N €N

@

such that

ne

[isirau forn

D. Uniform Integrability

77

Then {fi,..., fy} as a finite collection in L?(X, A,

according to Theorem D.13.

2) is p-th order uniformly integrable

Let us show that {fvi1, f+2, fwti, fia, --.} = (fin : 2 © N} is also p-th order

uniformly integrable. Let ¢ > 0. Then we have [

{X:|falP >€} Malas f x ifidu 0 there exists 5 > 0 such that f, |f |? du < eforeveryE ¢ &

with z(E) < 6.

Definition D.15. (Equi-integrability) Let (X, A, 44) be an arbitrary measure space and

consider L?(X, &, 4) where p € (0, 00). Let¥ = (fa : @ € A} C LP(X, A, x) where

A is an arbitrary index set. We say that the collection ¥ is p-th order equi-integrable if for every& > O there exists5 > 0 such that fy |\fal? dw < ¢ for alla € A whenever E € A and u(E) < 6. Theorem D.16. Let (X, 2, 12) be an arbitrary measure space and consider LP (X, A, 1)

where p © (0, 00). Let ¥ = { fy :@ € A} C LP(X, A, 1). IfF is a p-th order uniformly integrable collection then F is a p-th order equi-integrable collection.

Proof. Suppose F is p-th order uniformly integrable. Then by Theorem D.7, every ¢ > 0 there exists 2g € N such that

a

sup [ \felP au < &.2 acd J {X:| fo|P>£o} €

Then select§ =2p

@

> 0. Then for every E € A with 4(E) < 5, we have

— fimiran= f EMX:| fal? >£o} Ural? du + f EMX:| fal? £o} Lal? di + &9 u(E) 0 there exists § > 0 such that te \falPdu

for all a € A whenever E € & and z(E) collection. #

0 there exists 6 > 0 such that [ fal? du 0 such that f |fal? dp < iw E

for alla € A wheneverE € Mand w(E) < 6.

Then for all a € A whenever E € & and 4(E) < 6 we have

due = teal” f E ital? du < bPbPé =e. Picasa? E

This shows that {co fa : @ € A} is p-th order equi-integrable. 3.

Let us prove (c).

If g € LP°(X, &, 2) then we have |g|
0 such

that

fireau < seat

wheneverE € & and

(EZ) < 41.

Since {fa : a € A} is p-th order equi-integrable, for every « > 0 there exists 2 > 0 such that & [

E

fool? du < yal

for all a € A whenever EF € 2 and j4(E) < 53.

D. Uniform Integrability

7719

According to Lemma 16,7, we have |a + bl? < 2? {\al? + |p|? } and thus

fe + FI? 0 such that

[ E fal? du 0 such that fy | faldu,

whenever E € %jo,1) and 4,(£) < 6, Thus F is not equi-integrable.

a

0 such that f

E

\fa—f |? du

Now we have 4(X)

< oo.

0 and 6 > 0 given above there exists N € N such that

W{Xslfr-flzer} N,

Appendix

784 that is,

w{X:|f fl?ao— =e} N.

Thus we have {X:|fr—flP2e}

Ifa -flP?duN.

Then form > N we have

lis — FI = [ le - fl dp

-{ {X:lfn— fl? 0, we have lim sup n>00

| fn —f iB = 0. Then nonnegativity

of | fu — fl forn € N implies0 < liming || f, — f ||} < limsup || fu — f || = 0 and thus noo

Rn>Oo

wehave n>00 lim ||f,— f|2 =O andthen n> lim ||f, — fllp =.

2. (ii) > @. Assume that f ¢ L?(X, 2, w) and im Ifa — fllp = 0. Now

lim || f, — fllp =O implies that

n>00

lim

no

| f, — file = 0, that is, we have

tim, f fe - FP du =0.

n->00

According to Theorem D.14, this implies that (f, — f : n € N) is pth order uniformly integrable. Then by (d) of Theorem D.22, (f, — f + f : n € N) is p-th order uniformly

integrable, that is, (f, : n € N) is p-th order uniformly integrable. 3. (ii) > Gi). Assume that f € L?(X, A, 2) and im, fn — fllp =0.

According to Observation D.2, when p € [1, 00), || f |p is anorm on LP(X, &, w) and when p € (0, 1), pp(f, 8) is a metric on LPCX, &, p). Consider first the case p € [1, 00). By the triangle inequality of the norm || - ||» we have

fallp = fn -— F + filp S fe — Filp + Fle and thus

@

Ifallp — If llp < fa — fillp-

Interchanging the roles of f, and f in the argument above we have

Q)

Iflle — Wfallp = IF — fallp = Ifn — Filp-

D. Uniform Integrability

785

By (1) and (2) we have

and then

|Iallp — UFlp| < fn — Flys

Him |ILFally — I fllo| < im fa — fll =0.

