Measure and Integration: A First Course [1 ed.] 036734839X, 9780367348397

Table of contents :
Contents
Preface
Author
Note to the Reader
1 Review of Riemann Integral
2 Lebesgue Measure
3 Measure and Measurable Functions
4 Integral of Positive Measurable Functions
5 Integral of Complex Measurable Functions
6 Integration on Product Spaces
7 Fourier Transform
Bibliography
Index

Citation preview

Measure and Integration A First Course

Measure and Integration A First Course

M. Thamban Nair

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 c 2020 by Taylor & Francis Group, LLC

CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper International Standard Book Number-13: 978-0-367-34839-7 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a notfor-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Control Number: 2019950839 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Contents

Preface

vii

Author

ix

Note to the Reader

xi

1 Review of Riemann Integral 1.1 Definition and Some Characterizations . . . . . . . . . . . . 1.2 Advantages and Some Disadvantages . . . . . . . . . . . . . 1.3 Notations and Conventions . . . . . . . . . . . . . . . . . . .

1 1 8 11

2 Lebesgue Measure 2.1 Lebesgue Outer Measure . . . . . . . . . . . . . . . . . . . . 2.2 Lebesgue Measurable Sets . . . . . . . . . . . . . . . . . . . 2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 15 23 33

3 Measure and Measurable Functions 3.1 Measure on an Arbitrary σ-Algebra . . . . . . . . . . . 3.1.1 Lebesgue measure on Rk . . . . . . . . . . . . . . 3.1.2 Generated σ-algebra and Borel σ-algebra . . . . 3.1.3 Restrictions of σ-algebras and measures . . . . . 3.1.4 Complete measure space and the completion . . 3.1.5 General outer measure and induced measure . . 3.2 Some Properties of Measures . . . . . . . . . . . . . . . 3.3 Measurable Functions . . . . . . . . . . . . . . . . . . . 3.3.1 Probability space and probability distribution . . 3.3.2 Further properties of measurable functions . . . 3.3.3 Sequences and limits of measurable functions . . 3.3.4 Almost everywhere properties . . . . . . . . . . . 3.4 Simple Measurable Functions . . . . . . . . . . . . . . . 3.4.1 Measurability using simple measurable functions 3.4.2 Incompleteness of Borel σ-algebra . . . . . . . . 3.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . .

37 37 41 42 45 48 50 52 56 60 61 64 66 71 75 75 76

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v

vi

Contents

4 Integral of Positive Measurable Functions 4.1 Integral of Simple Measurable Functions . . . . . 4.2 Integral of Positive Measurable Functions . . . . . 4.2.1 Riemann integral as Lebesgue integral . . . 4.2.2 Monotone convergence theorem (MCT) . . 4.2.3 Radon-Nikodym theorem . . . . . . . . . . 4.2.4 Conditional expectation . . . . . . . . . . . 4.3 Appendix: Proof of the Radon-Nikodym Theorem 4.4 Problems . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

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81 81 88 95 97 103 104 105 111

5 Integral of Complex Measurable Functions 5.1 Integrability and Some Properties . . . . . . . . . . . 5.1.1 Riemann integral as Lebesgue integral . . . . . 5.1.2 Dominated convergence theorem (DCT) . . . . 5.2 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 H¨ older’s and Minkowski’s inequalities . . . . . 5.2.2 Completeness of Lp (µ) . . . . . . . . . . . . . . 5.2.3 Denseness of Cc (Ω) in Lp (Ω) for 1 ≤ p < ∞ . . 5.3 Fundamental Theorems . . . . . . . . . . . . . . . . 5.3.1 Indefinite integral and its derivative . . . . . . 5.3.2 Fundamental theorems of Lebesgue integration 5.4 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . .

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

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

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

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113 113 119 121 127 129 133 138 140 140 141 151 157

6 Integration on Product Spaces 6.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Product σ-algebra and Product Measure . . . . . . . . . . 6.3 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 6.4 Counter Examples . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 σ-finiteness condition cannot be dropped . . . . . . 6.4.2 Product of complete measures need not be complete 6.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . .

161 161 162 169 172 172 173 173

7 Fourier Transform 7.1 Fourier Transform on L1 (R) . . . . . . . . . 7.1.1 Definition and some basic properties . 7.1.2 Fourier transform as a linear operator 7.1.3 Fourier inversion theorem . . . . . . . 7.2 Fourier-Plancherel Transform . . . . . . . . . 7.3 Problems . . . . . . . . . . . . . . . . . . . .

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177 177 177 185 187 191 197

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Bibliography

199

Index

201

Preface

The concepts from the theory of measure and integration are vital to any advanced courses in analysis and its applications, specifically in the applications of functional analysis to other areas such as harmonic analysis, partial differential equations and integral equations, and in the theoretical investigations in applied mathematics. Therefore, an early introduction to such concepts becomes essential in the master’s program in mathematics. This book is an attempt toward that goal, requiring minimal background in mathematical analysis. It is essentially an updated version of the notes the author has been using for teaching courses on measure and integration for the last thirty years. The topics covered in this book are standard ones. However, the reader will definitely find that the presentation of the concepts and results differ from the standard texts, in the sense that it is more student-friendly. It starts with a short introduction on Riemann integration to motivate the necessity of the concept of integration of functions more general than those allowed in the theory of Riemann integration, and then, in Chapter 2, introduces the concept of Lebesgue measurable sets more general than the concept of intervals. Once we have this family of Lebesgue measurable sets, and the concept of a Lebesgue measure, it becomes almost obvious that one need not restrict the theory of integration to the subsets of the real line, but can be developed on any set together with a σ-algebra on it. Thus, the concept of a measure on a measurable space allows a theory of integration in a very general setting, which has immense potential for application to diverse areas of mathematics and its applications. The general theory of measure and integration is considered in Chapters 3, 4, and 5. Chapter 6 is concerned with the measure and integration on Cartesian product of measured spaces, namely, the product measure on product σ-algebra and integration of measurable functions on the product measure spaces. The final chapter, Chapter 7, on Fourier transform, is included only to show how the basic concepts of measure and integration are useful in proving results in another branch of analysis, which has lots of applications in partial differential equations and many engineering subjects. Since the probability space is a particular case of a measure space, at a few places, the implications of certain concepts to probability theory are also included, such as the concepts of random variable, distribution measure, distribution function, and conditional expectation.

vii

viii

Preface

Although the theory of integration is vast, the attempt in this book is to introduce the students to this modern subject in a simple and natural manner so that they can pursue the subject further with confidence, and also apply the concepts to other branches of mathematics such as those mentioned earlier. Thus, as the subtitle shows, the book is meant only as a first course on measure and integration. Advanced topics involving measures in the context of topological spaces and topological groups and so on are beyond the scope of this text. This book can be used for a one-semester course of about 45 lectures for the first- or second-semester of a master’s programme in mathematics. The book can also be used for the final year of a bachelor’s program, perhaps, omitting the last two chapters. To use this book for a course on measure and integration, no pre-requisite is assumed, except the mathematical maturity to appreciate and grasp concepts in analysis, though it is recommended that it be taught after a course on real analysis.

Acknowledgments: While teaching this course, as well as during the preparation of the notes, I have greatly benefited from the contributions of my students in terms of their questions in classes and also during the clarification of their doubts. One of those students, Rama Seshan, a research scholar from the electrical engineering department at IIT Madras, read the notes carefully and made suggestions from the students’ points of view. Dr. S. Sivanandan (IIT Delhi) and my former research scholar Dr. Ajoy Jana read some of the chapters and brought to my attention some typos and corrections. Dr. P. Sam Johnson (NIT Karnataka) and my research scholar Subhankar Mondal read all the chapters thoroughly and critically and suggested many corrections in the text. I am thankful to all of them. Finally, I thank my wife Sunita for her forbearance and encouragement. M. Thamban Nair

Author

M. Thamban Nair is professor of mathematics at the Indian Institute of Technology (IIT) Madras, Chennai, India. After completing his PhD thesis in 1984 at IIT Bombay, Mumbai (India), he did his post-doctoral research at the University of Grenoble (France), for a year under a French government scholarship. Following his return to India, he worked as a research scientist at IIT Bombay for a year. He taught at the Goa University for nearly a decade, and from December 1995 onward, he has been a faculty member at IIT Madras. He has also held visiting positions at the Australian National University, Canberra (Australia), University of Kaiserslautern (Germany), Sun Yat-sen University, Guangzhou (China), University of Saint-Etienne (France), Weierstrass Institute for Applied Analysis and Stochastics, Berlin (Germany), and University of Chemnitz (Germany). In addition, he has given many invited talks at various institutes in India and abroad. The broad area of Professor Nair’s research is in functional analysis and operator theory, more specifically, spectral approximation, the approximate solution of integral and operator equations, regularization of inverse and illposed problems. He has authored three books, Functional Analysis: A First Course (PHI-Learning, New Delhi), Linear Operator Equations: Approximation and Regularization (World Scientific, Singapore), and Calculus of One Variable (Ane Books, New Delhi), and co-authored Linear Algebra (Springer). He has published over 75 research papers in nationally and internationally reputed journals, including the Journal of Indian Mathematical Society, Proceedings of Indian Academy of Sciences, Proceedings of the American Mathematical Society, Journal of Integral Equations and Operator Theory, Mathematics of Computation, Numerical Functional Analysis and Optimization, Journal of Inverse and Ill-Posed Problems, and Inverse Problems. Professor Nair has received many awards for his academic achievements, including the C.L. Chandna Award of the Indo-Canadian Math Foundation for outstanding contributions in mathematics research and teaching for the year 2003, and the Ganesh Prasad Memorial Award of the Indian Mathematical Society for the year 2015.

ix

Note to the Reader

We shall use standard set-theoretic notations such as ∪, ∩, ⊆, ⊂, ∈ to denote ‘union’, ‘intersection’, ‘subset of’, ‘proper subset of’, ‘belong(s) to’, respectively. For sets S1 and S2 , the set {x ∈ S1 : x 6∈ S2 } is denoted by S1 \ S2 . If f is a function with domain S1 and codomain S2 , then we use the notation f : S1 → S2 , and it is also called a ‘map’ from S1 to S2 . Also, we use the following standard notations and symbols: N Z R C := ∀ ∃ ⇒ ⇐⇒ 7→

: : : : : : : : : :

set of all positive integers set of all integers set of all real numbers set of all complex numbers is defined by for all there exists or there exist implies or imply if and only if maps to

To mark the end of a proof (of a lemma, proposition, theorem, or corollary), we use the symbol , while the symbol ♦ is used to mark the ends of definitions, remarks, examples, and exercises. Numbering of definitions, results (lemmas, theorems, propositions, corollaries), remarks, examples, and exercises are done consecutively using three digits p.q.r, where p and q denote the chapter number and section number, respectively, and r denotes its actual occurrence.

xi

Chapter 1 Review of Riemann Integral

In this chapter, the definition and some basic results on the theory of Riemann integration are reviewed, and the limitations of Riemann integration are pointed out so as to convince the reader of the necessity for a more general integral.

1.1

Definition and Some Characterizations

Let f : [a, b] → R be a bounded function. The idea of a Riemann integral of f is to associate a unique number γ to f such that, in case f (x) ≥ 0 for all x ∈ [a, b], then γ can be thought of as the area of the region bounded by the graph of f , x-axis, and the lines with equations x = a and x = b. For its definition, first we consider a partition P of [a, b], that is, a finite set P := {xi : i = 0, 1, . . . , k} such that a = x0 < x1 < x2 < · · · < xk = b, usually written as P : a = x0 < x1 < x2 < · · · < xk = b, and consider the sums L(P, f ) :=

k X

mi ∆xi ,

U (P, f ) :=

i=1

k X

Mi ∆xi ,

i=1

where mi = inf{f (x) : xi−1 ≤ x ≤ xi },

Mi = sup{f (x) : xi−1 ≤ x ≤ xi }

and ∆xi = xi − xi−1 for i = 1, . . . , k. Clearly, for every partition P of [a, b], L(P, f ) ≤ U (P, f ). Note that if f (x) ≥ 0 for all x ∈ [a, b], then L(P, f ) is the total area of the rectangles with side lengths mi and xi − xi−1 , and U (P, f ) is the total area of the rectangles with side lengths Mi and widths xi − xi−1 , for i = 1, . . . , k. Thus, it is intuitively clear that the required area, say γ, under the graph of f must satisfy the relation: L(P, f ) ≤ γ ≤ U (P, f ) 1

2

Measure and Integration

for all partitions P of [a, b]. With this requirement in mind, we introduce the following definition. Definition 1.1.1 A bounded function f : [a, b] → R is said to be Riemann integrable on [a, b] if there exists a unique γ ∈ R satisfying L(P, f ) ≤ γ ≤ U (P, f ) for all partitions P of [a, b]. If such a γ exists, then it is called the Riemann integral of f and it is denoted by Z

b

f (x)dx.

♦

a

We shall see that every bounded function f : [a, b] → R having at most a finite number of discontinuities is Riemann integrable. However, every bounded function f on [a, b] need not be Riemann integrable, as the following example shows. Example 1.1.2 Let f : [0, 1] → R be the Dirichlet function, that is,  0 if x rational, f (x) = 1 if x irrational. Note that, for any partition P of [a, b], we have L(P, f ) = 0 and U (P, f ) = 1. Thus, for every number α ∈ [0, 1], we have L(P, f ) ≤ α ≤ U (P, f ) for every partition P of [0, 1]. In other words, α ∈ R satisfying L(P, f ) ≤ α ≤ U (P, f ) for all partitions P is not unique. Hence, f is not Riemann integrable. ♦ In the following, we shall denote the set of all partitions of [a, b] by P. Let f : [a, b] → R be a bounded function and let m = inf{f (x) : a ≤ x ≤ b},

M = sup{f (x) : a ≤ x ≤ b}.

Then, for any partition P = {xi : i = 0, 1, . . . , k} of [a, b], we have L(P, f ) =

k X

mi ∆xi ≥

i=1

k X

m∆xi = m(b − a)

i=1

and U (P, f ) =

k X i=1

Mi ∆xi ≤

k X

M ∆xi = M (b − a)

i=1

so that m(b − a) ≤ L(P, f ) ≤ U (P, f ) ≤ M (b − a).

Review of Riemann Integral

3

Thus, the sets {L(P, f ) : P ∈ P} and {U (P, f ) : P ∈ P} are bounded above and bounded below. Hence, L(f ) U (f )

:= :=

sup{L(P, f ) : P ∈ P}, inf{U (P, f ) : P ∈ P}

exist as real numbers. Using the quantities L(f ) and U (f ), we have the following characterization of Riemann integrability. Theorem 1.1.3 A bounded function f : [a, b] → R is Riemann integrable on [a, b] if and only if L(f ) = U (f ). Proof. Suppose f : [a, b] → R is Riemann integrable on [a, b], and let γ be the Riemann integral of f . Then, from the relations L(P, f ) ≤ γ ≤ U (P, f ) ∀P ∈ P

(∗)

we have L(f ) ≤ γ ≤ U (f ). Consequently, L(P, f ) ≤ L(f ) ≤ γ ≤ U (f ) ≤ U (P, f ) ∀P ∈ P. Now, since γ is the only number satisfying (∗), we have L(f ) = γ = U (f ). Conversely, suppose L(f ) = U (f ). Then we have L(P, f ) ≤ L(f ) = U (f ) ≤ U (P, f ) ∀P ∈ P. Thus, γ := L(f ) = U (f ) satisfies (∗). Further, if γ˜ ∈ R satisfies L(P, f ) ≤ γ˜ ≤ U (P, f ) ∀P ∈ P, then, from the definition of L(f ) and U (f ), it follows that L(f ) ≤ γ˜ ≤ U (f ). Hence, γ˜ = γ. Thus, we have proved that there exists a unique real number γ satisfying (∗). Remark 1.1.4 As you must have observed, the proof of Theorem 1.1.3 was very easy. We gave its proof in detail, mainly because of the fact that in standard text books, the Riemann integrability of a bounded function f : [a, b] → R is defined by requiring L(f ) = U (f ). ♦ Remark 1.1.5 The quantities L(P, f ) and U (P, f ) are known as lower Darboux sum and upper Darboux sum, respectively. Analogously, the quantities L(f ) and U (f ) are known as lower Darboux integral and upper Darboux inteRb gral, respectively. Therefore, in view of Theorem 1.1.3, the integral a f (x)dx in Definition 1.1.1 is also known as the Darboux-Riemann integral. ♦

4

Measure and Integration

Let P be a partition of [a, b]. A new partition of [a, b] obtained from P by adjoining additional points is called a refinement of P . Thus, if P = {xi : i = 1, . . . , k} is a partition of [a, b], P ∪ {t1 , . . . , t` }, with each tj satisfying xi−1 < tj < xi for some i ∈ {1, . . . , k}, is a refinement of P . If P1 and P2 are partitions of [a, b], we can consider a new partition P , which is a refinement of both P1 and P2 , by using all the partition points of P1 and P2 , taking repeated points only once. Such a partition is usually denoted by P1 ∪ P2 . Given any two partitions P and Q of [a, b], it can be seen that L(P, f ) ≤ L(P ∪ Q, f )

and U (P ∪ Q, f ) ≤ U (Q, f ).

(∗)

Thus, L(P, f ) ≤ U (Q, f ) for any partitions P and Q of [a, b]. Consequently, L(f ) ≤ U (f ). Exercise 1.1.6 Prove the relations in (∗) above.

♦

The characterization given in the following theorem is useful in deducing many properties of Riemann integral. Theorem 1.1.7 Let f : [a, b] → R be a bounded function. Then f is Riemann integrable if and only if for every ε > 0, there exists a partition P of [a, b] such that U (P, f ) − L(P, f ) < ε. Proof. Suppose f is Riemann integrable and let ε > 0 be given. By the definitions of L(f ) and U (f ), there exist partitions P1 and P2 of [a, b] such that L(f ) − ε/2 < L(P1 , f ) and U (P2 , f ) < U (f ) + ε/2. Let P = P1 ∪ P2 , the partition obtained by combining P1 and P2 . Then, we have L(f ) − ε/2 < L(P1 , f ) ≤ L(P, f ) ≤ U (P, f ) ≤ U (P2 , f ) < U (f ) + ε/2. Since L(f ) = U (f ) (cf. Theorem 1.1.3), it follows that U (P, f ) − L(P, f ) < [U (f ) + ε/2] − [L(f ) − ε/2] = ε. Conversely, suppose that for every ε > 0, there exists a partition P of [a, b] such that U (P, f ) − L(P, f ) < ε. Since L(P, f ) ≤ L(f ) ≤ U (f ) ≤ U (P, f ), we have U (f ) − L(f ) ≤ U (P, f ) − L(P, f ) < ε. This is true for every ε > 0. Hence, L(f ) = U (f ), and hence f is Riemann integrable.

Review of Riemann Integral

5

Here is an immediate consequence of the above theorem. Corollary 1.1.8 A bounded function f : [a, b] → R is Riemann integrable if and only if there exists a sequence (Pn ) of partitions of [a, b] such that U (Pn , f ) − L(Pn , f ) → 0

as

n → ∞,

and in that case the sequences (U (Pn , f )) and (L(Pn , f )) converge to the same Rb limit a f (x)dx. Exercise 1.1.9 Give proof for the above corollary.

♦

Next we give another characterization of Riemann integrability. For that purpose we introduce the following definition. Definition 1.1.10 Let P : a = x0 < x1 < · · · < xn = b be a partition of [a, b] and let T := {ti : i = 1, . . . , n} with ti ∈ [xi−1 , xi ], i = 1, . . . , n. The set T is called a tag set for P . Given a function f : [a, b] → R, the sum S(P, T, f ) :=

n X

f (ti )∆xi

i=1

is called the Riemann sum of f associated with (P, T ). The quantity |P | := max{xi − xi−1 : i = 1, . . . , n} is called the mesh of the partition P .

♦

Note that the Riemann sums may vary as the tag sets vary. It is obvious that, if f : [a, b] → R is a bounded function, then L(P, f ) ≤ S(P, T, f ) ≤ U (P, f ) for any partition P of [a, b] and for any tag set T for P . Therefore, by Theorem 1.1.7, we have the following result. Theorem 1.1.11 If f : [a, b] → R is Riemann integrable, then for every ε > 0, there exists a partition P such that Z b f (x)dx < ε S(P, T, f ) − a for every tag set T for P . Exercise 1.1.12 Supply details of the proof of the above theorem. In fact, the converse to Theorem 1.1.11 is also true.

♦

6

Measure and Integration

Theorem 1.1.13 Let f : [a, b] → R be a bounded function. Suppose there exists γ ∈ R such that for every ε > 0, there exists a partition P of [a, b] satisfying |S(P, T, f ) − γ| < ε Rb for every tag set T of P . Then f is Riemann integrable and γ = a f (x)dx. Proof. Let ε > 0 be given and let P : a = x0 < x1 < · · · < xk = b be as in the hypothesis of the theorem. Then we have γ − ε < S(P, T, f ) < γ + ε

(1)

for any tag set T corresponding to P . Let ui , vi ∈ [xi−1 , xi ] for i = 1, . . . , k be such that Mi − ε < f (ui )

and f (vi ) < mi + ε for i = 1, . . . , k.

(2)

Consider the tag sets T1 = {ui : i = 1, . . . , k} and T2 = {vi : i = 1, . . . , k} for the partition P . Then from (2), we have U (P, f ) − ε(b − a) < S(P, T1 , f ),

S(P, T2 , f ) < L(P, f ) + ε(b − a).

This, together with (1), implies U (P, f ) − ε(b − a) < γ + ε,

γ − ε < L(P, f ) + ε(b − a).

In particular, U (P, f ) − γ < ε[(b − a) + 1],

γ − L(P, f ) < ε[(b − a) + 1]

(3)

so that U (P, f ) − L(P, f ) < 2ε[(b − a) + 1]. Hence, by Theorem 1.1.7, f is Riemann integrable. Therefore, by the relations in (3), we have Z γ − ε[(b − a) + 1] < L(P, f ) ≤

b

f (x)dx ≤ U (P, f ) < γ < ε[(b − a) + 1]. a

Thus, Z γ −

a

b

f (x)dx < ε[(b − a) + 1]

for every ε > 0. Consequently, γ =

Rb a

f (x)dx.

If we know that f is Riemann integrable, then the following theorem is Rb better suited for obtaining approximations for a f (x) dx. For its proof, one may refer Ghorpade and Limaye [7].

Review of Riemann Integral

7

Theorem 1.1.14 Suppose f : [a, b] → R is a Riemann integrable function. Then for every ε > 0, there exists a δ > 0 such that Z S(P, T, f ) −

a

b

f (x)dx < ε

for any partition P of [a, b] with |P | < δ and for every tag set T for P . The conclusion in the above theorem is usually written as b

Z lim S(P, T, f ) =

|P |→0

f (x)dx. a

Here is an immediate consequence of Theorem 1.1.14. Corollary 1.1.15 Suppose f : [a, b] → R is a Riemann integrable function and (Pn ) is a sequence of partitions of [a, b] such that |Pn | → 0 as n → ∞. If Tn is a tag set for Pn for each n ∈ N, then Z S(Pn , Tn , f ) →

b

f (x)dx

as

n → ∞.

a

Looking at Theorem 1.1.13, one may ask the following question. If (Pn ) is a sequence of partitions of [a, b] such that lim |Pn | = 0 n→∞

and lim S(Pn , Tn , f ) = γ for some γ ∈ R, where Tn is a tag set n→∞ for Pn for each n ∈ N, then, is it true that f is Riemann integrable Rb and γ = a f (x)dx? The answer is in the negative as the following example shows. Example 1.1.16 Consider the Dirichlet function f : [0, 1] → R, of Example 1.1.2. That is,  0, x ∈ Q, f (x) := 1, x 6∈ Q. Let xi := i/n for i = 0, 1, . . . , n, and let ti be any rational point in the interval [xi−1 , xi ]. In this case we have lim |Pn | = 0 and S(Pn , Tn , f ) = 0 for all n ∈ N n→∞

so that lim S(Pn , Tn , f ) = 0. However, f is not Riemann integrable. n→∞

♦

8

Measure and Integration

1.2

Advantages and Some Disadvantages

We may observe, in view of Corollary 1.1.8, that if (Pn ) is a sequence of partitions of [a, b] such that (U (Pn , f )) and (L(Pn , f )) converge to the Rb same limit say γ, then f is Riemann integrable, and γ = a f (x)dx. If f : [a, b] → R is a continuous function, then using the uniform continuity of f , it can be shown that for any sequence (Pn ) of partitions of [a, b] satisfying |Pn | → 0 as n → ∞, we have U (Pn , f ) − L(Pn , f ) → 0 as n → ∞. Thus, by Corollary 1.1.8, we can conclude the following: Every continuous function f : [a, b] → R is Riemann integrable. The following results are also true; for their proofs, the reader may refer to Ghorpade and Limaye [7] or Rudin [13]. (a) Every bounded function f : [a, b] → R having at most a finite number of discontinuities is Riemann integrable. (b) Every monotonic function f : [a, b] → R is Riemann integrable. Thus, the set of all Riemann integrable functions is very large. In fact we have the following theorem, known as Lebesgue’s criterion for Riemann integrability, whose proof depends on some techniques involving the concept of oscillation of a function; refer to Delninger [4] for its proof. Lebesgue’s criterion for Riemann integrability: A bounded function f : [a, b] → R is Riemann integrable if and only if the set of points at which f is discontinuous is of measure zero. In the above, the terminology set of measure zero is used in the sense of the following definition. Definition 1.2.1 A set E ⊆ R is said to be of measure zero if for every ε > 0, there exists a countable family {In } of open intervals such that [ X E⊆ In and `(In ) < ε, n

n

where `(In ) is the length of the interval In .

♦

Example 1.2.2 We show that every countable subset of R is of measure zero. To see this, consider a countable set E = {an : n ∈ Λ}, where Λ is {1, . . . , k} for some k ∈ N or Λ = N. For ε > 0, let In := (an − ε/2n+1 , an + ε/2n+1 ),

n ∈ Λ.

Then E ⊆ ∪n∈Λ In

and

X n∈Λ

`(In ) =

X n∈Λ

(ε/2n ) ≤ ε.

♦

Review of Riemann Integral

9

Can an uncountable set be of measure zero? We shall answer this question affirmatively in the next chapter. Functions with only a finite or countably infinite number of discontinuities in [a, b] can be constructed easily. In fact, the following example shows that given any countable subset S of [a, b], there is a function f : [a, b] → R such that S is exactly the set of points in [a, b] at which f is discontinuous. Example 1.2.3 Let I = [a, b], S := {an : n ∈ N} ⊆ I and let f : I → R be defined by f (an ) = 1/n for all n ∈ N and f (x) = 0 for x ∈ I \ S. Clearly, this function is not continuous at any x ∈ S. We show that f is continuous at every x ∈ I \ S. Let x0 ∈ I \ S. Then f (x0 ) = 0. For ε > 0, we have to find a δ > 0 such that |x − x0 | < δ implies |f (x)| < ε. For δ > 0, let Jδ := (x0 − δ, x0 + δ) ∩ I. For ε > 0, let k ∈ N be such that 1/k < ε. Choose δ > 0 such that a1 , a2 , . . . , ak 6∈ Jδ . For instance, we may choose 0 < δ < min{|x0 − ai | : i = 1, . . . , k}. Then we have Jδ ∩ {a1 , a2 , . . .} ⊆ {ak+1 , ak+2 , . . .}. Hence, for x ∈ Jδ , we have either f (x) = 0 or f (x) = 1/n for some n > k. Thus, 1 |f (x)| ≤ < ε. k Thus we have proved that f is continuous at x0 . If we take S as the set of all rational numbers in I, then the function f is continuous at every irrational number in I and discontinuous at every rational number in I. ♦ Although the set of Riemann integrable functions on [a, b] is quite large, this class lacks some desirable properties. For example observe the following drawbacks of Riemann integrability and Riemann integration: (a) If (fn ) is a sequence of Riemann integrable functions on [a, b] and if fn (x) → f (x) as n → ∞ for every x ∈ [a, b], then it is not necessary that f is Riemann integrable. (b) Even if the function f in (a) is Riemann integrable, it is not necessary Rb Rb that a fn (x)dx → a f (x)dx as n → ∞. To illustrate the last two statements consider the following examples. Example 1.2.4 Let {r1 , r2 , . . .} be an enumeration of the set rational numbers in [0, 1]. For each n ∈ N, let  0 if x ∈ {r1 , . . . , rn }, fn (x) = 1 if x 6∈ {r1 , . . . , rn }.

10

Measure and Integration

Then each fn is Riemann integrable, as it is continuous except at a finite number of points. Note that fn (x) → f (x) as n → ∞ for every x ∈ [0, 1], where f : [0, 1] → R is the Dirichlet’s function, that is,  0 if x rational, f (x) = 1 if x irrational. We have seen in Example 1.1.2 that f is not Riemann integrable, which also follows from the Lebesgue’s criterion ofR Riemann integrability, since f is dis1 continuous everywhere. Thus, though 0 fn (x)dx = 1 for every n ∈ N and fn (x) → f (x) for every x ∈ [0, 1], we cannot even talk about the integral of f. ♦ Example 1.2.5 For n ∈ N, let fn : [0, 1] → R be defined by fn (x) = nχ(0,1/n] (x),

x ∈ [0, 1],

where χE denotes the characteristic function of E, that is,  1 if x ∈ E, χE (x) = 0 if x 6∈ E. Then we see that, for each x ∈ [0, 1], fn (x) → f (x) = 0 as n → ∞, but R1 Rb Rb f (x) dx = 1 for every n ∈ N. Hence a fn (x)dx 6→ a f (x)dx. ♦ 0 n Example 1.2.6 Consider fn : [0, 1] → R defined by  ne−nx if x ∈ (0, 1], fn (x) = 0 if x = 0. Then we have, for each x ∈ [0, 1], fn (x) → f (x) = 0 as n → ∞. Note that R1 R1 R1 f (x)dx = 1 − e−n → 1 as n → ∞. Thus, 0 fn (x)dx 6→ 0 f (x)dx. ♦ 0 n In Examples 1.2.5 and 1.2.6, we see that, although the sequence (fn (x)) converges for each x ∈ [0, 1], it is not uniformly bounded, that is, there does not exist an M > 0 such that |fn (x)| ≤ M for all x ∈ [a, b] and for all n ∈ N. In fact, in both the examples, the sequence (fn (1/n)) is unbounded. In this context, it is worth mentioning the following theorem, known as Arzela’s dominated convergence theorem, also called Arzela’s theorem. Theorem 1.2.7 (Arzela’s theorem) Suppose (fn ) is a sequence of Riemann integrable functions defined on [a, b] such that fn (x) → f (x) as n → ∞ for each x ∈ [a, b] for some Riemann integrable function f . If (fn ) is uniformly bounded, then Z

b

Z fn (x)dx →

a

b

f (x)dx a

as

n → ∞.

Review of Riemann Integral

11

For a recent elementary proof for the above theorem, one may refer to [8]. Note that, in Arzela’s theorem, we assumed that the limit function f is Riemann integrable. In this course we shall have a new type of integral, called the Lebesgue integral, which includes the Riemann integral, and derive Arzela’s theorem as a consequence of a more general result (see Theorem 5.1.12). In fact, the assumption of Riemann integrability of f is not required, but then, the integral of f is to be understood in the sense of the Lebesgue integral. In the next section we introduce certain notations and conventions which we shall use throughout the book.

1.3

Notations and Conventions

By countable family {An } of sets, we mean a family {An : n ∈ Λ} of sets An , n ∈ Λ, where either Λ = {1, 2, . . . , k} for some k ∈ N or Λ = N, the set of natural numbers. Then, ∪{An : n ∈ Λ} and ∩{An : n ∈ Λ} will be written as ∪n An and ∩n An , respectively. Also, for : n ∈ Λ} P a countable set {an ∈ R P of non-negative real numbers, the series n∈Λ an will be written as n an . By an interval we mean a subset I of R which is any of the following forms (a, b), [a, b), (a, b], [a, b] for any a, b ∈ R with a < b or of the forms (a, ∞), [a, ∞), (−∞, a), (−∞, a] for any a ∈ R. Here, for a ∈ R, (a, ∞) := {x ∈ R : a < x}, [a, ∞) := {x ∈ R : a ≤ x}, (−∞, a) := {x ∈ R : x < a}, (−∞, a] := {x ∈ R : x ≤ a}. The intervals (a, b), [a, b), (a, b], [a, b] are bounded intervals with end points a and b, and the intervals (a, ∞), [a, ∞), (−∞, a), (−∞, a], (−∞, ∞) are unbounded intervals. Length of an interval I is denoted by `(I). Thus, for a bounded interval I of end points a, b with a < b, `(I) is b − a. If I is an unbounded interval, then we sayP that its length is infinity, and we Pwrite `(I) = ∞. Also, for a divergent series n∈Λ an with an ≥ 0, we write n∈Λ an = ∞. Although ∞ is not a real number, we shall use the symbol ∞ as an extended real number. We have already used the symbol ∞ to denote the sum of a divergent series of non-negative real numbers. Similarly, we may use the symbol −∞ for the sum of a divergent series of non-positive real numbers.

12

Measure and Integration

˜ := R∪{∞, −∞}, called the set of all extended real Thus, we have the set R ˜ are defined as numbers. The order relations, addition and multiplication on R follows: For every a ∈ R, ∞ > a,

−∞ < a, ±∞ ± a = ±∞,  ±∞ if a > 0, ±∞ × a = ∓∞ if a < 0.

Further, ±∞ × 0 = 0,

±∞ ± ∞ = ±∞,

(±∞) × (±∞) = ∞.

However, ±∞ ∓ ∞ is not defined. In the above, addition and multiplication are commutative. Also, for any a ∈ R, we denote (a, ∞] := (a, ∞) ∪ {∞},

[−∞, a) := (−∞, a) ∪ {−∞},

[a, ∞] := [a, ∞) ∪ {∞},

[−∞, a] := (−∞, a] ∪ {−∞}.

˜ = R ∪ {−∞, ∞}, the We shall also use the notation [−∞, ∞] for the set R set of all extended real numbers. Intervals of the form (a, ∞] and [−∞, b) for a, b ∈ R are called neighbourhoods of ∞ and −∞, respectively. Throughout the text, we consider R with the usual metric, that is, d(x, y) := |x − y|,

x, y ∈ R.

Thus, a set A ⊆ R is open in R if and only if for each x ∈ A, there exists r > 0 such that (x − r, x + r) ⊆ A, and a set B ⊆ R is closed if and only if its complement is open in R, that is the set R \ B, is open. Further concepts based on open subsets of R will be introduced as and when they are required. Let S ⊆ R. Then (a) S is said to be bounded above if there exists b ∈ R such that s ≤ b for all s ∈ S, and in that case, b is called an upper bound of S; (b) S is said to be bounded below if there exists a ∈ R such that a ≤ s for all s ∈ S, and in that case, a is called a lower bound of S. (c) S is said to have a least upper bound b0 ∈ R if b0 is an upper bound of S and for any b < b0 , there exists s ∈ S such that b < s ≤ b0 . (d) S is said to have a greatest lower bound a0 ∈ R if a0 is a lower bound of S and for any a > a0 , there exists s ∈ S such that a0 ≤ s < a. It can be easily seen that if S ⊆ R has a least upper bound (respectively, greatest lower bound), then it is unique. An important property of R is its least upper bound property, stated as follows: Every subset of R which is bounded above has the least upper bound.

Review of Riemann Integral

13

Using the least upper bound property of R, it can be deduced that it has greatest lower bound property as well: Every subset of R which is bounded below has the greatest lower bound. If S ⊆ R is bounded above, then its least upper bound is also called the supremum of S, and it is denoted by sup(S). If S ⊆ R is bounded below, then its greatest lower bound is also called the infimum of S, and it is denoted by inf(S). If S is not bounded above, then we shall write sup(S) = ∞, and if S is not bounded below, then we shall write inf(S) = −∞. This convention is also adopted if S is a subset of [−∞, ∞]. ˜ Using the above definitions, supremum and infimum of a sequence in R ˜ can be defined as follows: Let (an ) be a sequence in R. Then we define sup an = sup{an : n ∈ N}, n

inf an = inf{an : n ∈ N}. n

Also, the limit supremum and limit infimum of (an ) are defined by lim sup an = inf sup an , n

k n≥k

lim inf an = sup inf an , n

k

n≥k

respectively. We observe that, if bk := sup an ,

ck := inf an n≥k

n≥k

for k ∈ N, then (bk ) is a decreasing sequence and (ck ) is an increasing sequence, ˜ Thus, we have and hence, they converge in R. lim sup an = lim sup an , n

k→∞ n≥k

lim inf an = lim inf an . n

k→∞ n≥k

˜ if lim sup an = lim inf n an , then we say that (an ) For a sequence (an ) in R, n ˜ and the common value, say a ∈ R, ˜ is called the limit of (an ), converges in R, and in that case we write lim an = a or an → a as n → ∞ or an → a.

n→∞

˜ Then the following statements can be easily Let (an ) be a sequence in R. verified. ˜ to a ∈ R if and only if for every ε > 0, there exists 1. (an ) converges in R N ∈ N such that an ∈ (a − ε, a + ε) for all n ≥ N . ˜ to ∞ if and only if for every r ∈ R, there exists 2. (an ) converges in R N ∈ N such that an ∈ (r, ∞] for all n ≥ N .

14

Measure and Integration ˜ to −∞ if and only if for every r ∈ R, there exists 3. (an ) converges in R N ∈ N such that an ∈ [−∞, r) for all n ≥ N . 4. If an ∈ R for all n ∈ N and (an ) converges in R, then (an ) converges in ˜ R. 5. If an ∈ R for all n ∈ N and (an ) diverges to ∞ (respectively, to −∞), ˜ to ∞ (respectively, to −∞). then it converges in R

Chapter 2 Lebesgue Measure

This chapter is a stepping stone for the subject of measure and integration. Here, we define the concept of an outer measure of a subset of the real line as a generalization of the concept of length of an interval, and then define the concept of measure of certain subsets of the real line which must satisfy certain intuitively desirable properties. The properties of the class of subsets which can be measured is a motivation for the concept of a general measure in the context of an arbitrary set which we shall consider in the subsequent chapters.

2.1

Lebesgue Outer Measure

In Chapter 1 we defined the concept of a set of measure zero for subsets of R while stating a characterization of Riemann integrability of bounded real valued functions defined on closed and bounded intervals (Definition 1.2.1). We have also observed that every countable subset of R is of measure zero. In particular, the set of rationals in any interval is of measure zero. So, we may ask the following questions: (1) If a subset E of R contains an open interval (of nonzero length), then can it be of measure zero? (2) If a subset E of R contains no open interval, is it of measure zero? (3) Can an uncountable subset E of R be of measure zero? Let us wait for a while to answer these questions. Given a set E ⊆ R, we shall denote by IE , the collection of all countable family {In } of open intervals which covers E, that is, E ⊆ ∪n In . Definition 2.1.1 The Lebesgue outer measure of E ⊆ R is defined as X m∗ (E) := inf `(In ), IE

n

15

16

Measure and Integration

where the infimum is taken over IE .