Thus we have lim, | fullp— IF llp| = 0, which is equivalent to iim, {ll fallp—fllp} =9,

thatis, lim Ilfall = Ilfllp-

Next consider the case p € (0, 1). By the triangle inequality of the metric ||f - gll>

where f, g € LP(X, 2, 4), we have

fn — OF < Ife — FIR + IF - OF and thus

(3)

fall - WI < Ife — FB.

Interchanging the roles of f, and f in the argument above we have

4

IFI5 — fall} SWF — fall = Ife — FIG.

By (3) and (4) we have

|i.fol — FS] < fa — FUR,

and then i i Him, |ILAallgPo ifi3|Pp < Jim

Thus we have im, |Il Fall? — a

fa

_

FilipP—=0.

= 0, which is equivalent to jim, {IfallZ—I FIZ} =0,

thatis, lim I fall> = IFIlZ, which is equivalentto lim || fully = If llo4. Gii) > (i). Assume that f € L7(X, 2, 4) and lim | fallp = If lp-

According Lemma 16.7, fora, b € Randp € (0, 00) wehave |a+b|? = 2?{|a|? +|b|?}.

Thus we have

(5)

2? {fal +1F1?} — Ifa — FIP = 0.

Since Am, Sn = f ae. on X, we have

©

slim [2?{Il? + If}

lhe — FP] =2PI71P ae. on x,

Note also that the assumption lim l|follp = Ifllp implies lim | fall? = ILfllZ, that is,

(7)

tim f fe | fal? du = [ IflP dy. iim,

786

Appendix

Applying Theorem 8.13 (Fatou’s Lemma) to the sequence of nonnegative functions given by (5) which converges 2?+!|f|? ae. on X according to (6), we have P+1) ep du < timint im i f P [2°(iit” Pp +12” PY }— _ ite —— 2] f|P au [artis im i |? =2 pt fini ip du-+timint ft lf, Ifa —— f?}du by (7) =—

a2 f ipidutimsup fife - fd x

n-co

JX

and then

®)

timsup fife noo

x

- fl? du 0. n> 00 x

Then (8) and (9) imply in, Ixelfa—f\? du =0, thatis, im, Wn - file = 0 and then

lim fn

flip =. 0

Definition D.26. Ler (X, 2, 2) be an arbitrary measure space. Consider the metric pp on

LP (X, A, 2) where p € (0, 00) defined by

()

Pe(f. 8) =IIf — gllp for f,g € LP(X, &, w) whenp € [1, 00).

2)

pp(f.8) = If — all for f.g € L?(X, %, w) whenp € (0, 1).

Let us call this metric pp on LP(X, &, 2) where p

€

(0,00) the standard metric on

LP(X, A, w). Let us call the topology on L?(X, A, 1) derived from the standard metric on L?(X, A, ) the standard metric topology on LF (X, A, p). Lemma D.27.

Let (X, 2, 4) be an arbitrary measure space and consider L? (X, &, 1)

where p € (0, 00). Let¥ C LP(X, M, 4) and let F be the closure of F in the standard metric

topology on LP (X, A, w). Then for every g € F there exists a sequence (fy :n € N) C ¥ such that

1° lim Ifa ~ gllp =0.

2° (fn in € N) converges to g in measure on X. 3° lim f, =g, ae. on X. n—>00

Proof. Let g € ¥. Then there exists a sequence (fy :n € N) C ¥ such that

@

lim, pC fn, 8) = 0.

D. Uniform Integrability

787

we have By the definition of pp in Definition D.26

(2)

im, fn —gllp=0@

tim, p(n, 8) =0e

tim

lA- si

-0.

Thus our sequence (f, : 2 € N) satisfies condition 1°. For C > 0 we have

@)

Wm stg= f Im eldu> f x

{X:|fa-8l=C}

fa — eld

= CPu{X 3 | fn — gl = Ch. Then applying the last member of (1), we have Jim, HX

: |fn—g8| = C} =0. This shows

that (f, : 2 € N) converges to g in measure on X. Thus our sequence (f, : n € N) satisfies condition 2°.

Then according to Theorem 6.24 (F. Riesz), condition 2° implies that there exists a subsequence (f,, : k € N) of (f, : m € N) that converges to g ae. on X. Relabel this subsequence as (f;, : 2 € N). Then we have a sequence (f, : n € N) satisfying conditions 1°, 2° and 3°.

Theorem D.28. Let (X, A, 2) be a finite measure space and consider LP(X, Q, ) where P € (0,00). Let ¥ c LP(X, A, 2) and let F be the closure of ¥ in the standard metric topology on L?(X, A, 4). If ¥ is a p-th order uniformly integrable collection then so is F. Proof. We prove this theorem by applying Theorem D.21 (Necessary and Sufficient Conditions for Uniform Integrability).

Let us represent F as ¥ = {fy :@ € A} and F as F = {gg : B ¢ B} whereB D A. To

show that ¥ is p-th order uniformly integrable, we show according to Theorem D.21 that

1° supper fx lgsl? du < 00. 2° Fis p-th order equi-integrable.

Since ¥ is p-th order uniformly integrable then according to Theorem D.21 we have

(D

M = sop f

acA/X

lfel? du 0 there exists § > 0 such that

@

[ |fel? du

We say that f is product-measurable if f is o(&y x --- X A,)-measurable. We say that f is factor-measurable if fx,,__.,.;,...,%,} #5 Bi-measurable for everyi = 1,...,m

Theorem E.5. Let (X1, 21), ..., (X1, Mn) be measurable spaces and consider the product

measurable space (X1 X «++ X Xn, (Ql X +++ xX U,)).

Let E € o(QMy x +++ x Ay).

Let f be a real-valued function defined on E. If f is o(Q, x --- X A,)-measurable on E, then its X;-section fix,,...,01,...,%9} 18 Wj-measurable on E{x1,...,@i,..., Xn} for every i=1,...,m Proof.

To show that ftx,,...,2;,...x9} 18 &;-measurable on E{x1,...,0@;,...,%n}, we show

that for everya € IR we have

a)

(Fist oveisnntn}) ((—00, ef) € 2h.

Now we have

(Firs vetestal) ((-00, 0]) = (F-1((-00, al) fa, «5 05s otek Since f is o (M1 x --- x %,)-measurable, we have f—!((—c0, a]) € a (M1 x --- x Ay). According to Theorem E.2 this implies

@)

(£71 (C00, a]) Morn,

07,666, an} € By.

Combining (2) and (3) we have (1). This completes the proof. Theorem E.6. Let (X1, 21), ..., (X1, Mn) be measurable spaces and consider the product

measurable space (X1x++ +x Xn, 0(Q1x-+-xMy)). LetE € o (Qi x-+-x Mn). Let f bea

real-valued function defined on E. If every X;-section of f, fic,....e1,...xn}> #9 Ai-measurable on E{xi,...,@;,...,%,} fori =1,...,n, then f iso(Qy x --- x A,)-measurable on E. Proof. To show that f is o (QM; x --- x 2{,,)-measurable on E, we show that for every a ¢ R we have

()

F7*(Cee, ai]) € o (Rr x --- x Bp).

Now since fixy,...,0;,...%} 18 2j-measurable on E{x1,...,0;,-..,%n} fori =1,...,", we

have

@

Ey = (Ferrntineta})

((-00, a]) € 2.

792

Appendix

Then we have n

F700, al) = Xn x + i=1

Kia % [(Fistennernnad) (C00, a)]

n

=() Xi x +++

i=1

=E,x:--x This proves (1).

Ki

x By x Xiq

EB, €Myx-:-x

Xx

A, Co(My

Xp x--- x An).

Kip

0 x Xn

Appendix F Functions of Bounded Oscillation The standard definition of a function of bounded variation on [a, b] C R relies on the fact that the domain of definition of the function is 1-dimensional. (See Definition 12.12 for instance.) Below we define a function of bounded oscillation on a closed box in R".

We show that

when n = 1 it is a function of bounded oscillation if and only if the function is a function of bounded variation.

[I] Partition of Closed Boxes in R” Definition F.1. A set D C R? is called an n-dimensional closed box in the topological space R" if it is represented with respect to an arbitrary orthogonal coordinate system (11,..-, Xn) € R" as:

D=[ai, a1 + 1] x +++ x [Guy Gn + cn); where —00 < a; < oo and cc; € (0,00) fori =1,...,n. Let &(D) = min{e;,..., ¢n} > 0. We call 8(D) the minimum width of the closed box D.

Observation F.2. Observe that if D C R” is an n-dimensional closed box in R” then its interior D° # # and moreover we have p?(D) = u?(D°) = c1 +++ cn € (0, 00). Definition F.3. Let {Dz; @ € A} be a collection of n-dimensional closed boxes. We say that

the collection in non-overlapping if (Da M De)° = O when oa" € A anda! # a”. Observe that if {D, : n € N} is anon-overlapping collection of closed boxes then we have

un( LJ Pn) = wt, neN

Terminological Convention.

neN

Let D be an n-dimensional closed box.

Let k = 1,..., 7.

When we say that a k-dimensional closed box E is contained in D we mean not only that E Cc D but also that E and D are parallel, that is, the edges of E are parallel to the edges of D. 793

794

Appendix

Definition F.4. Let D Cc R" be an n-dimensional closed box given by

D=[a, 6] x. x fa, 6]. Select points in R such that a®

— a


[fed — Fe-s)]. k=l

Since f is absolutely continuous on D, we have according to Theorem 13.17

(2)

f@) — f@)= f

De

fi du,.

Then substituting (2) in (1), we have

@

waEM=D|f rau vfff iriau, Hy, = frau. pit Hy, mn

ka 7 De

Since (3) holds for an arbitrary partition Pp of D, we have

Ve(f)= PpEPy sup VE(F, Po) < f D If'ldu,. This completes the proof.