♦

Note that m∗ (E) ≥ 0, and m∗ (E) can take the value ∞ as well. Definition 2.1.2 The map m∗ : E 7→ m∗ (E) from 2R , the power set of R, to [0, ∞] is called the Lebesgue outer measure on R. ♦ In due course, we shall refer to m∗ , simply as outer measure. One would expect that if A1 and A2 are any two disjoint subsets of R, then we must have m∗ (A1 ∪ A2 ) = m∗ (A1 ) + m∗ (A2 ). We shall see that, this is not necessarily true, and for the above to be true, we have to restrict the outer measure to a class of subsets of R, called Lebesgue measurable sets. Although, the class of Lebesgue measurable sets is going to be very large, it is important to know that there are non-Lebesgue measurable sets. Before, introducing this class of Lebesgue measurable sets, we observe some properties of the Lebesgue outer measure. Theorem 2.1.3 The following results hold. (i) If I is an open interval and A ⊆ I, then m∗ (A) ≤ `(I). (ii) m∗ (∅) = 0. (iii) If E is a countable subset of R, then m∗ (E) = 0. Proof. (i) Let A ⊆ I, where I is an open interval. Taking the singleton family {In } with In = I and n = 1, we obtain m∗ (A) ≤ `(I). (ii) For every ε > 0, we have ∅ ⊆ (−ε, ε). Hence by (i), m∗ (∅) ≤ 2ε. This is true for every ε > 0. Hence, m∗ (∅) = 0. (iii) First let E be a finite set, say E = {a1 , . . . , ak } ⊆ R. Then for every ε > 0, E ⊆ ∪ki=1 Ii , where Ii = (ai − ε, ai + ε). Hence, m∗ (E) ≤ 2kε. Since this is true for every ε > 0, m∗ (E) = 0. Next suppose that E is a countably infinite set, say E = {ai : i ∈ N}. Then taking In := (an − ε/2n+1 , an + ε/2n+1 ), we have E ⊆ ∪n∈N In so that m∗ (E) ≤

X n∈N

`(In ) =

X

(ε/2n ) ≤ ε.

n∈N

Since this is true for every ε > 0, we obtain m∗ (E) = 0. The following theorem is very useful in deriving many properties of the outer measure.

Lebesgue Measure

17

Theorem 2.1.4 Let E ⊆ R. For every S ε > 0, there exists a countable family {In } of open intervals such that E ⊆ n In and X `(In ) ≤ m∗ (E) + ε. n

Strict inequality occurs in the above if m∗ (E) < ∞. Proof. If m∗ (E) = ∞, then the conclusion is true trivially. So, assume that m∗ (E) < ∞. Then the results follow from the definition of infimum of a subset of R. By Theorem 2.1.4, for E ⊆ R, m∗ (E) = 0 if and only if for every ε > 0, there exists a countable family {In } of open intervals such that [ X E⊆ In and `(In ) ≤ ε. n

n

Thus, we can conclude: E is of measure zero if and only if its outer measure is zero.

Corollary 2.1.5 Let E ⊆ R. For every ε > 0, there exists an open set G in R such that E ⊆ G and m∗ (G) ≤ m∗ (E) + ε. Proof. Let ε > 0 be given. By TheoremS2.1.4, there exists a countable family {In } of open intervals such that E ⊆ n In and X `(In ) ≤ m∗ (E) + ε. n

Take G = n In . Then, by the definition of m∗ , m∗ (G) ≤ have m∗ (G) ≤ m∗ (E) + ε. S

P

n

`(In ). Thus, we

Remark 2.1.6 In view of Corollary 2.1.5, one may ask the following question: If E ⊆ R, then for every ε > 0, does there exist an open set G ⊃ E such that m∗ (G \ E) < ε?

We shall see, in Theorem 2.2.21, that the answer to the above question is in the affirmative only if E belongs to a particular class of subsets, called Lebesgue measurable sets. ♦

18

Measure and Integration

Notation: Sets considered in this chapter are subsets of R. Also, for E ⊆ R, we shall use the notation IE to denote the class of all countable family of open intervals {In } such that E ⊆ ∪n In . For subsets A and B of R, we define A + B = {x + y : x ∈ A, y ∈ B}. Also for E ⊆ R and a ∈ R, we denote, E + a = E + {a} so that E + a := {x + a : x ∈ E}. Theorem 2.1.7 Let A and B be subsets of R. Then the following results hold. (i) If A ⊆ B, then m∗ (A) ≤ m∗ (B). (ii) m∗ (A ∪ B) ≤ m∗ (A) + m∗ (B). (iii) If E ⊆ A and m∗ (E) = 0, then m∗ (A \ E) = m∗ (A). (iv) If E ⊆ R and x ∈ R, then m∗ (E + x) = m∗ (E). Proof. (i) Let A ⊆ B and let {In } ∈ IB . Then {In } ∈ IA . Hence X m∗ (A) ≤ `(In ). n

Now, taking infimum over all {In } ∈ IB , we have m∗ (A) ≤ m∗ (B). (ii) If either m∗ (A) or m∗ (B) is infinity, then the result holds. Next, assume that both m∗ (A) and m∗ (B) are finite. Hence, by Theorem 2.1.4, given ε > 0 there exist {In } and {Jn } in IA and IB , respectively, such that X X ε ε `(In ) < m∗ (A) + , `(Jn ) < m∗ (B) + . 2 2 n n ∞ Then, the collection {In }∞ n=1 ∪ {Jk }k=1 of open intervals covers A ∪ B. Therefore, X X m∗ (A ∪ B) ≤ `(In ) + `(Jn ) n

n

ε ε  ∗ + m (B) + ≤ m (A) + 2 2 = m∗ (A) + m∗ (B) + ε. 

∗

This is true for all ε > 0, so that m∗ (A ∪ B) ≤ m∗ (A) + m∗ (B). (iii) If E ⊆ A and m∗ (E) = 0, then by (i) and (ii), m∗ (A) ≤ m∗ (E) + m∗ (A \ E) = m∗ (A \ E) ≤ m∗ (A). Hence, m∗ (A \ E) = m∗ (A).

Lebesgue Measure

19

(iv) P Suppose E ⊆ R and x ∈ R. Given ε > 0, let {In } in IE be such that n `(In ) ≤ m∗ (E) + ε. Note that each In + x is an open interval with `(I + x) = `(I) and E + x ⊆ ∪n (In + x). Hence, X X m∗ (E + x) ≤ `(In + x) = `(In ) = m∗ (E) + ε. n

n

This is true for every ε > 0. Hence, m∗ (E + x) ≤ m∗ (E). Since E = (E + x) + (−x), it follows from the above that m∗ (E) ≤ m∗ (E + x). Thus the proof is complete. The property (i) in Theorem 2.1.7 is called the monotonicity property of m∗ , and the property (iv) is called the translation invariance of m∗ . Making use of the monotonicity of m∗ , we deduce the following. Corollary 2.1.8 Let E ⊆ R. Then there exists a set G which is a countable intersection of open sets in R such that E ⊆ G and m∗ (G) = m∗ (E). Proof. By Corollary 2.1.5, for each n ∈ N, there exists an open set Gn in R such that 1 E ⊆ Gn and m∗ (Gn ) ≤ m∗ (E) + . n T Take G = n Gn . Then, E ⊆ G and m∗ (E) ≤ m∗ (G) ≤ m∗ (Gn ) for every n ∈ N, so that 1 ∀ n ∈ N. m∗ (G) ≤ m∗ (E) + n Letting n tend to infinity, we obtain, m∗ (G) ≤ m∗ (E). Thus, we have proved m∗ (G) = m∗ (E). Definition 2.1.9 A subset of R is said to be a Gδ -set if it is a countable intersection of open sets, and a subset of R is said to be an Fσ -set if it is a countable union of closed sets. ♦ Since a subset of R is closed if and only if its complement is open, using De Morgan’s law, it follows that a subset of R is a Gδ -set if and only if its complement is an Fσ -set. Note that a Gδ -set need not be open and an Fσ -set need not be closed. For example, the set [0, 1) is neither open nor closed, but it is both a Gδ -set and an Fσ -set, as [0, 1) =

∞  \ n=1

−

1  ,1 , n

[0, 1) =

∞ h [ 0, n=1

In fact, every interval is both Gδ -set and Fσ -set.

n i . n+1

20

Measure and Integration In view of Definition 2.1.9, Corollary 2.1.8 can be stated as follows: For every E ⊆ R, ∃ a Gδ -set G ⊇ E such that m∗ (G) = m∗ (E).

The following corollary is immediate from Theorem 2.1.7 (ii). Corollary 2.1.10 For subsets A1 , . . . , An of R, m∗

n [

n  X Ak ≤ m∗ (Ak ).

k=1

k=1

More generally, we have the following theorem. Theorem 2.1.11 Suppose Ak ⊆ R for k ∈ N. Then m∗

∞ [

∞  X Ak ≤ m∗ (Ak ).

k=1

k=1

Proof. If m∗ (Ak ) = ∞ for some k ∈ N, then the result holds trivially. Hence, assume that m∗ (Ak ) < ∞ for every k ∈ N. Then, by Theorem 2.1.4, for each k ∈ N and ε > 0, there exists {Ik,n } ∈ IAk such that X `(Ik,n ) < m∗ (Ak ) + ε/2k . n

Since Ak ⊆ ∪n Ik,n , we have m∗

∞ [

k=1

Ak ⊆

S∞ S k=1

n Ik,n .

Therefore,

∞ ∞ ∞ X  X X X m∗ (Ak ) + ε. (m∗ (Ak ) + ε/2k ) = Ak ≤ `(Ik,n ) ≤ k=1 n

k=1

Thus, m∗

S∞

S

∞ k=1



Ak ≤

k=1

P∞

k=1

k=1

m∗ (Ak ) + ε. This is true for every ε > 0. Hence

the result follows. The property of m∗ in Theorem 2.1.11 is called the countable subadditivity of m∗ . Now we prove a result that we are waiting for. Theorem 2.1.12 The Lebesgue outer measure of any interval is its length. Proof. Let I be an interval. Case 1. Suppose I = [a, b] with −∞ < a < b < ∞: Let ε > 0 and let Iε := (a − ε, b + ε). Then, m∗ (I) ≤ m∗ (Iε ) ≤ b − a + 2ε.

Lebesgue Measure

21

This is true for all ε > 0. Hence, m∗ (I) ≤ b − a. Thus, it remains to show that m∗ (I) ≥ b − a. For this, it is enough to show that X b−a≤ `(In ) ∀ {In } ∈ II , (∗) n

because, in that case we can take infimum over all such {In } ∈ II and obtain b − a ≤ m∗ (I). So, let {In } ∈ II . If `(In ) = ∞ for some n ∈ N, then (∗) holds trivially. So, we may assume that each In is of finite length. By the compactness of I, there exists a finite sub-collection {In1 , , . . . , Ink } of {In } such that I ⊆ ∪ki=1 Ini . Let Ini := (ai , bi ) for i ∈ {1, . . . , k}. We may assume, without loss of generality, that (ai , bi ) ∩ [a, b] 6= ∅ and ai+1 < bi for i ∈ {1, . . . , k − 1}. Then k X i=1

`(Ini ) =

k−1 k X X (bi − ai+1 ) ≥ bk − a1 ≥ b − a, (bi − ai ) = bk − a1 + i=1

i=1

so that b−a≤

k X i=1

`(Ini ) ≤

X

`(In ).

n

Thus, we have proved (∗). Case 2. Suppose I is an interval of finite length whose end points a and b with a < b, which is not necessarily a closed interval: Then for sufficiently small ε > 0, we have [a + ε, b − ε] ⊆ I ⊆ [a − ε, b + ε]. Hence, m∗ ([a + ε, b − ε]) ≤ m∗ (I) ≤ m∗ ([a − ε, b + ε]) so that by Case 1, b − a − 2ε ≤ m∗ (I) ≤ b − a + 2ε. Since this is true for every ε > 0, it follows that m∗ (I) = b − a. Case 3. Suppose I is of infinite length: In this case, for every M > 0 there exists a closed interval IM of length M such that IM ⊆ I. By Case 1, m∗ (IM ) = M . Thus, M = m∗ (IM ) ≤ m∗ (I). Since this is true for every M > 0, m∗ (I) = ∞. By Theorem 2.1.3 (iii), outer measure of every countable subset of R is zero, and by Theorem 2.1.12, every interval is of positive measure. Hence, we can conclude:

22

Measure and Integration Every interval is an uncountable set.

Can an uncountable set be of measure 0? The answer is in the affirmative as the following example of Cantor’s ternary set or simply Cantor set shows. Example 2.1.13 (Cantor ternary set) Let us first recall how the Cantor ternary set is constructed. Consider the unit interval [0, 1]. Let C1 be the set obtained after removing its “middle third” J1 := ( 31 , 23 ) from C0 := [0, 1]. That is h 1i h2 i C1 = 0, ∪ ,1 . 3 3 Next, let C2 be the set obtained from C1 after removing the “middle thirds” from each of the two subintervals in C1 . Let the removed set be J2 . Thus, h 1i h2 1i h2 7i h8 i C2 = 0, ∪ , ∪ , ∪ ,1 . 9 9 3 3 9 9 Continue this procedure to obtain C3 , C4 , and so on. At the nth stage, we obtain Cn = Cn−1 \ Jn , where Jn is the union of the middle thirds of each of the subintervals in Cn−1 . Now the Cantor set C is defined by C=

∞ \

Cn .

n=1

Note that C1 ⊃ C2 ⊃ C3 ⊃ · · · . and m∗ (C1 ) = 2/3, m∗ (C2 ) = (2/3)2 , m∗ (C3 ) = (2/3)3 , etc., and more generally,  2 n m∗ (Cn ) = , n ∈ N. 3 Hence, k \   2 k m∗ (C) ≤ m∗ Cn = m∗ (Ck ) = ∀ k ∈ N. 3 n=1 Thus, m∗ (C) = 0. To see that C is an uncountable P∞set, first we recall that every number a ∈ [0, 1] can be written as a series n=1 a3nn , where an ∈ {0, 1, 2} for n ∈ N. It is possible that a number a ∈ [0, 1] can have two different representations; one with only a finite number of terms and the other with an infinite number of terms. For example, the number 13 can be written as 1 1 0 0 0 = + 2 + 3 + 4 + ··· 3 3 3 3 3

and

1 2 2 2 = 2 + 3 + 4 + ··· . 3 3 3 3

Lebesgue Measure Similarly, the number

2 3

23

is represented as

0 2 0 2 0 = + + 2 + 3 + ··· 3 3 3 3 3

2 2 2 1 2 = + 2 + 3 + 4 + ··· . 3 3 3 3 3 Pk More generally, if x ∈ (0, 1] has a finite sum representation, say x = j=1 for some k ∈ N with ak 6= 0, then ak ∈ {1, 2} and hence and

aj 3j

∞

ak − 1 X 2 ak = + 3k 3k 3k+j j=1 so that x has an infinite expansion as well. Now, let [0, 1]. Then a ∈ C if and only if a ∈ Cn for every n. Recall that each Cn consists of 2n disjoint closed intervals, and the 2n+1 closed intervals of Cn+1 is obtained by removing the middle one-third from each of the 2n intervals of Cn . Hence, it is seen that an is either 0 or 2. Hence we obtain C=

∞ nX o bn : b ∈ {0, 2}, n ∈ N . n 3n n=1

Thus, C is in one-one correspondence with the family of all sequences (bn ) with bn ∈ {0, 2}, and hence, C is an uncountable set. ♦

2.2

Lebesgue Measurable Sets

Now investigate the question whether the equality m∗ (A1 ∪ A2 ) = m∗ (A1 ) + m∗ (A2 ).

(1)

holds for any two disjoint subsets A1 and A2 of R. Suppose for a moment that (1) is true for any two disjoint sets A1 and A2 of R. Then we also have m

∗

n [



Ai =

i=1

n X

m∗ (Ai )

(2)

i=1

for any pairwise disjoint sets A1 , . . . , An . Now, consider a denumerable disjoint family {An }∞ n=1 , that is, An ∩ Am = ∅ whenever n 6= m. Then using the equality (2) above, monotonicity (Theorem 2.1.7 (i)), and the subadditivity (Theorem 2.1.11) of m∗ , we obtain n X i=1

m∗ (Ai ) = m∗

n [ i=1

∞ ∞  [  X Ai ≤ m∗ Ai ≤ m∗ (Ai ) i=1

i=1

24

Measure and Integration

for every n ∈ N, so that n X

∗

∗

m (Ai ) ≤ m

∞ [

i=1



Ai ≤

i=1

∞ X

m∗ (Ai )

i=1

for every n ∈ N. Taking limit as n → ∞, we obtain m∗

∞ [

∞  X Ai = m∗ (Ai ).

i=1

(3)

i=1

We now show that (3) does not hold for certain denumerable disjoint family {An }∞ n=1 of subsets of R, and consequently, there are disjoint subsets A1 and A2 for which (1) does not hold. For this, first we prove the following. Theorem 2.2.1 There exists a set E ⊂ [0, 1] such that if {r1 , r2 , . . .} is an enumeration of the rational numbers in [−1, 1] and En := E + rn for n ∈ N, then {En } is a disjoint family satisfying 1 ≤ m∗

∞ [

 En ≤ 3.

n=1

Proof. Consider the relation ∼ on R by defining x ∼ y ⇐⇒ x − y ∈ Q. It can be easily seen that ∼ is an equivalence relation on R. Hence, R is the disjoint union of equivalence classes. Let E be the subset of [0, 1] such that its intersection with each equivalence class is a singleton set. Such a set E exists by using the axiom of choice on the collection E := {[x] ∩ [0, 1] : x ∈ [0, 1]}, where [x] is the equivalence class of x. We note that if x ∼ y, then the rational number r := x − y satisfies −1 ≤ r ≤ 1. Let {r1 , r2 , . . .} be the set of all rational numbers in [−1, 1]. Let En := E + rn for n ∈ N. Then {En } is a disjoint family. Indeed, if En ∩ Em 6= ∅ for some n 6= m, then there exist x, y ∈ E such that x + rn = y + rm so that x ∼ y; consequently, x = y so that rn = rm , which is not possible. Since E ⊆ [0, 1] and rn ∈ [−1, 1], we have En = E + rn ⊆ [−1, 2] ∀ n ∈ N. S∞

Hence, n=1 En ⊆ [−1, 2]. Also, for each x ∈ [0, 1], there S exists y ∈ E such ∞ that x ∼ y so that x ∈ En for some n ∈ N. Hence, [0, 1] ⊆ n=1 En . Thus, [0, 1] ⊆

∞ [

En ⊆ [−1, 2]

n=1

so that, using the monotonicity of m∗ , the required inequality follows.

Lebesgue Measure

25

From the above theorem, we deduce the following. Theorem 2.2.2 Let E and {En } be as in Theorem 2.2.1. Then m∗ (E) > 0 and ∞ ∞ [  X En < m∗ (En ). m∗ n=1

n=1 ∗

Proof. By the sub-additivity of m and by Theorem 2.2.1, we have 1 ≤ m∗

∞ [ n=1

∞  X En ≤ m∗ (En ). n=1

From this, using the fact that m∗ (En ) = m∗ (E) for all  n ∈ N, it follows that S∞ ∗ ∗ m (E) > 0. Therefore, the relation 1 ≤ m n=1 En ≤ 3 in Theorem 2.2.1 S  P ∞ ∞ ∗ shows that m∗ n=1 En = n=1 m (En ) is not possible. By the above theorem, the relation (3), and hence the relation (1) does not hold for some disjoint family of sets involved. Thus, we have also proved the following result. Theorem 2.2.3 There exist disjoint subsets A1 and A2 of R such that m∗ (A1 ∪ A2 ) < m∗ (A1 ) + m∗ (A2 ). The above discussion motivates us to look for a family of sets in which the relation (1), and hence (2) and (3) hold for all possible disjoint family of sets involved. Definition 2.2.4 A set E ⊆ R is said to be Lebesgue measurable if m∗ (A) = m∗ (A ∩ E) + m∗ (A ∩ E c ) for every A ⊆ R.

♦

Notation: We denote the set of all Lebesgue measurable sets by M. The proof of the following theorem is easy and hence it is left as an exercise (Problem 10). Theorem 2.2.5 The following results hold. (i) E ∈ M ⇐⇒ m∗ (A) ≥ m∗ (A ∩ E) + m∗ (A ∩ E c ) (ii) E ∈ M ⇒ E c ∈ M. (iii) ∅ ∈ M,

R ∈ M.

(iv) m∗ (E) = 0 ⇒ E ∈ M.

∀ A ⊆ R.

26

Measure and Integration Since countable sets are of zero outer measure, Theorem 2.2.5 implies: M contains all countable sets. In particular, Q ∈ M and Qc ∈ M.

Theorem 2.2.6 Let A1 , A2 be subsets of R such that A1 ∩ A2 = ∅. If either A1 or A2 belongs to M, then m∗ (A1 ∪ A2 ) = m∗ (A1 ) + m∗ (A2 ). Proof. Suppose A1 ∩ A2 = ∅. Assume A1 ∈ M. Then m∗ (A1 ∪ A2 ) = m∗ ((A1 ∪ A2 ) ∩ A1 ) + m∗ ((A1 ∪ A2 ) ∩ Ac1 ). Since (A1 ∪ A2 ) ∩ A1 = A1 and (A1 ∪ A2 ) ∩ Ac1 = A2 , we obtain the result. The following corollary is immediate from the above theorem. Corollary 2.2.7 If {A1 , . . . , An } is a disjoint family in M, then m∗

n [

n  X Ai = m∗ (Ai ).

i=1

i=1

The proof of the following theorem is along the same lines as we have deduced (3) from (2). However, we give its detailed proof. Theorem 2.2.8 Let {An : n ∈ N} be a disjoint family in M. Then m∗

∞ [

∞  X An = m∗ (An ).

n=1

n=1

Proof. If {An } is a finite family, then by Corollary 2.2.7, m∗

n [ i=1

n  X Ai = m∗ (Ai ) i=1

for every n ∈ N. Hence, by using the monotonicity of m∗ , we have ∞ X n=1

m∗ (An ) ≥ m∗

∞ [ n=1

n n  [  X An ≥ m∗ Ai = m∗ (Ai ) i=1

i=1

for all n ∈ N. Letting n tend to infinity, the result follows. In view of Theorem 2.2.8 and Theorem 2.2.2, we have the following corollary. Corollary 2.2.9 The set E in Theorem 2.2.1 does not belong to M. Definition 2.2.10 Those subsets of R which do not belong to M are called non-measurable sets. ♦

Lebesgue Measure

27

Corollary 2.2.9 shows that there are non-measurable sets. The following theorem shows that non-measurable sets are, in fact, plenty. Theorem 2.2.11 Let A ⊆ R be such that m∗ (A) > 0. Then there exists a subset E ⊆ A such that E 6∈ M. Proof. First let us assume that A is a bounded set. We follow the arguments used in the proof of Theorem 2.2.2. Consider the equivalence relation ∼ on R as in the proof of Theorem 2.2.2, that is, x ∼ y ⇐⇒ x − y ∈ Q. Let E be the subset of A such that its intersection with each equivalence class [x] with x ∈ A is a singleton set. Such a set E exists by using the axiom of choice on the collection E := {[x] ∩ A : x ∈ A}, where [x] is the equivalence class of x. Let a = inf A and b = sup A. Then A ⊆ [a, b]. We note that if x ∼ y, then the rational number r := x − y satisfies a − b ≤ r ≤ b − a. Let {r1 , r2 , . . .} be the set of all rational numbers in [a − b, b − a]. Let En := E + rn for n ∈ N. Then {En } is a disjoint family and, for every x ∈ E and n ∈ N, 2a − b ≤ x + rn ≤ 2b − a so that ∞ [

A⊆

En ⊆ [2a − b, 2b − a].

n=1

Hence, 0 < m(A) ≤ m∗

∞ [

 En ≤ 3(b − a).

n=1 ∗

∗

∗

Since m (En ) = m (E +rn ) = m (E) for all n ∈ N, as in the proof of Theorem 2.2.2, ∞ X

m∗ (En ) = lim

k→∞

n=1

S

k X n=1

m∗ (En ) =



0 if m∗ (E) = 0, ∞ if m∗ (E) > 0.



P∞ ∞ ∗ Hence, m∗ n=1 En 6= n=1 m (En ). In particular E 6∈ M. Next, suppose that A is not bounded. Then for every α > 0, Aα := A ∩ [−α, α] is a bounded set. Since A = ∪α>0 Aα and m∗ (A) > 0, there exists β > 0 such that m∗ (Aβ ) > 0. By the arguments in the above paragraph, there exists Eβ ⊆ Aβ ⊆ A such that Eβ 6∈ M. This completes the proof. We have seen that M contains all countable subsets of R. Since the Cantor ternary set is of zero outer measure, it also belongs to M. Now we show that M contains a lot more subsets of R.

28

Measure and Integration

Theorem 2.2.12 Let A1 , A2 ∈ M. Then A1 ∪ A2 ∈ M. More generally, if {A1 , . . . , An } ⊆ M, then ∪ni=1 Ai ∈ M. Proof. Let A ⊆ R. We have to show that

m∗ (A) ≥ m∗ A ∩ (A1 ∪ A2 ) + m∗ A ∩ Ac1 ∩ Ac2 .

(1)

Note that A ∩ (A1 ∪ A2 ) = (A ∩ A1 ) ∪ (A ∩ A2 ∩ Ac1 ), so that
m∗ A ∩ (A1 ∪ A2 ) ≤ m∗ (A ∩ A1 ) + m∗ (A ∩ A2 ∩ Ac1 ). Therefore, the right-hand side of (1) is less than or equal to
m∗ (A ∩ A1 ) + m∗ (A ∩ A2 ∩ Ac1 ) + m∗ A ∩ Ac1 ∩ Ac2 .

(2)

Now, since A2 ∈ M, we get
m∗ (A ∩ A2 ∩ Ac1 ) + m∗ A ∩ Ac1 ∩ Ac2 = m∗ (A ∩ Ac1 ). Thus, the expression in (2) is less than or equal to m∗ (A ∩ A1 ) + m∗ (A ∩ Ac1 ) which is equal to m∗ (A), since A1 ∈ M. Thus,

m∗ A ∩ (A1 ∪ A2 ) + m∗ A ∩ Ac1 ∩ Ac2 ≤ m∗ (A) which completes the proof of (1). The last part follows by repeated application of the first part. Corollary 2.2.13 If A1 , A2 ∈ M, then A1 \ A2 ∈ M. Proof. Let A1 , A2 ∈ M. Since A1 \ A2 = A1 ∩ Ac2 = (Ac1 ∪ A2 )c , the result follows from Theorem 2.2.12, a complement of any element in M is an element in M. By Theorem 2.2.12 and Corollary 2.2.13, M is closed under finite unions and complementations. Next we show that M is closed under countable unions as well. For this purpose we shall make use of the following lemma which is more general than Corollary 2.2.7. Lemma 2.2.14 Let {A1 , . . . , An } be a disjoint family in M. Then for any A ⊆ R, n n   X [ m∗ A ∩ Ai = m∗ (A ∩ Ai ). i=1

i=1

Lebesgue Measure

29

Proof. Let n = 2. Since A1 ∈ M, m∗ (A ∩ (A1 ∪ A2 )) = m∗ (A ∩ (A1 ∪ A2 ) ∩ A1 ) + m∗ (A ∩ (A1 ∪ A2 ) ∩ Ac1 ). But, A ∩ (A1 ∪ A2 ) ∩ Ac1 = A ∩ A2 .

A ∩ (A1 ∪ A2 ) ∩ A1 = A ∩ A1 , Hence, we have

m∗ (A ∩ (A1 ∪ A2 )) = m∗ (A ∩ A1 ) + m∗ (A ∩ A2 ). Thus, the result is proved for n = 2. The result for general n follows by induction. S∞ Theorem 2.2.15 If En ∈ M for n ∈ N, then n=1 En ∈ M. S Proof. Let E = n En , where En ∈ M for n ∈ N, and let A ⊆ R. We have to show that m∗ (A) ≥ m∗ (A ∩ E) + m∗ (A ∩ E c ) . (1) S We write E as a disjoint union E = n An where A1 = E1 and for n ≥ 2, An = En \ ∪n−1 i=1 Ei . Let Fn = ∪ni=1 Ai . Note that Fn ∈ M, by Theorem 2.2.12 and Corollary 2.2.13. Hence, m∗ (A) ≥ m∗ (A ∩ Fn ) + m∗ (A ∩ Fnc ) . (2) Now, Lemma 2.2.14 implies m∗ (A ∩ Fn ) =

n X

m∗ (A ∩ Ai )

i=1

and the relation we obtain

Fnc

∗

⊇ E implies m (A ∩ Fnc ) ≥ m∗ (A ∩ E c ). Thus, from (2) c

m∗ (A) ≥

n X

m∗ (A ∩ Ai ) + m∗ (A ∩ E c ).

i=1

This is true for all n ∈ N. Letting n tend to infinity, m∗ (A) ≥

∞ X

m∗ (A ∩ Ai ) + m∗ (A ∩ E c ).

i=1

Therefore, using the subadditivity of m∗ , m∗ (A) ≥

∞ X

m∗ (A ∩ Ai ) + m∗ (A ∩ E c )

i=1

≥ m∗

∞ [

 (A ∩ Ai ) + m∗ (A ∩ E c )

i=1

= m∗ (A ∩

∞ [

Ai ) + m∗ (A ∩ E c )

i=1

= m∗ (A ∩ E) + m∗ (A ∩ E c ) .

30

Measure and Integration

Thus, (1) is proved. M is closed under countable unions and complementations.

In view of Theorem 2.2.15, the property of m∗ stated in Theorem 2.2.8 is called the countable additivity of m∗ on M. T∞ Corollary 2.2.16 If En ∈ M for n ∈ N, then n=1 En ∈ M. S c T∞ ∞ c Proof. By De Morgan’s law, we have n=1 En = . Hence, the n=1 En result follows by Theorem 2.2.15 and the fact that E ∈ M implies E c ∈ M. Definition 2.2.17 The function m∗ restricted to M is called the Lebesgue measure on R, and it is denoted by m. For E ∈ M, m(E) := m∗ (E) is called the Lebesgue measure of E. ♦ Let us list some of the important properties of the family M that we have proved: • ∅ ∈ M; • E ∈ M ⇒ E c ∈ M; [ [ X • {Ei } disjoint family in M ⇒ Ei ∈ M & m∗ ( Ei ) = m∗ (Ei ). i

i

i

Remark 2.2.18 We may observe that, for a set A ∈ M, the family MA := {E ⊆ A : E ∈ M} also has the properties listed in the above box by replacing M by MA . It is to be observed that, in this case, E c appearing in the box corresponds to the complement of E ⊆ A in A, that is, the set A \ E. ♦ The following theorem shows that the class M is very large. Theorem 2.2.19 The following results hold. (i) For any a ∈ R, the intervals (a, ∞), [a, ∞), (−∞, a) and (−∞, a] belong to M. (ii) For any a, b ∈ R with a < b, the intervals (a, b), [a, b), (a, b], and [a, b] belong to M. (iii) Open subsets of R belong to M. (iv) Closed subsets of R belong to M.

Lebesgue Measure

31

(v) Every Gδ -set belongs to M. (vi) Every Fσ -set belongs to M. Proof. We first prove that (a, ∞) ∈ M for any a ∈ R, and then deduce other results by using some of the properties of m∗ . So, let a ∈ R and E = (a, ∞). Let A ⊆ R and ε > 0. By Theorem 2.1.4, there exists a countable family {In } of open intervals such that [ X A⊆ In , `(In ) ≤ m∗ (A) + ε. n

n

Note that A∩E ⊆

[ (In ∩ (a, ∞)),

A ∩ Ec ⊆

n

In0

[ (In ∩ (−∞, a]). n

:= In ∩ (−∞, a], we have X X m∗ (A ∩ E) + m∗ (A ∩ E c ) ≤ m∗ (In0 ) + m∗ (In00 )

Hence, taking

:= In ∩ (a, ∞) and

In00

n

n

X [m∗ (In0 ) + m∗ (In00 )].

=

n

Note that In0 and In00 are intervals such that In0 ∩ In00 = ∅ and In0 ∪ In00 = In so that m∗ (In0 ) + m∗ (In00 ) = `(In0 ) + `(In00 ) = `(In ). Thus, we have proved that m∗ (A ∩ E) + m∗ (A ∩ E c ) ≤

X

`(In ) ≤ m∗ (A) + ε.

n

This is true for any ε > 0, so that we get m∗ (A ∩ E) + m∗ (A ∩ E c ) ≤

X

`(In ) ≤ m∗ (A).

n

Hence, (a, ∞) = E ∈ M. Next we observe that for any a ∈ R, [a, ∞) =

∞  \ n=1

(−∞, a) = R \ [a, ∞),

a−

 1 ,∞ , n

(−∞, a] = R \ (a, ∞),

and for any a, b ∈ R with a < b, (a, b) = (a, ∞) ∩ (−∞, b),

[a, b] = [a, ∞) ∩ (−∞, b].

Thus, every interval listed in (i) and (ii) belongs to M. Also, using the facts that M is closed under countable unions, countable intersections and complementation, and the fact that every open subset of R is a countable union of open intervals, the results listed in (iii)-(vi) follow.

32

Measure and Integration

Now, we prove a companion result to Corollary 2.1.5 and Corollary 2.1.8, which also answers the question raised in Remark 2.1.6. First, let us observe the following result. Proposition 2.2.20 Let E ∈ M. Then for every ε > 0, there exists an open set G in R such that E ⊆ G and m∗ (G \ E) ≤ ε. Proof. Let ε > 0 be given. By Corollary 2.1.5, there exists an open set G in R such that E ⊆ G and m∗ (G) ≤ m∗ (E) + ε. Since E and G are in M and G \ E = G ∩ E c is also in M. Therefore, m∗ (E) + m∗ (G \ E) = m∗ (G) ≤ m∗ (E) + ε. Since m∗ (E) < ∞, we obtain m∗ (G \ E) ≤ ε. Theorem 2.2.21 Let E ⊆ R. Then the following are equivalent: (i) E ∈ M. (ii) For every ε > 0, there exists an open set G in R containing E such that E ⊆ G and m∗ (G \ E) ≤ ε. (iii) There exists a Gδ -set G in R such that E ⊆ G and m∗ (G \ E) = 0. Proof. (i) ⇒ (ii): Suppose E ∈ M and ε > 0 be given. If m∗ (E) < ∞, then by Proposition 2.2.20, there exists an open set G in R such that E ⊆ G and m∗ (G \ E) ≤ ε. For the case when m∗ (E) is not necessarily finite, we ∗ write E = ∪∞ n=1 En , where m (En ) < ∞ for every n ∈ N. For example, we can take En := E ∩ [−n, n] for n ∈ N. Since m∗ (En ) < ∞ for each n ∈ N, again by Proposition 2.2.20, there exists open set Gn ⊇ En such that m∗ (Gn \ En ) < ε/2n . Taking G = ∪∞ n=1 Gn , we have G is open, G ⊇ E and ∞ ∞ G \ E = [∪∞ n=1 Gn ] \ [∪n=1 En ] ⊆ ∪n=1 (Gn \ En ).

Therefore, m∗ (G \ E) ≤

∞ X

m∗ (Gn \ En ) ≤

n=1

∞ X ε = ε. 2n n=1

Thus, (ii) holds. (ii) ⇒ (iii): Assume (ii). Then, for every n ∈ N, there exists an open set Vn in R such that E ⊆ Vn Let G =

T

n

and m∗ (Vn \ E) ≤

1 . n

Vn . Then E ⊆ G ⊆ Vn and G \ E ⊆ Vn \ E for all n ∈ N so that m∗ (G \ E) ≤

1 n

∀ n ∈ N.

Letting n tend to infinity, we obtain m∗ (G \ E) = 0. Thus, (iii) holds.

Lebesgue Measure

33

(iii) ⇒ (i): Assume (iii). Then there exists a Gδ -set G in R such that E ⊆ G and m∗ (G \ E) = 0. Therefore, G \ E ∈ M. We know that G ∈ M so that by Corollary 2.2.13, E = G \ (G \ E) ∈ M. This completes the proof. Corollary 2.2.22 Let E ⊆ R. Then E ∈ M if and only if there exist a Gδ -set G and an Fσ -set F such that F ⊆ E ⊆ G and m∗ (G \ F ) = 0. Proof. Suppose E ∈ M. Then by Theorem 2.2.21, there exists a Gδ -set G such that E ⊆ G and m∗ (G \ E) = 0. Also, by taking E c in place of E, there exists a Gδ -set H such that E c ⊆ H and m∗ (H \ E c ) = 0. Then F := H c is an Fσ -set satisfying F = Hc ⊆ E

and E \ F = E \ H c = E ∩ H = H \ E c

so that m∗ (E \ F ) = m∗ (H \ E c ) = 0. Since G \ F = (G \ E) ∪ (E \ F ) we obtain m∗ (G \ F ) = 0. Conversely, suppose that there exists a Gδ -set G and an Fσ -set F such that F ⊆ E ⊆ G and m∗ (G \ F ) = 0. In particular, G \ E ⊆ G \ F so that m∗ (G \ E) = 0 and hence G \ E ∈ M. Therefore, by Corollary 2.2.13, E = G \ (G \ E) ∈ M. Using Theorem 2.2.21, we obtain the following theorem. The details of its proof are left as an exercise. Theorem 2.2.23 Let E ⊆ R. Then the following are equivalent: (i) E ∈ M. (ii) For every ε > 0, there exists a closed set F in R such that F ⊆ E and m∗ (E \ F ) ≤ ε. (iv) There exists an Fσ -set F in R such that F ⊆ E and m∗ (E \ F ) = 0.

2.3

Problems

1. Let X be a set and A be a family of subsets of X. Show that A is an algebra on X if and only if X ∈ A, and A, B ∈ A implies A \ B ∈ A and A ∪ B ∈ A.

34

Measure and Integration 2. Prove that, in Definition 2.1.1, m∗ (E) remains the same if we take IE to be the collection of all countable family {In } of intervals of finite length, not necessary open intervals. [Hint: Given any open interval I and ε > 0, there exist intervals I1 and I2 containing one or both end points such that I1 ⊆ I ⊆ I2 and |`(I) − `(Ii )| < ε for i = 1, 2.] 3. Prove that, for any E ⊆ R, m∗ (E) = inf{m∗ (G) : G ⊇ E, G open}. [Hint: Use Corollary 2.1.5.] 4. Show that, if E ⊆ A and m∗ (E) = 0, then m∗ (A ∪ E) = m∗ (A). 5. Show that, every interval is a Gδ -set and an Fσ -set. 6. Prove that if E is a countable subset of R, then R \ E is a Gδ set, and in particular, R \ Q is a Gδ set. 7. From Theorem 2.1.11, deduce that outer measure of every countable set is 0. 8. Deduce Corollary 2.1.8 from Theorem 2.1.11. 9. If E is a subset of an interval I such that m∗ (E) = 0, then prove that I \ E is dense in I. [Hint: Note that, if J ⊆ I is an interval which does not contain any point from I \ E, then J ⊆ E and hence m∗ (J) = 0.]