#

Remark F.21. Theorem F.20 is not valid if the function f is a function of bounded variation but not absolutely continuous on D. For example consider the Cantor function t defined on [0,1]. Now t is real-valued, increasing and continuous on [0, 1] with 7(0) = 0 and z(1) = 1. Thus z is a BV function on [0, 1] and V(r) = 1. On the other hand, ¢’ exists and is equal to 0 a.e. on [0, 1] so that Sion |’|dp, =O041= Vd (x).

This page intentionally lett blank

Bibliography [1] L. Ambrosio and P. Tilli, Selected Topics on “Analysis in Metric Spaces", Scuola Normale Superiore, Pisa, 2000. [2] H. Bauer, Wahrscheinlichkeitstheorie und Grundziige der Maftheorie, 3rd ed., Walter de Gruyter, Berlin, 1978. [3] P. Billingsley, Probability and Measure,

1979.

2nd ed., John Wiley and Sons, New York,

[4] A. M. Bruckner, J. B. Bruckner and B. S. Thomson, Real Analysis, Prentice-Hall, Upper Saddle River, New Jersey, 1997. [5] R. C. Buck, Advanced Calculus, 3rd ed., McGraw-Hill, New York, 1978.

[6] R. M. Dudley, Real Analysis and Probability, Wadsworth & Brooks/Cole, 1989. [7] L. C. Evans and R. F. Gariepy, Measure Theory and Fine Property of Functions, CRC Press, Boca Raton, 1992. [8] K. J. Falconer, The Geometry of Fractal Sets, Cambridge University Press, UK, 1985. [9] G. B. Folland, Real Analysis, John Wiley and Sons, New York, 1985. [10] P. Halmos, Measure Theory, Van Nostrand, New York, 1950.

[11] E. Hewitt and K. Stromberg, Real and Abstract Analysis, Springer Verlag, New York, 1969. [12] P. Mattila, Geometry of Sets and Measures in Euclidean Spaces, Cambridge University Press, UK, 1995. [13] L P. Natanson, Theorie der Funktionen einer reellen Verdnderlichen, Akademie Verlag,

Berlin, 1961.

[14] C. A. Rogers, Hausdorff Measures, Cambridge University Press, UK, 1970, 1998. [15] H. L. Royden, Real Analysis, 3rd ed., Macmillan, New York, 1988. [16] W. Rudin, Principles of Mathematical Analysis, 2nd ed., McGraw-Hill, New York,

1964.

803

804

Bibliography

[17] W. Rudin, Real and Complex Analysis, 3rd ed., McGraw-Hill, New York, 1987. [18] E. C. Titchmarsh, The Theory of Functions, 2nd ed., Oxford University Press, UK,

1939,

Index C, D, L, O, P, R, and T are abbreviations for Corollary, Definition, Lemma, Observation,

Proposition, Remark, and Theorem respectively. A * * * * * * * *

* * * * *

absolute continuity of Lebesgue-Stieltjes measure .. .. D22.16, p.532; T22.21, p.535 absolutely continuous function .... D13.1, p.283; D22.13, p.531 absolutely continuous measure ... D114, p.236 additive set function vee DAS, pl adjoint of a linear transformation . 029.9, p.737 algebra .. . D1.1,p3 almost uniform convergence .... 6.10, p.109 approximation by continuous functions inR... . 76.36, p.124; T6.39, p.127 GRY owe cee eee ee ener eee renee eee senaeseng enna beeen ane 124.22, p.630 in a locally compact Hausdorff space . T19.37, p.483; T19.38, p.486 approximate identity in convolution ... . D23.52, p.592 approximation by step functions .. .. D6.29, p.119 area of a spherical hypersurface .. .. T26.21, p.695 atom in a measurable space .... .». D1.35, p.19 in a measure space .. . D1.35, p.19 average function .. .. D25,3, p.644

B * Baire category theorem .. * Banach space

«ee T1547, p.362 .. D15.8, p.341

* Banach-Steinhaus theorem ...

.. T15.51, p.365

* Banach-Zarecki theorem .. T13.8, * bi-Lipschitz mapping . -. D28.7, FDIA ee ... D15.29, * Borel-Cantelli lemma ..................cecccceseecesneeceeeceeeeeeeseeeeeeeeeenenes T6.6, * Borel measurability of limits and derivatives Vimits occ cece cece ene neneeeneecsense eee peceeteaeeeesseessecuereneeeeene TB.2, derivatives : * Borel measurable function . * Borel measure .... . D19.15, 805

p.287 p.721 p.353 p.106 p.763

p.470

806

*

* * * * * * * *

Index

inner regular .... outer regular .... Borel measure on R’ . inner regular .... outer regular . regular .. signed ... Borel measure space on R” Borel outer measure .. Borel regular outer measure .... Borel regularity of Lebesgue outer measure onR .. on R? . Borel set ..... Borel o-algebra ..... Borel’s theorem on approximation by continuous functions . bounded convergence theorem

* bounded linear functional .