10. Prove Theorem 2.2.5. 11. Suppose {A1 , . . . , An } is a disjoint family  of subsets  ofPR such that n ∗ Sn ∗ A1 , . . . , An−1 belong to M. Then show that m i=1 Ai = i=1 m (Ai ). [Hint: Theorem 2.2.6.] f = M ∪ {E0 }, where E0 6∈ M. Prove that if {An } is a countable disjoint 12. Let M   f then m∗ S An = P m∗ (An ). family in M, n n [Hint: Use Problem 11.] 13. Justify the statement: There exists a countably infinite disjoint family N of subsets of R such that N ∩ M = ∅. [Hint: Theorem 2.2.11.] 14. Deduce Corollary 2.2.7 from Theorem 2.2.8. 15. Let E ⊆ R be such that m∗ (E) < ∞. Prove that the following are equivalent: (a) E ∈ M. (b) There exists a Gδ set G ⊇ E and E0 ⊆ R with m∗ (E0 ) = 0 such that E = G \ E0 . (c) There exists an Fσ set F ⊆ E and F0 ⊆ R with m∗ (F0 ) = 0 such that E = F ∪ F0 . 16. If A, B ∈ M with A ⊆ B and if m(A) < ∞, then show that m(B \ A) = m(B) − m(A). 17. If A, B ∈ M, prove that m(A ∪ B) + m(A ∩ B) = m(A) + m(B).

Lebesgue Measure

35

18. Let A, B ∈ M such that A ⊂ B and m(B) = m(A) < ∞. If C ⊆ R is such that A ⊆ C ⊆ B, then prove that C ∈ M and m(C) = m(A). 19. If E ∈ M, then prove that m(E) = sup{m(K) : K ⊆ E, K compact}. [Hint: You may use Problem 3. Recall that a subset of R is compact if and only if it is closed and bounded.] 20. Find an open dense subset A of R with m(A) < ∞. [Hint: Use density of the set of rational numbers.] 21. Show that for every ε > 0, there exists an open dense subset A of R such that m(A) < ε. [Hint: Use arguments similar to those used in solving Problem 20.] 22. Find a closed subset B of R with empty interior and with m(B) = ∞. [Hint: Use Problem 21.] 23. Find a subset E of [0, 1] which is a countable union of closed nowhere dense sets such that m(E) = 1. [Hint: Use Problem 20. Recall that a set A is said to be nowhere dense if its closure has an empty interior.] 24. Verify the statement in Remark 2.2.18. 25. Supply details of the proof of Theorem 2.2.23.

Chapter 3 Measure and Measurable Functions

In the last chapter we defined the concept of a measure of certain subsets of the real line. These subsets include most of the usual sets that we deal with in analysis, such as intervals, open sets, closed sets, and so on. We have also seen, that not every subset of the real line can be measured in a satisfactory manner. However, the sets that can be measured follow certain nice set theoretic rules. In this chapter we build the measure theory on an arbitrary set, by making use of the essential properties that measurable subsets of the real line satisfy.

3.1

Measure on an Arbitrary σ-Algebra

Consider the class M of all Lebesgue measurable subsets of R and m the Lebesgue measure on R. Recall the following properties: (1) R ∈ M, (2) E ∈ M implies E c ∈ M, (3) En ∈ M for n ∈ N implies

S

n

En ∈ M, and

(4) {En } is a countable disjoint family in M implies [  X m En = m(En ). n

n

Motivated by the above properties of the family M of all measurable sets of R and the Lebesgue measure m on R, we introduce the following two definitions.

37

38

Measure and Integration

Definition 3.1.1 Let X be a set and A be a family of subsets of X. Then A is called a σ-algebra on X if (a) X ∈ A, (b) A ∈ A implies Ac := X \ A ∈ A, S∞ (c) An ∈ A for n ∈ N implies n=1 An ∈ A. The pair (X, A) is called a measurable space, and members of A are called measurable sets. ♦ Remark 3.1.2 In view of (a) and (b) in Definition 3.1.1, if A is a σ-algebra, then ∅ ∈ A. Therefore, the denumerable union in (c) implies the finite unions as well. If we replace the countable union in the condition (c) in the above definition by finite unions, then the resulting family is called an algebra. Clearly, every σ-algebra is an algebra, and the converse is not true. We shall use the concept of an algebra in Chapter 6. ♦ Definition 3.1.3 Let (X, A) be a measurable space. Then a function µ : A → [0, ∞] is called a measure on (X, A) if (a) µ(∅) = 0, and S  P (b) µ n An = n µ(An ) for any countable disjoint family {An } in A. The triple (X, A, µ) is called a measure space. For A ∈ A, µ(A) is called the measure of A. ♦ Let us first make an easy observation. Theorem 3.1.4 Let X be a set and A be a family of subsets of X. Then A is a σ-algebra on X if and only if X ∈ A and the following two conditions are satisfied. (i) A, B ∈ A implies A \ B ∈ A. T (ii) An ∈ A for n ∈ N implies n An ∈ A. Proof. For subsets A, B, A1 , A2 , . . . of X, we have A \ B = A ∩ B c = (Ac ∪ B)c ,

\

An =

[

n

Ac = X \ A,

[ n

An =

Acn

c

,

(1)

n

\

Acn

c

.

(2)

n

If A is a σ-algebra, then (i) and (ii) follow from (1). Conversely, if X ∈ A and A satisfies (i) and (ii), then by (2), A is a σ-algebra.

Measure and Measurable Functions

39

Convention: In due course, we adopt the following convention: (a) If the σ-algebra A is understood from the context, then instead of saying that “(X, A) is a measurable space”, we may say that “X is a measurable space”. (b) If the σ-algebra A and the measure µ are understood from the context, then instead of saying that “(X, A, µ) is a measure space”, we may say that “X is a measure space”. (c) If the σ-algebra A is understood from the context, then instead of saying that “µ is a measure on the measurable space (X, A)”, we may say that “µ is a measure on X”. Remark 3.1.5 A measure as in Definition 3.1.3 is also called a positive measure, since there are notions such as signed measures and complex measures. In this book we do not consider such measures. Hence, we use the terminology measure for the positive measure. ♦ Theorem 3.1.6 Let (X, A, µ) be a measure space and A, B ∈ A. Then (i) A ⊆ B implies µ(A) ≤ µ(B), (ii) A ⊆ B and µ(A) < ∞ imply µ(B \ A) = µ(B) − µ(A). Proof. Suppose A, B ∈ A such that A ⊆ B. Then B is the disjoint union of A and B \ A. By Theorem 3.1.4, B \ A ∈ A, and by the definition of the measure, µ(B) = µ(A) + µ(B \ A). From this, (i) and (ii) follow. The property (i) in Theorem 3.1.6 is called the monotonicity property of µ. Definition 3.1.7 Let (X, A, µ) be a measure space. (a) If µ(X) = 1, then µ is called a probability measure. (b) If µ(X) < ∞, then µ is called a finite measure, and (X, A, µ) is called a finite measure space. S∞ (c) If there exists Xn ∈ A, n ∈ N such that X = n=1 Xn and µ(Xn ) < ∞ for each n ∈ N, then µ is called a σ-finite measure, and (X, A, µ) is called a σ-finite measure space.  The assertions in the following examples may be verified by the reader.

40

Measure and Integration

Example 3.1.8 The triple (R, M, m) is a measure space, and it is a σ-finite measure space. ♦ Example 3.1.9 For A ∈ M, MA defined as in Remark 2.2.18 is a σ-algebra, and the map mA defined by mA (E) := m(E),

E ∈ MA ,

is a σ-finite measure on (A, MA ).

♦

Example 3.1.10 For any set X, the family {∅, X} and the power set 2X are σ-algebras, and they are the smallest and largest (in the sense of set inclusion), respectively, of all σ-algebras on X. ♦ Example 3.1.11 Given any measurable space (X, A), µ defined by µ(A) = 0 for all A ∈ A is a measure on (X, A). This measure is called the zero measure on X. ♦ Example 3.1.12 Let (X, A, µ) be a measure space. (i) For every α > 0, the map E 7→ αµ(E) defines a measure on (X, A). (ii) If 0 < µ(X) < ∞, then the map E 7→ on (X, A).

µ(E) µ(X)

is a probability measure ♦

Example 3.1.13 Let X be a non-empty set and A = {∅, X}. Let µ be defined by µ(∅) = 0 and µ(X) = 1. Then (X, A, µ) is a measure space. ♦ Example 3.1.14 Let X be any set, A = 2X , the power set of X, and µ be defined by  # E if E is a finite set µ(E) = ∞ if E is an infinite set. Here, E # denotes the number of elements in E. Then µ is a measure on (X, A). This measure is called the counting measure on X. Note that the counting measure on X is (i) finite if and only if X is a finite set and (ii) σ-finite if and only if X is a countable set.

♦

Example 3.1.15 Let X be any set, A be the power set of X, x0 ∈ X, and µx0 be defined by  1 if x0 ∈ E µx0 (E) = 0 if x0 ∈ 6 E. Then (X, A, µx0 ) is a measure space, and the measure µx0 is called the Dirac measure on X centered at x0 . Dirac measure µx0 is also denoted by δx0 . ♦ Let us have an illustration of the Dirac measure: Think of a thin wire of negligible weight. Suppose a bead of weight 1 unit is kept

Measure and Measurable Functions

41

at some point x0 on the wire. Then the weight of a part, say E, of the wire is 1 if x0 belongs to E, and the weight is zero if x0 does not belong to E. Example 3.1.16 Let X be a set, A be the power set of X, xi ∈ X and wi ≥ 0 for i ∈ {1, . . . , k}. For E ⊆ X, let ∆E = {i : xi ∈ E}, and let µ be defined by X wi . µ(E) = i∈∆E

Then (X, A, µ) is a measure space. Note that µ is a probability measure if and Pk only if i=1 wi = 1. ♦ Analogous to the illustration of the Dirac measure we have the following corresponding to Example 3.1.16: Think of a thin wire of negligible weight. Suppose beads of weights w1 , . . . , wk are kept at points x1 , . . . , xk , respectively, P on the wire. Then the weight of a part, say E, of the wire is i∈∆E wi , where ∆E = {i : xi ∈ E}. Another illustration of Example 3.1.16: Let us assume that the wire costs nothing, but the beads at x1 , . . . , xk cost rupees (Indian currency) w1 , . . . , wk , respectively. Then the Pcost of a part, say E, of the wire together with the beads on it is i∈∆E wi rupees, where ∆E = {i : xi ∈ E}. An important example of a measure which is closely related to the Lebesgue measure on R is the Lebesgue measure on Rk for k ∈ N with k ≥ 2.

3.1.1

Lebesgue measure on Rk

Before giving the definition, let us introduce a few notations: Let k ∈ N with k ≥ 2. Then, for intervals I1 , . . . , Ik in R, sets of the form D := I1 × · · · × Ik , are the analogues of intervals in Rk . Such sets are called k-cells. Given a k-cell D := I1 × · · · × Ik , the number `(D) = `(I1 ) × · · · × `(Ik ) represents the length or area or volume or hyper-volume according as k is 1 or 2 or 3 or greater than 3, respectively. If I1 , . . . , Ik are open intervals, then the k-cell D := I1 × · · · × Ik is called an open k-cell.

42

Measure and Integration

Definition 3.1.17 For E ⊆ Rn , the Lebesgue outer measure of E is defined as X m∗k (E) := inf `(Dn ), n

where the infimum is taken over the collection of all countable family {Dn } of open k-cells such that E ⊆ ∪n Dn . ♦ It can be shown, as in the case of Lebesgue outer measure m∗ , that 1. m∗k (∅) = 0; 2. A ⊆ Rk , B ⊆ Rk , A ⊆ B ⇒ m∗k (A) ≤ m∗k (B); S∞ P∞ 3. An ⊆ Rk , n ∈ N ⇒ m∗k ( n=1 An ) ≤ n=1 m∗k (An ). As in the case of R, a set E ⊆ Rk is said to be Lebesgue measurable if m∗k (A) = m∗k (A ∩ E) + m∗k (A ∩ E c ) for all A ⊆ Rk . Further, the family Mk of all Lebesgue measurable subsets of Rk is a σ-algebra, and the map E 7→ mk (E) := m∗k (E) is a measure on Mk , called the Lebesgue measure on Rk .

3.1.2

Generated σ-algebra and Borel σ-algebra

We know that an arbitrary family of subsets of a non-empty set X need not be a σ-algebra. However, an arbitrary family of subsets of X is associated with a unique σ-algebra in a certain way. First let us prove the following. Theorem 3.1.18 Intersection of any family of σ-algebras on a set X is again a σ-algebra on X. Proof. Let {Aα : α ∈ Λ} be a family T of σ-algebras on a set X, where Λ is some index set. We show that A := α∈Λ Aα is also a σ-algebra on X: Since X ∈ Aα for every α ∈ Λ, we have X ∈ A. Now, let A ∈ A. Then A ∈ Aα for every α ∈ Λ. Since each Aα is a σ-algebra, Ac ∈ Aα for every α ∈ Λ. Hence, Ac ∈ A. Next, let {An } be a countable family in A. Then each An ∈ Aα for every α ∈ Λ. Again, since each Aα is a σ-algebra, ∪n An ∈ Aα for every α ∈ Λ. Therefore, ∪n An ∈ A. Thus, A is a σ-algebra. Theorem 3.1.19 Let X be a set and S be a family of subsets of X. Then the intersection of all σ-algebras containing S is a σ-algebra on X, and it is the smallest σ-algebra on X containing S. Proof. Let F be theTfamily of all σ-algebras on X containing S. Then by Theorem 3.1.18, AS = A∈F A is a σ-algebra on X. Since S ⊆ A for every A ∈ F, S ⊆ AS . Clearly AS ⊆ A for every A ∈ F. Hence, AS is the smallest (in terms of set inclusion) σ-algebra containing S.

Measure and Measurable Functions

43

Definition 3.1.20 Let X be a set and S be a family of subsets of X. The smallest σ-algebra containing S is called the σ-algebra generated by S, and it is denoted by AS . ♦ Observe that, if S1 and S2 are families of subsets of a set X such that S1 ⊆ S2 , then AS1 ⊆ AS2 . On the sets Rk , other than the σ-algebra Mk , there is another natural σ-algebra, the σ-algebra generated by all open subsets of Rk . The same is the case with any topological space. Definition 3.1.21 Let Y be a topological space with topology T . Then the σalgebra generated by T is called the Borel σ-algebra on Y , and the members of the Borel σ-algebra are called Borel sets. ♦ The Borel σ-algebra on a topological space Y will be denoted by BY . Notation: If Y = Rk , then we shall denote BRk by Bk . We shall also denote B1 by B. Note that m|B1 is a measure on (R, B1 ). Also, by the observations after Definition 3.1.17, mk |Bk is a measure on (R, Bk ). For the sake of completion of presentation, we recall that, a topological space is a pair (Y, T ) consisting of a set Y together with a topology T , that is a family T of subsets of Y , satisfying the following properties: (1) {∅, X} ⊆ T , (2) A, B ∈ T ⇒ A ∩ B ∈ T , [ (3) S ⊆ T ⇒ A∈T. A∈S

Recall also that, given a topological space (Y, T ), the members of the topology T are called open sets in Y , and a set A ⊆ Y is called a closed set if Ac is an open set. If the topology T on Y is understood from the context, then instead of saying “(Y, T ) is a topological space”, we say that “Y is a topological space”. Let us now recall some concepts from the theory of metric spaces: Let (Ω, d) be a metric space. Then, by an open ball centered at x ∈ Ω, we mean a set of the form B(x, r) := {u ∈ Ω : d(x, u) < r} for some r > 0. A point x ∈ Ω is said to be an interior point of a set A ⊆ Ω if A contains an open ball centered at x, and the set A is said to be an open set if every point in A is an interior point. It can be easily shown that the family of all open subsets of Ω is a topology on Ω, called the topology induced by the metric d. A subset D of Ω is said to be dense in Ω if for every x ∈ Ω, every open ball centered at x contains some point from D. If Ω contains a countable dense subset, then it is said to be a separable metric space. It can be shown that

44

Measure and Integration

if Ω is a separable metric space, then every nonempty open subset of Ω is a countable union of open balls. Example 3.1.22 Let (Ω, d) be a separable metric space. Then the family S := {B(x, r) : x ∈ Ω, r > 0} generates the Borel σ-algebra BΩ . Indeed, S ⊆ BΩ so that AS ⊆ BΩ . Since Ω is a separable metric space, every open subset of Ω is a countable union of open balls, AS contains all open sets, so that BΩ ⊆ AS . ♦ As a particular case of the above example, we see that S := {(a, b) : a, b ∈ R with a < b} generates the Borel σ-algebra B. In fact, we have the following theorem. Theorem 3.1.23 Each of the following family of sets generate the Borel σalgebra B on R: S S1 S2 S3 S4 S5 S6 S7

:= := := := := := := :=

{(a, b) : a, b ∈ R with a < b}, {[a, b) : a, b ∈ R with a < b}, {(a, b] : a, b ∈ R with a < b}, {[a, b] : a, b ∈ R with a < b}, {(a, ∞) : a ∈ R}, {(−∞, b) : b ∈ R}, {[a, ∞) : a ∈ R}, {(−∞, b] : b ∈ R}.

Proof. We have already observed that AS = B. Now, let a, b ∈ R with a < b. Then, [a, b) =

∞  \ n=1

1  a− ,b , n

(a, b] =

∞  \

a, b +

n=1

1 , n

∞  \

[a, b] =

n=1

1 1 a− ,b+ . n n

Hence, S1 , S2 , S3 are subfamilies of AS = B. Consequently, AS1 , AS2 , AS3 are subfamilies of B. 2 < b − a. Then Also, for a, b ∈ R with a < b, let m ∈ N be such that m 1 1 a + n < b − n for every n ≥ m and we have (a, b) =

∞ h [ 1  a+ , b , n n=m

(a, b) =

∞  [ n=m

a, b−

1i , n

(a, b) =

∞ h [ 1 1i a+ , b− . n n n=m

Therefore, S is a subfamily of each of the σ-algebras AS1 , AS2 , AS3 . Consequently, B = AS is a subfamily of each of AS1 , AS2 , AS3 . Thus, we have proved that BR is the same as each of AS , AS1 , AS2 , AS3 .

Measure and Measurable Functions

45

Further, for a, b ∈ R, since the intervals (a, ∞) and (−∞, b) are open sets, S4 and S5 are subfamilies of B; consequently, AS4 and AS5 are subfamilies of B. Also, since [a, ∞) =

∞ [

[a, n),

(−∞, b] =

n=1

∞ [

(−n, b],

n=1

S6 and S7 are subfamilies of AS1 and AS2 , respectively. But, we have already proved that AS1 = B = AS2 . Hence, S6 and S7 are also subfamilies of BR . Next, for a, b ∈ R with a < b, we observe that (a, b] = (a, ∞) ∩ (−∞, b],

[a, b) = [a, ∞) ∩ (−∞, b).

Since (c, ∞) ∈ S4 , (−∞, c] ∈ S7 , [c, ∞) ∈ S6 , (−∞, c) ∈ S5 , for any c ∈ R, we obtain (a, b] = (a, ∞) ∩ (−∞, b] = (a, ∞) ∩ (R \ (b, ∞)) ∈ AS4 , (a, b] = (a, ∞) ∩ (−∞, b] = (R \ (−∞, a]) ∩ (−∞, b] ∈ AS7 , [a, b) = [a, ∞) ∩ (−∞, b) = [a, ∞) ∩ (R \ [b, ∞)) ∈ AS6 , [a, b) = [a, ∞) ∩ (−∞, b) = (R \ (−∞, a)) ∩ (−∞, b) ∈ AS5 . Consequently, B = AS2 ⊆ AS4 and BR = AS2 ⊆ AS7 , BR = AS1 ⊆ AS6 and BR = AS1 ⊆ AS5 . This completes the proof. Since M contains all open sets and B is the smallest σ-algebra containing all open sets, we have B ⊆ M. We shall see that B is a proper subfamily of M (Theorem 3.4.9).

3.1.3

Restrictions of σ-algebras and measures

Recall from Example 3.1.9 that if A ∈ M, then MA is a σ-algebra on A and the map mA : E 7→ m(E) is a measure on (A, MA ). The σ-algebra MA and the measure mA can be thought of as restrictions of M and m, respectively. Analogously, we can define restrictions of a general σ-algebra to a measurable set and a measure to a sub-σ-algebra. In this connection, the following theorem can be proved easily. Theorem 3.1.24 Let (X, A, µ) be a measure space, X0 ∈ A and A0 be a σ-algebra on X0 such that A0 ⊆ A. Then µ0 defined by µ0 (E) = µ(E),

E ∈ A0 ,

is a measure on (X0 , A0 ). Definition 3.1.25 Let (X, A, µ) be a measure space, X0 ∈ A and A0 be a σ-algebra on X0 such that A0 ⊆ A. Then the measure defined in Theorem 3.1.24 is called the restriction of µ to A0 , and it is denoted by µ|A0 . ♦

46

Measure and Integration

Remark 3.1.26 In Theorem 3.1.24 and Definition 3.1.25, X0 can be the same as X, but A0 can be different from A. ♦ Definition 3.1.27 The restriction of the Lebesgue measure on Rk to the Borel σ-algebra Bk , namely, mk |Bk , is called the Borel measure on Rk . ♦ In Theorem 3.1.24, A0 can be any σ-algebra on X0 such that A0 ⊆ A. For example, it can be {X0 , ∅}. One may ask the following question: Given a measurable space (X, A) and X0 ∈ A, what would be the largest σ-algebra on X0 contained in A? Here is an answer to it. Theorem 3.1.28 Suppose (X, A) is a measurable space and X0 ∈ A. Then A0 = {A ∩ X0 : A ∈ A} is a σ-algebra on X0 , and it is the largest σ-algebra on X0 contained in A. Proof. Clearly X0 ∈ A0 . Let E ∈ A0 . Then there exists A ∈ A such that E = A ∩ X0 , and X0 \ E = X0 \ (A ∩ X0 ) = X0 ∩ (A ∩ X0 )c . Since (A ∩ X0 )c ∈ A, we have X0 \ E ∈ A0 . Next, let {En } be a countably infinite family in A0 . Let An ∈ A such that En = An ∩ X0 for n ∈ N. Then ∞ [ n=1

S∞

En =

∞ [

(An ∩ X0 ) =

n=1

∞ [

 An ∩ X0 .

n=1

S∞

Since n=1 An ∈ A, we have n=1 En ∈ A0 . Now, we show that A0 is the largest σ-algebra on X0 such that A0 ⊆ A. Suppose A˜0 is a σ-algebra on X0 such that A˜0 ⊆ A. Then for every E ∈ A˜0 , we have E ∈ A so that E = E ∩ X0 ∈ A0 . Thus, A˜0 ⊆ A0 . Definition 3.1.29 Let (X, A) be a measurable space and X0 ∈ A. (a) The σ-algebra A0 given in Theorem 3.1.28 is called the restriction of the σ-algebra A to X0 . (b) If µ is a measure on (X, A) and µ0 is the restriction of µ to A0 , then the measure space (X0 , A0 , µ0 ) is called the restriction of the measure space (X, A, µ) to X0 . ♦ One may observe that, given a measurable space (X, A) and X0 ∈ A, the σ-algebra restricted to X0 is the same as {E ⊆ X0 : E ∈ A} (Problem 18). Notation: In due course, we shall denote the restriction of a σ-algebra A to a measurable set X0 ∈ A by AX0 , whereas the corresponding restriction of

Measure and Measurable Functions

47

the measure µ will be denoted by the same notation µ; but we say that µ is a measure on AX0 or, by abusing the terminology, µ is a measure on X0 . In the context of the Lebesgue measurable space (R, M, m), a question of interest would be the following: Does there exist a σ-algebra A on R properly containing M such that the restriction of the Lebesgue outer measure m∗ to A is a measure?

The answer is in the negative. We deduce this from the following theorem, which is, in fact, a characterization of Lebesgue measurability. Theorem 3.1.30 A subset E of R is Lebesgue measurable if and only if m∗ (I) ≥ m∗ (I ∩ E) + m∗ (I ∩ E c ) for every open interval I. Proof. Let E ⊆ R. Clearly, if E is Lebesgue measurable, then m∗ (I) ≥ m∗ (I ∩ E) + m∗ (I ∩ E c )

(1)

holds for every open interval I. Conversely, suppose (1) holds for every open interval I. We have to show that m∗ (A) ≥ m∗ (A ∩ E) + m∗ (A ∩ E c )

(2)

for every A ⊆ R. Let ε > 0 be given. By the definition of m∗ , there exists a countable family {In } of open intervals such that [ X A⊆ In and `(In ) ≤ m∗ (A) + ε. (3) n

n

Hence, using (1) and (3), m∗ (A ∩ E) + m∗ (A ∩ E c ) ≤ m∗ (

[

In ∩ E) + m∗ (

n

[

In ∩ E c )

n

X ≤ [m∗ (In ∩ E) + m∗ (In ∩ E c )] n

≤

X

m∗ (In ) ≤ m∗ (A) + ε.

n

Thus, m∗ (A ∩ E) + m∗ (A ∩ E c ) ≤ m∗ (A) + ε for every ε > 0. Hence, (2) holds for every A ⊆ R.

48

Measure and Integration

Theorem 3.1.31 Suppose there is a σ-algebra A on R such that M ⊆ A and m∗ |A is a measure on A. Then A = M. Proof. We have to prove that A ⊆ M. For this, let E ∈ A. Since A contains intervals and m∗ is a measure on A, we have m∗ (I) = m∗ (I ∩ E) + m∗ (I ∩ E c ) for every open interval I. Therefore, by Theorem 3.1.30, we have E ∈ M. Thus, A ⊆ M.

3.1.4

Complete measure space and the completion

Definition 3.1.32 Let (X, A, µ) be a measure space. We say that µ is a complete measure on (X, A), if for every A ∈ A with µ(A) = 0, E ⊆ A implies E ∈ A. If µ is a complete measure on (X, A), then we say that (X, A, µ) is a complete measure space, and A is complete with respect to the measure µ. ♦ A measure space (X, A, µ) is not complete if and only if there exists A ∈ A and E ⊆ A such that µ(A) = 0 and E ∈ 6 A.

Example 3.1.33 Recall that (see Theorem 2.2.5 (iv)), if A ⊆ R such that m∗ (A) = 0, then A ∈ M. Hence, it follows that (R, M, m) is complete. We shall see that (R, B, m) is not complete (Theorem 3.4.9). ♦ Example 3.1.34 Let X = {a, b, c} and A = {∅, X, {a}, {b, c}}. Let µ on A be defined by µ(∅) = µ({b, c}) = 0

and µ(X) = µ({a}) = 1.

Then µ is a measure on (X, A); it is not complete, as µ({b, c}) = 0,

{b} ⊆ {b, c} but {b} 6∈ A.

♦

Example 3.1.35 For any nonempty set X, we know that A = {∅, X} is a σ-algebra and µ defined by µ(∅) = 0 and µ(X) = 0 is a measure on A. Then we see that µ is complete if and only if X is a singleton set. ♦ Theorem 3.1.36 Let (X, A, µ) be a measure space. Let N := {E ⊆ X : ∃ B ∈ A with E ⊆ B and µ(B) = 0} and A˜ := {A ∪ E : A ∈ A, E ∈ N }.

Measure and Measurable Functions

49

Then the following are true. (i) A˜ is a σ-algebra on X containing A; (ii) µ ˜ defined on A˜ by µ ˜(A ∪ E) = µ(A)

for A ∈ A, E ∈ N

˜ with µ is a complete measure on (X, A) ˜(A) = µ(A) for every A ∈ A; ˜µ (iii) (X, A, ˜) is the smallest complete measure space containing (X, A, µ), ˆµ in the sense that if (X, A, ˆ) is a complete measure space with A ⊆ Aˆ ˆ and µ ˆ(A) = µ(A) for every A ∈ A, then A˜ ⊆ A. ˜ In particular, X ∈ A. ˜ Proof. (i) First we observe that A ⊆ A˜ and N ⊆ A. Suppose A ∈ A and E ∈ N . Note that (A ∪ E)c = Ac ∩ E c . Let B ∈ A with E ⊆ B and µ(B) = 0. Then E c ⊇ B c , so that E c = B c ∪ (E c \ B c ) = B c ∪ (E c ∩ B). Thus, (A ∪ E)c = Ac ∩ E c = Ac ∩ [B c ∪ (E c ∩ B)] = [Ac ∩ B c ] ∪ [Ac ∩ (E c ∩ B)]. Note that Ac ∩ B c ∈ A and Ac ∩ (E c ∩ B) ⊆ B with µ(B) = 0. Hence, c ˜ (A S∞∪ E) ∈ A. Next,˜ let An ∈ A and En ∈ N for n ∈ N. We have to show that n=1 (An ∪ En ) ∈ A. Note that ∞ [

(An ∪ En ) =

∞ [

∞   [ En . An ∪

n=1

n=1

n=1

Since En ⊆ Bn for some Bn ∈ A with µ(Bn ) = 0, we have ∞ [ n=1

En ⊆

∞ [

Bn with

n=1

∞ [

∞ ∞ [  X Bn ≤ Bn ∈ A and µ µ(Bn ) = 0.

n=1

S∞

n=1

n=1

S∞

˜ Thus, we have proved Therefore, n=1 En ∈ N so that n=1 (An ∪ En ) ∈ A. ˜ that A is a σ-algebra on X containing A. (ii) Since ∅ ∈ A, µ ˜(∅) = µ(∅) = 0. Next, let An ∈ A and En ∈ N for n ∈ N be such that {An ∪ En } is a disjoint family. Then, {An } and {En } are disjoint families and hence ∞ [  µ ˜ (An ∪ En )

∞ ∞  [  [  = µ ˜ An ∪ En

n=1

n=1

n=1

∞ ∞ [  X = µ An = µ(An ) n=1

=

∞ X n=1

˜ Thus, µ ˜ is a measure on A.

n=1

µ ˜(An ∪ En ).

50

Measure and Integration

For showing the completeness of µ ˜, let F ⊆ A ∪ E for some A ∈ A and ˜ Note that E ∈ N with µ ˜(A ∪ E) = 0. We have to show that F ∈ A. F = F ∩ (A ∪ E) = (F ∩ A) ∪ (F ∩ E), where F ∩ A ⊆ A with µ(A) = µ ˜(A ∪ E) = 0 so that F ∩ A ∈ N . Also, F ∩ E ⊆ E with E ∈ N so that F ∩ E ∈ N . Since ˜ N ⊆ A˜ and A˜ is a σ-algebra, F = (F ∩ A) ∪ (F ∩ E) ∈ A. ˆµ (iii) Suppose that (X, A, ˆ) is a complete measure space with A ⊆ Aˆ and µ ˆ(A) = µ(A) for every A ∈ A. Since A ⊆ Aˆ and µ ˆ is a complete measure, ˆ Thus, using the fact that Aˆ is a σ-algebra, we also obtain we have N ⊆ A. ˆ A˜ ⊆ A. Definition 3.1.37 Given a measure space (X, A, µ), the complete measure ˜µ space (X, A, ˜) obtained as in Theorem 3.1.36 is called the completion of (X, A, µ). ♦ Theorem 3.1.38 The σ-algebra M of Lebesgue measurable sets is the completion of the Borel σ-algebra B. Proof. Recall from Corollary 2.2.22 that for every E ∈ M, there exist a Gδ -set G ⊇ E and an Fσ -set F ⊆ E such that m(G \ F ) = 0. We know that Gδ -sets and Fσ -sets are Borel sets. Note that E = F ∪ (E \ F ), where E \ F ⊆ G \ F with m(G \ F ) = 0. Thus M is the completion of the Borel σ-algebra B. We shall see in Theorem 3.4.9, using the concept of a measurable function, that (R, B, m) is not complete, and B ( M.

3.1.5

General outer measure and induced measure

Recall that the Lebesgue measure m on R is obtained by restricting the Lebesgue outer measure m∗ to the σ-algebra M. Likewise, we now describe a procedure of obtaining a measure on a set X by first defining a general outer measure µ∗ on all subsets of X and then restricting it to a general class of measurable sets constructed out of µ∗ . Definition 3.1.39 Let X be a set. A function µ∗ : 2X → [0, ∞] is called an outer measure on X if the following three conditions are satisfied: (a) µ∗ (∅) = 0, (b) A ⊆ B ⇒ µ∗ (A) ≤ µ∗ (B),

Measure and Measurable Functions (c) µ∗

S

n



An ≤

P

n

51

µ∗ (An ) for every countable family {An } in 2X . ♦

Definition 3.1.40 Let µ∗ be an outer measure on a set X. Then a set E ⊆ X is said to be µ∗ -measurable if for every A ⊆ X, µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E c ).

♦

The proof of the following theorem is exactly as in the case of Lebesgue measurable sets (see Problem 13). Theorem 3.1.41 (Carath´ eodory’s theorem) Let µ∗ be an outer measure on a set X and let A be the family of all µ∗ -measurable subsets of X. Then A is a σ-algebra, and µ := µ∗ |A , the restriction of µ∗ to A, is a complete measure on A. Remark 3.1.42 We have seen that m∗ defined on 2R satisfies the properties (a), (b), (c) in Definition 3.1.39, and hence m∗ is an outer measure on R, and Lebesgue measurable subsets of R are nothing but m∗ -measurable sets according to Definition 3.1.40, and Lebesgue measure is the complete measure as in Theorem 3.1.41. We could have started our theory in the abstract setting as above and obtained M and Lebesgue measure as a particular case and proved its special properties. Instead, we preferred to consider Lebesgue outer measure and Lebesgue measurable sets first, as it was felt that, our approach would serve as a better motivation for the abstract theory. ♦ Given an algebra A0 on a set X, an outer measure can be constructed on X by using a set function which vanishes on empty set. More precisely, we have the following theorem. Its proof is easy; the reader may supply the details (Problem 16). Theorem 3.1.43 Let A0 be an algebra on a set X and µ0 : A0 → [0, ∞] be such that µ0 (∅) = 0. Then µ∗ : 2X → [0, ∞] defined by ∗

µ (A) = inf

∞ nX

µ0 (En ) : A ⊆

n=1

[

En with En ∈ A0 , n ∈ N

o

n

is an outer measure on X. Once we get an outer measure µ∗ out of an algebra A0 and a set function µ0 as in the above theorem, a natural question is whether sets in the algebra A0 are µ∗ measurable. The answer is in the affirmative if the function µ0 has countable additivity property on A0 , as the following theorem shows. The reader may see its proof in Folland [6]. Theorem 3.1.44 Let A0 be an algebra on a set X and let µ0 : A0 → [0, ∞] be such that (a) µ0 (∅) = 0;

52

Measure and Integration

(b) for any disjoint family {An :P n ∈ N} of sets in A0 such that ∪∞ n=1 An ∈ ∞ ∞ A0 , we have µ0 (∪n=1 An ) = n=1 µ0 (An ). Then µ∗ : 2X → [0, ∞] defined by ∗

µ (E) := inf

∞ nX

∞ µ0 (An ) : {An }∞ n=1 ⊆ A0 , E ⊆ ∪n=1 An

o

n=1

is an outer measure on X and every set in A0 is µ∗ -measurable. Further, µ∗ (E) = µ0 (E) for every E ∈ A0 . It can be seen that the family of all finite disjoint unions of intervals of the from (a, b] for a, b ∈ R with a < b is an algebra on R, and this algebra satisfies all the properties in Theorem 3.1.44 (Problem 15).

3.2

Some Properties of Measures

Here are two theorems which are simple consequences of the countable additivity of the measure. Theorem 3.2.1 Let (X, A, µ) be a measure space and An ∈ A be such that An ⊆ An+1 for all n ∈ N. Then ∞ [  Ai µ(An ) → µ

as

n → ∞.

i=1

S∞ S∞ Proof. We write i=1 Ai as a disjoint union i=1 Ei by taking E1 = A1 and Ei = Ai \ Ai−1 for i = 2, 3, . . .. Then n n ∞ ∞ ∞ [  [  [  X X Ei . µ(Ei ) = lim µ µ Ai = µ Ei = µ(Ei ) = lim i=1

But,

Sn

i=1

i=1

n→∞

i=1

i=1

n→∞

i=1

Ei = An . Hence, ∞ [  µ Ai = lim µ(An ). n→∞

i=1

This completes the proof. Theorem 3.2.2 Let (X, A, µ) be a measure space and An ∈ A such that An ⊇ An+1 for all n ∈ N and µ(Ak ) < ∞ for some k ∈ N. Then ∞  \ µ(An ) → µ Ai i=1

as

n → ∞.

Measure and Measurable Functions 53 T∞ T∞ Proof. Since i=1 Ai = i=k Ai for any k ∈ N, we may assume without loss of generality that µ(A1 ) < ∞, instead of the assumption µ(Ak ) < ∞ for some k ∈ N. Now, let Bn = A1 \ An , n ∈ N. Then Bn ⊆ Bn+1 for all n ∈ N, so that by the Theorem 3.2.1, ∞ [  µ Bi = lim µ(Bn ). n→∞

i=1

S∞ T∞ But, i=1 Bi = A1 \ i=1 Ai . Therefore, since µ(A1 ) < ∞, by Theorem 3.1.6 (ii), we have ∞ ∞ ∞ [    \  \ µ Bi = µ A1 \ Ai = µ(A1 ) − µ Ai , i=1

i=1

i=1

µ(Bn ) = µ(A1 \ An ) = µ(A1 ) − µ(An ).
T∞ Thus, µ(An ) → µ i=1 Ai as n → ∞. It is to be mentioned that the conclusion in Theorem 3.2.2 need not hold if µ(An ) = ∞ for every n ∈ N. To see this, consider the example of (R, M, m) and An = [n, ∞) for n ∈ N. In this case, we have An ⊇ An+1 for every T∞ T∞n ∈ N, m(An ) = ∞ for every n ∈ N and n=1 An = ∅. Thus, m(An ) 6→ m( i=1 Ai ). Also, in Theorem 3.2.2 , the assumption µ(Ak ) < ∞ for some k ∈ N is not a necessary condition. To see this, consider (R, M, m) and An = [1 − n1 , ∞) for n ∈ N. Note that µ(An ) = ∞ and An ⊇ T An+1 for
every n ∈ N, and T ∞ ∞ A = [1, ∞). Thus, we have m(A ) → m A n n=1 n i=1 i . Now, let us introduce some set theoretic notions. Definition 3.2.3 A sequence (An ) of sets is called (a) monotonically increasing if An ⊆ An+1 for every n ∈ N, (b) monotonically decreasing if An ⊇ An+1 for every n ∈ N.