* * * * *

bounded linear mapping bounded oscillation bounded variation ... box in R® BV function ....

c * * * *

Cantor ternary set .. Cantor-Lebesgue function . Carathéodory condition ... Cauchy sequence with respect to a norm ... with respect to convergence in measure .

* Cavalieri’s formula

...

* Cavalieri’s principle .. * chain rule for differentials ... for Dini derivates ...

for measurable mappings for Radon-Nikodym derivatives ... * change of variable of integration inR ... inR" .. * closed graph theorem .... * complete extension of a measure space ..

.-.. D19.15, - D19.15, . D25.36, . D25.36, . D25.36, . D25.36, . D25.36, . D24.14,

p.470 p.470 p.664 p.664 p.664 p.664 p.664 p.625

. D2.14, p.35 vee D215, p.35

. L3.21, p.53 - 124.11, p.624 . D1.16, p.10 - D1.16, p.10 76.36, p.124 .». T7.16, p.143 «++. D15.29, p.353

. D15.29, DE10. sees D12.12, .... DFA, .. D12.12,

p.353 p.796 p.274 p.793 p.274

T15.41, p.360 D29.17, p.743 . p.88; DA.14, p.759 +. p.90; p.288; p.539 D2.2, p.29 -. D15.8, p.341 .- D6.25, p.116 . 123.67, p.608

w+. Prob. 22.4, p.547 p.674 L14.9, p.330

.. 71.40, p.21 sees T11.23, p.252 ++ p.308; T13.32, p.311 --- 126.9, p.683 . 715.61, p.373 DS5.1, p.99

Index

807

* complete measure space ...

... D1.34, p.18

* complete o-algebra

.».. D1.34, p.18

* completion

of ac-algebra

of a measure space .. of a product measure space .. of Borel measure space on R . of Borel measure space on R® * condition (L)

-.. D5.2, p.99

-- D5.1, p.99 . 123.23, p.564 .-+. 75.7, p.102 . 124.15, p.625 » Prob. 13.10, p.319

* condition (N) ......:.ccscsscscaeccccseacnsenesscnececneapecesnecereeapsaesecaersene D13.6, p.286

* * * * * * * *

convergence almost everywhere D6.1, p.104; 16.22, p.114; T6.24, _ D6.14, p.111; T6.22, p.114; T6.24, convergence in measure . convergence in norm .. . D15.8, p.341; T16.25, p.403; T16.28, p.405; T16.30, convex COMbINAatION 20.0... 0.ccesseeveneceesenseeneeaeeneceecesneestensenseaee D27.15, convex function .. D14.1, convex hull .. . D27.21, convex set ... «. D27.15, convexity condition D14.1,

* convolution of functions

* * * * *

p.706 p.323 p.708 p.706 p.323

. D23.30, p.569

countably additive set function . D1.19, p.11 countably subadditive set function .. a D119, p.11 counting measure . D17.17, p.439 covering class of sets .. «++» D2.20, p.38 covering sequence Of SCtS ..........scceecesecnseenveerseersaeeeeecesvensenesenssens D2.20, p.38

D * * * *

Darboux sum decomposition of an increasing function decreasing sequence of sets .. density of a set ... Lebesgue ... symmetric .... * derivative of a function * * * * * * * *

p.115 p.115 p.406

of a set function

D7.19, p.146 . T22.29, p.546 we. DLS, p.4 . D25.25, p.658 . D25.25, p.658 «. D25.25, p.658 D12.1, p.258 . D25.19, p.655

diameter of a set .... D27.1, p.699 difference set of a set .. - D3.28, p.59 differentiable function D12.1, p.258 differentiable mapping .. se. p.673 differential of a mapping ... «s+ p.-673 differentiation under the integral sign . . P23.37, p.574 digital expansion of real MUMDETS ............0.cceecseceesceeccereneeseensvenreneseneees p.751 digital representation of real numbers . DA,3, p.753

* dilation of a set ...

* Dini derivate

..» D312, p.48

-. D125, p.263

808 * dominated convergence theorem .... under convergence in measure . * dual ofa normed linear space .. * 3-cover

E * * * * * * *

Egoroff’s theorem envelope of a function .. equal almost everywhere equicontinuity ...... equi-integrability ... equivalent measures equivalent norms .

* essential bound ...

Index

.. 19.20, p.188 19.22, p.191

w p.359

.. D27.3, p.699

-. T6.12, p.110 D7.24, p.150 «-. D4,17, p.81 . Prob, 16.16, p.427 .. DD.15, p.777 . D11.24, p.252 . D15.11, p.342 . D16.35, p.410

* essential supremum . D16.36, p.410 * extended complex-valued function . p.392 * extended real number system ..........csccceseeesesereseesecesceeeeeetenscnseneceeneeenees p.ll F * * * *

Fy-set factor-measurability of a function ... factor-measurability of a set .... Fatou’s lemma ... generalized .. under convergence in measure .