♦

We S may observe that, for any family {Aα : α ∈ Λ} of subsets of a set X, the set α∈Λ Aα is an upper bound of {Aα : α ∈ Λ} with respect to the partial order of “inclusion” of sets T on the power set of X. It is, in fact, the least upper bound. Similarly, the set α∈Λ Aα is a lower bound of {Aα : α ∈ Λ}, and it is the greatest lower bound. Thus, we may define [ \ sup Aα = Aα , inf Aα = Aα . α∈Λ

α∈Λ

α∈Λ

α∈Λ

For a monotonically increasing sequence (An ) of subsets of a set X, the set S ∞ i=1 Ai can be considered as the limitTof (An ), and for a monotonically de∞ creasing sequence (An ) of sets, the set i=1 Ai can be considered as the limit of (An ). Motivated by these considerations, we have the following definition.

54

Measure and Integration

Definition 3.2.4 Let (An ) be a sequence of subsets of a set X. Then we define the following: ∞ [ ∞ \

lim sup An := inf sup An = n→∞

k∈N n≥k

∞ \ ∞ [

lim inf An := sup inf An = n→∞

An ,

k=1 n=k

k∈N n≥k

An .

k=1 n=k

The set lim sup An is called the limit superior of (An ), and the set lim inf An n→∞

n→∞

is called the limit inferior of (An ). If lim sup An = lim inf An , then we say n→∞

n→∞

that the limit of (An ) exists, and the common set is called the limit of (An ), denoted by lim An . ♦ n→∞

For a sequence (An ) of subsets of a set X, the following results can be verified easily: 1. If (An ) is monotonically increasing, then lim An exists and n→∞

lim An =

n→∞

∞ [

Ai .

i=1

2. If (An ) is monotonically decreasing, then lim An exists and n→∞

lim An =

n→∞

∞ \

Ai .

i=1

It is also true that, for x ∈ X, • x ∈ lim sup An ⇐⇒ x ∈ An for infinitely many n ∈ N, n→∞

• x ∈ lim inf An ⇐⇒ x ∈ An for all but finitely many n ∈ N. n→∞

Thus, Theorem 3.2.1 and Theorem 3.2.2 can be restated as follows: If (X, A, µ) is a measure space and (An ) is a sequence in A which is either monotonically increasing or monotonically decreasing with µ(A1 ) < ∞, then µ( lim An ) = lim µ(An ). n→∞

n→∞

What can we say if (An ) is a general sequence of sets from A? Theorem 3.2.5 Let (X, A, µ) be a measure space and (An ) be a sequence of sets in A. Then we have the following.

Measure and Measurable Functions

55

(i) µ(lim inf An ) ≤ lim inf µ(An ). n→∞

n→∞

S∞ (ii) If µ( n=1 An ) < ∞, then µ(lim sup An ) ≥ lim sup µ(An ). n→∞

n→∞

(iii) If lim An and lim µ(An ) exist, then n→∞

n→∞

µ( lim An ) ≤ lim µ(An ). n→∞

n→∞

S∞ (iv) If lim An exists and if µ( n=1 An ) < ∞, then lim µ(An ) exists and n→∞

n→∞

µ( lim An ) = lim µ(An ). n→∞

n→∞

S

Proof. For k ∈ N, let Bk = n≥k An and Ck = Bk ⊇ Bk+1 and Ck ⊆ Ck+1 for all k ∈ N. (i) By Theorem 3.2.1,

T

n≥k

An . Then, we have

∞ [  µ(lim inf An ) = µ Ck = lim µ(Ck ). n→∞

k=1

k→∞

But, µ(Ck ) ≤ µ(Ak ). Therefore, lim µ(Ck ) = lim inf µ(Ck ) ≤ lim inf µ(Ak )

k→∞

k→∞

k→∞

so that we obtain the S∞required inequality. (ii) Suppose µ( n=1 An ) < ∞, i.e., µ(B1 ) < ∞. Then, by Theorem 3.2.2, ∞ \  µ(lim sup An ) = µ Bk = lim µ(Bk ). n→∞

k=1

k→∞

But, µ(Bk ) ≥ µ(Ak ). Therefore, lim µ(Bk ) = lim sup µ(Bk ) ≥ lim sup µ(Ak )

k→∞

k→∞

k→∞

so that we obtain the required inequality. (iii) This follows from (i). (iv) By (ii), lim sup µ(An ) ≤ µ(lim sup An ) = µ( lim An ) = µ(lim inf An ) n→∞

n→∞

n→∞

n→∞

and by (i), µ(lim inf An ) ≤ lim inf µ(An ). Hence, n→∞

n→∞

lim sup µ(An ) ≤ µ( lim An ) ≤ lim inf µ(An ). n→∞

n→∞

n→∞

Thus, lim µ(An ) exists and µ( lim An ) = lim µ(An ). n→∞

n→∞

n→∞

56

3.3

Measure and Integration

Measurable Functions

Suppose that f is a real valued function defined on a subset E of R. Then, we know by definition that, f is continuous on E if and only if for every open set G of R, the set f −1 (G) is open in E, that is, f −1 (G) = E ∩ V for some open subset V of R. Here, the topology on R is considered to be the usual topology. Now, let A be either the Borel σ-algebra B on R or the σ-algebra M of all Lebesgue measurable subsets of R. For E ∈ A, let AE be the restriction of A to E. Thus, f continuous ⇒ f −1 (G) ∈ AE for every open set G of R. Since AE contains sets which are not open, the converse of the above statement is not true. In other words, there can exist open sets G in R such that f −1 (G) ∈ AE , but f −1 (G) is not open. For example, for the function f : R → R defined by  1, x > 0, f (x) = 0, x ≤ 0, we have  (0, ∞),    (−∞, 0], f −1 (G) = R,    ∅,

1 ∈ G, 0 6∈ G, 0 ∈ G, 1 6∈ G, {0, 1} ⊆ G, {0, 1} ∩ G = ∅.

for every open set G ⊆ R. Thus, taking E = R, f −1 (G) ∈ B1 for every open set G ⊆ R, but the function is not continuous. In view of the above observations, we introduce the following definition. Definition 3.3.1 Let (X, A) be a measurable space and let K be either R or C. Then f : X → K is said to be a measurable function with respect to the σ-algebra A if f −1 (G) ∈ A for every open set G ⊆ K. A measurable function f : X → K is said to be a real measurable function if K = R, and it is said to be a complex measurable function if K = C. ♦ Instead of considering real or complex measurable functions, we may also have occasion to deal with functions with values in a general topological space. Thus, we have the following definition. Definition 3.3.2 Let (X, A) be a measurable space and (Y, T ) be a topological space. A function f : X → Y is said to be measurable with respect to the pair (A, T ) if f −1 (G) ∈ A for every G ∈ T . ♦

Measure and Measurable Functions

57

Convention: If the σ-algebra A and the topology T are understood from the context, then we shall simply say “f is a measurable function” instead of saying that “f is a measurable function with respect to the pair (A, T ).” We may observe: Let A and A0 be σ-algebras on a set X. If f is measurable with respect to A, then f is measurable with respect to A0 if and only if A ⊆ A0 .

Definition 3.3.3 For a given topological space (Y, T ), a function f : R → Y is called (i) Lebesgue measurable if f is measurable with respect to the σ-algebra M on R, and (ii) Borel measurable if f is measurable with respect to the Borel σalgebra B on R. ♦ As we have already mentioned earlier, we shall see in Theorem 3.4.9 that B ( M. Therefore, a Lebesgue measurable function need not be Borel measurable. Some of the topological spaces that we frequently use in the study of measure and integration are the following: (a) R with usual topology, (b) C with usual topology, (c) Rk with usual topology, (d) [−∞, ∞] := R ∪ {∞, −∞}, the extended real line, with topology generated by open subsets of R (under usual topology) and intervals of the ˜ := [−∞, ∞], form (a, ∞] and [−∞, b) for a, b ∈ R. Thus, for G ⊆ R ˜ if and only if for every x ∈ G there exists G is open in R an interval I of the form (a, b), (a, ∞] or [−∞, b) such that x ∈ I ⊆ G, where a, b ∈ R. ˜ then we say that f is an Definition 3.3.4 In the Definition 3.3.2, if Y = R, extended real valued measurable function. ♦ ˜ we can infer the following: By the definition of the topology on R, ˜ := [−∞, ∞] is a countable union of intervals of 1. Every open subset of R the form (a, b), [∞, b), (a, ∞], where a, b ∈ R. ˜ and {−∞} and 2. The intervals [−∞, ∞) and (−∞, ∞] are open sets in R, ˜ {∞} are closed sets in R.

58

Measure and Integration Using the property of any σ-algebra, we can state the following. Let (X, A) be a measurable space and (Y, T ) be a topological space. Suppose that there is a subfamily T0 of T with the property that every open set in T is a countable union of sets from T0 . Then, a function f : X → Y is measurable with respect to (A, T ) if and only if f −1 (B) ∈ A for every B ∈ T0 .

˜ C, and Rk have the property stated in the The topological spaces R, R, above box with T0 as specified below: (a) For Y = R, T0 is the family of all open intervals. ˜ T0 is the family of all intervals of the form (a, b), (a, ∞], or (b) For Y = R, [−∞, b). (c) For Y = C, T0 is the family of all open balls in Y , that is family of all sets of the form {z ∈ C : |z − ξ| < r} for ξ ∈ C and r > 0. (d) For Y = Rk , T0 is the family of all open balls or open k-cells. Here, an open ball in Rk is a set of the form {x ∈ Rk : |x − u| < r}, P 1/2 k 2 for u ∈ Rk and r > 0, where |x − u| := |x − u | , for i i i=1 x = (x1 , . . . , xk ) and u = (u1 , . . . , uk ) in Rk , and an open k-cell is a set of the form I := I1 × · · · × Ik , where I1 , . . . , Ik are open intervals in R. In due course, we shall also use the following definition. Definition 3.3.5 Let X be a measurable space, Y be a topological space. Then a function f : X → Y is said to be measurable on a measurable set E if f |E is measurable with respect to the restricted σ-algebra AE . ♦ An important class of functions in measure and integration is the class of all characteristic functions. Definition 3.3.6 Given a set X, the characteristic function of a subset E of X is the function χE : X → R defined by  1, x ∈ E, χE (x) = ♦ 0, x 6∈ E. Theorem 3.3.7 Let X be a measurable space and E ⊆ X. Then χE is measurable if and only if E is measurable. Proof. We observe that for an open subset G of R,  E, 1 ∈ G, 0 6∈ G,    c 1 6∈ G, 0 ∈ G, E , −1 χE (G) = X, 1 ∈ G, 0 ∈ G,    ∅, 1 6∈ G, 0 6∈ G. Thus, χE is measurable if and only if E is measurable.

Measure and Measurable Functions

59

The proof of the following theorem is easy, and hence it is left as an exercise (see Problem 35). Theorem 3.3.8 Suppose f : X → Y is a measurable function. (i) If E ∈ A, then f is measurable on E. ˜ then for any c ∈ R, cf is measurable. (ii) If Y is either R or C or R, Theorem 3.3.9 Let X be a measurable space, Y be a set and f : X → Y . Then the following results hold. (i) S := {E ⊆ Y : f −1 (E) ∈ A} is a σ-algebra on Y . (ii) If Y is a topological space, then f is measurable if and only if S contains BY , the Borel σ-algebra on Y . Proof. (i) The fact that S is a σ-algebra on Y follows from the relations f −1 (∅) = ∅, f −1

f −1 (Y ) = X, ∞ [ n=1

f −1 (Ac ) = [f −1 (A)]c ,

∞  [ An = f −1 (An ) n=1

for subsets A, An of Y for n ∈ N. (ii) Suppose that Y is a topological space. By (i), S is a σ-algebra. If f is measurable, then S contains all open sets. Since BY is the smallest σ-algebra containing all open sets, S contains BY as well. Converse is obvious as BY contains all open sets in Y . As an immediate consequence of Theorem 3.3.9, we have the following theorem. Theorem 3.3.10 Let X be a measurable space and Y be a topological space. Then, a function f : X → Y is measurable if and only if for every B ∈ BY , f −1 (B) ∈ A. In view of the above theorem, we have the following definition of measurability in a general context. Definition 3.3.11 Let (X1 , A1 ) and (X2 , A2 ) be two measurable spaces. A function f : X1 → X2 is said to be measurable with respect to the pair (A1 , A2 ) of σ-algebras if f −1 (B) ∈ A1 for every B ∈ A2 . ♦ The following theorem shows how a measure on one measurable space can induce a measure on another measurable space using a measurable function in the sense discussed in Definition 3.3.11. This theorem is important in the context of probability theory.

60

Measure and Integration

Theorem 3.3.12 Let (X1 , A1 ) and (X2 , A2 ) be measurable spaces and let f : X1 → X2 be measurable in the sense of Definition 3.3.11. Let µ be a measure on (X1 , A1 ). Then ν : A2 → [0, ∞] defined by ν(A) = µ(f −1 (A)),

A ∈ A2 ,

is a measure on (X2 , A2 ). Proof. Clearly, ν(∅) = µ(f −1 (∅)) = 0. Next, let {An } be a countable disjoint family in A2 . Then [ [ f −1 ( An ) = f −1 (An ) n

n

and {f −1 (An )} is also a disjoint family. Hence, [ X X [ ν( An ) = µ( f −1 (An )) = µ(f −1 (An )) = ν(An ). n

n

n

n

This completes the proof. As a particular case of the above theorem, we may state the following theorem. Theorem 3.3.13 Let (X, A, µ) be a measure space and f : X → R be a measurable function. Then ν : B1 → [0, ∞] defined by ν(A) = µ(f −1 (A)),

A ∈ B1 ,

is a measure on (R, B1 ). Definition 3.3.14 The measure ν defined as in Theorem 3.3.12 is called the measure induced by f and µ. ♦

3.3.1

Probability space and probability distribution

In the context of probability theory, where a measure is a probability measure, the σ-algebra is called a σ-field, and the set on which the σ-field is considered is called a sample space. Measurable sets are called events, measure spaces are called probability spaces, and measurable functions are called random variables. Let Ω be a sample space, F a σ-field of events in Ω, µ a probability measure on (Ω, F), and f : Ω → R a random variable. Then, by Theorem 3.3.13, ν : BR → [0, ∞] defined by ν(B) = µ(f −1 (B)),

B ∈ BR ,

is a probability measure. It is called the probability distribution or simply the distribution of the random variable f . The function F : R → R defined by F (t) = µ(f −1 (−∞, t]),

t ∈ R,

Measure and Measurable Functions

61

is called the distribution function or cumulative distribution function induced by the random variable f . Note that F (t) = µ({ω ∈ Ω : f (ω) ≤ t}),

t ∈ R.

It is often written as F (t) = µ(f ≤ t),

t ∈ R.

Note that F is monotonically increasing and non-negative. It can be shown that F is a measurable function (Exercise). A function g : R → [0, ∞) is called a probability density function of the random variable f , if the improper R∞ integral −∞ g(t)dt exists and Z

t

F (t) =

g(s)ds,

x ∈ R.

−∞

In Chapter 5, we shall give a characterization of F for which such a probability density function exists.

3.3.2

Further properties of measurable functions

Recall that the measurability of a function from a measurable space to a topological space is defined using open sets in the topological space. However, for a real or an extended real valued function, the measurability can be characterized using a very special type of sets as in Theorem 3.1.23. Theorem 3.3.15 Let (X, A) be a measurable space and f : X → [−∞, ∞]. Then, the following are equivalent: (i) f is measurable. (ii) {x ∈ X : f (x) > a} ∈ A

∀a ∈ R.

(iii) {x ∈ X : f (x) ≥ a} ∈ A

∀a ∈ R.

(iv) {x ∈ X : f (x) < a} ∈ A

∀a ∈ R.

(v) {x ∈ X : f (x) ≤ a} ∈ A

∀a ∈ R.

˜ Now, for Proof. Recall that for any a ∈ R, (a, ∞] is an open set in R. a ∈ R, we observe that {x ∈ X : f (x) > a} = f −1 ((a, ∞]), {x ∈ X : f (x) ≥ a} = f −1 ([a, ∞]), {x ∈ X : f (x) < a} = f −1 ([−∞, a)), {x ∈ X : f (x) ≤ a} = f −1 ([−∞, a]).

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Measure and Integration

Note also that [a, ∞] =

∞ \

a−

n=1

˜ \ [a, ∞], [−∞, a) = R

 1 ,∞ , n

[−∞, a] =

∞ \ 

− ∞, a +

n=1

1
, n

and for any a, b ∈ R with a < b, (a, b) = (a, ∞] ∩ [−∞, b). From these observations together with the properties of a σ-algebra, the equivalence of (i)-(v) follows. Since every open subset of R2 is a countable union of open 2-cells, the proof of the following theorem is obvious. Theorem 3.3.16 f : X → R2 is measurable if and only if f −1 (I1 × I2 ) is a measurable set for every open 2-cell I1 × I2 in R2 . From this we deduce the following theorem. Theorem 3.3.17 Let X be a measurable space and f, g be real measurable functions on X. Then the function h : X → R2 defined by h(x) = (f (x), g(x)),

x ∈ X,

is measurable. Proof. Since every open subset of R2 is a countable union of open rectangles, it is sufficient to show that for open intervals I1 and I2 , h−1 (I1 × I2 ) is measurable. Note that h−1 (I1 × I2 )

= {x ∈ X : h(x) ∈ I1 × I2 } = {x ∈ X : f (x) ∈ I1 and g(x) ∈ I2 } = f −1 (I1 ) ∩ g −1 (I2 ).

Since f and g are measurable functions, f −1 (I1 ) ∩ g −1 (I2 ) is a measurable set. Consequently, h is a measurable function. Theorem 3.3.18 Let X be a measurable space, and Y and Z be topological spaces. If f : X → Y is measurable and g : Y → Z is continuous, then g ◦ f : X → Z is measurable. Proof. Suppose f : X → Y is measurable and g : Y → Z is continuous. Let V be an open set in Z. By continuity of g, the set g −1 (V ) is open in Y , and by measurability of f , f −1 (g −1 (V )) is measurable in X. Hence, from the identity (g ◦ f )−1 (V ) = f −1 (g −1 (V )) the set (g ◦ f )−1 (V ) is measurable in X.

Measure and Measurable Functions

63

As corollaries to the above theorem we have the following two theorems. Theorem 3.3.19 Let X be a measurable space. (i) If f and g are real measurable functions on X and λ ∈ R, then f + g, f g and λf are real measurable functions. (ii) If f and g are real measurable functions on X, then f + ig is a complex measurable function. (iii) If f is a complex measurable function on X, then the functions Ref , Imf, |f |, defined by [Ref ](x) := Ref (x),

[Imf ](x) = Imf (x),

|f |(x) := |f (x)|,

respectively, for x ∈ X, are real measurable functions on X. (iv) If f and g are complex measurable functions on X and λ ∈ C, then f + g, f g and λf are complex measurable functions on X. Proof. (i) By Theorem 3.3.17, the function x 7→ (f (x), g(x)) from X to R2 is measurable. Now, the results follow from Theorem 3.3.18 by making use of the following facts: (a) The functions (α, β) 7→ α + β, (α, β) 7→ αβ from R2 to R and the function α 7→ λα from R to R are continuous. (b) The function (α, β) 7→ α + iβ from R2 to C is continuous. (c) The functions z 7→ Rez, z 7→ Imz and z 7→ |z| from C to R are continuous. Remark 3.3.20 From (i) and (iv) in Theorem 3.3.19, it follows that the set VR of all real measurable functions and the set VC of all complex measurable functions are vector spaces over R and C, respectively. Since R ⊆ C, VC is a vector space over R as well. ♦ Remark 3.3.21 In Theorem 3.3.19, we avoided the case of extended real valued functions, because f + g need not be defined at some points. Also, so far we have not considered any topology on [−∞, ∞]×[−∞, ∞]. However, for non-negative extended real valued measurable functions f and g, we shall show the measurability of f + g, after considering the notion of simple measurable functions. ♦ We shall have occasion to use the following definition and the subsequent theorem. Definition 3.3.22 Given a function f : X → [−∞, ∞] defined on a set X, its positive part, negative part and absolute value or modulus, denoted by f + , f − , and |f |, respectively, are defined by f + (x) = max{f (x), 0}, f − (x) = max{−f (x), 0}, |f | = max{f (x), −f (x)}

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Measure and Integration

for x ∈ X.

♦

We may observe that f = f + − f −,

|f | = f + + f − .

Theorem 3.3.23 Let f : X → [−∞, ∞]. If f is measurable, then f + , f − and |f | are measurable. In case f is real valued, then f is measurable if and only if f + and f − are measurable. Proof. Suppose f is measurable. By Theorem 3.3.8 (ii), the function −f is measurable. Now, measurability of f + , f − , and |f | follows from Theorem 3.3.25, since f + = sup{f, 0},

f − = sup{−f, 0},

|f | = sup{f, −f }.

Next, suppose that f is real valued and f + , f − are measurable. Then by Theorem 3.3.19 (i), f = f + − f − is also measurable. In the second part of the above theorem, we assumed that f is real valued. In a later theorem (see Theorem 3.4.8), we shall drop this condition.

3.3.3

Sequences and limits of measurable functions

Analogous to the cases of real sequences, we define supremum, infimum, limit superior, and limit inferior of sequences of functions. Definition 3.3.24 Let (fn ) be a sequence of functions from a set X to [−∞, ∞]. Then supremum, infimum, limit superior, limit inferior of (fn ), denoted by sup fn , n

inf fn , n

lim sup fn ,

lim inf fn , n

n

are defined by (supn fn )(x) = supn fn (x), (inf n fn )(x) = inf n fn (x), , (lim supn fn )(x) = lim supn fn (x), (lim inf n fn )(x) = lim inf n fn (x) respectively, for x ∈ X.

♦

We may observe that if lim sup fn = lim inf fn , then lim fn (x) exists for n

n

n

every x ∈ X, and lim fn (x) = lim sup fn (x) = lim inf fn (x) n

for every x ∈ X.

n

n

Measure and Measurable Functions

65

Theorem 3.3.25 Let (fn ) be a sequence of extended real valued measurable functions on X. Then sup fn ,

inf fn , n

n

lim sup fn ,

lim inf fn

n

n

are extended real valued measurable functions. Proof. For a ∈ R, we note that {x ∈ X : sup fn (x) > a} = n

[

{x ∈ X : fn (x) > a},

n

{x ∈ X : inf fn (x) < a} = n

[ {x ∈ X : fn (x) < a}. n

By Theorem 3.3.15, {x ∈ X : fn (x) > a} and {x ∈ X : fn (x) < a} are measurable sets. Hence, by the properties of the σ-algebra, sup fn and inf fn n

n

are also measurable. Since lim sup fn = inf sup fn , k n≥k

n

lim inf fn = sup inf fn , n

k

n≥k

it also follows that lim sup fn and lim inf fn are measurable functions. n

n

Theorem 3.3.26 Let (fn ) be a sequence of real valued measurable functions on X such that for each x ∈ X, there exists Mx > 0 satisfying |fn (x)| ≤ Mx for all n ∈ N. Then supn fn , inf n fn , lim supn fn , lim inf n fn are real valued measurable functions. Proof. Since for each x ∈ X, there exists Mx > 0 satisfying |fn (x)| ≤ Mx for all n ∈ N, the functions supn fn , inf n fn , lim supn fn , lim inf n fn are real valued. Also, by Theorem 3.3.25, these functions are measurable as well. Definition 3.3.27 Suppose (fn ) is a sequence of functions defined on a mea˜ := [−∞, ∞]. surable space X with values in Y which is either of R, C or R Then we say that (fn ) converges pointwise on a subset E of X if for each x ∈ E, the sequence (fn (x)) converges in Y , and in that case we write fn → f

pointwise on E,

where f : E → Y is defined by f (x) = lim fn (x), x ∈ E. The function f is n→∞

called the pointwise limit of (fn ). If fn → f pointwise on X, we simply say, fn → f pointwise. ♦ As a corollary to the above two theorems, we obtain the following result. Theorem 3.3.28 Let (fn ) be a sequence of real or extended real valued or complex measurable functions on X. If fn → f pointwise on X, then f is measurable.

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Measure and Integration

Proof. If fn ’s are either extended real valued (respectively, real valued), then the result is a consequence of Theorem 3.3.25 (respectively, Theorem 3.3.26). Next, suppose that fn ’s are complex valued. Then each fn can be written as fn = gn + ihn , where gn = Re (fn ) and hn = Im (fn ). By Theorem 3.3.19 (iii), gn and hn are real measurable functions. Since fn → f pointwise, it can be easily seen that (gn ) and (hn ) converge pointwise to g := Re (f ) and h := Im (f ), respectively. Again, by Theorem 3.3.26, g and h are real measurable functions. Hence, by Theorem 3.3.19 (ii), f = g + ih is complex measurable.

3.3.4

Almost everywhere properties

In measure theory, there are weaker forms of convergence than pointwise convergence. One such convergence is the almost everywhere convergence, which is “almost” pointwise convergence. Definition 3.3.29 Let (X, A, µ) be a measure space and (fn ) be a sequence of measurable functions defined on a set E ∈ A taking values in Y which is ˜ We say that (fn ) converges almost everywhere on E either of R, C, or R. if the set E0 := {x ∈ E : (fn (x)) converges} belongs to A and µ(E \ E0 ) = 0, and in that case, we also say that (fn ) converges to f almost everywhere on E and write fn → f a.e. on E, where f : E → Y is a function satisfying f (x) = lim fn (x), x ∈ E0 . n→∞

♦

We shall, sometimes, use the terminology “(fn (x)) converges for almost all x ∈ E” and “fn (x) → f (x) for a.a. x ∈ E” to mean “(fn ) converges a.e. on E” and “fn → f a.e. on E”, respectively. Remark 3.3.30 In Definition 3.3.29 we have defined f only on the set E0 = {x ∈ E : fn (x) converges}. Thus, while saying that fn → f a.e., we are free to define f on E \ E0 . By Theorem 3.3.28, we know that, f is measurable on E0 . However, arbitrary extension of f to E need not be measurable. We shall see soon that if either µ is complete or by extending f to all of E by defining f = 0 on E \ E0 , then f is measurable on E. ♦ Example 3.3.31 (i) For each n ∈ N, let fn : [0, 1] → R be defined by  (−1)n if x ∈ Q, fn (x) = 1 if x 6∈ Q. Note that for every x ∈ [0, 1] \ Q, (fn (x)) converges to 1 and for x ∈ [0, 1] ∩ Q, (fn (x)) diverges. Since m(Q) = 0, (fn ) converges almost everywhere on [0, 1] to the constant function 1, but it does not converge pointwise on [0, 1].

Measure and Measurable Functions (ii) Let gn : [0, 1] → R be defined by  1 gn (x) = (−1)n

67

if x ∈ Q, if x 6∈ Q

for each n ∈ N. In this case, (gn ) neither converges pointwise nor converges a.e. on [0, 1]. ♦ The concept of almost everywhere can be defined in a more general context. Definition 3.3.32 Let (X, A, µ) be a measure space, E ∈ A and P be a property on the elements of X. Then P is said to hold almost everywhere on E if the set E0 := {x ∈ E : P holds at x} belongs to A and µ(E \ E0 ) = 0, and, in that case, we write P holds a.e. on E. The fact “P holds a.e. on X” is simply written as “P holds a.e.” ♦ Thus, if f is a function defined on a measure space (X, A, µ) with values ˜ or C, then f = 0 a.e. on E if the set E0 := {x ∈ E : f (x) = 0} in R or R belongs to A and µ(E \ E0 ) = 0. Note that if f is measurable and E ∈ A, then E0 ∈ A so that f = 0 a.e. on E if and only if µ({x ∈ E : f (x) 6= 0}) = 0. Example 3.3.33 Consider the following functions from R to R, where R is with Lebesgue measure:    1 if x ≤ 0, 1 if x ∈ Q, 1 if x = 0, f (x) = g(x) = h(x) = 0 if x > 0, x if x 6∈ Q x if x 6= 0. We observe that (i) f is continuous except at 0, so that f is continuous a.e., (ii) g(x) = x except for x ∈ Q, so that g(x) = x for a.a. x ∈ R, and (iii) h(x) = x except for x = 0, so that g(x) = x for a.a. x ∈ R.

♦

Remark 3.3.34 Note that, in Example 3.3.33, f is continuous a.e., but there is no continuous function which is equal to this function a.e., whereas g and h are equal to continuous functions a.e., with h is continuous a.e., but g is not continuous a.e. ♦ Now, in view of the concept of almost everywhere convergence, we may ask whether we can modify Theorem 3.3.28 by relaxing the pointwise convergence to almost everywhere convergence. In this regard we have the following theorem. Theorem 3.3.35 Let (fn ) be a sequence of extended real valued measurable functions on a measure space (X, A, µ) and fn → f a.e. on X. If µ is a complete measure, then f is measurable.

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Measure and Integration

Proof. Let E = {x ∈ X : fn (x) → f (x)}. By assumption, E ∈ A and µ(E c ) = 0. We have to prove that f −1 (G) ∈ A for any open set G in [−∞, ∞]. So, let G be an open subset of [−∞, ∞]. Note that f −1 (G) = {x ∈ E : f (x) ∈ G} ∪ {x ∈ E c : f (x) ∈ G}. Since fn → f pointwise on E, by Theorem 3.3.28, the set {x ∈ E : f (x) ∈ G} is in AE , and since {x ∈ E c : f (x) ∈ G} ⊆ E c and µ(E c ) = 0, by the completeness of µ, {x ∈ E c : f (x) ∈ G} ∈ A. Thus, f −1 (G) ∈ A is the union of two sets in A. If the measure µ in the above theorem is not complete, then it is not necessary that f is measurable. Here is a simple illustration of this fact. Example 3.3.36 Suppose (X, A, µ) is an incomplete measure space. Then there exists A ∈ A with µ(A) = 0 and E ⊆ A such that E 6∈ A. Let f : X → R be defined by   3 if x ∈ E, 2 if x ∈ A \ E, f (x) =  1 if x ∈ X \ A. Then, f is not a measurable function, since f −1 ({3}) = E 6∈ A. If we define g = χX\A for n ∈ N, then each g is measurable and g = f a.e. Defining fn = g for every n ∈ N, we have fn → g a.e. ♦ Note that, if (fn ) is as in the above example and if we define h(x) = 1 for every x ∈ X, then fn → h a.e., where h is measurable and f = h a.e. This is, in fact, true for any sequence of measurable functions which converges a.e. as the following theorem shows. Theorem 3.3.37 Let (fn ) be a sequence of extended real valued measurable functions on a measure space X. If (fn ) converges a.e. on X, then there exists a measurable function f such that fn → f a.e. on X. For its proof we shall make use of a special case of the following result, which we call pasting lemma. Lemma 3.3.38 (Pasting lemma) Let X be a measurable space, Y be a topological space, E ∈ A, and f1 : E → Y and f2 : E c → Y be measurable with respect to the σ-algebras AE and AE c , respectively. Then f : X → Y defined by  f1 (x), x ∈ E, f (x) = f2 (x), x ∈ E c is a measurable function. Proof. Let G be an open set in Y . Then {x ∈ X : f (x) ∈ G} = {x ∈ E : f (x) ∈ G} ∪ {x ∈ E c : f (x) ∈ G} = {x ∈ E : f1 (x) ∈ G} ∪ {x ∈ E c : f2 (x) ∈ G}.

Measure and Measurable Functions

69

Since f1 and f2 are measurable with respect to the restricted σ-algebras AE and AE c , respectively, we have {x ∈ E : f1 (x) ∈ G} ∈ AE ⊆ A,

{x ∈ E : f2 (x) ∈ G} ∈ AE c ⊆ A.

Hence, {x ∈ X : f (x) ∈ G} ∈ A for every open set G in Y ; consequently, f is measurable. As an immediate corollary to the pasting lemma (Lemma 3.3.38), we have the following. Corollary 3.3.39 Let X be a measure space, Y be a topological space, and g : X → Y be measurable on a measurable set E, where µ(E c ) = 0. Then there exists a measurable function f : X → Y such that f = g a.e. Proof. Take f1 = g and f2 = 0 in Lemma 3.3.38. Proof of Theorem 3.3.37. Let E := {x ∈ X : (fn (x)) converges}. By assumption, E ∈ A with µ(E c ) = 0. Therefore, by Theorem 3.3.28, the function f0 : E → Y defined by f0 (x) = limn→∞ fn (x), x ∈ E, is measurable with respect to the restricted σ-algebra AE . Let  f0 (x), x ∈ E, f (x) = 0, x ∈ Ec. Then, by pasting lemma (Lemma 3.3.38), f is measurable. Since fn → f pointwise on E and µ(E c ) = 0, fn → f a.e. on X. We know that almost everywhere convergence is weaker than pointwise convergence (see Example 3.3.31) and pointwise convergence is weaker than uniform convergence. However, every sequence of functions which converges almost everywhere also converges uniformly on every set of finite measure except on a set of arbitrarily small measure. This is the so-called Egoroff ’s theorem stated below. Theorem 3.3.40 (Egoroff ’s theorem) Let (X, A, µ) be a finite measure space and (fn ) be a sequence of extended real valued measurable functions on X which converges a.e. to a measurable function f . Then for every ε > 0, there exists E ∈ A such that µ(X \ E) < ε and fn → f uniformly on E. Proof. Let X0 := {x ∈ X : (fn (x)) converges }. Since (fn ) converges a.e. on X, we know that X0 ∈ A and µ(X \ X0 ) = 0. Note that fn → f pointwise on X0 . Thus, for each x ∈ X0 and for each m ∈ N, there exists kx,m ∈ N such that |fn (x) − f (x)| < 2−m ∀ n ≥ kx,m . S∞ Therefore, X0 = k=1 Fm,k , where Fm,k := {x ∈ X0 : |fn (x) − f (x)| < 2−m

∀ n ≥ k}.

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Measure and Integration

Note that Fm,k ⊆ Fm,k+1

∀k ∈ N

so that µ(Fm,k ) → µ(∪∞ j=1 Fm,j ) = µ(X0 )

as k → ∞.

In particular, since µ(X0 ) < ∞, µ(X0 \ Fm,k ) = µ(X0 ) − µ(Fm,k ) → 0

as k → ∞.

Hence, for a given ε > 0 and m ∈ N, there exists km such that µ(X0 \ Fm,k ) < ε2−m

∀ k ≥ km

so that µ(X0 \ Fm,km ) < ε2−m Let E =

∩∞ m=1 Fm,km .

∀ m ∈ N.

Then

∞ ∞  [  X (X0 \ Fm,km ) ≤ µ(X0 \ E) = µ µ(X0 \ Fm,km ) < ε. m=1

m=1

Note that if x ∈ E, then x ∈ Fm,km for all m ∈ N so that |fn (x) − f (x)| < 2−m

∀ n ≥ km .

Thus fn → f uniformly on E. Further, since µ(X\X0 ) = 0, we have µ(X\E) = µ(X0 \ E) < ε. The following example shows that the conclusion in Egoroff’s theorem need not hold if µ is not a finite measure. Example 3.3.41 Let R be with Lebesgue measure and fn := χ(n,n+1] for n ∈ N. Clearly, fn → 0 pointwise. Let 0 < ε < 1. Suppose there exists E ⊆ R such that m(R \ E) < ε and

sup |fn (x)| → 0

as n → ∞.

(∗)

x∈E

Then there exists N ∈ N such that supx∈E |fn (x)| < 1/2 for all n ≥ N . Since fN = 1 on (N, N + 1], we obtain E ∩ (N, N + 1] = ∅. Consequently, (N, N +1] ⊆ R\E. This forces m(R\E) ≥ 1, which contradicts the assumption that m(R \ E) < ε < 1. Thus, there does not exist E ⊆ R satisfying (∗). ♦ We shall use Egoroff’s theorem to deduce the following. Theorem 3.3.42 Let (X, A, µ) be a finite measure space and (fn ) be a sequence of extended real valued measurable functions on X which converges a.e. to a measurable function f . Then, for every ε > 0, there exists N ∈ N such that µ({x ∈ X : |fn (x) − f (x)| > ε}) < ε ∀ n ≥ N.

Measure and Measurable Functions

71

Proof. Let ε > 0 be given. Then, by Egoroff’s theorem, there exists E ∈ A, E ⊆ A such that µ(A \ E) < ε and fn → f uniformly on E. Hence, there exists N ∈ N such that |fn (x) − f (x)| < ε ∀ x ∈ E

and ∀ n ≥ N.

This implies that, for n ≥ N , {x ∈ X : |fn (x) − f (x)| > ε} ⊆ X \ E. Since X \ E = (X \ A) ∪ (A \ E) and since µ(X \ A) = 0 and µ(A \ E) < ε, we have µ({x ∈ X : |fn (x) − f (x)| > ε}) < ε ∀ n ≥ N. This completes the proof. Definition 3.3.43 A sequence (fn ) of extended real valued measurable functions on a measure space (X, A, µ) is said to converge in measure to a measurable function f if for every ε > 0, there exists N ∈ N such that µ({x ∈ X : |fn (x) − f (x)| > ε}) < ε ∀ n ≥ N.

♦

Thus, Theorem 3.3.42 shows that if fn → f a.e., then (fn ) converges in measure to f. The concept of convergence in measure is extensively used in probability theory.

3.4

Simple Measurable Functions

In this section, we consider a class of functions which are more general than characteristic functions, and which plays a very important role in the theory of measure and integration. Definition 3.4.1 Let X be a nonempty set. A function ϕ : X → R is called a simple function if it takes only a finite number of values. ♦ If α1 , . . . , αn are the distinct values of a simple function ϕ : X → R, then ϕ can be represented as n X ϕ= αi χEi , i=1

where Ei = {x ∈ X : ϕ(x) = αi }, i = 1, . . . , n. The above representation of ϕ is called its canonical representation. Pn Note that if ϕ = i=1 αi χEi is the canonical representation of a simple function ϕ, then {E1 , . . . , En } is a decomposition of X, that is, X = ∪ni=1 Ei

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Measure and Integration

and {E1 , . . . , En } is a disjoint family. If one of the values of P ϕ is 0 and if n α1 , . . . , αn are the nonzero distinct values of ϕ, then also ϕ = i=1 αi χEi is the canonical representation of ϕ, but X=

n [

Ei

with E0 = {x ∈ X : ϕ(x) = 0}.

i=0

Suppose a1 , . . . , an in R and Ai ⊆ X, i = 1, . . . , n. Then we see that the function ϕ : X → R defined by ϕ=

n X

ai χAi

i=1

takes only a finite number of values, and hence, it is a simple function. Thus, ϕ : X → R is a simple function if and only if there Pn are a1 , . . . , an in R and Ai ⊆ X, i = 1, . . . , n, such that ϕ = i=1 ai χAi . Recall that a function f : R → R is called a step function if there are disjoint intervals I1 , . . . , In and a1 , . . . , an in R such that f=

n X

a i χI i .

i=1

Thus, every real valued step function defined on R is a simple function. Theorem 3.4.2 Let (X, A) be a measurable Pn space and ϕ be a simple function on X with canonical representation ϕ = i=1 αi χEi . Then ϕ is measurable if and only if Ei ∈ A for i = 1, . . . , n. Proof. Suppose ϕ is a measurable function. Then, each Ei is a measurable set as it is the inverse image of the closed (singleton) set {αi }. Conversely, if each Ei is a measurable set, then by Theorem 3.3.19 (i), ϕ is a measurable function. The next theorem is one of the most important results in the theory of measure and integration, and it will help us in inferring many properties of a measurable function. Theorem 3.4.3 Let (X, A) be a measurable space and f : X → [0, ∞] be a measurable function. Then there exists a sequence (ϕn ) of simple measurable functions on X such that (i) 0 ≤ ϕn ≤ ϕn+1 for every n ∈ N and (ii) ϕn → f pointwise on X.

Measure and Measurable Functions

73

In fact, the sequence (ϕn ) defined by n

ϕn =

n2 X i−1 i=1

2n

χEi,n + nχFn ,

n ∈ N,

satisfies the requirements (i) and (ii) above, where Fn := {x ∈ X : f (x) ≥ n} and n i−1 i o Ei,n := x ∈ X : n ≤ f (x) < n , i = 1, . . . , n2n . 2 2 Proof. Note that, for each n ∈ N, {Ei,n : i = 1, . . . , n2n } ∪ {Fn } is a disjoint family of measurable sets. Let x ∈ X. If f (x) = ∞, then ϕn (x) = n for all n ∈ N, so that ϕn (x) → f (x) as n → ∞. In case 0 ≤ f (x) < ∞, then there exists k ∈ N such that f (x) ≤ k. Hence, for every n ≥ k, x ∈ Ei,n for some i ∈ {1, 2, . . . , n2n }, and in that case |f (x) − ϕn (x)| ≤

1 . 2n

Therefore, in this case also, ϕn (x) → f (x) as n → ∞. Next suppose that x ∈ Ei,n for some n ∈ N and for some i ∈ {1, 2, . . . , n2n }. Then ϕn (x) = (i − 1)/2n , and   i−1 i−1 1 , + n+1 . ϕn+1 (x) ∈ 2n 2n 2 Thus, ϕn (x) ≤ ϕn+1 (x). If x ∈ Fn , then ϕn (x) = n and ϕn+1 (x) ∈ {n +

i : i = 0, 1, . . . , 2n+1 }. 2n+1

Thus, we get ϕn (x) ≤ ϕn+1 (x) for every x ∈ X and for every n ∈ N. We have the following theorem as an immediate corollary to the above theorem. Theorem 3.4.4 Let f be a real measurable function on a measurable space (X, A). Then there exists a sequence of simple measurable functions on X which converges to f pointwise. Proof. Write f as f = f + − f − . Then, by Theorem 3.4.3, there exist increasing sequences (ϕn ), (ψn ) of non-negative simple measurable functions on X which converge pointwise to f + and f − , respectively. Also, for each n ∈ N, ϕn − ψn is a simple measurable function and (ϕn − ψn ) converges pointwise to f .

74

Measure and Integration We shall also make use of Theorem 3.4.3 to prove the following theorem.

Theorem 3.4.5 Let (X, A, µ) be a (possibly incomplete) measure space and ˜ A, ˜µ let (X, ˜) be its completion. Let f : X → R be a measurable function with ˜ Then there exists a measurable function g : X → R with respect respect to A. to A such that f = g a.e. Proof. Let N := {F ⊆ X : ∃ B ∈ A with F ⊆ B and µ(B) = 0}. Then, ˜ there exists A ∈ A and F ∈ N such that by Theorem 3.1.36, for every E ∈ A, E = A ∪ F , so that χE = χA a.e. Note that χE is measurable with respect to A˜ and χA is measurable with respect to A. Similarly, for any simple function ˜ there exists a simple function ψ ϕ, which is measurable with respect to A, Pk measurable with respect to A such that ϕ = ψ a.e. In fact, if ϕ = i=1 αi χEi , wherePEi = Ai ∪ Fi with Ai ∈ A and Fi ∈ N for i = 1, . . . k, then we may take k ψ = i=1 αi χAi . If we take F := ∪ki=1 Fi , then we see that ϕ = ψ on X \ F , where µ(F ) = 0. ˜ Next, suppose that f : X → R is a measurable function with respect to A. By Theorem 3.4.4, there exists a sequence (ϕn ) of simple measurable functions ˜ such that ϕn → f pointwise. By using the considerations in the last on (X, A) paragraph, there exists a sequence (ψn ) of simple measurable functions on Pkn (n) (X, A) such that ϕn = ψn a.e. for every n ∈ N. Now, if ϕn = i=1 αi χ (n) , E

(n) Ei

(n) Ai

where = we may take ψn =

(n) (n) ∪F with Ai ∈ A Pkni (n) i=1 αi χA(n) . Let

and

(n) Fi

i

∈ N for i = 1, . . . kn , then

i

A :=

kn ∞ [ [ n=1 i=1

(n)

Ai ,

F :=

kn ∞ [ [

(n)

Fi

.

n=1 i=1

Note that X \ A ⊆ F and ϕn = ψn on A for all n ∈ N, where µ(F ) = 0. Since ϕn → f pointwise on X, we have f (x) = lim ψn (x) for x ∈ A. Since n→∞ each ψn is measurable with respect to A, by Theorem 3.3.28, the function f0 , which is the restriction of f to A, is measurable with respect to the restricted σ-algebra AA . Now, let  f0 (x) for x ∈ A, g(x) := 0 for x ∈ X \ A. Then, by pasting lemma (Lemma 3.3.38), g is measurable with respect to A. Note that g = f a.e. As an immediate corollary of the above theorem we have the following. Theorem 3.4.6 If f : R → R is a Lebesgue measurable function, then there exists a Borel measurable function g : R → R such that f = g a.e. with respect to the Lebesgue measure on (R, M).

Measure and Measurable Functions

3.4.1

75

Measurability using simple measurable functions

From Theorem 3.4.3, we deduce the following theorem. Theorem 3.4.7 Let f and g be extended real valued non-negative measurable functions on a measurable space (X, A). Then f + g is measurable. More generally, if (fn ) is a sequence of P extended real valued non-negative ∞ measurable functions on (X, A), then f := n=1 fn is measurable. Proof. By Theorem 3.4.3, there exist increasing sequences (ϕn ) and (ψn ) of non-negative simple measurable functions on X which converge pointwise to f and g, respectively. Also, for each n ∈ N, ϕn + ψn is a simple measurable function and (ϕn + ψn ) converges pointwise to f + g. Hence, by Theorem 3.3.28, f + g is measurable. Now, let (fn ) be a sequence of extended real valued non-negative measurPk able functions, and for each k ∈ N, let gk := n=1 fn . Then, by thePfirst part, ∞ each gk is measurable, and hence, again, by Theorem 3.3.28, f := n=1 fn is measurable, since it is the pointwise limit of the sequence (gk ) of measurable functions. Recall that, in Theorem 3.3.23, we have proved that if f is real valued, then it is measurable if and only if f + and f − are measurable. We have this result for extended real valued function as well, as a consequence of Theorem 3.4.3. Theorem 3.4.8 Let f be an extended real valued function defined on a measurable space X. Then f is measurable if and only if f + and f − are measurable. Proof. We have already proved in Theorem 3.3.23 that if f is measurable, then both f + and f − are measurable. Conversely, suppose that f + and f − are measurable. By Theorem 3.4.3, there exist increasing sequences (ϕn ) and (ψn ) of non-negative simple measurable functions which converge to f + and f − , respectively. Then, (ϕn −ψn ) converges pointwise to f := f + −f − . Hence, by Theorem 3.3.28, f is measurable. We end this chapter by proving one of the results which we promised at the end of Section 3.1.2.

3.4.2

Incompleteness of Borel σ-algebra

We have seen (see Theorem 3.1.38) that M is the completion of the Borel σ-algebra B1 on R. Now, we show that B1 is not complete with respect to the Lebesgue measure m. For this, first we recall from Example 2.1.13 that if the Cantor ternary set C satisfies m∗ (C) = 0; in particular, C ∈ M. Since C is a countable intersection of finite unions of closed intervals, C is a closed set and C ∈ B.

76

Measure and Integration

Theorem 3.4.9 The Lebesgue measure on the Borel σ-algebra B is not complete. In particular, B is a proper subfamily of M. Proof. Let C be the Cantor ternary set considered in Example 2.1.13. Since C ∈ B1 and m∗ (C) = 0, it is enough to identify a subset of C which is not in B. For this, first we represent each x ∈ [0, 1] in binary expansion as x=

∞ X ϕn (x) , 2n n=1

with ϕn (x) ∈ {0, 1} such that (ϕn (x)) is not an eventually constant sequence. Now, define f : [0, 1] → R by f (0) = 0 and for 0 < x ≤ 1, f (x) =

∞ X 2ϕn (x) . 3n n=1

Note that f is an injective function and its range is contained in the Cantor set C. We observe that, for each n ∈ N, ϕn = χEn , where En is a finite union of subintervals of [0, 1]. Hence, each ϕn is a Lebesgue measurable function. Therefore, by Theorem 3.4.7, f is also a Lebesgue measurable function. Now, let E0 ⊆ [0, 1] be a non-Lebesgue measurable set, and let F0 = f (E0 ). Thus, F0 ⊆ C and m(C) = 0 so that F0 ∈ M. Now, if F0 ∈ B, then by the Lebesgue measurability of f (see Theorem 3.3.10), E0 = f −1 (F0 ) ∈ M; arriving at a contradiction. Thus, we have proved that the subset F0 of C does not belong to B. The particular case follows, since F0 ∈ M \ B.

3.5

Problems

1. Let X be a set and A be a family of subsets of X. Show that A is a σ-algebra if and only if A is non-empty and conditions (b) and (c) in Definition 3.1.1 are satisfied. [Hint: Using (b) and (c), show A ∈ A implies ∅ ∈ A.] 2. Show that the condition (a) in Definition 3.1.3 can be replaced by “∃ A0 ∈ A such that µ(A0 ) < ∞.” 3. Show that an algebra A0 on a set X is a σ-algebra if and only if A0 is closed under countable increasing unions, that is, An ∈ A0 , n ∈ N and An ⊆ An+1 S for all n ∈ N imply n An ∈ A0 . [Hint: For S {An : n ∈SN} ⊆ A0 , construct a disjoint family {Bn : n ∈ N} ⊆ A0 such that n An = n An .]

Measure and Measurable Functions

77

4. Let (X, A, µ) be a measure space. Prove that for every A, B ∈ A, µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B). [Hint: A ∪ B = (A ∩ B) ∪ (A \ A ∩ B) ∪ (B \ A ∩ B), a disjoint union.] 5. Let (X, A, µ) be a measure space. Prove that, if α is a non-negative real number, then the function E 7→ αµ(E) is also a measure on (X, A). 6. If µ1 , . . . , µk are measures on a measurable space (X, P A) and if α1 , . . . , αk are non-negative real numbers, then prove that E 7→ ki=1 αi µi (E) is a measure on (X, A). 7. Let X be an uncountable set and A ⊆ 2X such that A ∈ A if and only if either A or Ac is countable. Define µ : A → [0, ∞] such that  0 if A is countable, µ(A) = ∞ if A is uncountable. Show that A is a σ-algebra and µ is a measure on (X, A). ˙[Hint: For showing S An ∈ A whenever An ∈ A for every n ∈ N, first show n∈N

that A, B ∈ A implies A ∪ B ∈ A.] 8. Let X be a set and X0 be a subset of X. Show that the family of all those sets E such that either E ⊆ X0 or X \ E ⊆ X0 is a σ-algebra. 9. Let X be a set and X be the family of all countable subsets of X. Show that A := {E ⊆ X : E ∈ X or E c ∈ X } is a σ-algebra. 10. Let (X, A, µ) be a measure space and X0 ∈ A. Define µ0 : A → [0, ∞] by µ0 (A) = µ(A ∩ X0 ) for every A ∈ A. Show that µ0 is a measure on X. Deduce that, if x0 ∈ X and {x0 } ∈ A such that µ({x0 }) = 1, then µ0 is the Dirac measure at x0 . 11. Let (X, A, µ) be a measure space. Prove that µ is σ-finite if and only S if there exists a countable disjoint family {An } ⊆ A such that X = n An and µ(An ) < ∞ for every n ∈ N. [Hint: For {An : n ∈ N} ⊆ SA, construct {Bn : n ∈ N} ⊆ A such that S Bn ⊆ Bn+1 for all n ∈ N and n An = n An .] 12. Prove Theorem 3.1.19 first, and deduce Theorem 3.1.18. 13. Let X be a set. Let A, µ∗ and µ be as in Section 3.1.5. Prove the following: (a) A is a σ-algebra on X. (b) µ := µ∗ |A is a measure on A. (c) If A ⊆ X and µ∗ (A) = 0, then A ∈ A. [Hint: Follow the procedure of construction of M and m from m∗ ] 14. The measure µ in Problem 13 is complete - Why? 15. Show that the family of all finite disjoint unions of intervals of the form (a, b] for a, b ∈ R with a < b is an algebra A0 on R, and it satisfies the properties in Theorem 3.1.44. 16. Supply the details of the proof of Theorem 3.1.43.

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Measure and Integration

17. Let A0 be a σ-algebra on a set X and µ0 : A0 → [0, ∞] such that µ0 (∅) = 0. Let µ∗ : 2X → [0, ∞] be the outer measure generated by µ0 , that is, X [ µ∗ (A) := inf{ µ0 (An ) : {An } ⊆ A0 , A ⊆ An }. n

n

Show that, for E ⊆ X, the following are equivalent: (a) µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E c ) for every A ∈ 2X . (b) µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E c ) for every A ∈ A0 . [Hint: Use arguments as in the proof of Theorem 3.1.30.] 18. Let A be a σ-algebra on X and X0 ∈ A. Show that {E ⊆ X0 : E ∈ A} is a σ-algebra, which is the same as AX0 , the restriction of A to X0 . 19. Let S be a family of subsets of a set X and let AS be the σ-algebra generated by S. Show that AAS = AS . 20. Let A be as in Problem 7 or Problem 9. If S is the family of all single subsets of X, then show that A = AS . 21. Prove that the σ-algebra generated by the family of all closed subsets of a topological space Y is the Borel σ-algebra on Y . [Hint: A set E is closed if and only if E c is open.] 22. Show that the Borel σ-algebra on R2 is generated by the family S = {I × R : I open interval} ∪ {R × I : I open interval}. [Hint: Every open rectangle is of the form (I × R) ∩ (R × J) for some open intervals I and J.] 23. Let (An ) and (Bn ) be sequences of subsets of a set X such that there exists k ∈ N with An = Bn for all n ≥ k. Show that lim inf An = lim inf Bn n→∞

[Hint: If A0n = is increasing.]

n→∞

S∞

j=n

and

lim sup An = lim sup Bn . n→∞

Aj and A00n =

T∞

j=n

n→∞

Aj , then (A0n ) is decreasing and (A00n )

24. Let (An ) be a sequence of subsets of a set X and x ∈ X. prove that (a) x ∈ lim sup An ⇐⇒ x ∈ An for infinitely many n ∈ N, n→∞

(b) x ∈ lim inf An ⇐⇒ x ∈ An for all but finitely many n ∈ N. n→∞

25. Let (xn ) be a sequence in R which converges to some a ∈ R. Prove that, if lim {xn } exists, and if a ∈ 6 {xn : n ∈ N}, then lim {xn } 6= {a}. Here, {x} n→∞

n→∞

denotes the singleton set containing x. 26. From Problem 25 deduce that limn→∞ { n1 } does not exist. 27. Let (En ) be a sequence of subsets of a set X and E ⊆ X. Prove that lim En = E ⇐⇒ lim χEn (x) = χE (x)

n→∞

n→∞

∀ x ∈ X.

Measure and Measurable Functions

79

28. Show that there is no E ⊆ R such that lim χ{1/n} (x) = χE (x) for every n→∞

x ∈ R, but lim χ{1/n} (x) = χ{0} (x) for every x ∈ R \ {0}. n→∞

29. For fn : X → R with n ∈ N, show that {x ∈ X : lim fn (x) = 0} = n→∞

∞ [ ∞ \ ∞ \

{x ∈ X : |fn (x)|
0, there exists N ∈ N and E ∈ A such that µ(E) < ε and |fn (x) − f (x)| < ε for every n ≥ N and for every x ∈ X \ E. 48. Let (X, A, µ) be a measure space and (fn ) be a sequence of real measurable functions on X which converges in measure to f . Show that there exists a subsequence (fkn ) of (fn ) such that fkn → f a.e.

Chapter 4 Integral of Positive Measurable Functions

In this chapter we introduce one of the most important concepts in measure theory, namely, the integral of non-negative measurable functions, and prove many important results, including the most celebrated monotone convergence theorem. The definition of the concept of the integral is motivated in a natural manner; first defining the integral of a simple measurable function motivated by the Riemann integral of a step function, and then using the fact that any non-negative measurable function f is a pointwise limit of a monotonically increasing sequence of non-negative simple measurable functions. The method used for the motivation of the definition is replicated in proving the monotone convergence theorem as well.

4.1

Integral of Simple Measurable Functions

Throughout this chapter, (X, A, µ) is a measure space. P n Suppose ϕ : [a, b] → R is a step function given by ϕ = i=1 αi χIi , where Sn {Ii : i = 1, . . . , n} is a disjoint family of intervals such that [a, b] = i=1 Ii . Then we know that the Riemann integral of ϕ is given by Z

b

ϕ(x)dx = a

n X

αi `(Ii ).

i=1

Recall that anyP simple measurable function ϕ on X has the canonical n representation ϕ = i=1 αi χAi , which is akin to a step function. Hence, motivated by the definition of the Riemann integral of a step function, we define the integral of a simple measurable function. Definition 4.1.1 Let ϕ be a non-negative Pn simple measurable function on X with the canonical representation ϕ = i=1 αi χAi (see Theorem 3.4.2). Then R the integral of ϕ over X with respect to µ, denoted by X ϕdµ, is defined 81

82

Measure and Integration

by Z ϕ dµ := X

n X

αi µ(Ai ).

♦

i=1

Note that if A is a measurable set, then χA is a simple measurable function and Z χA dµ = µ(A). X

Remark 4.1.2 Let ϕ be a simple measurable Pn function on a measurable space X with the canonical representation ϕ = i=1 αi χAi . (a) InPDefinition 4.1.1, we assumed that ϕ is non-negative, for otherwise n the sum i=1 αi µ(Ai ) may not be well-defined, since it can happen that for some i 6= j, µ(Ai ) = µ(Aj ) = ∞ and αi = −αj 6= 0 so that we end up in having expression of the form ∞ − ∞. R (b) If ϕ is identically zero, then clearly, X ϕdµ = 0. If ϕ ≥ 0 is not identically zero, and if it takes the value 0, then the term corresponding to this value need not be written in the canonical representation and in the expression for the integral of ϕ. R (c) If ϕ ≥ 0, and αi > 0 and µ(Ai ) = ∞ for some i ∈ {1, . . . , n}, then ϕ dµ = ∞. For example, if the measure space is (R, M, m) and if ϕ = χ[0,∞) , X then Z ϕ dm = µ([0, ∞)) = ∞. ♦ R

In view of the above remark, we have the following definition. Definition 4.1.3 Let ϕ be a non-negativeR simple measurable function on X. Then ϕ is P said to be integrable on X if X ϕ dµ < ∞. ♦ n If ϕ = i=1 αi χAi is the canonical representation of a non-negative simple Pn measurable function ϕ, then for E ∈ A, χE ϕ = i=1 αi χAi ∩E is the canonical representation of χE ϕ (see Problem 44 in Chapter 3) so that Z χE ϕ dµ = X

n X

αi µ(Ai ∩ E).

i=1

This observation motivates the following definition. Definition 4.1.4 If ϕ is a non-negative simple measurable function on X and if E ∈ A, then we define Z Z ϕ dµ = χE ϕ dµ, E

X

and it is called the integral of ϕ over E with respect to µ.

♦

Integral of Positive Measurable Functions

83

Example 4.1.5 Let X = [a, b], A = M[a,b] , the Lebesgue σ-algebra restricted to [a, b], and let µ be the Lebesgue measure. Let A = [a, b] ∩ Q and B = [a, b] ∩ Qc . Then we have Z Z χA dµ = µ(A) = 0 and χB dµ = µ(B) = b − a. X

X

Recall that the functions χA and χB are not Riemann integrable.

♦

Exercise 4.1.6 Let E ∈ A. Let µE be the measure induced by µ on (E, AE ), where AE is the restriction of the σ-algebra R A to E.RProve that, for any nonnegative simple measurable function ϕ, E ϕdµ = E ϕ|E dµE , where ϕ|E is the restriction of ϕ to E. ♦ Convention: In the following, when we speak about measurable functions, it is meant that they are defined on a measurable space (X, A), and when we speak about integrals, they are with respect to a measure µ on (X, A). In due course, we shall use the following properties of the characteristic functions: (1) For disjoint sets A and B, χA∪B = χA + χB . (2) For any sets A and B, χA∩B = χA χB . (3) If A ⊆ B, then χB\A = χB − χA . (4) For any sets A and B, χA∪B = χA + χB − χA χB . In the above (1) and (2) can be seen easily; (3) follows from (1) by observing that B = A ∪ (B \ A), and (4) is a consequence of (1) and (2), since A ∪ B = (A \ A ∩ B) ∪ (A ∩ B) ∪ (B \ A ∩ B). Recall that a function ϕ : X → R is a simple measurable function Pn if and only if there are βi ∈ R and Bi ∈ A for i = 1, . . . , n such that ϕ = i=1 βi χBi . So, it is natural to ask whether Z ϕ dµ = X

k X

βi µ(Bi )

i=1

is true. The answer is in the affirmative. Towards proving this, we first prove the following lemma. Lemma 4.1.7 Let {B1 , . . . , Bk } be a disjoint family of measurable sets and let β1 , . . . , βk be non-negative real numbers (not necessarily distinct). Let ψ := Pk i=1 βi χBi . Then Z k X ψ dµ = βi µ(Bi ). X

i=1

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Measure and Integration

Proof. In case β1 , . . . , βk are distinct, the given representation of ψ is its canonical representation and hence the proof follows from the definition. Next assume that some of β1 , . . . , βk are repeated. Suppose α1 , . . . , αn are the distinct numbers among β1 , . . . , βk . For each i ∈ {1, . . . , n}, let ∆i := {j ∈ {1, . . . , k} : βj = αi }. Then it is clear that ψ=

k X

βj χBj =

j=1

n  X X i=1

βj χBj



=

n X i=1

j∈∆i

X

αi

χBj =

n X

αi χAi ,

i=1

j∈∆i

S where Ai = j∈∆i Bj . Since {B1 , . . . , Bk } is aPdisjoint family, {A1 , . . . , An } is also a disjoint family, and hence, µ(Ai ) = j∈∆i µ(Bj ) for i = 1, . . . , n. Pn Therefore, ψ = i=1 αi χAi is the canonical representation of ψ, and hence, Z ψ dµ = X

n X

αi µ(Ai ) =

i=1

=

n X

αi

i=1

n X X

βj µ(Bj ) =

i=1 j∈∆i

X

µ(Bj )

j∈∆i k X

βi µ(Bi ).

i=1

This completes the proof. Example 4.1.8 Let X = [a, b], A = M[a,b] , the Lebesgue σ-algebra restricted to [a, b], and let µ be the Lebesgue measure. Let ϕ be a non-negative step R function. By Lemma 4.1.7, X ϕ dµ is the Riemann integral of ϕ. ♦ Example 4.1.9 Let X be a finite set, say X = {x1 , . . . , xk }, A = 2X and µ be the counting measure. Then every function f : X → [0, ∞) can be represented as k X f= f (xi )χ{xi } . i=1

Then, by Lemma 4.1.7,

R X

f dµ =

Pk

i=1

f (xi ).

♦

Example 4.1.10 Using the result in Example 4.1.9, every finite sum of nonnegative real numbers can be represented as an integral. This can be seen as follows: Let a1 , . . . , an be in [0, ∞). Let X = {1, . . . , n}, A = 2X and µ be the counting measure on X. Define f : X → R by f (j) = aj ,

j ∈ {1, . . . , n}.

Then f is a simple function and it has the representation f=

n X j=1

aj χ{j} .

Integral of Positive Measurable Functions 85 R Pn Hence, by Lemma 4.1.7, X f dµ = j=1 aj . We shall see, in due course, that every countable sum of non-negative real numbers can also be represented as an integral. ♦ Remark 4.1.11 Let X be a countably infinite set, Pnsay X = {x1 , x2 , . . .}. Consider a function f : X → [0, ∞) and fn := i=1 f (xi )χ{xi } for each n ∈ N. Then the following can be verified easily (Exercise): (i) fn (x) → f (x) for each x ∈ X. (ii) If µ is the counting measure on X, then Z fn dµ → X

∞ X

as n → ∞.

f (xi )

i=1

More generally, we have the following: Let (X, A, µ) be a measure space, {En : n ∈ N} be a disjoint family of measurable sets, and (αn ) be a sequence of non-negative real numbers. Then the function ∞ X (∗) αn χEn f := n=1

is wellPdefined taking values in {0, α1 , α2 , . . .}. As in the previous case, if n fn := i=1 αi χEi , then f (x) = lim fn (x), n→∞

and Z fn dµ = X

n X

αi µ(Ei ) →

∞ X

x ∈ X,

αi µ(Ei )

as n → ∞.

i=1

i=1

R We shall formally define integral R X f dµ ofP a general non-negative mea∞ surable function f on X and obtain X f dµ = n=1 αn µ(En ) whenever f is as in (∗). ♦ In Lemma 4.1.7, we assumed that {B1 , . . . , Bk } is a disjoint family. We shall drop this assumption by making use of the following result. Theorem 4.1.12 Let ϕ and ψ be non-negative simple measurable functions on X and c be a non-negative real number. Then Z Z Z (ϕ + ψ) dµ = ϕ dµ + ψ dµ, X

X

Z

X

Z c ϕ dµ = c

X

ϕ dµ. X

Pn Pm Proof. Let ϕ := i=1 αi χAi and ψ := j=1 βj χBj be the canonical representations of ϕ and ψ, respectively. Let Eij = Ai ∩ Bj for i ∈ {1, . . . , n} and j ∈ {1, . . . , m}.

86

Measure and Integration

Since

Sn

i=1

Ai = X =

Sm

Bj , we have

j=1

Ai = Ai ∩

m [

m  [ Bj = (Ai ∩ Bj ),

j=1

Bj = Bj ∩

n [

i ∈ {1, . . . , n},

j=1 n  [ Ai = (Ai ∩ Bj ),

i=1

j ∈ {1, . . . , m}.

i=1

Thus, ϕ=

n X

n X

α i χAi =

i=1

ψ=

m X

m X

βj χBj =

j=1

ϕ+ψ =

αi

i=1

m X

χAi ∩Bj



=

j=1

βj

j=1

n X

n X m X

αi χAi ∩Bj ,

i=1 j=1

χAi ∩Bj



=

i=1

m X n X

βj χAi ∩Bj ,

j=1 i=1

n X m X (αi + βj )χAi ∩Bj ,

cϕ =

n X m X

cαi χAi ∩Bj .

i=1 j=1

i=1 j=1

Now, since {Ai ∩ Bj : i = 1, . . . , n; j = 1, . . . , m} is a disjoint family of measurable sets, by Lemma 4.1.7, we have Z (ϕ + ψ)dµ = X

m n X X (αi + βj )µ(Ai ∩ Bj ) i=1 j=1

=

=

=

n X m X

αi µ(Ai ∩ Bj ) +

i=1 j=1 n m X X

αi

i=1 n X

µ(Ai ∩ Bj ) +

j=1

αi µ(Ai ) +

and Z cϕ dµ = X

n X m X

µ(Ai ∩ Bj )

βj

i=1

βj µ(Bj )

Z ϕdµ +

ψdµ, X

cαi µ(Ai ∩ Bj ) = c

i=1 j=1

m X

∩ Bj )

j=1

Z X

βj µ(Ai i=1 j=1 m n X X j=1

i=1

=

n X m X

n X m X

Z αi µ(Ai ∩ Bj ) = c

i=1 j=1

Thus, the proof is complete. From the above theorem, we deduce a few corollaries.

ϕ dµ. X

Integral of Positive Measurable Functions

87

Corollary 4.1.13 If ϕ1 , . . . , ϕn are simple non-negative measurable functions on X, then Z X n n Z  X ϕi dµ = ϕi dµ. X

i=1

i=1

X

Proof. This is immediate from Theorem 4.1.12. The next corollary is a generalization of Lemma 4.1.7, and it is the result that answers affirmatively the question raised prior to the statement of Lemma 4.1.7. Corollary 4.1.14 Let B1 , . . . , Bk be measurable sets and β1 , . . . , βk be nonnegative real numbers. Then Z X k X

k  X βi χBi dµ = βi µ(Bi ).

i=1

i=1

Proof. Since ϕi := βi χBi is a simple non-negative measurable function for each i = 1, . . . , k, the result follows from Corollary 4.1.13. Corollary 4.1.15 Let ϕ and ψ be non-negative simple measurable functions on X such that ϕ ≤ ψ. Then Z Z ϕdµ ≤ ψdµ. X

Further, if

R X

X

ϕ dµ < ∞, then Z Z Z (ψ − ϕ)dµ = ψdµ − ϕdµ. X

X

X

Proof. We write ψ = ϕ + (ψ − ϕ) and observe that ψ − ϕ is a non-negative simple measurable function. Hence, by Theorem 4.1.12, Z Z Z ψ dµ = ϕ dµ + (ψ − ϕ) dµ. (∗) X

R

X

X

R

R Since XR(ψ − ϕ) dµ ≥ 0, weR have X ϕdµ ≤R X ψdµ. From (∗), it also follows R that, if X ϕ dµ < ∞, then X (ψ − ϕ)dµ = X ψdµ − X ϕdµ. Corollary 4.1.16 Let (ϕn ) be a sequence of non-negative simple measurable functions on R  X. Suppose 0 ≤ ϕn ≤ ϕn+1 for all n ∈ N. Then the sequence ϕ dµ converges in [0, ∞]. X n Proof. In view of Corollary 4.1.15, the sequence cally increasing, and hence it converges in [0, ∞].

R

 ϕ dµ is monotonin X

88

Measure and Integration

Next theorem shows that each non-negative simple measurable function is associated with a measure on (X, A) in a natural way. Corollary 4.1.17 Let ϕ be a non-negative simple measurable function on X. Then Z ν(E) := ϕ dµ, E ∈ A, E

defines a measure on A. Proof. Clearly, ν(∅) = 0. Let {En } be a disjoint P family of measurable sets, S and let E = n En . We have to show that ν(E) = n ν(En ). Pk Let ϕ = i=1 αi χAi for some non-negative reals α1 , . . . , αk and measurable sets A1 , . . . , Ak . Then χE ϕ =

k X

αi χE χAi =

i=1

k X

αi χE∩Ai .

i=1

Hence, by Lemma 4.1.7, Z

Z ϕ dµ =

E

χE ϕ dµ = X

k X

αi µ(E ∩ Ai ).

(1)

i=1

Note that, for each i = 1, . . . , k, h [  i [  X En ∩ Ai = µ µ(E ∩ Ai ) = µ (En ∩ Ai ) = µ(En ∩ Ai ), n

n

(2)

n

since {En ∩ Ai : n ∈ N} is a disjoint family in A. Now, (1) and (2) imply that Z ν(E)

=

ϕdµ = E

=

i=1

XZ n

k X X

En

ϕdµ =

αi µ(En ∩ Ai ) =

n

X

k XX n

αi µ(En ∩ Ai )

i=1

ν(En ).

n

This completes the proof that ν is a measure. Definition 4.1.18 The measure ν defined in Corollary 4.1.17 is called the measure induced by ϕ and µ. ♦

4.2

Integral of Positive Measurable Functions

Let f : X → [0, ∞] be a measurable function. By Theorem 3.4.3, there exists an increasing sequence (ϕn ) of non-negative simple measurable functions

Integral of Positive Measurable Functions

89

which converges to f pointwise. In view of Corollary 4.1.16, we may define Z Z f dµ := lim ϕn dµ. n→∞

X

X

For this definition to be meaningful, it is necessary to prove that the above limit is independent of the sequence (ϕn ), as long as it is increasing and converging pointwise to f . That is, we have to show that, if (ϕn ) and (ψn ) are increasing sequences of non-negative simple measurable functions converging to f pointwise, then Z Z ψn dµ. ϕn dµ = lim lim n→∞

X

n→∞

X

This is a consequence of the following theorem. Theorem 4.2.1 Let f : X → [0, ∞] be a measurable function and (ϕn ) be an increasing sequence of non-negative simple measurable functions on X which converges to f pointwise. Then Z Z lim ϕn dµ = sup ϕdµ, n→∞

X

ϕ∈Sf

X

where Sf is the set of all non-negative simple measurable functions ϕ satisfying ϕ ≤ f. Proof. By the definition of Sf , we have ϕn ∈ Sf for all n ∈ N. ThereZ Z R fore, ϕn dµ ≤ sup ϕdµ for all n ∈ N. By Corollary 4.1.16, ( X ϕn dµ) X

ϕ∈Sf

X

converges in [0, ∞]. Hence, Z

Z ϕn dµ ≤ sup

lim

n→∞

X

ϕ∈Sf

ϕdµ. X

To show the other way inequality, let ϕ ∈ Sf . We are going to show that Z Z r ϕdµ ≤ lim ϕn dµ (1) X

n→∞

X

for every r ∈ (0, 1), so that by letting r → 1, the result will follow. Now, to prove (1), let r ∈ (0, 1) and for each n ∈ N, let En = {x ∈ X : rϕ(x) ≤ ϕn (x)}. S∞ Clearly, En ∈ A and En ⊆ En+1 for every n ∈ N. Further, X = n=1 En . To see this, let x ∈ X. If ϕ(x) = 0, then x ∈ En for all n ∈ N. Next, suppose ϕ(x) > 0. Then r ϕ(x) < ϕ(x) ≤ f (x). Since ϕn (x) → f (x), there exists k ∈ N such that r ϕ(x) ≤ ϕk (x) ≤ f (x). Hence, x ∈ Ek .

90

Measure and Integration Now, let ν be the measure induced by ϕ, that is, Z ν(E) = ϕdµ, E ∈ A. E

Since rϕ ≤ ϕn on En , Z

Z rϕdµ ≤

r ν(En ) = En

Z ϕn dµ ≤

En

ϕn dµ.

(2)

X

By Theorem 3.2.1, lim ν(En ) = ν(X). Hence, (2) leads to n→∞

Z

Z ϕdµ = r ν(X) = r lim ν(En ) ≤ lim

r

n→∞

X

n→∞

ϕn dµ. X

Thus, we have proved (1), and the proof is complete. Notation: Given an extended real valued non-negative measurable function f , we shall use the notation Sf to denote the set of all non-negative simple measurable functions ϕ satisfying ϕ ≤ f . Motivated by Theorem 4.2.1, we define the integral of a non-negative measurable function as follows. Definition 4.2.2 Let f : X → [0, ∞] be a measurable function. Then the integral of f over X is defined as Z Z f dµ := sup ϕ dµ. ϕ∈Sf

X

X

If E ∈ A, then the integral of f over E is defined as Z Z f dµ := χE f dµ. E

♦

X

Definition 4.2.3 A measurable function f : X → [0, ∞] is said to be integrable if Z f dµ < ∞. ♦ X

Example 4.2.4 Let (an ) be a sequence of non-negative real numbers. Let X = N, A = 2N and µ be the counting measure on N. Define f : X → R by f (j) = aj , j ∈ N. P∞ Then f can be represented as f = j=1 aj χ{j} . Note that f = lim

n→∞

n X j=1

aj χ{j} = lim ϕn , n→∞

Integral of Positive Measurable Functions 91 Pn where ϕn = j=1 aj χ{j} . Clearly, (ϕn ) is an increasing sequence of nonnegative simple measurable functions which converges to f pointwise. Hence, by Theorem 4.2.1, we have Z Z n ∞ X X f dµ = lim ϕn dµ = lim aj = aj . n→∞

X

n→∞

X

j=1

j=1

Thus, every countable sum of non-negative real numbers can be represented as an integral. ♦ Theorem 4.2.5 Let f and g be extended real valued non-negative measurable functions on X such that f ≤ g. Then Z Z f≤ g. X

X

Proof. Since f ≤ g, we have Sf ⊆ Sg . Hence, Z Z Z Z f = sup ϕ ≤ sup ϕ= g, X

ϕ∈Sf

X

ϕ∈Sg

X

X

completing the proof. Theorem 4.2.6 Let f and g be extended real valued non-negative measurable functions on X and c ≥ 0. Then f + g and cf are measurable functions, and Z Z Z (f + g)dµ = f dµ + g dµ, X X X Z Z cf = c f. X

X

Proof. We have already proved (see Theorem 3.4.7) that f + g and cf are measurable functions. In fact, by Theorem 3.4.3, there exist increasing sequences (ϕn ) and (ψn ) of non-negative simple measurable functions on X which converge pointwise to f and g, respectively, so that (ϕn + ψn ) and (cϕn ) are increasing sequences of non-negative simple measurable functions which converge pointwise to f + g and c f , respectively. Hence, by Theorem 3.3.28, f + g and c f are measurable functions. Therefore, by Theorem 4.2.1 and Theorem 4.1.12, Z Z (f + g)dµ = lim (ϕn + ψn )dµ n→∞ X X Z  Z = lim ϕn dµ + ψn dµ n→∞ X Z X Z = lim ϕn dµ + lim ψn dµ n→∞ X n→∞ X Z Z = f dµ + g dµ, X

X

92

Measure and Integration

and

Z

Z

Z

c f dµ = lim

cϕn dµ = lim c

n→∞

X

n→∞

X

Z ϕn dµ = c

X

f dµ. X

This completes the proof. Corollary 4.2.7 Let f and g be extended Rreal valued non-negative measurable functions on X such that f ≤ g on X. If X f dµ < ∞, then Z Z Z (g − f )dµ = g dµ − f dµ. X

X

X

R

Proof. Suppose X f dµ < ∞. Since g = f + (g − f ), where both f and g − f are non-negative, by Theorem 4.2.6, we have Z Z Z gdµ = f dµ + (g − f ) dµ. X

Now, since

R X

X

f dµ < ∞, we obtain

X

R X

(g − f ) dµ =

R X

gdµ −

R X

f dµ.