* finite measure

* finitely additive set function . * finitely subadditive set function .. * Fubini’s theorem ......... * Fubini-Tonelli theorem ..

* fundamental theorem of calculus ... G * Gs-set * gamma function ....

.-. D1.18, p.10 .... DE.4, p.791 ... DE.1, p.789 78.13, p.168 «.. T9.19, p.187 .». T9.21, p.190 . D1.30, p.17

.». D119, p.11 D1.19, p.11 123.18, p.560; T23.20, p.562; T23.27, p.567 +» T23,19, p.561; T23.28, p.568 p.300

D1.18, p.10 «++, 026.12, p.689

H * Hahn decomposition ............cccssecsseeseeesseveeeecesvene D10.12, p.219; D10.13, p.219

* Hahn-Banach theorem teal case .., complex case ... * Hardy-Littlewood maximal function * Hardy-Littlewood maximal theorem * Hausdorff dimension

. 715.64, . 715.70, .-. D25.7, . T25.11, ws. D27.34,

p.374 p.380 p.647 p.650 p.715

Index

809

* Hausdorff measure of a unit ball .... * Hausdorff measure of a unit cube . * Hausdorff measure ....

* * * * * *

Hilbert norm ... Hilbert space ... Hopf extension theorem . Hilder condition .... Hélder mapping Hilder’s inequality forp € (0, 1) andg € (—o0, 0) ... for p,q € (1,00) ......

for p = 1landg =o0.... I * * * * * * * * * * *

image measure improper Riemann integral ... increasing sequence of sets ... indefinite integral in R” . indefinite integral ... inner product space . inner product ... integrable . integration by image measure integration by parts .............. integration with respect to a signed measure .

* intermediate value theorem for a derivative

* isodiametric inequality .. F USOMEMY 2.0... cece eeec eee c eee eseeeneeen nace seca J * Jensen’s inequality .. * Jordan curve length parametric representation. rectifiable * Jordan decomposition of a BV function * Jordan decomposition of a signed measure * jump of a real-valued increasing function ..

L *L' nom... *1? nom... * LP nom... * gP

«+» 129.8, p.735 129.8, p.735 .. D27.7, p.701

D15.25, . D15.25, -. 720.5, Prob. 16.7, D28.3,

p.350 p.350 p.506 p.425 p.719

.. T16.54, p.421 . 716.14, p.396

.. T16.40, p.411

wee T1.44, p.22 .» R9.29, p.195 DL.5, p.4 . D25.22, p.657 . D13.12, p.289 - D15.19, p.349 - D15.19, p.349 D7.4, p.132; D8.1, p.159; D9.1, p.177 ... 19.34, p.201 . 713.29, p.307 . D10.32, p.228 T12.3, p.260

129.6, p.734 e ae ces ecne se censenstanse se D28.7, p.721

«.. 114.16, p.335 .». D28.14, p.724 ++. D28.20, p.726 . D28.14, p.724

... D28.20, p.726 vee DI2.19, p.277; T12.18, p.277 . D10.17, p.221; T10.21, p.223 «++ p.263

vee wee vee ..

T1515, T1515, T16.23, DI7.19,

p.345 p.345 p.401 p.441

810 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

* * * *

Index

Lebesgue measure on R” .. D24.8, p.622 Lebesgue o-algebra on R . .-. D3.1, p.42 Lebesgue o-algebra on R® D24.8, p.622 Lebesgue decomp. of a signed measure .... D11.5, p.236; T11.10, p.238; T11.13, p.243 Lebesgue decomposition of an increasing function. . T13.20, p.296 Lebesgue density of a set. . D25.25, p.658 Lebesgue density theorem . ww . €25.32, p.662 Lebesgue differentiation theorem for increasing function . T12.10, p.269 Lebesgue differentiation theorem for indefinite integral . .. T13.15, p.291 Lebesgue differentiation theorem ... . 125.17, p.654 Lebesgue inner measure onR .. ..-. D3.31, p.60 Lebesgue integrable .. D7.4, p.132; D8.1, p.159; D9.1, p.177 Lebesgue integral ..... D7.4, p.132; D8.1, p.159; D9.1, p.177 Lebesgue measurable function . «. D4.1, p.72 Lebesgue measurable set in R . D3.1, p.42 Lebesgue measurable space on R .. D3.1, p.42 Lebesgue measurable space on R” .. .. D24.8, p.622 Lebesgue measure on R ... D3.1, p42 ... D3.1, p42 Lebesgue measure space on R . .. D248, p.622 Lebesgue measure space on R” .. D3.1, p.42 Lebesgue outer measure on R .... Lebesgue outer measure on R* D24.5, p.620 Lebesgue point .. D25.13, p.652 Lebesgue set D25.13, p.652 Lebesgue’s dominated convergence theorem T9.20, p.188; T9.22, p.191 Lebesgue’s theorem on convergence in measure .. 16.22, p.114 Lebesgue’s theorem on integral of the derivative .... wee T12.10, p.269 Lebesgue-Radon-Nikodym theorem .. . T11.10, p.238 Lebesgue-Stieltjes measure .. D22.5, p.527 absolute continuity ..

singularity Lebesgue-Stieltjes outer measure ... limit inferior ofa sequence of sets .. limit ofa sequence of sets ......... limit superior ofa sequence of sets . linear functional linear transformation of the Lebesgue integral on R ..... of the Lebesgue integral on R” . of the Lebesgue measure on RR” Lipschitz condition . Lipschitz mapping . locally compact space locally integrable ...