Corollary 4.2.8 Let f be an extended real valued non-negative measurable function on X. Let (ϕn ) be a decreasing sequence of non-negative simple meaR surable functions which converges pointwise to f and X ϕ1 dµ < ∞. Then Z Z f dµ = lim ϕn dµ. X

n→∞

X

Proof. By the hypothesis, (ϕ1 − ϕn ) is an increasing sequence of nonnegative simple measurable functions which converges pointwise to ϕ1 − f . Note that ϕ1 − f ≥ 0. Hence, by Theorem 4.2.1 (taking ϕ1 − ϕn and ϕ1 − f in place of ϕn and f , respectively), Z Z (ϕ1 − ϕn )dµ. (ϕ1 − f )dµ = lim n→∞

X

X

R

Since f ≤ ϕn ≤ ϕ1 for every n ∈ N and X ϕ1 dµ < ∞, by Theorem 4.2.5 and Corollary 4.2.7, we have Z Z Z ϕ1 dµ − f dµ = (ϕ1 − f )dµ X X X Z = lim (ϕ1 − ϕn )dµ n→∞ X Z Z = ϕ1 dµ − lim ϕn dµ. X

Thus,

R X

f dµ = limn→∞

R X

ϕn dµ.

n→∞

X

Integral of Positive Measurable Functions

93

Notation: OnceR the measure µ under is understood, we may R R R consideration R write E f dµ as E f and X f dµ as X f or f . In the following theorem, we list some of the properties of the integral. Theorem 4.2.9 Let f and g be extended real valued non-negative measurable functions defined on X, c ∈ [0, ∞) and E ∈ A. Then we have the following. R R (i) f ≤ g ⇒ E f ≤ E g. R R (ii) A, B ∈ A, A ⊆ B ⇒ A f ≤ B f . R R R (iii) A, B ∈ A, A ∩ B = ∅ ⇒ A∪B f = A f + B f . R R (iv) E c f = c E f . R (v) f (x) = 0 ∀ x ∈ E ⇒ E f = 0. R (vi) µ(E) = 0 ⇒ E f = 0. Z Z (vii) f = sup ϕ. E

ϕ∈Sf

E

Proof. (i) Suppose f ≤ g. Then χE f ≤ χE g so that, by Theorem 4.2.5, Z Z Z Z f= χE f ≤ χE g = g. E

X

X

E

(ii) Let A, B ∈ A be such that A ⊆ B. Then, χA f ≤ χB f . Hence, (i) implies that Z Z Z Z f= χA f ≤ χB f = f. A

X

X

B

(iii) Let A, B ∈ A be such that A ∩ B = ∅. Then χA∪B = χA + χB so that, by Theorem 4.2.6, Z Z Z   f = χA∪B f = χA + χB f A∪B ZX Z X Z Z = χA f + χB f = f+ f. X

X

A

B

(iv) This follows from Theorem 4.2.6 by taking χE fRin place R of f . (v) If f (x) = 0 for all x ∈ E, then χE f = 0. Hence, E f = X χE f = 0. (vi) Suppose µ(E) = 0 and ϕ ∈ SχE f . Then we have Z Z ϕ = χE ϕ and ϕ= χE ϕ = 0. X

X

Hence, Z

Z f=

E

Z χE f =

X

sup ϕ∈Sχ

E

ϕ = 0. f

X

94

Measure and Integration

R R (vii) Let ϕ R∈ Sf . Then, χE ϕ ≤ χE f so that by (i), X χE ϕ ≤ X χE f . R Thus, supϕ∈Sf E ϕ ≤ E f . To show the reverse inequality, let ϕ ∈ SχE f . Then ϕ ∈ Sf and ϕ = χE ϕ. Hence, Z Z Z Z f= χE f = sup ϕ = sup χE ϕ. E

ϕ∈Sχ

X

E

f

ϕ∈Sχ

X

E

f

X

Also, we have SχE f ⊆ Sf . Hence, Z

Z f=

E

ϕ∈Sχ

Z χE ϕ ≤ sup

sup E

ϕ∈Sf

X

f

Z χE ϕ = sup

X

ϕ∈Sf

ϕ. E

This completes the proof. Theorem 4.2.10 Let f : X → [0, ∞] be a measurable function on X. Then Z f = 0 ⇐⇒ f = 0 a.e. on X. X

Proof. Let A = {x ∈ X : f (x) 6= 0} and B = {x ∈ X : f (x) = 0}. Then A ∩ B = ∅ and A ∪ B = X. Suppose f = 0 a.e. Then, µ(A) = 0. Hence, by parts (iii), (v), and (vi) in Theorem 4.2.9, we have Z Z Z f= f+ f = 0. X

Conversely, suppose that A=

∞ [ n=1

An

R X

A

B

f = 0. Observe that

n 1o where An := x ∈ X : f (x) ≥ . n

We prove that µ(An ) = 0 for every n ∈ N so that µ(A) = 0. Note that χAn f ≥ n1 χAn for every n ∈ N. Hence, by Theorem 4.2.9 (i), we have Z Z Z 1 0= f≥ f= χAn f ≥ µ(An ) ∀ n ∈ N. n X An X Therefore, µ(An ) = 0 for every n ∈ N, and thus the proof is over. Theorem 4.2.11 Suppose f : X → [0, ∞] is integrable. Then (i) µ({x ∈ X : f (x) ≥ n}) → 0 as n → ∞, (ii) µ({x ∈ X : f (x) = ∞}) = 0. In particular f takes values in [0, ∞) a.e.

Integral of Positive Measurable Functions

95

Proof. Let An := {x : f (x) ≥ n}, n ∈ N. By Theorem 4.2.9 (i), we have Z Z Z Z f≥ f= f χ An ≥ nχAn = nµ(An ) X

An

X

X

for all n ∈ N. Hence, 1 µ(An ) ≤ n

Z f →0

as n → ∞.

X

This proves (i). To see (ii), note that A := {x ∈ X : f (x) = ∞} =

∞ \

An .

n=1

Since µ(A) ≤ µ(An ) for all n ∈ N and µ(An ) → 0 as n → ∞, we obtain µ(A) = 0. Thus, (ii) is also proved. This, in particular, shows that f takes values in [0, ∞) a.e. Remark 4.2.12 For an integrable non-negative measurable function f , not only that µ(An ) → 0, where An := {x ∈ X : f (x) ≥ n} as in Theorem 4.2.11 (i), we also have nµ(En ) → 0 (Problem 9). Another stronger version of Theorem 4.2.11 (i) when µ is a finite measure is Theorem 4.2.21. ♦

4.2.1

Riemann integral as Lebesgue integral

We know that if ϕ is a non-negative step function on [a, b], then it is R Lebesgue measurable on [a, b], and X ϕ dµ is the Riemann integral of ϕ (see Example 4.1.8), that is, Z

b

Z ϕ dµ =

X

ϕ(x)dx. a

Is it true for any non-negative Riemann integrable function on [a, b]? The answer is in the affirmative. Theorem 4.2.13 If f : [a, b] → [0, ∞) is a Riemann integrable function, then f is measurable with respect to the σ-algebra M[a,b] and Z

b

Z f (x)dx =

a

f dm. [a,b]

Proof. Suppose f : [a, b] → [0, ∞) is a Riemann integrable function. Let (Pn ) be a partition of [a, b] such that Pn+1 is a refinement of Pn and |Pn | → 0 as n → ∞. Then we know that Z b Z b L(Pn , f ) → f (x)dx U (Pn , f ) → f (x)dx a

a

96

Measure and Integration

as n → ∞, where L(Pn , f ) and U (Pn , f ) denote the lower sum and upper sum corresponding to the partition Pn . Note that if (n)

(n)

Pn : a = x0

(n)

(n)

Ii

(n)

= inf f (x),

mi

(n)

< · · · < xkn = b,

< x1

(n)

= sup f (x),

Mi

(n)

(n)

= [xi−1 , xi ),

(n)

x∈Ii

x∈Ii

for i = 1, . . . , kn , then L(Pn , f ) :=

kn X

(n)

(n)

mi (xi

(n)

− xi−1 ),

U (Pn , f ) :=

i=1

kn X

(n)

(n)

Mi (xi

(n)

− xi−1 ).

i=1

Thus, taking ϕn =

kn X

(n)

mi χ

I

i=1

, (n)

kn X

ψn =

(n)

Mi χ

I

i

i=1

we have Z

,

b

Z ϕn = L(Pn , f ) →

[a,b]

(n) i

f (x)dx

(1)

f (x)dx.

(2)

a

and Z

Z

b

ψn = U (Pn , f ) → [a,b]

a

as n → ∞. Since Pn+1 is a refinement of Pn , we have ϕn ≤ ϕn+1 ≤ f ≤ ψn+1 ≤ ψn

∀ n ∈ N.

Thus, (ϕn ) and (ψn ) converge pointwise, say to ϕ and ψ, respectively. Then, ψ − ϕ ≥ 0 and, by Theorem 3.3.28, ϕ and ψ are measurable, and for every n ∈ N, ϕn ≤ ϕn+1 ≤ ϕ ≤ f ≤ ψ ≤ ψn+1 ≤ ψn ∀ n ∈ N. R  R  This implies that ϕ is an increasing sequence, ψ is a den n [a,b] [a,b] creasing sequence, and for every n ∈ N, Z Z Z Z ϕn ≤ ϕ≤ ψ≤ ψn . (3) [a,b]

[a,b]

[a,b]

[a,b]

Further, by Theorem 4.2.1 and Corollary 4.2.8, Z Z Z lim ϕn = ϕ, lim n→∞

[a,b]

Now, by (1) and (2), Z Z (ψn − ϕn ) = [a,b]

n→∞

[a,b]

[a,b]

Z ψn − [a,b]

Z ψn =

[a,b]

 ϕn → 0

ψ. [a,b]

as n → ∞.

(4)

Integral of Positive Measurable Functions 97 R Therefore, by (3), [a,b] (ψ − ϕ)dm = 0. Hence, by Theorem 4.2.10, ϕ = ψ a.e. As ϕ ≤ f ≤ ψ, we obtain f = ϕ a.e. (see Problem 41 in Chapter 3). Since the Lebesgue measure is complete, f is measurable (by Theorem 3.3.35), and by (1) and (4), we have Z Z Z Z b f (x)dx = lim ϕn dm = ϕdm = f dm. n→∞

a

[a,b]

[a,b]

[a,b]

This completes the proof.

4.2.2

Monotone convergence theorem (MCT)

Suppose (fn ) is a sequence of extended real valued non-negative measurable functions on X such that fn → f pointwise for some measurable R R function f . A natural question is whether we have the convergence X fn → X f . The answer is: not necessarily. To see this let us consider an example. Example 4.2.14 Consider the measure space (X, A, µ) = ([0, 1], M[0,1] , m) and fn := n χ(0, 1 ] , n ∈ N. n R Then, we have lim fn (x) = 0 for every x ∈ X. Note that X fn = 1 for n→∞

all n ∈ N. Thus, fn ≥ 0 on [0, 1] for all nR ∈ N, and R taking f (x) = 0 for all x ∈ [0, 1], we have fn → f pointwise, but X fn 6→ X f . ♦ However, if (fn ) is monotonically increasing, which converges to f pointR R wise, then we do have the convergence X fn → X f . This is the celebrated monotone convergence theorem (MCT). Theorem 4.2.15 (MCT) Let (fn ) be a sequence of extended real valued non-negative measurable functions defined on X such that (i) fn ≤ fn+1 for all n ∈ N and (ii) (fn ) converges pointwise on X. Let f (x) = lim fn (x), x ∈ X. Then f is measurable and n→∞

Z

Z fn →

X

f

n → ∞.

as

X

Proof. Since fn (x) → f (x) for every x ∈ X, by Theorem 3.3.28, the function f : X → [0, ∞] is measurable. Further, we have fn ≤ fn+1 ≤ f

∀ n ∈ N,

and by Theorem 4.2.9 (i), Z Z Z fn ≤ fn+1 ≤ f X

X

X

∀ n ∈ N.

98

Measure and Integration R

Hence, limn→∞

X

fn exists and Z

Z fn ≤

lim

n→∞

f.

X

X

Following the same lines of arguments as in the proof of Theorem 4.2.1 with the sequence (fn ) in place of (ϕn ) we obtain Z Z Z f := sup ϕdµ ≤ lim fn . ϕ∈Sf

X

n→∞

X

X

For the sake of completion, let us imitate the arguments here briefly: Let ϕ ∈ Sf , and for 0 < r < 1 and n ∈ N, let En = {x ∈ X : rϕ(x) ≤ fn (x)}. S∞ Then (En ) is an increasing sequence in A and X = n=1 En . Consider the measure induced by ϕ, that is, Z ν(E) = ϕdµ, E ∈ A. E

Then we have Z

Z

Z

rϕdµ ≤

r ν(En ) = En

fn dµ ≤ En

fn dµ. X

Taking limit, Z

Z ϕdµ = r ν(X) = lim r ν(En ) ≤ lim

r

n→∞

X

n→∞

fn dµ. X

Since this is true for every r ∈ (0, 1) and for every ϕ ∈ Sf , Z Z Z f := sup ϕdµ ≤ lim fn dµ. X

ϕ∈Sf

n→∞

X

X

This completes the proof of the theorem. Remark 4.2.16 Traditionally, MCT (Theorem 4.2.15) is proved first and then the result in Theorem 4.2.1 is observed. We followed the reverse path as it is felt that Theorem 4.2.1 is a good motivation for defining the integral of non-negative measurable functions. ♦ Corollary 4.2.17 Let (fn ) be a sequence P∞ of extended real valued non-negative measurable functions on X and f := n=1 fn . Then for every E ∈ A, Z f= E

∞ Z X n=1

E

fn .

Integral of Positive Measurable Functions 99 Pn Proof. Let gn = i=1 fi for n ∈ N. Then 0 ≤ gn ≤ gn+1 for every n ∈ N and gn → f pointwise on X. Hence, by MCT (Theorem 4.2.15) and Theorem 4.2.6, Z Z n Z ∞ Z X X f = lim gn = lim fi = fi . n→∞

X

n→∞

X

X

i=1

X

i=1

This completes the proof. Corollary 4.2.18 Let f : X → [0, ∞] be a measurable function, and let S∞ {En : n ∈ N} be a disjoint family in A. Let E = n=1 En . Then Z f dµ = E

∞ Z X

f dµ.

En

n=1

Proof. We observe that χE f = χS∞

i=1

f = lim χSn Ei n→∞

i=1

f = lim Ei

n X

n→∞

χEi f,

i=1

Pn where ( i=1 χEi f ) is an increasing sequence of non-negative measurable functions. Hence, by MCT (Theorem 4.2.15), Z

Z X n

Z f dµ =

χE f dµ = lim

E

n→∞

X

=

n Z X

lim

n→∞

i=1

X

χEi f =

X i=1 ∞ Z X i=1

χEi f

X

χEi f =

∞ Z X i=1

f.

Ei

This completes the proof. Corollary 4.2.19 Let f : X → [0, ∞] be a measurable function and let Z ν(E) := f dµ, E ∈ A. E

Then ν is a measure on X. Further, if g : X → [0, ∞] is any measurable function, then Z Z gdν = gf dµ. (∗) X

X

Proof. Clearly, ν(∅) = 0. Let {En : n ∈ N} be a disjoint family in A. Then, by Corollary 4.2.18, we have ν

∞ [



Z

Ei =

i=1

Thus, ν is a measure.

f dµ = S

i

Ei

∞ Z X i=1

Ei

f dµ =

∞ X i=1

ν(Ei ).

100

Measure and Integration

Next, we observe that the relation (∗) in the theorem holds if g is a characteristic function of a measurable set. Indeed, if g = χE for some measurable set E, then Z Z gdν =

χE dν = ν(E)

X

and

X

Z

Z gf dµ =

X

Z χE f dµ =

X

f dµ E

so that, by the definition of ν, we obtain (∗). Now, using the property of the integral, (∗) holds for all simple non-negative measurable functions as well. Since any measurable function g : X → [0, ∞] is a pointwise limit of an increasing sequence of simple non-negative measurable functions, the proof of (∗) can be completed by invoking Theorem 4.2.1 or MCT (Theorem 4.2.15). Definition 4.2.20 The measure ν defined in Corollary 4.2.19 is called the measure induced by f and µ. ♦ Here is another application of MCT. Theorem 4.2.21 Let f : X → [0, ∞] be measurable and µ be a finite measure. Let An := {x ∈ X : f (x) ≥ n} for n ∈ N. Then Z f dµ < ∞ ⇐⇒ X

∞ X

µ(An ) < ∞.

n=1

Proof. Let Bn = {x ∈ X : n < f (x) ≤Sn + 1} for n ∈ N0 := N ∪ {0}. Then ∞ we see that Bn = An \ An+1 and X = n=0 Bn , where {Bn : n ∈ N0 } is a disjoint family in A and A0 = X. Note that Z nµ(Bn ) ≤ f dµ ≤ (n + 1)µ(Bn+1 ∀ n ∈ N0 . Bn

Hence, by Corollary 4.2.18, ∞ X

Z nµ(Bn ) ≤

f dµ ≤ X

n=0

∞ X

(n + 1)µ(Bn+1 .

n=0

R P∞ Thus, X f dµ < ∞ if and only if n=0 nµ(Bn ) < ∞. Since X is of finite measure, µ(Bn ) = µ(An ) − µ(An+1 ) for all n ∈ N0 , so that ∞ X

nµ(Bn =

n=1

This completes the proof.

∞ X n=1

n[µ(An ) − µ(An+1 ) =

∞ X n=1

µ(An ).

Integral of Positive Measurable Functions

101

In MCT we assumed that the sequence (fn ) of non-negative measurable functions is monotonically increasing. What can we say, if this condition is dropped? Corollary 4.2.22 (Fatou’s lemma) Let (fn ) be a sequence of extended real valued non-negative measurable functions on a measure space (X, A, µ). Then Z Z (lim inf fn ) ≤ lim inf fn . n

X

n

X

Proof. For each k ∈ N, let gk = inf fn . Then gk ≤ gk+1 for all k ∈ N, and n≥k

lim gk = f := lim inf fn . Hence, by MCT (Theorem 4.2.15), n

k→∞

Z

Z gk =

lim

k→∞

X

f. X

R R But, since gk ≤ fk for all k ∈ N, we have X gk ≤ X fk so that Z Z Z Z f = lim gk = lim inf gk ≤ lim inf fk . k→∞

X

k

X

X

k

X

This completes the proof. In Fatou’s lemma, strict inequality can hold if the sequence (fn ) is not monotonically increasing. The following two examples illustrate this. Example 4.2.23 Consider the functions fn as in Example 4.2.14, that is, (X, A, µ) = ([0, 1], M[0,1] , m) and fn := n χ(0, 1 ] , n

n ∈ N.

We have seen that fn ≥ 0 on [0, 1] for all n ∈ N, lim fn (x) = 0 for every n→∞ R x ∈ X and X fn = 1 for all n ∈ N. Also, taking f (x) = 0 for all x ∈ [0, 1], we have fn → f pointwise. Note that Z Z (lim inf fn ) dµ = ( lim fn ) dµ = 0, X

n

X n→∞

Z lim inf n

fn dµ = 1. X

Thus, strict inequality holds in Fatou’s lemma. Note that, (fn ) is not monotonically increasing. ♦ Example 4.2.24 Consider the measure space (X, A, µ) = (R, M, m) and fn := χ[n,∞) ,

n ∈ N.

102

Measure and Integration R Then, we have lim fn (x) = 0 for every x ∈ R and X fn = ∞ for all n ∈ N. n→∞ Hence, Z Z (lim inf fn ) dµ = X

( lim fn ) dµ = 0,

n

X n→∞

Z fn dµ = ∞.

lim inf n

X

Thus, in this example, strict inequality holds in Fatou’s lemma. Note that, (fn ) is not monotonically increasing. ♦ In MCT we assumed that fn ≤ fn+1 for every n ∈ N, that is, for each x ∈ X, fn (x) ≤ fn+1 (x) for every n ∈ N. The following theorem shows that the same conclusion as in MCT holds if the above condition is replaced by fn ≤ fn+1

a.e. for each n ∈ N.

Theorem 4.2.25 (MCT) Let (fn ) be a sequence of extended real valued nonnegative measurable functions defined on X such that fn ≤ fn+1 a.e. for every n ∈ N. Then (fn ) converges a.e. to a measurable function f and Z Z fn = f. lim n→∞

X

X

Proof. T Let E = {x ∈ X : fn (x) ≤ fn+1 (x) for every n ∈ N}. Then we ∞ have E = n=1 En , where En := {x ∈ X : fn (x) ≤ fn+1 (x)} for each n ∈ N. By the assumption that fn ≤ fn+1 a.e., we have En ∈ A and µ(Enc ) = 0 for each n ∈ N. Hence, ∞ ∞ [  X µ(E c ) = µ Enc ≤ µ(Enc ) = 0. n=1

n=1

Let g(x) = lim fn (x) for x ∈ E. Then g is measurable with respect to AE . n→∞

Since fn ≤ fn+1 on E and lim fn (x) = g(x) for every x ∈ E, by MCT n→∞

(Theorem 4.2.15), Z

Z gdµ = lim

n→∞

E

fn dµ. E

By pasting lemma (Lemma 3.3.38), the function f defined by  g(x), x ∈ E, f (x) = 0, x ∈ Ec is measurable. Since µ(E c ) = 0 and µ(Enc ) = 0 for every n ∈ N, we have Z Z Z Z f dµ = f dµ + f dµ = gdµ, X

Z fn dµ = X

Ec

E

Z fn dµ + E

E

Z

Z fn dµ =

Ec

fn dµ. E

Integral of Positive Measurable Functions

103

Thus, Z

Z f dµ =

X

Z gdµ = lim

E

n→∞

Z fn dµ = lim

E

n→∞

fn dµ. X

This completes the proof.

4.2.3

Radon-Nikodym theorem

Note that if µ and ν are as in Corollary 4.2.19, then for every E ∈ A, µ(E) = 0 ⇒ ν(E) = 0. Definition 4.2.26 Suppose µ and ν are measures on a measurable space (X, A). Then ν is said to be absolutely continuous with respect to µ if for every A ∈ A, µ(E) = 0 ⇒ ν(E) = 0, and this fact is written as ν −∞. k k Thus, −∞ < β0 < 0. Hence, there exists B0 ∈ A such that B0 ⊆ A0 and β0 := inf{λ(B) −

λ(B0 ) −

1 β0 µ(B0 ) < . k 2

108

Measure and Integration

Next, let A1 := A0 \ B0 . Then, A1 ⊆ A0 and, since β0 < 0, λ(A1 ) −

1 µ(A1 ) k

= λ(A0 \ B0 ) −

1 µ(A0 \ B0 ) k

1 1 µ(A0 )] − [λ(B0 ) − µ(B0 )] k k 1 β0 > [λ(A0 ) − µ(A0 )] − k 2 1 > [λ(A0 ) − µ(A0 )] > 0. k [λ(A0 ) −

=

Thus, A1 ∈ S. We may also observe that, if B ∈ A and B ⊆ A1 , then λ(B) − k1 µ(B) ≥ β20 . For otherwise, there exists B ∈ A such that B ⊆ A1 and λ(B) − k1 µ(B) < β20 so that the set B ∪ B0 , which is a subset of A0 , satisfies λ(B ∪ B0 ) −

1 µ(B ∪ B0 ) k

1 1 µ(B)] + [λ(B0 ) − µ(B0 )] k k

=

[λ(B) −


0, (d) βn+1 ≥ β0 /2n+1 . T Define A∞ := n∈N An . Then, we have λ(A∞ ) = lim λ(An )

and µ(A∞ ) = lim µ(An ).

n→∞

n→∞

Hence, by (c) above, 1 1 µ(A∞ ) ≥ λ(A0 ) − µ(A0 ) > 0. k k

λ(A∞ ) −

Let B ∈ A be such that B ⊆ A∞ . Then for each n ∈ N, B ⊆ An+1 and, by (d) above, 1 λ(B) − µ(B) ≥ βn+1 ≥ β0 /2n+1 ≥ 0. k Hence, A∞ ∈ G. Thus, we have shown that there exists Aˆ ∈ G. ˆ then χ χ = χ so that Now, let h := f + k1 χAˆ . Observe that, if B ⊆ A, B B A ˆ Z Z Z 1 f dµ + λ(B) = ν(B), hdµ = f dµ + µ(B) ≤ k B B B ˆ then and if B ⊆ X \ A, Z

Z f dµ ≤ ν(B).

hdµ = B

B

Hence, for any B ∈ A, Z Z Z hdµ = hdµ + B

ˆ B∩A

hdµ ≤ ν(B ∩ A) + ν(B \ A) = ν(B).

ˆ B∩(X\A)

ˆ > 0. Thus, we have shown that h ∈ F. Since Aˆ ∈ G, we also have µ(A) Step (2) (Existence part when µ and ν are σ-finite measures): Suppose µ and ν are σ-finite measures, not necessarily finite, on (X, A) such that R ν 0} = ∞ n=1 {x ∈ X : f (x) > 1/n}.] R 9. Let f : X → C be measurable such that X |f |p dµ < ∞ for some p > 0. Let An = {x ∈ X : |f (x)| ≥ n}. Show that µ(An ) = o(1/np ), i.e., lim np µ(En ) = 0. n→∞ R p [Hint: k := {x ∈ X : k ≤ |f (x)| < k +1} and observe that X |f | dµ = P∞ RWrite B S∞ p |f | dµ and A = B .] n k k=0 B k=n k

10. Let (fn ) be a sequence such that fn → f R of non-negative measurable functions R pointwise and lim X fn dµ exists and equal to X f dµ < ∞. Prove that n→∞ R R lim E fn dµ = E f dµ for every E ∈ A. n→∞

[Hint: Use Fatou’s lemma for fn χE and fn − fn χE .] 11. Using Fatou’s lemma (Corollary 4.2.22), derive the following: If (fn ) is a sequence of non-negative measurable functions Rsuch that fn → f pointwise and R fn ≤ f for every n ∈ N, then X f dµ = lim X fn dµ. n→∞

12. RJustify the statement: R n If f : R → [0, ∞) is a continuous function, then f dm = limn→∞ −n f (x)dx. R

112

Measure and Integration

√ 13. Let f (x) = 1/ x for 0 < x ≤ 1 and  n, √ fn (x) = 1/ x, R

if if

0 < x ≤ 1/n2 , 1/n2 < x ≤ 1.

R1 n→∞ 0

fn (x)dx, and find its value.  1/x2 , if 14. Let f (x) = 1/x2 for 1 ≤ x < ∞ and fn (x) = 0, if Show that Z n Z Z dx fn dm = f dm = lim , 2 n→∞ [1,∞) 1 x [1,∞) Show that

(0,1]

f dm = lim

1 < x ≤ n, n < x < ∞.

and find its value. 15. Let p ∈ R and f (x) = 1/xp for 0 < x ≤ 1. Show that  1 Z , if p < 1, 1−p f dm = ∞, if p ≥ 1. (0,1] 16. Let p ∈ R and f (x) = 1/xp for 1 ≤ x < ∞. Show that  1 Z , if p > 1, p−1 f dm = ∞, if p ≤ 1. [1,∞) 17. Let f (x) = 1/(1 + x2 ) for 1 ≤ x < ∞. Show that its value.

R [1,∞)

f dm is finite and find

18. Let (X, A, µ) be a finite measure space and f ≥ 0 be a measurable function on X. Let g(t) := µ({x ∈ X : f (x) ≤ t}), t ∈ R. Prove that there exists a measure ν on the Borel σ-algebra on R such that ν((−∞, t]) = g(t) for all t ∈ R. [Hint: Define ν(B) := µ{x ∈ X : f (x) ∈ B} for B ∈ B.]

Chapter 5 Integral of Complex Measurable Functions

In the last chapter, we considered integral of measurable functions which are extended real valued and non-negative. In this chapter, we extend the concept of integral from non-negative real valued measurable functions to real or complex valued measurable functions. Also, we shall prove one of the most important theorems in the theory of measure and integration, namely, the dominated convergence theorem (DCT), and derive many important and useful consequences of this theorem.

5.1

Integrability and Some Properties

Throughout this chapter we consider a measure space (X, A, µ). Recall (see Definition 3.3.22) that if f is a real valued function defined on X, then f = f + − f −. Here f + and f − are the positive part and negative part of f , defined by f − (x) := max{−f (x), 0},

f + (x) := max{f (x), 0},

respectively, for x ∈ X. So, it is natural to extend the concept of integral to real measurable functions as in the following definition. Definition 5.1.1 Suppose f is a real measurable function on X.R Then the integral of f over X with respect to the measure µ, denoted by X f dµ, is defined by Z Z Z + f dµ = f dµ − f − dµ X X X R R provided at least one of X f + dµ and X f − dµ is finite. ♦ Since |f | = f + + f − , we have Z Z Z |f | dµ = f + dµ + f − dµ X

X

X

113

114

Measure and Integration

and hence, R R f + dµ and X Rf − dµ are Rfinite if andRonly if X |f | dµ is finite, and in that case X f dµ = X f + dµ − X f − dµ. R

X

R Note that X |f | dµ is defined in the case of complex measurable functions f as well. Hence, we introduce the following definition. Definition 5.1.2 A real or R complex measurable function f on X is said to be integrable (over X) if X |f |dµ < ∞. ♦ Notation and convention: The set of all integrable complex measurable functions on X will be denoted by L(X, A, µ)

or L(µ)

or L(X).

If Ω is a Lebesgue measurable subset of R and f ∈ L(Ω), where the σ-algebra on Ω is the restriction of M to Ω, then f is called a Lebesgue integrable function on Ω. If f is a real measurable function, then by extending the co-domain of f to C, f can be thought of as a complex valued measurable function. Hence, we have the following: A real measurableR function Rf is integrable R if and only if f ∈ L(µ), and in that case, X f dµ = X f + dµ − X f − dµ.

Suppose f is a complex measurable function on X. Since f = Ref + iImf , we have |Ref | ≤ |f |, |Imf | ≤ |f |, |f | ≤ |Ref | + |Imf |. Hence, f ∈ L(µ) ⇐⇒ Ref ∈ L(µ)

and

Imf ∈ L(µ).

Thus, we have the following definition. Definition 5.1.3 If f ∈ L(µ), then the integral of f over X with respect to µ is defined by Z Z Z f dµ := Ref dµ + i Imf dµ. X

X

X

If E ∈ A, then the integral of f over E with respect to µ is defined by Z Z f dµ := χE f dµ. ♦ E

X

Integral of Complex Measurable Functions 115 R RNotation and convention: The integral X f dµ is sometimes written as f (x)dµ(x). Once the measure is understood from the context, we shall also X use the notation Z Z f or f X

R for the integral X f dµ. If Ω is a Lebesgue measurable subset of R and f ∈ L(Ω), then the integral of f over Ω is called its Lebesgue integral. Let us prove some simple properties of functions in L(µ). Theorem 5.1.4 If f and g are in L(µ) and c ∈ C, then f + g and cf are in L(µ), and Z Z Z Z Z (f + g) = f+ g, cf = c f. X

X

X

X

X

Proof. Suppose f and g are in L(µ) and c ∈ C. Since |f + g| ≤ |f | + |g| and |cf | = |c| |f |, both f + g and cf belong to L(µ). Now to prove the equalities of the integrals, first we consider real and complex cases separately: Case 1: f and g are real valued and c ∈ R. In this case, note that f + g = (f + − f − ) + (g + − g − ). Thus, (f + g)+ − (f + g)− = (f + − f − ) + (g + − g − ), so that (f + g)+ + f − + g − = (f + g)− + f + + g + . Hence, by Theorem 4.2.6, Z Z Z Z Z Z (f + g)+ + f− + g− = (f + g)− + f+ + g+ . X

X

X

X

X

X

Since each integral on both sides of the above equation is finite, we obtain, Z Z Z Z Z Z (f + g)+ − (f + g)− = f+ − f− + g+ − g− . X

X

X

X

X

X

Thus, Z

Z (f + g) =

X

Z f+

X

g. X

Next, note that cf = (c+ − c− )(f + − f − ) = (c+ f + + c− f − ) − (c+ f − + c− f + )

116

Measure and Integration

so that (cf )+ − (cf )− = (c+ f + + c− f − ) − (c+ f − + c− f + ) and hence (cf )+ + (c+ f − + c− f + ) = (cf )− + (c+ f + + c− f − ). Again, by Theorem 4.2.6, we have Z Z Z Z Z Z + + − − + − + + (cf ) + c f + c f = (cf ) + c f + c− f − . X

X

X

X

X

X

Since each integral on both sides of the above equation is finite, we obtain, Z Z Z cf = (cf )+ − (cf )− X X X Z Z Z Z + + + − − − = c f − c f + c f − c− f + X X X X Z Z hZ i hZ i = c+ f+ − f − − c− f+ − f− X X X X Z hZ i + − + − = (c − c ) f − f X X Z = c f. X

Case 2: f and g are complex valued and c ∈ C. In this case, we observe that f + g = Re (f + g) + iIm (f + g) = [Re f + Re g] + i[Im f + Im g]. Hence, using Case 1, Z Z Z (f + g) = [Re f + Re g] + i [Im f + Im g] X X X Z Z Z h i hZ i = Re f + Re g + i Im f + Im g X X X X Z Z hZ i hZ i = Re f + i Im f + Re g + i Im g X X X Z X Z = f+ g. X

X

Let c = α + iβ with α, β ∈ R. Then we have cf = (α + iβ)(Re f + i Im f ) = (α Re f − β Im f ) + i(α Im f + β Re f ).

Integral of Complex Measurable Functions

117

Now, using Case 1, we get Z Z Z cf = (α Re f − β Im f ) + i (α Im f + β Re f ) X X X Z Z Z Z = α Re f − β Im f + iα Im f + iβ Re f X X X Z ZX = (α + iβ) Re f + i(α + iβ) Im f X X Z hZ i = c Re f + i Im f X Z X f. = c X

This completes the proof. Corollary 5.1.5 Let f, g ∈ L(µ) be real valued such that f ≤ g on X. Then Z Z f≤ g. X

X

Proof. By Theorem 5.1.4, Z Z Z Z g= [(g − f ) + f ] = (g − f ) + f. X

X

Since g − f ≥ 0, we have

R X

X

(g − f ) ≥ 0. Hence,

X

R X

f≤

R X

g.

Theorem 5.1.6 If f ∈ L(µ), then Z Z f ≤ |f |. X

X

R R Proof. Let f ∈ L(µ), and let θ ∈ R be such that X f = X f eiθ . Then, using Theorem 5.1.4, Z Z Z Z Z −iθ −iθ f = e−iθ f= e f= Re(e f ) + i Im (e−iθ f ). X

X

X

X

X

From this, we obtain X Im (e−iθ f ) = 0, and hence, Z Z Z Z −iθ −iθ + f = Re(e f ) = [Re(e f )] − [Re(e−iθ f )]− X X X X Z Z −iθ + ≤ [Re(e f )] ≤ |f |. R

X

X

This completes the proof. The following corollary is immediate from the above theorem.

118

Measure and Integration

Corollary 5.1.7 If f is measurable and f = 0 on E ∈ A, then

R E

f dµ = 0.

Theorem 5.1.8 The following results hold. (i) Suppose f and g are complex measurable functions such that f = 0 a.e. and g = 0 a.e. Then f + g = 0 a.e. R (ii) Suppose f ∈ L(µ) is such that E f = 0 for all E ∈ A. Then f = 0 a.e. Proof. (i) We know that, if h is a complex measurable function, then h = 0 a.e. if and only if the set Eh := {x ∈ X : h(x) 6= 0} is of measure zero. Note that Ef +g ⊆ Ef ∪ Eg . Hence, µ(Ef +g ) ≤ µ(Ef ) + µ(Eg ) = 0 so that µ(Ef +g ) = 0. (ii) First observe that it is enough to prove for the case of a real valued f . So, let f be a real valued measurable function. Let E = {x ∈ X : f (x) ≥ 0}. By hypothesis, Z Z f+ = f =0 X

so that by Theorem 4.2.10, f by (i), f = f + − f − = 0 a.e.

+

E

= 0 a.e. Similarly, we have f − = 0 a.e. Hence,

Remark 5.1.9R Theorem 5.1.4 shows that L(µ) is a vector space over C and the R map f 7→ X f dµ is a linear functional on L(µ). Further, the map f 7→ |f |dµ is a seminorm on L(µ). X By a seminorm on a real or complex vector space V , we mean a function p : V → R such that p(u + v) ≤ p(u) + p(v),

p(αv) = |α|p(v)

for all u, v ∈ V and for all scalar α. It can be easily shown that if p is a seminorm on a vector space V , then p(0) = 0

and p(v) ≥ 0 ∀ v ∈ V.

To have an example of a seminorm other than the one given above, let us consider the vector space B(Ω) of all bounded real or complex valued functions defined on a nonempty set Ω. Then, for each ω ∈ Ω, the map f 7→ f (ω) is a linear functional on B(Ω) and the map f 7→ |f (ω)| is a seminorm on B(Ω). Now, in view of Theorem 4.2.10 and Theorem 5.1.8, we have Z |f |dµ = 0 ⇐⇒ f = 0 a.e. X

So, let Z := {f ∈ L(µ) : f = 0 a.e.}.

Integral of Complex Measurable Functions

119

By Theorem 5.1.8, Z is a subspace of L(µ). Consider the quotient space L(µ) := L(µ)/Z. By abusing the notation, we shall denote the equivalence class of f ∈ L(µ) by the same notation f . Thus, the map Z f 7→ |f | dµ X

defines a norm on L(µ). By a norm on a real or complex vector space V , we mean a seminorm p : V → R which also satisfies the condition v ∈ V,

p(v) = 0

⇒

v = 0.