* Lusin’s condition (N) .

.. D22.16, p.532; T22.21, p.535

. 122.24, p.538 D22.5, p.527 DL, p.5 .. DL.8, p.5 . D1.6, p.5 «. D1S.26, p.351

«+. T9.32, . 724.32, «e+ 124.31, . Prob. 13.10, p.. 319; DC.1, . D28.3, D19.4, D25.1,

p.199 p.638 p.638 p.769 p.719 p.466 p.643

.. D13.6, p.286

Index

811

# Lusin’s theorem ..........ccecececsrereterscsceterscscucseseneeees 76.39, p.127; T19.37, p.483

M * mapping bi-Lipschitz differentiable .0.........ccccccc cee eseeseceeeea ees enseeee ness Holder Lipschitz * maximal function for a measure * maximal function ... * maximal theorem for a measure * measurable function * measurable mapping .. * measurable space * measurable with respect to an outer measure ...

.- D28.7, p.721 e se sae eans e e es neane p.673

.. D28.3, p.719 .. D28.3, p.719 . D25.38, p.665 .. D25.7, p.647 . 725.40, p.665

D4.1, p.72 .-- D138, p.20 - D1.29, p.17 D2.2, p.29

* measure .... a. D124, p.13 finite ... - D1.30, p.17 o-finite .. +. D1.30, p.17 * measure space .. «. D1.30, p.17 * metric * metric outer measure .. 7 .. D2.17, p.35 * MTC SPACE .......ceecceecevennseeeseeeneensecnsenecenecseeseneeseenseeeceneseneeee D1.16, p.10 * Minkowski’s inequality FOL P = 00 ooo ieccsecesscrsenecceesaseesscnssenseeeseseeeeeassensneereensager 716.41, p.412 forp € (0, 1) ... .-.. T16.55, p.421 forp € [1, 00) .. - T16.17, p.398 * Minkowski’s inequality for integrals ... .. 123.69, p.610 * monotone class

* monotone class generated by a class of sets * monotone convergence theorem .. for measurable sets .. generalized ........... * monotone sequence of sets * monotone set function ... * u*-measurable set N * n-dimensional Borel measure space ... * negative part of a signed measure . * negative set ..... * non Lebesgue measurable set

* non overlapping intervals * non-singular linear transformation ..

... D23.10, p.553 . D23.13, p.554

. T8.5, p.162 . T1.26, p.14 19.17, p.186

. DLS, p.4

. D119, pd

... D2.2, p.29

.». D24.14, p.625 . D10.22, p.223 « D10.9, p.218 3.20, p.52; T3.45, p.68

D13.1, p.283

-. D15.5, p.340

812

Index

* normed linear space ... complete * null set in a measure space in a signed measure space . Oo * * * * * * * * * *

* * P * * * * * * * * * * * * * * * * * * * * *

oscillation open mapping theorem .. orthogonal transformation . 0ce oscillation of a function . outer measure ..... based on a set function on a covering class .. Borel regular .. Lebesgue . metric

regular .... o-finite ...

p-digital expansion of real numbers p-digital representation of real numbers .... p-th order integrable p-th order equi-integrability . p-th order uniform integrability partition of a box ... partition of unity . polar coordinates . positive homogeneity o! positive homogeneity ...... positive linear functional positive measure space positive measure ......... positive part of a signed measure . positive set ........cceceeeeee positively separated sets ... product-measurability of a function product-measurability of a set . product measurable space . product measure space product measure

.-. D15.5, p.340 .-. D15.8, p.341 «+. D132, p.18 .. D10.9, p.218

eee e nena esse

.+-» DE8, p.795 . 715.57, p.370 e e e ense e e ne e e p.640

. DC2, p.769 D2.1, p.29 . 72.21. p.38

w+. D2.14, p.35 . D2.15, p.35 D3.1, p.42 . D2.17, p.35

..-- D2,10, p.33 -- D2.12, p.34 «DA, «» DA, « D164, DD.15, . DD5,

p.753 p.753 p.393 p.777 p.772

. DF4, p.794 . T19.12, p.468

.. P26.10, p.685 . €24.33, p.639 . T3.18, p.51; D15.5, p.340 . D19.26, p.475 «+. p.212 «» p.212 . D10.22, p.223 D10.9, p.218 . D2.17, p.35 . DE, p.791 . DE.1, p.789 D23.1, p.548 D23.1, p.548 ... D23.1, p.548

Index

813

R * Radon measure * Radon-Nikodym derivative

* * * * * * *

Radon-Nikodym theorem .. rectifiable curve ........ regular outer measure ... regularity of Hausdorff measure .. regularity of Lebesgue outer measure on R Riemann integrable . Riemann integral . improper ...

* Riemann sum .....

* Riesz-Markoff theorem .. * Riesz representation theorem on L? forp € (1, 00) .... on LP forp € [1, 00)

....

».. D19.19, p.472 .. D111, p.235

. 11.14, p.244; T11.16, p.245 - D28.20, p.726

. D2.10, p.33 . 727.27, p.110 . 13.21, p.53 .. D718, p.145 «. D7.18, p.145 «. R9.29, p.195 .» D7.18, p.145

.. T19.34, p.479 .- 118.7, p.461

T18.6, p.457

* Riesz’s theorem on convergence in measure . .. 16.24, p.115 * right-continuous modif ication of increasing function ...............0000008 D22.1, p.525 s * * * * * * *

s-dimensional Hausdorff measure ... Schwarz’s inequality .... section of a product set .. semi-differentiable .. semi-integrable semialgebra semicontinuous function ...

* sequence of sets decreasing .... increasing .. monotone ....

* set function additive countably additive .. countably subadditive . finitely additive ...... finitely subadditive monotone ..... subadditive

* shrink nicely * signed measure ... * sipned measure space ....

D27.7, « 715.22, p.349; C16.15, D23.8, D12.1, D7 4, p.132; D8.1, p.159; D9.1, D21.1, «+. D15.79, .. D15.85,

p.701 p.397 p.552 p.258 p.177 p.516 p.386 p.389

... DLS, p.4 . DLS, p.4 w

DIS, p.4

«+ D1.19, . D119, - D119, - D119, . D119,

p.11 p11 p.11 p.11 p.11

. D1.19, p.11 . D119, p.11

D25.16, -- D10.1, D10.1, .. D10.29,

p.654 p.212 p.212 p.227

.. D10.29, p.227

814

* * * * * * * * * * * * * * *

Index

simple function singular function . singular measures ... singularity of Lebesgue-Stieltjes measure . spherical coordinates ... Steiner symmetrization . step function ........... subadditive set function . sup NOIM .........6.. support of a function .. symmetric derivative of a measure .. symmetry with respect to a hyperplane .. o-algebra generated by a collection of subsets . g-algebra . O-COMPACE SEL oie eecccesecceeeeeteseecc seen een eeneeee nese

* g-finite measure .... * g-finite outer measure

..

T * testing set * Tonelli’s theorem ...

* total oscillation of a function ... * total variation of a function

of a signed measure . * totally disconnected .. * translate of a set * translation invariance

Lebesgue integral in R Lebesgue integral in R” .... Lebesgue measure space on R* . Lebesgue outer measure on R? . Lebesgue measure space on R .. Lebesgue outer measure on R .. * translation invariant o-algebra . * translation invariant measure * truncation of a function ....

U * uniform absolute continuity of the integral ... * uniform boundedness theorem . * uniform integrability

- D10.29, p.227 .--. D7.1, p.131 . D13.10, p.289; D22.13, p.531 . D10.16, p.221 .. 122.24, p.538 . P26.11, p.686 729.5, p.732 D6.29, p.119 - D1.19, p.11 . 715.15, p.345 .-. D19.7, p.467 - D25.19, p.655 D29.4, p.731 D1.12, p.7 «. DL3, p.3 nene eae eaen es e cetene eaese e p.465 .-» D130, p.17 «+. D212, p.34

vee D1.30, p.17

D2.2, p.29 123.17, p.558; T23.26, p.567

DE 10, p.796

wi. D12.12, p.274

«+». D10.22, p.223 « D29.15, p.742 D3.12, p.48

19.31, p.197 . 124.28, p.633 .. 124.27, p.633 . 124.26, p.632 .. T3.16, p.49 . 13.15, p.49 . D3.14, p.49 . D3.14, p49 .». D8.15, p.169

-. T8.20, p.172; T9.26, p.193 . 715.49, p.364 ... DD.S, p.772

Index

815

. 715,15, p.345; T19.50, p.491; D23.45, p.585 * Urysohn’s lemma ... Vv * Vitali cover .. * Vitali covering theorem . * volume of a ball in R” ... Ww * weak convergence in L? forp € (1, 00)

see eeeeeeee p.729

... T19.11, p.467

.. D127, p.266 ... T12.9, p.266 . 726.13, p.690 ...D16.46, p.414 . D16.31, p.407 .. D16.46, p.414

Y * Young’s convolution theorem ............ssceesecessecesseceaeesensesensenersee 123.44, p.582 * Young’s inequality ..........ccscssessecessersvenceensaseceeeeseensseseeecsesaee 116.12, p.395 Z