The usual notation for a norm is k · k. As another example of a norm other than the one given above, let us consider again the vector space B(Ω) of all bounded real or complex valued functions defined on a set Ω. Then, the map f 7→ kf k := sup |f (x)| x∈Ω

is a norm on B(Ω). In later sections, we shall deal with some other spaces with norms. ♦

5.1.1

Riemann integral as Lebesgue integral

Recall from Theorem 4.2.13 that if f : [a, b] → [0, ∞) is a Riemann integrable function, then it is Lebesgue integrable, that is, integrable over [a, b] with respect to the Lebesgue measure on [a, b], and Z b Z f (x)dx = f dm. a

[a,b]

Now, we show that this result still holds if f is a real valued Riemann integrable function. So, suppose that f : [a, b] → R is a Riemann integrable function. Then, we may recall from the theory of Riemann integration that |f | is also Riemann integrable and Z Z b b f (x)dx ≤ |f (x)|dx. a a Let us observe that f+ =

1 (|f | + f ), 2

f− =

1 (|f | − f ). 2

120

Measure and Integration

Hence, both f + and f − are Riemann integrable, and Z b Z Z b  1 b + f (x)dx = |f (x)|dx + f (x)dx , 2 a a a Z b Z Z b  1 b f − (x)dx = f (x)dx . |f (x)|dx − 2 a a a

(1)

(2)

By Theorem 4.2.13, f + and f − are Lebesgue measurable and Z b Z Z b Z f + (x)dx = f + dm, f − (x)dx = f − dm. a

[a,b]

a

[a,b]

Therefore, f is Lebesgue integrable, and Z Z Z f dm = f + dm − [a,b]

[a,b]

Z =

f − dm

[a,b]

b

f + (x)dx −

a

Z

b

f − (x)dx.

a

Now, using relations (1) and (2), we obtain Z Z b f dm = f (x)dx. [a,b]

a

Thus, we have proved the following theorem. Theorem 5.1.10 If f : [a, b] → R is a Riemann integrable function, then it is Lebesgue integrable, and Z b Z f (x)dx = f dm. a

[a,b]

It is to be remarked that the concept of Riemann integral can be extended to complex valued functions as well. Suppose f is a complex valued bounded function defined on an interval [a, b]. Then f is said to be Riemann integrable if Ref and Imf are Riemann integrable, and in that case we define the Riemann integral of f by Z b Z b Z b f (x)dx = Ref (x)dx + i Imf (x)dx. a

a

a

Thus, we can also assert the following. Theorem 5.1.11 If f : [a, b] → C is a Riemann integrable function, then it is Lebesgue integrable, and Z b Z f (x)dx = f dm. a

[a,b]

Integral of Complex Measurable Functions

121

Notation: If the measure space is R with Lebesgue measure m, and if J is an interval with end points a and b, where a < b, then for every complex Rb Rb measurable function f on J, we use the notation a f dm or a (f (x)dx for R Ra R f dm and define b f dm = − J f dm. J

5.1.2

Dominated convergence theorem (DCT)

Recall that, in the monotone convergence theorem (MCT), we required the sequence (fn ) of functions to be non-negative and monotonically increasing. Now, we shall prove convergence results without these assumptions; but using some other conditions. First, let us prove the following theorem, which is analogous to a theorem in the theory of Riemann integration (see, e.g., [2]). Theorem 5.1.12 Let (fn ) be a sequence of complex measurable functions which converges uniformly to a function f on X and let µ(X) < ∞. Then Z lim |fn − f |dµ = 0. n→∞

X

Further, if f ∈ L(µ), then fn is integrable for all large enough n, and Z Z lim fn dµ = f dµ. n→∞

X

X

Proof. Let ε > 0 be given and let N ∈ N be such that |fn (x) − f (x)| ≤ ε ∀ n ≥ N

∀ x ∈ X.

Then, Z

Z |fn − f |dµ ≤

X

εdµ = εµ(X) ∀ n ≥ N. X

Z |fn − f | = 0.

Thus, lim

n→∞

X

Note that fn − f ∈ L(µ) for all n ≥ N . Hence, if f ∈ L(µ), then fn = f + (fn − f ) ∈ L(µ) for all n ≥ N and, by Theorem 5.1.4 and Theorem 5.1.6, Z Z Z Z fn dµ − = (fn − f ) dµ ≤ f dµ |fn − f | dµ ≤ εµ(X) X

X

X

for all n ≥ N . Hence, lim

R

n→∞ X

fn dµ =

X

R X

f dµ.

Now, we would like to relax some of the conditions in Theorem 5.1.12 on (fn ) and X. The resulting theorem is the dominated convergence theorem (DCT), one of the most important results in the theory of integration. Theorem 5.1.13 (DCT) Suppose (fn ) is a sequence of complex measurable functions such that

122

Measure and Integration

(a) (fn ) converges pointwise on X and (b) there exists g ∈ L(µ) satisfying |fn | ≤ |g| for all n ∈ N. Let f (x) := lim fn (x), x ∈ X. Then fn and f are integrable, n→∞

Z

Z |fn − f | = 0

lim

n→∞

and

Z

lim

fn =

n→∞

X

X

f. X

Proof. Since |fn | ≤ |g| for some g ∈ L(µ) and (fn (x)) converges for every x ∈ X, it follows that |f | ≤ |g| so that fn ∈ L(µ) and f ∈ L(µ). Also, we have |fn − f | ≤ |fn | + |f | ≤ 2|g|

∀ n ∈ N.

Thus, 2|g| − |fn − f | ≥ 0 for all n ∈ N, and 2|g| − |fn − f | → 2|g| pointwise as n → ∞. Hence, by Fatou’s lemma, Z Z 2|g| = lim inf (2|g| − |fn − f |) n X X Z ≤ lim inf (2|g| − |fn − f |) n X Z Z |fn − f |. = 2|g| − lim sup n

X

X

Thus, Z

Z n

|fn − f | ≤ 0.

|fn − f | ≤ lim sup

0 ≤ lim inf

n

X

X

Z |fn − f | exists and it is equal to 0. Now, by Theorem

Consequently, lim

n→∞

X

5.1.4 and Theorem 5.1.6, Z Z Z Z = (fn − f ) dµ ≤ fn dµ − |fn − f | dµ → 0 f dµ X

X

as n → ∞, so that lim

X

X

R

n→∞ X

fn =

R X

f.

The following example shows that the condition (b) in the dominated convergence theorem cannot be dropped. Example 5.1.14 For each n ∈ N, let fn : R → R be defined by  1/n if 0 < x < n, fn (x) = 0 otherwise. R Then, (fn ) converges to the zero function uniformly. But R fn dm = 1 for every n ∈ N. Note that, (fn ) is not dominated by an integrable function. This example also shows that the condition µ(X) < ∞ in Theorem 5.1.12 cannot be dropped. ♦

Integral of Complex Measurable Functions

123

The conclusions in DCT hold good if we replace pointwise convergence by convergence almost everywhere provided we define f appropriately. Theorem 5.1.15 (DCT) Suppose (fn ) is a sequence of complex measurable functions such that (fn ) converges a.e., and there exists g ∈ L(µ) satisfying |fn | ≤ |g| a.e. for all n ∈ N. Then there exists a measurable function f such that fn → f a.e., fn and f are integrable, Z Z Z lim |fn − f | = 0 and lim fn = f. n→∞

n→∞

X

X

X

Proof. Let A := {x ∈ X : fn (x) → f (x)} T∞ and for each n ∈ N, let Bn := {x ∈ X : |fn (x)| ≤ |g(x)|}. Let E = A ∩ ( n=1 Bn ). By hypothesis, we have µ(Ac ) = 0 and µ(Bnc ) = 0 for every n ∈ N. Hence, µ(E c ) = 0. Let ( lim fn (x), x ∈ E, n→∞ f (x) := 0, x ∈ X \ E. Then, by pasting lemma (Lemma 3.3.38), fn → f a.e., and the conclusion follows by applying Theorem 5.1.13 on (E, AE , µE ) and using the fact that integrals over E c is 0. The following theorem is an immediate consequence of DCT, and hence we omit its proof. Theorem 5.1.16 (Bounded convergence theorem) (BCT) Suppose (fn ) is a sequence of complex measurable functions which converges pointwise to a function f on X. If µ(X) < ∞ and (fn ) is uniformly bounded, that is, there exists M > 0 such that |fn (x)| ≤ M for all x ∈ X and for all n ∈ N, then fn and f are integrable, Z Z Z lim |fn − f | = 0 and lim fn = f. n→∞

n→∞

X

X

X

Example 5.1.17 Let fn (x) =

nx , 1 + n 2 x2

x ∈ [0, 1].

We observe that fn (x) → 0 for each x ∈ [0, 1] and 0 ≤ fn (x) ≤

1 2

∀ x ∈ [0, 1], n ∈ N.

Hence, by Theorem 5.1.16, Z 1 Z fn (x)dx = 0

fn dm → 0.

[0,1]

Note that the convergence of (fn ) to the zero function is not uniform, and hence, Theorem 5.1.12 cannot be applied. ♦

124

Measure and Integration

Now, we prove a theorem analogous to Corollary 4.2.17. Theorem 5.1.18 (fn ) is a sequenceP of complex measurable funcR P∞Suppose ∞ tions such that n=1 X |fn | converges. Then n=1 fn converges a.e. to an integrable function and ! Z ∞ ∞ Z X X fn = fn . X

n=1

n=1

X

R P∞ P∞ R Proof. By Corollary 4.2.17, we have XR P n=1 |fn | = n=1 X |fn |, and ∞ hence, by the assumption in the theorem, |f | < ∞. Therefore, by n n=1 P X P∞ ∞ Theorem 4.2.11, n=1 |fn | is finite a.e. In particular, n=1 fn converges a.e. Let ∞ n X X |fn |. fj and h := gn := n=1

j=1

P∞

Thus, (gn ) converges to g := n=1 fn a.e. and |gn | ≤ h for all n ∈ N with h ∈ L(µ). Hence, by DCT (Theorem 5.1.15), Z lim

n→∞

Z gn dµ =

X

g dµ = X

Z X ∞ X

 fn .

n=1

But, Z lim

n→∞

Thus,

gn = lim X

n→∞

Z X n X

j=1

n Z ∞ Z  X X fj = lim fj = fj . n→∞

j=1

X

j=1

X

R  P∞ X

 P ∞ R f n=1 n = j=1 X fj and the proof is completed.

Now we consider some more consequences of DCT. Theorem 5.1.19 Let f ∈ L(µ). For n ∈ N, let  f (x), |f (x)| ≤ n, fn (x) = n, |f (x)| > n. Then, for each n ∈ N, fn is a bounded measurable function and Z |f − fn |dµ → 0 as n → ∞. X

Proof. Clearly, |fn | ≤ n and |fn | ≤ |f | for all n ∈ N. In particular, each fn is a bounded measurable function. Also, fn → f pointwise. Indeed, since f is complex valued, for each x ∈ X, there exists k ∈ N such that |f (x)| ≤ k, so R that fn (x) = f (x) for all n ≥ k. Therefore, by DCT, X |f − fn |dµ → 0.

Integral of Complex Measurable Functions

125

The implication of the above theorem can be stated as follows: For every f ∈ L(µ), there R exists a sequence (fn ) of bounded functions in L(µ) such that lim X |fn − f |dµ = 0. n→∞

The following theorem is an immediate consequence of Theorem 5.1.19. Theorem 5.1.20 If f ∈ L(µ), then for every ε > 0, there exists a bounded g ∈ L(µ) such that Z |f − g|dµ < ε. X

Recall that if f is a measurable function and E ∈ A with µ(E) =R 0, then |f |dµ = 0. Suppose µ(E) 6= 0, but µ(E) is small. Can we say that E |f | dµ E small? Not necessarily, as the following example shows. R

Example 5.1.21 Let f : R → R be defined by  1/x2 , x 6= 0, f (x) = 0, otherwise. Let An = {x ∈ R : 0 < x < 1/n}, n ∈ N. Then we have m(An ) = 1/n and Z Z Z n2 χAn dm = n2 m(An ) = n. |f |dm = χAn |f |dm ≥ An

R

Thus, m(An ) → 0, but

R An

R

|f |dm → ∞.

♦

Such a situation does not arise if f ∈ L(µ). Theorem 5.1.22 Let f ∈ L(µ). Then for every ε > 0, there exists δ > 0 such that Z A ∈ A, µ(A) < δ ⇒ |f |dµ < ε. A

Proof. Let ε > 0 be given. If f is bounded, say, |f | ≤ M0 for some M0 > 0, then for every A ∈ A, Z |f |dµ ≤ M0 µ(A) A

so that

ε µ(A) < M0

Z ⇒

|f |dµ < ε. A

Now, let us consider the general case of f ∈ L(µ).R In this case, by Theorem 5.1.20, there exists a bounded g ∈ L(µ) such that X |f − g|dµ < ε/2. Hence, for A ∈ A, Z Z Z Z ε |f |dµ ≤ |g|dµ + |f − g|dµ ≤ |g|dµ + . 2 A A A A

126

Measure and Integration

Let M > 0 be such that |g| ≤ M . Then we have Z Z ε ε |f |dµ ≤ |g|dµ + ≤ M µ(A) + . 2 2 A A R Thus, µ(A) < ε/2M implies A |f |dµ < ε. Corollary 5.1.23 Let f ∈ L(µ). If (An ) is a sequence of sets in A such that R µ(An ) → 0 as n → ∞, then An |f |dµ → 0 as n → ∞. Proof. Let (An ) be a sequence of sets in A such that µ(An ) → 0 as n → ∞. Let ε > 0 be given. By Theorem 5.1.22, there exists δ > 0 such that Z |f |dµ < ε. (∗) A ∈ A, µ(A) < δ ⇒ A

Since µ(An ) → 0, there exists N ∈ N such that µ(An ) < δ Hence, by (∗) above,

R An

∀ n ≥ N.

|f |dµ < ε for all n ≥ N .

Corollary 5.1.24 Let f ∈ L(µ) and An = {x ∈ X : |f (x)| > n} for n ∈ N. R Then An |f |dµ → 0 as n → ∞. Proof. By Theorem 4.2.11 (i), R we know that µ(An ) → 0 as n → ∞. Hence, by Corollary 5.1.23, we obtain An |f |dµ → 0 as n → ∞. R Recall from Theorem 5.1.8 (ii) that, if f ∈ L(µ) is such that E f = 0 for all E ∈ A, then f = 0 a.e. Now, in the case when X is an interval and µ is the Lebesgue measure, then we obtain the same conclusion with measurable sets replaced by subintervals. Theorem 5.1.25 Let J be an interval and let f ∈ L(J). For any given a ∈ J, Rx if a f dm = 0 for every x ∈ J, then f = 0 a.e. on J. R Proof. First we observe that I f dm = 0 for every interval I ⊆ J. Indeed, for every open interval with end points c, d, where c < d, Z d Z d Z c f dm = f dm − f dm = 0. c

R

a

a

Next, we show that G∩J f dm = 0 for every open set G ⊆ R: Let G be an open set in R. We know that, G can be written as a countable S∞ S∞ disjoint family of open intervals, say G = n=1 In . Then G ∩ J = n=1 Jn , where Jn := In ∩ J and {Jn } is a countable disjoint family of intervals or empty sets. Without loss of generality, we may assume that Jn 6= ∅ for every n ∈ N. Let fn = χJn f, n ∈ N.

Integral of Complex Measurable Functions Then we have

Z

Z f dm = 0 ∀ n ∈ N.

fn dm = G∩J

Let hk :=

Pk

n=1

127

(1)

Jn

fn , k ∈ N. Then hk → f pointwise on G ∩ J and |hk | ≤

k X

|fn | =

n=1

k X

χJn |f | ≤ |f |

∀ k ∈ N.

n=1

Hence, by DCT and (1), Z

Z f dm = lim

G∩J

k→∞

hk dm = lim

k→∞

G∩J

k Z X n=1

fn dm = 0.

(2)

G∩J

Now, let E be any measurable subset of J. Then, for every n ∈ N there exists an open set Gn ⊇ E such that m(Gn \ E) < 1/n. Note that Gn ∩ J = E ∪ [(Gn \ E) ∩ J], where E and (Gn \ E) ∩ J are disjoint measurable sets of finite measure. Hence, by (2), Z Z Z f dm = f dm − f dm Gn ∩J (Gn \E)∩J E Z Z = f dm ≤ |f |dm. (Gn \E)∩J (Gn \E)∩J R Since m(Gn \E) → 0, by Corollary 5.1.23, (Gn \E)∩J |f |dm → 0. Therefore, we R can conclude that E f dm = 0. This is true for every measurable set E ⊆ J. Therefore, by Theorem 5.1.8(ii), f = 0 a.e.

5.2

Lp Spaces

Let (X, A, µ) be a measure space. For 1 ≤ p < ∞, we denote by Lp (X, A, µ) the set of all complex measurable functions f on X such that |f |p is integrable. In short, we may denote this set by Lp (µ) or by Lp (X). Thus, for a complex measurable function f on X, Z f ∈ Lp (µ) ⇐⇒ |f |p dµ < ∞. X

Definition 5.2.1 A measurable function f on X is said to be an essentially bounded function if there exists Mf > 0 such that |f | ≤ Mf a.e. on X. ♦

128

Measure and Integration

The set of all essentially bounded functions is denoted by L∞ (X, A, µ) or simply by L∞ (µ) or L∞ (X). Thus, for a complex measurable function f on X, f ∈ L∞ (µ) ⇐⇒ ∃ Mf > 0 such that |f | ≤ Mf a.e. on X. For a complex measurable function f on X, we denote Z 1/p kf kp := |f |p dµ if 1 ≤ p < ∞, X

and kf k∞ := inf{Mf > 0 : |f | ≤ Mf a.e. on X}. Note that, kf kp can be ∞. Thus, for 1 ≤ p ≤ ∞, and for any complex measurable function f on X, f ∈ Lp (µ) ⇐⇒ kf kp < ∞. Let Zp := {f ∈ Lp (µ) : f = 0 a.e.}. Theorem 5.2.2 For 1 ≤ p ≤ ∞, Lp (µ) is a vector space over C and Zp is a subspace of Lp (µ). Proof. Let 1 ≤ p ≤ ∞. Note that for f, g ∈ Lp (µ), |f + g| ≤ |f | + |g| ≤ 2 max{|f |, |g|}. Hence, kf + gk∞ ≤ 2 max{kf k∞ , kgk∞ } and for 1 ≤ p < ∞, |f + g|p ≤ 2p max{|f |p , |g|p } so that

Z

|f + g|p dµ ≤ 2p max

X

nZ

|f |p dµ,

X

Z

o |g|p dµ .

X

Hence, f + g ∈ Lp (µ). It is easy to see that αf ∈ Lp (µ) for every f ∈ Lp (µ) and α ∈ C. Using these, all axioms of a vector space can be verified. It can also be verified that, for f, g ∈ Lp (µ), f = 0 a.e. and g = 0 a.e. imply f + g = 0 a.e., which we already know by Theorem 5.1.8, and αf = 0 a.e. for every α ∈ C. Hence, Zp is a subspace of Lp (µ). Consider the quotient space Lp (µ) := Lp (µ)/Zp . Notation: As in the case of the space L(µ) considered in Remark 5.1.9, we use the same symbol f for f ∈ Lp (µ) and for the corresponding equivalence class [f ]. If Ω is a Lebesgue measurable subset of R, then we shall denote the

Integral of Complex Measurable Functions

129

spaces Lp (Ω, MΩ , m) and Lp (Ω, MΩ , m) by Lp (Ω) and Lp (Ω), respectively. Note that the spaces L(µ) and L(µ) introduced earlier are the spaces L1 (µ) and L1 (µ), respectively. We shall show that f 7→ kf kp is a norm on Lp (µ) (c.f. Remark 5.1.9) and the corresponding metric is complete.

5.2.1

H¨ older’s and Minkowski’s inequalities

One of the crucial inequalities required is the Young’s inequality: Lemma 5.2.3 (Young’s inequality) Let a, b be non-negative real numbers and p, q ∈ (1, ∞) be such that p1 + 1q = 1. Then ab ≤

bq ap + . p q

Proof. We know that the function ϕ : R → R defined by ϕ(x) = ex ,

x ∈ R,

is convex, that is, for every x, y ∈ R and 0 < λ < 1, ϕ(λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y). Taking λ = 1/p we have 1 − λ = 1/q and x

y

ep+q ≤

ex ey + . p q x

y

Now, taking x > 0 and y > 0 such that a = e p and b = e q , that is, x = ln(ap ) q p and y = ln(bq ), we obtain ab ≤ ap + bq . Remark 5.2.4 Observe that the Young’s inequality for p = 2 and q = 2 follows from the identity (a − b)2 = a2 + b2 − 2ab. ♦ In fact, Young’s inequality considered in Lemma 5.2.3 is a particular case of the Jensen’s inequality. Theorem 5.2.5 (Jensen’s inequality) Let f be a real valued integrable function on a probability space (X, A, µ), that is, with µ(X) = 1, and I be an open interval such R that f (x) ∈ I for all x ∈ X. Let g : I → R be a convex function. Then X f dµ ∈ I and Z  Z g f dµ ≤ (g ◦ f ) dµ. X

X

Proof. Since f (x) ∈ I for all x ∈ X and µ(X) = 1, it follows (see Corollary R 5.1.5) that X f dµ ∈ I. For a, b, c ∈ I with a < b < c, we have b = (1 − λ)a + λc with λ =

b−a . c−a

130

Measure and Integration

By the convexity of g, g(b) ≤ (1 − λ)g(a) + λg(c), so that g(b) − g(a) ≤ λ[g(c) − g(a)].

(1)

g(b) − g(a) g(c) − g(a) ≤ . b−a c−a

(2)

Thus,

Writing µ = 1 − λ =

c−b c−a ,

(1) implies

g(b) − g(a) ≤ (1 − µ)[g(c) − g(a)] so that µ[g(c) − g(a)] ≤ g(c) − g(b), that is, g(c) − g(b) g(c) − g(a) ≤ . c−a c−b

(3)

The relations (2) and (3) give g(b) − g(a) g(c) − g(b) ≤ . b−a c−b

(4)

Note that the right-hand side of (4) is independent of a. For a fixed b, let β be the supremum of the left-hand side of (4) as a varies such that a < b. Then we have g(c) − g(b) g(b) − g(a) ≤β≤ (5) b−a c−b for all a ∈ I with a < b. The above relations can also be written as g(b) − g(a) g(b) − g(c) ≤β≤ , b−a b−c

(6)

which is also true for all c ∈ I with b < c. The relations in (5) and (6) show that g(b) − g(t) ≤ β(b − t) ∀ t ∈ I. R Now, taking b = X f dµ and t = f (x) for x ∈ X, we obtain Z  Z  g f dµ − g(f (x)) ≤ β f dµ − f (x) . X

X

On integration, taking into account that µ(X) = 1, we get Z Z  Z Z  g f dµ − (g ◦ f )dµ ≤ β f dµ − f dµ = 0. X

X

Thus, we obtain the required inequality.

X

X

Integral of Complex Measurable Functions

131

To see that the Young’s inequality is a particular case of Jensen’s inequality, let x1 , . . . , xn be real numbers and λ1 , . . . , λn be non-negative real numbers such that λ1 + · · · + λn = 1. Then, considering the convex function g(t) = et , X = {x1 , . . . , xn } with measure µ on the power set of X by defining µ({xi }) = λi , i = 1, . . . , n, f : X → R by f (x) = x, x ∈ X, and taking ai = exi , i = 1, . . . , n, we obtain Z g

 n X  Pn f dµ = g xi λi = e i=1 xi λi = aλ1 1 · · · aλnn ,

X

i=1

Z (g ◦ f )dµ =

n X

X

i=1

g(xi )λi =

n X

λi ai .

i=1

Thus, by Jensen’s inequality, for non-negative real numbers a1 , . . . , an and λ1 , . . . , λn with λ1 + · · · + λn = 1, we have aλ1 1 · · · aλnn ≤ λ1 a1 + · · · + λn an . Taking n = 2, p = 1/λ1 , q = 1/λ2 , a = aλ1 1 , b = aλ2 2 we obtain the Young’s inequality in Lemma 5.2.3. Recall the notation Z 1/p p kf kp := |f | dµ for 1 ≤ p < ∞, X

for complex measurable functions f on X. In the following, for 1 ≤ p ≤ ∞, we take q ∈ [1, ∞] such that 1 1 + =1 p q with the convention that 1/∞ = 0. Theorem 5.2.6 (H¨ older’s inequality) Let f and g be complex measurable functions on X and 1 ≤ p ≤ ∞. Then Z |f g|dµ ≤ kf kp kgkq , X

where q is such that

1 p

+

1 q

= 1.

Proof. For the case p = 1 and p = ∞, it is easy to see that the inequality holds. Hence, assume that 1 < p < ∞. Then 1 < q < ∞. First we observe that if one of kf kp and kgkq is zero or infinity, then the inequality holds. Hence, we assume that 0 < kf kp < ∞ and 0 < kgkq < ∞. For x ∈ X, taking a = |f (x)|/kf kp and b = |g(x)|/kgkq in the Young’s inequality, we have |f (x)g(x)| |f (x)|p |g(x)|q ≤ + . kf kp kgkq pkf kpp qkgkqq

132

Measure and Integration

Now, taking integrals over X, we get R R Z |g|q |f |p 1 1 1 X X = + = 1. |f g| ≤ p + kf kp kgkq X pkf kp qkgkqq p q R Hence, X |f g| ≤ kf kp kgkq , which completes the proof. Theorem 5.2.7 (Minkowski’s inequality) Let f and g be complex measurable functions on X and 1 ≤ p ≤ ∞. Then kf + gkp ≤ kf kp + kgkp . Proof. We observe that the inequality for p = 1 and p = ∞ follows from the inequality |f + g| ≤ |f | + |g|. So, let 1 < p < ∞ and q > 1 be such that 1 1 p + q = 1. We note that the inequality holds if kf + gkp = 0. Hence, assume that kf + gkp 6= 0. Note that Z Z |f + g|p dµ = |f + g|p−1 |f + g|dµ X X Z Z p−1 ≤ |f + g| |f |dµ + |f + g|p−1 |g|dµ. X

X

By H¨ older’s inequality, we have Z Z 1/q |f + g|p−1 |f | ≤ kf kp |f + g|(p−1)q = kf kp kf + gkp/q p , X

Z

X

|f + g|p−1 |g| ≤ kgkp

Z

X

|f + g|(p−1)q

1/q

= kgkp kf + gkp/q p .

X

Thus, kf +

gkpp

Z =

  |f + g|p dµ ≤ kf kp + kgkp kf + gkp/q p .

X p/q

Now canceling out kf + gkp , as kf + gkp 6= 0, we obtain the required inequality. Now, the proof of the following theorem is immediate. Theorem 5.2.8 For 1 ≤ p ≤ ∞, Lp (µ) is a vector space over C and f 7→ kf kp is a norm on Lp (µ). In particular, (f, g) 7→ kf − gkp is a metric on Lp (µ).

Integral of Complex Measurable Functions

133

Remark 5.2.9 If X = N, A = 2N and µ is the counting measure, then every complex valued function on X is measurable. Hence, in this case, for any p ∈ [1, ∞], we have Zp = {0} and Lp (µ) = Lp (µ) = `p (N). Here, `∞ (N) is the space of all bounded sequences in C for 1 ≤ p < ∞, Pand ∞ `p (N) is the space of all sequences (an ) in C such that n=1 |an |p < ∞. Also, if X = {1, . . . , k} for some k ∈ N, A = 2X and µ is the counting measure, then every complex valued function on X is measurable, and hence, for any p ∈ [1, ∞], Zp = {0} and Lp (µ) = Lp (µ) = Ck .

5.2.2

♦

Completeness of Lp (µ)

We prove that the metric on Lp (µ) induced by the norm k · kp is complete. Before that, let us observe the following result which will be used in the sequel. Lemma 5.2.10 Let 1 ≤ p < ∞ and let (fn ) be a Cauchy sequence in Lp (µ) having a subsequence which converges almost everywhere to a measurable function f . Then f ∈ Lp (µ) and (fn ) converges to f in Lp (µ). Proof. Suppose (fn ) is a Cauchy sequence in Lp (µ) with 1 ≤ p < ∞ having a subsequence (fnk ) which converges a.e. to a measurable function f . Then for each n ∈ N, |fnk − fn | → |f − fn | a.e. on X. Therefore, by Fatou’s lemma, Z

|f − fn |p ≤ lim inf k

X

Z

|fnk − fn |p .

X

Now, let ε > 0 be given. Since (fn ) is a Cauchy sequence in Lp (µ) and nk ≥ k, there exists N ∈ N such that kfnk − fn kp < ε for all n, k ≥ N . Thus, Z X

|f − fn |p ≤ lim inf k

Z

|fnk − fn |p < εp

∀ n ≥ N,

X

showing that f ∈ Lp (µ) and kf − fn kp → 0 as n → ∞. We shall also make use of the following result. Theorem 5.2.11 Let Ω be any nonempty set. Then the map (f, g) 7→ kf − gk∞ := sup |f (t) − g(t)|, t∈Ω

is a complete metric on B(Ω).

f, g ∈ B(Ω),

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Measure and Integration

Proof. Let (fn ) be a Cauchy sequence in B(Ω). Then for every t ∈ Ω, the sequence (fn (t)) is a Cauchy sequence of complex numbers. Hence, (fn (t)) converges for every t ∈ Ω. Let f (t) := lim fn (t), n→∞

t ∈ Ω.

We show that f ∈ B(Ω) and kfn − f k∞ → 0. Now, let k ∈ N be such that kfn − fk k∞ ≤ 1 for all n ≥ k. Then we have |fn (t)| ≤ kfn − fk k∞ + kfk k∞ < 1 + kfk k∞ for all n ≥ k and for all t ∈ Ω. Hence, |f (t)| ≤ max{1 + kfk k∞ , kf1 k∞ , kf2 k∞ , . . . , kfk k∞ }

∀t ∈ Ω

showing that f ∈ B(Ω). Next, let ε > 0 be given. Let N ∈ N be such that kfn − fm k∞ < ε for all n, m ≥ N . Let t ∈ Ω and let m ≥ N . Then we have |fn (t) − fm (t)| ≤ kfn − fm k∞ < ε for all n ≥ N . In particular, |fn (t) − fm (t)| < ε for all n ≥ N . Letting n tend to ∞, |f (t) − fm (t)| ≤ ε. Then, for every n ≥ N , we have |f (t) − fn (t)| ≤ |f (t) − fm (t)| + |fm (t) − fn (t)| < 2ε. Since N is independent of t, it follows that kf − fn k∞ < 2ε for all n ≥ N . Thus, the Cauchy sequence (fn ) converges in B(Ω). Theorem 5.2.12 For 1 ≤ p ≤ ∞, the metric (f, g) 7→ kf − gkp on Lp (µ) is complete. Proof. First we consider the case 1 ≤ p < ∞: Let (fn ) be a Cauchy sequence in Lp (µ). By Lemma 5.2.10, it is enough to show that (fn ) has a subsequence which converges a.e. Since (fn ) is a Cauchy sequence, for each i ∈ N, there exists ni ∈ N such that kfn − fni kp < 1/2i for all n ≥ ni . Without loss of generality we may assume that ni ≤ ni+1 for all i ∈ N. Then we have kfni+1 − fni kp < 1/2i for all i ∈ N. Note that fnk = fn1 +

k−1 X

(fni+1 − fni ).

i=1

P∞ Thus, it is enough to show that i=1 (fni+1 − fni ) converges a.e. on X. Now, let k ∞ X X gk := |fni+1 − fni |, g := |fni+1 − fni |. i=1

i=1

Integral of Complex Measurable Functions

135

Then gk (x) → Zg(x) as kZ→ ∞ for every x ∈ X, and by monotone convergence theorem, lim gkp = g p . But, k→∞

X

X

Z X

gkp

1/p

= kgk kp ≤

k X

kfni+1 − fni kp ≤ 1.

i=1

R Hence,P X g p < ∞. Therefore, by Theorem 4.2.11 (ii), g(x) < ∞ a.e. on X. ∞ Thus i=1 (fni+1 − fni ) converges a.e. on X, completing the proof for the case 1 ≤ p < ∞. Next we consider the case p = ∞. Let (fn ) be a Cauchy sequence in L∞ (µ). Then the sets Ak := {x ∈ X : |fk (x)| > kfk k∞ } and Bm,n := {x ∈ X : |fn (x) − fm (x)| > kfn − fm k∞ } are of measure zero for every k, m, n ∈ N. Hence, the set [   [  E := Ak ∪ Bm,n k

m,n

is of measure zero, and for each x ∈ Ω := E c , |fk (x)| ≤ kfk k∞ for all k ∈ N, and |fn (x) − fm (x)| ≤ kfn − fm k∞ ∀ m, n ∈ N. Since (fn ) is Cauchy in L∞ (µ), it follows that (f˜n ) with f˜n = fn |Ω is a Cauchy sequence in B(Ω) with respect to k · k∞ . By Theorem 5.2.11, B(Ω) is complete with respect to k · k∞ . Hence, there exists f˜ ∈ B(Ω) such that kf˜n − f˜k∞ → 0 as n → ∞. Defining f (x) = f˜(x) for x ∈ Ω and f (x) = 0 for x 6∈ Ω, it follows that f ∈ L∞ (µ), and kfn − f k∞ → 0 as n → ∞. Note that, as part of the proof of the above theorem, we have proved the following. Proposition 5.2.13 If 1 ≤ p < ∞, then every Cauchy sequence in Lp (µ) has a subsequence which converges almost everywhere. In particular, if (fn ) is a sequence in Lp (µ) which converges to f in Lp (µ), then (fn ) has a subsequence which converges to f a.e. Before closing this subsection, let us observe certain relations between Lp spaces. Theorem 5.2.14 If µ(X) < ∞, then for 1 ≤ p ≤ r ≤ ∞, Lr (µ) ⊆ Lp (µ).

136

Measure and Integration

Proof. Let µ(X) < ∞ and f be a complex measurable function on X. Suppose 1 ≤ p < ∞. Then we have Z |f |p dµ ≤ kf kp∞ µ(X). X

In particular, L∞ (µ) ⊆ Lp (µ) 1 p

for every p ∈ [1, ∞).

1 q

Also, for p, q ∈ (1, ∞) with + = 1, by H¨older’s inequality, Z Z 1/p |f | dµ ≤ |f |p dµ [µ(X)]1/q . X

X

Thus, Lp (µ) ⊆ L1 (µ)

for every p ∈ [1, ∞).

Next, let 1 < p < r < ∞. Again by H¨ older’s inequality, we have Z Z 1/s |f |p dµ ≤ |f |ps dµ [µ(X)]1/t X

X

for any s, t ∈ (1, ∞) with 1s + 1t = 1. Taking s = r/p, we have 1/t = 1 − p/r so that Z p/r Z |f |r dµ [µ(X)]1−p/r . |f |p dµ ≤ X

X r

p

Thus, L (µ) ⊆ L (µ). Exercise 5.2.15 Suppose 1 ≤ s ≤ p ≤ ∞. Show that, if f ∈ Lp (µ) is such that µ({x ∈ X : f (x) 6= 0}) < ∞, then f ∈ Ls (µ). ♦ If µ(X) = ∞, then it is not necessary to hold the relations Lp (µ) ⊆ Lr (µ) or Lr (µ) ⊆ Lp (µ) (see Problem 12). However, we have the following result. Theorem 5.2.16 If 1 ≤ p ≤ r ≤ ∞, then `p (N) ⊆ `r (N). Further, if p < r, then the above inclusion is proper. Proof. Suppose (an ) ∈ `p (N) for 1 ≤ p < ∞. Then, (an ) converges to 0. In particular, (an ) is a bounded sequence. Hence, `p (N) ⊆ `∞ (N) for every p with 1 ≤ p < ∞. Next, let 1 ≤ p < r < ∞ and (an ) ∈ `p (N). Let M > 0 be such that |an | ≤ M for all n ∈ N. Then ∞ X n=1 p

|an |r =

∞ X n=1 r

|an |p |an |r−p ≤ M r−p

∞ X

|an |p .

n=1

This shows that ` (N) ⊆ ` (N). To see the last part, let 1 ≤ p < r < ∞, P∞ 1 P∞ 1 1 and consider the sequence (1/n p ). Since n=1 n diverges and n=1 nr/p converges, we have (1/n1/p ) ∈ `r (N) \ `p (N). Also, if an = 1 for all n ∈ N, then (an ) ∈ `∞ (N) \ `p (N) for any p ∈ [1, ∞).

Integral of Complex Measurable Functions

137

More generally, for any denumerable set X with counting measure µ, we have the proper inclusions Lp (µ) ⊂ Lr (µ) whenever 1 ≤ p < r ≤ ∞ (see Problem 13). We close this subsection with another application of DCT in connection with Lp functions. We shall also make use of this result in the context of Fourier transform in Chapter 7. We may recall from calculus (see [10]) thatRif f : R → R is Riemann n integrable on [−n, n] for every n ∈ N and if lim −n |f (x)|dx exists, then the n→∞ R∞ integral −∞ f (x)dx, called an improper integral of f , is defined by Z ∞ Z n f (x)dx = lim f (x)dx. n→∞

−∞

−n

An analogous result is true in the context of general integral also. Proposition 5.2.17 Let E ∈ A. For each n ∈ N, let En ∈ A be such that µ(En ) < ∞ and χEn → 1 pointwise. Let 1 ≤ p < ∞, f ∈ Lp (µ) and fn := χEn f for n ∈ N. Then, for each n ∈ N, fn ∈ Ls (µ) ∩ Lp (µ) for all s ∈ [1, p] and Z |f − fn |p dµ → 0 as n → ∞. X Z Z 1 In particular, if f ∈ L (µ), then f dµ = lim f dµ. n→∞

X

En

Proof. Let n ∈ N and 1 ≤ p < ∞. Then Z Z Z |fn |p dµ = |f |p dµ ≤ |f |p dµ = kf kpp X

En

X

p

so that fn ∈ L (µ). Also, since µ({x ∈ X : fn (x) 6= 0}) ≤ µ(En ) < ∞, it can be seen that fn ∈ Ls (µ) for every s ∈ [1, p] (see Exercise 5.2.15). Further, Z Z p |f − fn | dµ = (1 − χEn )|f |p dµ, X

X

(1 − χEn )|f | → 0 pointwise and (1 − χEn )|f |p ≤ |f |p with |f |p ∈ L1 (µ). Therefore, by DCT (Theorem 5.1.13), Z |f − fn |p dµ → 0 as n → ∞. p

X

If f ∈ L1 (µ), then we have Z Z Z Z f dµ − f dµ = (f − fn )dµ ≤ |f − fn |dµ. X

Since

R X

En

X

|f − fn |dµ → 0 as n → ∞, we obtain

X

R X

f dµ = lim

R

n→∞ En

f dµ.

Remark 5.2.18 Let 1 ≤ p < ∞. Clearly, for every s ∈ [1, p], Ls (µ) ∩ Lp (µ) is a subspace of Lp (µ). Proposition 5.2.17 shows that Ls (µ) ∩ Lp (µ) is dense in Lp (µ). ♦

138

5.2.3

Measure and Integration

Denseness of Cc (Ω) in Lp (Ω) for 1 ≤ p < ∞

We may recall that the space C[a, b] of all complex valued continuous functions defined on [a, b] is a complete metric space with respect to the metric d(f, g) := sup |f (x) − g(x)|,

f, g ∈ C[a, b].

x∈[a,b]

It can be easily shown that C[a, b] is not complete with respect to the metric Z b dp (f, g) := |f (x) − g(x)|p dx, f, g ∈ C[a, b], a

for 1 ≤ p < ∞. Note that C[a, b] ⊆ Lp [a, b]. So, it is natural to ask whether C[a, b] is dense in Lp [a, b]. We shall answer this affirmatively in a slightly more general context. Definition 5.2.19 Let Ω be a (Lebesgue) measurable subset of R. A function f : Ω → C is said to be of compact support if there exists a compact subset K of Ω with f (x) = 0 for all x 6∈ K. ♦ The set of all continuous functions f : Ω → C with compact support is denoted by Cc (Ω). It can be verified easily that Cc (Ω) is a vector space over C. Clearly, if Ω itself is compact, then Cc (Ω) is C(Ω), the space of all continuous functions on Ω. In particular, Cc ([a, b]) = C[a, b]. The question is whether every f ∈ Lp (Ω) can be approximated by a sequence of functions from Cc (Ω). The answer is known to be affirmative if Ω is a locally compact subset of R, that is, if for each x ∈ Ω, there exists an open set in Ω containing x, whose closure in Ω is compact. Proof of this result, in this generality, is beyond the scope of this book; one may refer to Rudin [14]. However, for certain special cases, the result can be proved rather easily. In the following we shall prove the result when Ω is a countable disjoint union of intervals. In particular, Ω can be an open set in R or an interval of any of the following forms: [a, b], [a, ∞), (−∞, b]. First we prove a result for characteristic functions of intervals. Proposition 5.2.20 Let Ω be a measurable subset of R and J ⊆ Ω be an interval of finite length. Let 1 ≤ p < ∞. Then for every ε > 0, there exists fε ∈ Cc (Ω) such that Z |χJ − fε |p dm < ε.

Ω

Proof. Let a and b be the end points of J with a < b. Let δ > 0 be such that a + 4δ < b so that a + 2δ < b − 2δ and [a + δ, b − δ] ⊆ J. Thus, we have a < a + δ < a + 2δ < b − 2δ < b − δ < b. Let gδ be the trapezoidal function defined on R such that it is (a) 1 on [a + 2δ, b − 2δ],

Integral of Complex Measurable Functions

139

(b) linear on [a + δ, a + 2δ] and [b − 2δ, b − δ], and (c) 0 on (−∞, a + δ) and (b − δ, ∞). Thus, gδ : R → R is given by  1    1 δ (x − a − δ) gδ (x) = 1 (b − δ − x)    δ 0

if if if if

x ∈ [a + 2δ, b − 2δ], x ∈ [a + δ, a + 2δ], x ∈ [b − 2δ, b − δ]. x 6∈ [a + δ, b − δ].

Then we see that gδ ∈ Cc (Ω) with gδ (x) = 0 for all x 6∈ [a + δ, b − δ]. Hence, Z

|χJ (x) − gδ (x)|p dx =

Z

a+2δ

2

J

a+δ 1

Z =

2δ

1 [1 − (x − a − δ)]p dx δ

y p dy

0

=

2δ/(p + 1).

Now, for a given ε > 0, we may take δ > 0 such that on Ω by  gδ (x), x ∈ J, fε (x) = 0, x 6∈ J. R Then we have fε ∈ Cc (Ω) and Ω |χJ − fε |p dm < ε.

2δ p+1

< ε and define fε

Theorem 5.2.21 Let 1 ≤ p < ∞ and Ω be a countable disjoint union of intervals. Then Cc (Ω) is dense in Lp (Ω). Proof. Let f ∈ Lp (Ω) and ε > 0. It is enough to prove for the case when f is a non-negative real valued function. In that case, we know, by Theorem 3.4.3, that there is an increasing sequence (ϕn ) of non-negative simple measurable functions on Ω such that ϕn → f pointwise. We observe that, for each n ∈ N, ϕn ∈ Lp (Ω),

|f − ϕn |p ≤ 2p |f |p ,

f − ϕn → 0

pointwise.

Hence, by DCT, Z Ω

|f − ϕn |p dm =

Z

(f − ϕn )p dm → 0.

Ω

Thus, given ε > 0, there exists a non-negative simple measurable function ϕ such that kf − ϕkp < ε. Hence, it is enough to prove that, corresponding to the above ϕ, there exists g ∈ Cc (Ω) such that kϕ − gkp < ε. Note that the simple function ϕ is of Pk the form ϕ = i=1 αi χAi for some αi ∈ [0, ∞) and measurable sets Ai ⊆ Ω with m(Ai ) < ∞ for i = 1, . . . , k. Hence, it is enough to prove that for every

140

Measure and Integration

measurable set E ⊆ Ω with m(E) < ∞, there exists g ∈ Cc (Ω) such that kχE − gkp < ε. So let E be a measurable subset of Ω. We know, by Theorem 2.2.21, that there exists an open set G ⊆ R such that E ⊆ G and m(G) ≤ m(E) + ε,

m(G \ E) < ε.

Let G0 = G ∩ Ω. Then E ⊆ G0 and we have χG0 = χE + χ(G0 \E) so that kχG0 − χE kpp =

Z Ω

|χG0 − χE |p dm =

Z Ω

χ(G0 \E) dm = m(G0 \ E) < ε.

Since G is a countable disjoint union of intervals of finite length, by the asS sumption on Ω, G0 can be written as G0 = n In , where {In } is a countable disjoint family of intervals of finite length. Since m(G) < ∞, ∞ X

[  In = m(G0 ) < ∞. `(In ) = m n

n=1

Let k ∈ N be such that

P∞

n=k+1

Gk ⊆ G0 ,

`(In ) < ε and Gk =

m(G0 \ Gk ) =

∞ X

Sk

n=1 In .

Then

`(In ) < ε.

n=k+1

Hence, kχG0 − χGk kpp =

Z Ω

|χG0 − χGk |p dm =

Z Ω

χ(G0 \Gk ) dm = m(G0 \ Gk ) < ε.

Thus, by Minkowski’s inequality, kχE − χGk kp ≤ kχE − χG0 kp + kχG0 − χGk kp < 2ε1/p . Pk But, χGk = n=1 χIn . Hence, it is enough to prove that for each interval J ⊆ Ω of finite length, there exists g ∈ Cc (Ω) such that kχJ − gkp < ε. This is proved in Proposition 5.2.20.

5.3 5.3.1

Fundamental Theorems Indefinite integral and its derivative

We may recall from calculus that if f : [a, b] → R is a Riemann integrable function, then an indefinite integral of f is the function g : [a, b] → R defined

Integral of Complex Measurable Functions by

141

x

Z

x ∈ [a, b],

f (t)dt,

g(x) = c + a

for some c ∈ R. We know that g is continuous, and if f is continuous, then g is differentiable and g 0 = f . A natural question is: What can we say about g if f is known to be only Lebesgue integrable? In this section we investigate this issue when f is a real or complex valued Lebesgue integrable function. We denote by K the field R of real numbers or the field C of complex numbers, and we denote by L[a, b] the set of all K-valued functions defined on [a, b] which are integrable with respect to the Lebesgue measure on [a, b]. Definition 5.3.1 For f ∈ L[a, b] and c ∈ K, the function g : [a, b] → K defined by Z x g(x) = c + f dm, x ∈ [a, b], a

is called an indefinite integral of f .

♦

As in the case of Riemann integral, we have the following theorem. Theorem 5.3.2 Let f ∈ L[a, b] and g : [a, b] → K be defined by Z x g(x) = f dm, x ∈ [a, b]. a

Then g is continuous. Proof. Let x ∈ [a, b] and (xn ) be a sequence in [a, b] such that xn → x as n → ∞. Then, using the notation χn for χ[a,xn ] , we have Z xn Z b g(xn ) = f dm = χn f dm. a

a

Note that χn f is measurable, |χn f | ≤ |f | and χn f → χn f pointwise. Hence, by DCT, Z b Z b g(xn ) = χn f dm → χn f dm = g(x) a

a

as n → ∞. Thus, g is a continuous function.

5.3.2

Fundamental theorems of Lebesgue integration

We observe that R x if f ∈ L[a, b] is real valued and non-negative, then the function g(x) := a f dm, x ∈ [a, b], is monotonically increasing. Hence, if f is real valued, then writing Z x Z x Z x f dm = f + dm − f − dm, a

a

a

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Measure and Integration

we see that g is a difference of two monotonically increasing functions. In this context, let us state the following theorem whose proof is relegated to the Appendix to this chapter (Section 5.4). Theorem 5.3.3 Let ϕ : [a, b] → R be a monotonically increasing function. Then ϕ is differentiable a.e., ϕ0 is non-negative and Lebesgue measurable, and Z

b

ϕ0 dm ≤ ϕ(b) − ϕ(a).

a

Making use of the above theorem, we prove the following theorem, which is known as a fundamental theorem of Lebesgue integration (FTLI). Theorem 5.3.4 (FTLI-1) Let f ∈ L[a, b] and g : [a, b] → K be defined by Z x g(x) = f dm, x ∈ [a, b]. a 0

Then g is differentiable a.e., g ∈ L[a, b] and g 0 = f a.e. Proof. Without loss of generality, we may assume that f is real valued. Since f ∈ L[a, b], both f + and f − are non-negative integrable functions, and hence, the functions g1 and g2 defined by Z x Z x g1 (x) = f + dm, g2 (x) = f − dm, x ∈ [a, b], a

a

are non-negative, monotonically increasing, and g = g1 − g2 . Hence, by Theorem 5.3.3, g1 and g2 are differentiable a.e., g10 and g20 are non-negative and Lebesgue measurable, and b

Z

gi0 dm ≤ gi (b) − gi (a)

for i = 1, 2.

a

Therefore, g is differentiable a.e., |g 0 | ≤ g10 + g20 and Z a

b

|g 0 |dm ≤

Z a

b

g10 dm +

Z

b

g20 dm ≤ g1 (b) − g1 (a) + g2 (b) − g2 (a).

a

Thus, g 0 ∈ L[a, b]. Now, we show that g 0 = f a.e. We split the proof into two cases: Case (1): f is bounded. Let M > 0 be such that |f (x)| ≤ M for all x ∈ (a, b). Let (tn ) be a sequence of real numbers in (0, 1] such that tn → 0 as n → ∞. We know that g(x + tn ) − g(x) → g 0 (x) tn

a.e.

Integral of Complex Measurable Functions

143

Let us extend the function f to [a, b + 1] by defining its value as 0 on (b, b + 1] and designate the extended function by the same notation f . Note that for x ∈ (a, b), and for all n ∈ N such that x + tn ∈ [x, b], Z g(x + tn ) − g(x) 1 x+tn f dm. fn (x) := = tn tn x Then we have |fn | ≤ M and fn → g 0 almost everywhere. Hence, by DCT, for every y ∈ [a, b], Z y Z y g 0 dm. fn dm → a

a

Note that, for y ∈ [a, b], using the translation invariant property of the Lebesgue measure, Z y Z 1 y fn dm = [g(x + tn ) − g(x)] dm tn a a Z y+tn  Z y 1 = g(x) dm − g(x) dm tn a+tn a Z y+tn  Z a+tn Z y 1 = g(x) dm − g(x)dm − g(x) dm tn y y a Z y+tn  Z a+tn 1 = g(x) dm − g(x) dm . tn y a Since g is continuous, we know from real analysis that Z Z 1 a+tn 1 y+tn g(x) dm → g(y), g(x) dm → g(a) = 0 tn y tn a as n → ∞. Hence, Z y Z 0 g dm = lim n→∞

a

Thus, Z

y

Z

a

y

f dm ∀ y ∈ [a, b].

fn dm = g(y) = a

y

(g 0 − f )dm = 0 ∀ y ∈ [a, b].

a

Therefore, by Theorem 5.1.25, g 0 = f a.e. Case (2): f is not necessarily bounded. Without loss of generality, assume that f is non-negative. For n ∈ N, let fn be as in Theorem 5.1.19, that is,  f (x), f (x) ≤ n, fn (x) = n, f (x) > n. Then fn → f pointwise and f − fn ≥ 0 for all n ∈ N. Let Z x Z x gn (x) := fn dm and hn (x) := (f − fn )dm, a

a

x ∈ [a, b],

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Measure and Integration

for each n ∈ N. Since, for each n ∈ N, gn and hn are monotonically increasing functions on [a, b], by Theorem 5.3.3, they are differentiable almost everywhere and their derivatives gn0 and h0n are non-negative and measurable almost everywhere. Also, by Case (1) above, gn0 = fn a.e. for each n ∈ N. Note that g = gn + hn so that g is differentiable a.e., g is monotonically increasing and g 0 = gn0 + h0n = fn + h0n ≥ fn

a.e.

0

Since fn → f pointwise, we have g ≥ f a.e. Therefore, again by Theorem 5.3.3, Z b Z b g(b) = g(b) − g(a) ≥ g 0 dm ≥ f dm = g(b). a

Hence,

Rb a

0

a

0

(g − f )dm = 0. Since g − f ≥ 0 a.e., we have g 0 = f a.e.

Recall again from calculus that if f : [a, b] → R is a Riemann integrable function and if f = g 0 for some differentiable function g : [a, b] → R, then Z x g(x) = g(a) + f (t)dt ∀ x ∈ [a, b]. a

In this connection, we may ask the following question: If a function g : [a, b] → K is differentiable almost everywhere with g 0 ∈ L[a, b], then do we have the relation Z x g(x) = g(a) + g 0 dm ∀ x ∈ [a, b]? a

Well, it is too much to expect, in view of the fact that an indefinite integral of an integrable function is continuous, and a function which is differentiable a.e. need not be continuous. For instance, consider g := χ[a,c] : [a, b] → R for c = (a + b)/2. Then g is differentiable almost everywhere with g 0 = 0. Note that Z x g(a) + g 0 dm = 1 ∀ x ∈ [a, b] whereas g(x) = 0 ∀ x ∈ [c, b]. a

At this point, we prove that the function g defined in Theorem 5.3.4 is not only continuous, differentiable a.e. and satisfies g 0 = f a.e., but it is also absolutely continuous, as per the following definition. Definition 5.3.5 A function ϕ : [a, b] → K is said to be absolutely continuous if for every ε > 0, there exists δ > 0 such that for every family {Ii : i = 1, . . . , n} of non-overlapping subintervals of [a, b], n X i=1

`(Ii ) < δ

⇒

n X

|ϕ(xi ) − ϕ(yi )| < ε,

i=1

where xi and yi are the end points of Ii , i = 1, . . . , n.

♦

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145

The following statements can be easily verified. 1. If ϕ is a Lipschitz function, that is, if there exists K > 0 such that |ϕ(x)− ϕ(y)| ≤ K|x − y| for all x, y ∈ [a, b], then ϕ is absolutely continuous. 2. If ϕ is absolutely continuous, then ϕ is uniformly continuous. Theorem 5.3.6 Let f ∈ L[a, b] and g : [a, b] → K be defined by Z x g(x) = f dm, x ∈ [a, b]. a

Then g is absolutely continuous. Proof. Let {Ii : i = 1, . . . , n} be a family of non-overlapping subintervals of [a, b] with left and right end points of Ii as xi and yi , respectively. Then we have Z n Z yi n n Z yi X X X ≤ |f |dm = |f |dm, f dm |g(xi ) − g(yi )| = i=1

i=1

xi

i=1

xi

A

Sn Pn where A = i=1 Ii . Note that m(A) = i=1 `(Ii ). Hence, by Theorem 5.1.22, for every ε > 0, there exists δ > 0 such that n X

`(Ii ) < δ

n X

⇒

i=1

|g(xi ) − g(yi )| < ε.

i=1

Thus, g is absolutely continuous. In view of the above theorem, a more appropriate question should be the following: If a function g : [a, b] → K is differentiable a.e. with g 0 ∈ L[a, b] and if g is absolutely continuous, then do we have the relation Z x g(x) = g(a) + g 0 dm ∀ x ∈ [a, b] ? a

We answer this question affirmatively. Theorem 5.3.7 Suppose g : [a, b] → K is absolutely continuous and differentiable a.e. with g 0 ∈ L[a, b]. Then Z x g(x) = g(a) + g 0 dm ∀ x ∈ [a, b]. a

For its proof, we make use of the following lemma, whose proof is given in the Appendix to this chapter (Section 5.4).

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Measure and Integration

Lemma 5.3.8 Suppose ϕ : [a, b] → K is absolutely continuous and ϕ0 exists a.e. If ϕ0 = 0 a.e., then ϕ is a constant function. Remark 5.3.9 We shall see that the condition that ϕ : [a, b] → K is absolutely continuous itself would imply that ϕ0 exists a.e. (see Theorem 5.3.14 and Theorem 5.3.16). ♦ Proof of Theorem 5.3.7. Let Z x h(x) = g 0 dm,

x ∈ [a, b].

a

By Theorem 5.3.6, h is absolutely continuous and, by Theorem 5.3.4, h is differentiable a.e. and h0 = g 0 a.e. Thus, the function g − h is absolutely continuous, differentiable a.e. and (g − h)0 = 0 a.e. Therefore, by Lemma 5.3.8, g − h is a constant function. In particular, g(x) − h(x) = g(a) − h(a) so that Z x g(x) − g(a) = h(x) − h(a) = g 0 dm. a

This completes the proof. Now, we show that the assumption in Theorem 5.3.7 that g is differentiable a.e. with g 0 ∈ L[a, b] is redundant. Thus, in fact, we have the following theorem in place of Theorem 5.3.7, which is again known as a fundamental theorem of Lebesgue integration. Theorem 5.3.10 (Fundamental theorem of Lebesgue integration) Suppose g : [a, b] → K is absolutely continuous. Then g is differentiable a.e., g 0 ∈ L[a, b] and Z x

g 0 dm

g(x) = g(a) +

∀ x ∈ [a, b].

a

For its proof we need some preparatory results. First a definition. Definition 5.3.11 A function ϕ : [a, b] → K is said to be of bounded variation if there exists M > 0 such that P Va,b (ϕ) :=

n X

|ϕ(xi ) − ϕ(xi−1 )| ≤ M

i=1

for every partition P : a = x0 < x1 < · · · < xn = b of [a, b]. The quantity P Va,b (ϕ) := sup Va,b (ϕ) P

is called the total variation of ϕ, where supremum is taken over all partitions P of [a, b]. ♦ Remark 5.3.12 Suppose γ : [a, b] → C is a continuous function. If γ is of bounded variation, then one may define the length of the curve γ as its total variation. ♦

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147

The following results can be seen easily. 1. Every monotonically increasing function ϕ : [a, b] → R is of bounded variation and its total variation is ϕ(b) − ϕ(a) (see Problem 23). 2. Every characteristic function on [a, b] is of bounded variation. As an example of a function which is not of bounded variation, consider the function (see Problem 24) ϕ : [0, 1] → R defined by  sin(1/x), x > 0, ϕ(x) = 1, x = 0. It is also important to observe that a continuous function need not be of bounded variation. For example, consider the function ϕ : [0, 1] → R defined by  x sin(1/x), x > 0, ϕ(x) = 0, x = 0. Then it can be shown that (Problem 25) f ∈ C[0, 1] but not of bounded variation. However, if ϕ : [a, b] → K is a Lipschitz function, then ϕ is of bounded variation. Theorem 5.3.13 Let f ∈ L[a, b] and g : [a, b] → K be defined by Z x g(x) = f dm, x ∈ [a, b]. a

Then g is of bounded variation and Va,b (g) ≤

Rb a

|f |dm.

Proof. Let a = x0 < x1 < · · · < xn = b be a partition of [a, b]. Then n X

n Z X |g(xi ) − g(xi−1 )| =

i=1

≤

xi

xi−1 i=1 Z n x i X

f dm

|f |dm

i=1 Z b

xi−1

|f |dm.

= a

Since f ∈ L[a, b], it follows that g is of bounded variation. Theorem 5.3.14 Every absolutely continuous function on [a, b] is of bounded variation.

148

Measure and Integration

Proof. Let ϕ : [a, b] → K be an absolutely continuous function. Hence, for a given ε > 0, there exists δ > 0 such that for every family {Ii : i = 1, . . . , n} of non-overlapping subintervals of [a, b], n X

⇒

`(Ii ) < δ

i=1

n X

|ϕ(xi ) − ϕ(yi )| < ε,

i=1

where xi and yi are the end points of Ii , i = 1, . . . , n. Consider a partition P : a = t0 < t1 < · · · < tm = b of [a, b]. Let a = a0 < a1 < · · · < ak = b be another partition of [a, b] such that ai −ai−1 < δ for i = 1, . . . , k. Now, consider the combination of the partitions {ti }m i=0 and {ai }ki=0 , say {si }`i=0 , where any two repeated points are taken only once. Then we have ` m X X |ϕ(si ) − ϕ(si−1 )|. |ϕ(ti ) − ϕ(ti−1 )| ≤ i=1

i=1

For each i ∈ {1, . . . , k}, let si,1 , . . . , si,ni be the points from {si }`i=0 that lie in [ai−1 , ai ]. Then we have ni X

|ϕ(si,j ) − ϕ(si,j−1 )| < ε

j=1

for each i ∈ {1, . . . , k}, so that ` X i=1

|ϕ(si ) − ϕ(si−1 )| =

ni k X X

|ϕ(si,j ) − ϕ(si,j−1 )|
0, there exists I ∈ I such that x∈I

and `(I) < ε.

♦

Lemma 5.4.2 (Vitali covering lemma) Let E ⊆ R with m∗ (E) < ∞. If I is a Vitali cover of E, then for every ε > 0, there exist pairwise disjoint intervals I1 , . . . , In in I such that m∗ (E \ ∪ni=1 Ii ) < ε. Proof. Let ε > 0 be given and let G be an open set such that E ⊆ G and m∗ (G) < m∗ (E) + ε. Without loss of generality assume that the intervals in I are closed. Since I is a Vitali cover of E, we can also assume that I ⊆ G for every I ∈ I. For some n ∈ N, consider pairwise disjoint intervals I1 , . . . , In in I. If m∗ (E \ ∪ni=1 Ii ) < ε, then we are done. Otherwise, choose In+1 ∈ I as follows: Let κn := sup{`(I) : I ∈ I, I ∩ Ij 6= ∅, j = 1, . . . , n}. Choose In+1 ∈ I such that `(In ) ≥ `(In+1 ) > κn /2. Since {In } is a disjoint family of intervals and since their union is contained in G, we have ∞ X n=1

∗ `(In ) = µ∗ (∪∞ n=1 In ) ≤ m (G) < ∞.

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Measure and Integration

Hence, there exists N ∈ N such that ∞ X

`(In ) < ε/5.

n=N +1 N N We show that m∗ (E \ ∪N i=1 Ii ) < ε. Let x ∈ E \ ∪i=1 Ii . Since E \ ∪i=1 Ii ⊆ N N G \ ∪i=1 Ii , there exists Ix ∈ I such that Ix ⊆ G \ ∪i=1 Ii . In particular, Ix ∩ Ij = ∅, j = 1, . . . , N . Also, there exists some n > N such that Ix ∩ In 6= ∅. For, otherwise, κn > `(Ix ) for all n > N which is not possible, since κn → 0 as n → ∞. Note that

Ix ∩ In 6= ∅

⇒

`(Ix ) ≤ κn ≤ 2`(In+1 ) ≤ 2`(In ).

Therefore, Ix ⊆ In + 2`(In )[−1, 1] = Jn , say. Thus, ∞ E \ ∪N i=1 Ii ⊆ ∪n=N +1 Jn , P∞ where `(Jn ) ≤ 5`(In ), so that m∗ (E \ ∪N i=1 Ii ) ≤ 5 n=N +1 `(In ) < ε.

The above lemma is used in the proof of the next theorem. Theorem 5.4.3 If f : [a, b] → R is monotonically increasing, then it is differentiable a.e. Proof. For each x ∈ [a, b], define the quantities f (x + h) − f (x) , h h→0+ f (x) − f (x − h) D− f (x) = lim sup , h h→0+ f (x + h) − f (x) D+ f (x) = lim inf , + h h→0 f (x) − f (x − h) . D− f (x) = lim inf h h→0+ D+ f (x)

=

lim sup

Let A = {x ∈ [a, b] : D+ f (x) > D− f (x)}, B = {x ∈ [a, b] : D− f (x) > D+ f (x)}. Note that x 6∈ A ∪ B implies D+ f (x) ≤ D− f (x),

D− f (x) ≤ D+ f (x)

so that D− f (x) ≤ D− f (x) ≤ D+ f (x) ≤ D+ f (x) ≤ D− f (x)

Integral of Complex Measurable Functions

153

which implies that all the four quantities D− f (x), D− f (x), D+ f (x), D+ f (x) are the same, and hence f is differentiable at x. In other words, f is differentiable on [a, b] \ A ∪ B. Hence, it is enough to show that A and B are sets of measure zero. Observe that, x ∈ A ⇐⇒ ∃ s, t ∈ Q such that D+ f (x) > s > t > D− f (x). S Hence, A = s,t∈Q As,t , where As,t := {x ∈ [a, b] : D+ f (x) > s > t > D− f (x)}. Thus, it is enough to prove that As,t is of measure 0 for each s, t ∈ Q. Now, let ε > 0 be given. Then, by the definition of m∗ (·), there exists an open set V ⊆ R such that As,t ⊆ V,

m∗ (V ) ≤ m∗ (As,t ) + ε.

Also, by the definition of D− f , for each x ∈ As,t , there exists hx > 0 such that [x − h, x] ⊆ V, f (x) − f (x − h) < th ∀ h ∈ (0, hx ]. Let IV := {[x − h, x] : h ∈ (0, hx ], x ∈ As,t }. We observe that IV is a Vitali cover of As,t . Hence, by Vitali covering lemma (Lemma 5.4.2), there exist disjoint intervals I1 , . . . , In in IV such that m∗ (As,t \ ∪nj=1 Ij ) < ε. Writing Ij := [xj − hj , xj ], j = 1, . . . , n, we have n n X X [f (xj ) − f (xj − hj )] ≤ t hj ≤ tm∗ (V ) < t[m∗ (As,t ) + ε]. j=1

j=1

Now, let G = As,t ∩ (∪nj=1 Ij◦ ), where S ◦ denotes the interior of the set S. Then for every y ∈ G, there exists ky > 0 and jy ∈ {1, . . . , n} such that [y, y + k] ⊆ Ijy ,

f (y + k) − f (y) > sk

∀k ∈ (0, ky ].

We observe that IG := {[y, y + k] : k ∈ (0, ky ], y ∈ G} is a Vitali cover of G. Hence, there exist disjoint intervals J1 , . . . , Jm in IG such that m∗ (G \ ∪m i=1 Ji ) < ε.

154

Measure and Integration

Let Ji = [yi , yi + ki ], i = 1, . . . , m. Then we have m m X X [f (yi + ki ) − f (yi )] > s ki . i=1

i=1

Since each Ji is contained in some Ij , we have m X [f (yi + ki ) − f (yi )]

=

i=1

n X

X

[f (yi + ki ) − f (yi )]

j=1 {i:Ji ⊆Ij }

=

n X [f (xj ) − f (xj − hj )] j=1


m (G) − ε > m (As,t ) − 2ε.

i=1

Applying this on (∗), we obtain s[m∗ (As,t ) − 2ε] < s

m X

ki < t[m∗ (As,t ) + ε].

i=1

Hence, s[m∗ (As,t ) − 2ε] < t[m∗ (As,t ) + ε] for every ε > 0 so that s m∗ (As,t ) ≤ t m∗ (As,t ). Since t < s, it follows that m∗ (As,t ) = 0. We have shown that m∗ (A) = 0. Following analogous arguments, we obtain ∗ m (B) = 0. This completes the proof.

Integral of Complex Measurable Functions

155

For both the theorems above, we did not use any measure theoretic arguments except the fact that if {In } is a countable disjoint family of intervals, then ∞ X m∗ (∪∞ I ) = `(In ). n n=1 n=1

Making use of Theorem 5.4.3, we prove Theorem 5.3.3. We shall also make use of Fatou’s lemma, though the statement of the theorem is quite measuretheoretic-free. Proof of Theorem 5.3.3. By Theorem 5.4.3, ϕ is differentiable a.e. Let E ⊆ [a, b] be such that ϕ0 (x) exists for all x ∈ E and m∗ ([a, b] \ E) = 0. Let  0 ϕ (x), x ∈ E, f (x) = 0, x ∈ [a, b] \ E. Define ϕ(x) = ϕ(b) for x ∈ [b, b + 1] and for n ∈ N, let fn (x) = n[ϕ(x + 1/n) − ϕ(x)],

x ∈ [a, b].

Note that each fn is non-negative, and fn → f pointwise. Hence, f is measurable; in particular, ϕ0 is measurable. Also we have Z

b

Z

ϕ(x)dx

a

Z

a b+1/n

Z ϕ(y)dy − n

= n a+1/n

Z

b+1/n

ϕ(y)dy − n

ϕ(b)dy = ϕ(b),

b

Z

ϕ(y)dy.

b+1/n

Z ϕ(y)dy = n b

a+1/n

Z ϕ(y)dy ≥ n

n

a+1/n

a

b+1/n

n

ϕ(y)dy Z

b

Z

b

a

= n But

b

Z ϕ(x + 1/n)dx − n

fn (x)dx = n a

b

a+1/n

ϕ(a)dy = ϕ(a).

a

a

Hence, Z

b

fn (x)dx ≤ ϕ(b) − ϕ(a). a

Therefore, by Fatous’ lemma, Z

b

Z f (x)dx ≤ lim inf

a

Thus, the proof is complete.

n

b

fn (x)dx ≤ ϕ(b) − ϕ(a). a

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Measure and Integration

Recall from Theorem 5.3.14 that every absolutely continuous function is of bounded variation and hence by Theorem 5.3.16 it is differentiable a.e. Now, as a consequence of Vitali covering lemma, we prove Lemma 5.3.8. Proof of Lemma 5.3.8. Let ϕ : [a, b] → K be absolutely continuous and ϕ0 = 0 a.e. Let c ∈ (a, b]. We show that f (c) = f (a). Now, let ε > 0 be given, and let δ > 0 be as in the definition of absolute continuity of ϕ (see Definition 5.3.5). Let E := {x ∈ [a, c] : f 0 (x) = 0}. Then for each x ∈ E, there exists αx ∈ [x, c] such that |ϕ(y) − ϕ(x)| < ε(y − x) whenver x < y < αx .

(1)

Then I := {[x, y] : x ∈ E, x < y < αx } is a Vitali cover of E. Hence, by Vitali covering lemma (Lemma 5.4.2), there exists a finite subfamily {[xi , yi ] : i = 1, . . . , k} of I such that m∗ (E \

k [

[xi , yi ]) < δ.

i=1

Since m∗ ([a, c] \ E) = 0, from the above inequality, we obtain k   [ m∗ [a, c] \ [xi , yi ] < δ, i=1

Pk

equivalently, (c − a) − i=1 (yi − xi ) < δ. Thus, taking y0 = a and xk+1 = c, Pk we have i=0 (xi+1 − yi ) < δ. Therefore, by the absolute continuity of ϕ, k X

|ϕ(xi+1 ) − ϕ(yi )| < ε.

(2)

i=1

Also, by (1), we have k X

|ϕ(yi ) − ϕ(xi )| < ε

i=1

k X (yi − xi ) ≤ ε(c − a).

(3)

i=1

The relations (2) and (3) imply k k X X |ϕ(c) − ϕ(a)| = (ϕ(xi+1 ) − ϕ(yi )) + (ϕ(yi ) − ϕ(xi )) i=1

=

k X

|ϕ(xi+1 ) − ϕ(yi )| +

i=1


0. Therefore, we have ϕ(c) = ϕ(a), and the proof is complete.

Integral of Complex Measurable Functions

5.5

157

Problems R

1. If f ∈ L(µ) such that

X

f ≥ 0, then show that

R X

f=

R X

Ref.

2. Suppose fR is a real measurable function. Prove that, if f ≥ 0 a.e. or if f ≤ 0 a.e., then X f dµ is meaningful in the sense of Definition 5.1.1. Further, prove the following: R R (a) If f ≥ 0 a.e., then X f dµ = X f + dµ. R R (b) If f ≤ 0 a.e., then X f dµ = − X f − dµ. 3. If f is a bounded complex measurable function and µ(X) < ∞, then prove that f ∈ L(µ). 4. Prove Corollary 5.1.7. 5. Show that L(µ) is a vector space over C, and f 7→ on L(µ).

R X

f is a linear functional

6. Suppose µ(X) < ∞ and (fn ) is a uniformly bounded sequence of measurable functions. Prove the following: (a) fn ∈ L1 (µ) for every n ∈ N. R R (b) If fn → f a.e., then f ∈ L1 (µ) and X fn dµ → X f dµ. R 7. Show that the set N := {f ∈ L(µ) : X |f | = 0} is a subspace of the vector R space L(µ), and the map [f ] 7→ X |f | is a norm on the quotient space L(µ)/N . R 8. For f ∈ L(R, m), show that the integral R f (x)e−itx dx exists for each t ∈ R R and the function t 7→ R f (x)e−itx dx is continuous and bounded on R. [Hint: Use DCT.] R R 9. If f ∈ L(µ) such that X f = X |f |, then show that there exists c ∈ C such that f (x) = c|f (x)| for almost all x ∈ X. R R [Hint: Write XRf as α X f for some complex number α with |α| =R1, and use the facts (i) g ≥ 0 implies g is real valued a.e., and (ii) g ≥ 0 and g = 0 implies g = 0 a.e.] R 10. Prove that [f ] 7→ X |f |dµ is a norm on the space L(µ) := L(µ)/Z, where Z = {f ∈ L(µ) : f = 0 a.e.}. 11. Justify the statement: Suppose (fn ) is a sequence of Riemann integrable functions on [a, b] such that fn → f pointwise on [a, b] and if (fn ) is uniformly Rb R bounded, then f is Lebesgue integrable, and we have a fn (x) dx → X f. 12. Let  1   , x f (x) =   0,

x ∈ [1, ∞), and x 6∈ [1, ∞)

 1    √ , x g(x) =    0,

Show that f ∈ L2 (R) \ L1 (R) and g ∈ L1 (R) \ L2 (R).

x ∈ (0, 1), x 6∈ (0, 1)

158

Measure and Integration

13. Let X be a denumerable set with counting measure µ on X. Show that for 1 ≤ p < r ≤ ∞, Lp (µ) ⊆ Lr (µ). 14. Prove that for 1 ≤ p < ∞, the metric d(·, ·) defined by Z b |f (x) − g(x)|p dx, f, g ∈ C[a, b], dp (f, g) := a

on C[a, b] is not complete. 15. Prove that for 1 ≤ p < ∞, the space P[a, b] of all polynomial functions on [a, b] is dense in C[a, b] with respect to the metric (f, g) 7→ kf − gkp . [Hint: Use the fact that every function in C[a, b] is a uniform limit of a sequence of polynomials, and that kf − gkp ≤ kf − gk∞ (b − a)1/p .] 16. For p, q ∈ [1, ∞], let (fn ) and (gn ) be sequences in Lp (µ) and Lq (µ), respectively, such that kfn − f kp → 0 and kgn − gkq → 0 for some f ∈ Lp (µ) and g ∈ Lq (µ). Prove that kfn gn − f gk1 → 0. 17. Suppose p, q, r ∈ (1, ∞] are such that r1 = p1 + 1q . If f ∈ Lp (µ) and g ∈ Lq (µ), then prove that f g ∈ Lr (µ) and kf gkr ≤ kf kp kgkq . [Hint: Write

1 p/r

+

1 q/r

= 1 and apply H¨ older’s inequality.]

for 0 ≤ θ ≤ 1. 18. Let p, q, r ∈ (1, ∞] be such that p ≥ r, q ≥ r and r1 = pθ + 1−θ q If f ∈ Lp (µ) ∩ Lq (µ), then prove that f ∈ Lr (µ) and kf kr ≤ kf kθp kf k1−θ . q [Hint: Write inequality.]

1 p/rθ

+

1 q/r(1−θ)

= 1 and r = rθ + r(1 − θ) and apply H¨ older’s

19. Let f ∈ L1 ((0, 2π]). Using the notation introduced in the beginning of Section 5.3.1, define the n-th Fourier coefficient of f by Z 2π 1 fˆ(n) := f (x)e−inx dm(x) for n ∈ Z. 2π 0 Also, extending 2π-periodically, define fτ (x) := f (x − τ ) for x, τ ∈ R. Prove the following: (a) |fˆ(n)| ≤

1 kf k1 2π

(b) |fˆ(n)| ≤

1 kf 4π

for all n ∈ Z.

− fπ/n k1 for all n ∈ Z.

(c) The function τ 7→ fτ is continuous from R to L1 ((0, 2π]). (Hint: Cc (0, 2π] is dense in L1 ((0, 2π]) and function in Cc (0, 2π]) are uniformly continuous.) (d) |fˆ(n)| → 0 as |n| → ∞ so that |fˆ(n)| = o(1). (e) If f is absolutely continuous, then |fˆ(n)| ≤ that |fˆ(n)| = o(1/|n|).

1 ˆ0 |f (n)| |n|

for all n ∈ Z so

[Hint: Theorem 5.3.10.] 20. Let f ∈ L1 ((0, 2π]) be absolutely continuous. Using the notations in Problem 1 ˆ0 19, prove that |fˆ(n)| ≤ |n| |f (n)| for all n ∈ Z so that |fˆ(n)| = o(1/|n|). [Hint: Theorem 5.3.10.]

Integral of Complex Measurable Functions

159

21. Let f : [a, b] → R be continuous and differentiable on (a, b), and there exists M > 0 such that |f 0 (x)| ≤ M for all x ∈ (a, b). Show that f is absolutely continuous. 22. If f : [a, b] → R is an absolutely continuous and f (x) 6= 0 for all x ∈ [a, b], then show that 1/f is also absolutely continuous. 23. Show that every monotonically increasing function F : [a, b] → K is of bounded variation and its total variation is F (b) − F (a).  sin(1/x), x > 0, 24. Show that f : [0, 1] → R defined by f (x) = is not of 1, x=0 bounded variation.  x sin(1/x), x > 0, 25. Show that f : [0, 1] → R defined by f (x) = is contin0, x=0 uous, but not of bounded variation. 26. Suppose f : [a, b] → K is of bounded variation and |f | ≥ c > 0 on [a, b]. Show that 1/f is of bounded variation on [a, b]. 27. Suppose f and g are functions of bounded variation on [a, b]. Show that f g is also of bounded variation on [a, b]. 28. Suppose f : [a, b] → R is continuous and differentiable on (a, b), and there exists M > 0 such that |f 0 (x)| ≤ M for all x ∈ (a, b). Show (without using Problem 21) that f is also of bounded variation on [a, b]. Rb 29. Show that, if ϕ : [a, b] → C is absolutely continuous, then Va,b (ϕ) ≤ a |ϕ0 |dm. 30. Write the steps involved in proving Theorems 5.3.18 and 5.3.19.

Chapter 6 Integration on Product Spaces

So far we have been concerned with measurable functions of one variable and their integration. In this chapter we consider measurable functions of more than one variable, that is, functions on measurable spaces of the form X := X1 × X2 × · · · × Xk with appropriate σ-algebras and measures on them. We shall restrict our study for the case of k = 2. Thus, the idea is to construct a new σ-algebra and a measure on X1 × X2 using the measure spaces X1 and X2 , and see how integration on the product space is related to the integration on the component spaces. For this purpose we shall use most of the concepts and basic theorems introduced in the previous chapters.

6.1

Motivation

Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be measure spaces. We would like to have a σ-algebra A1 ⊗ A2 on X1 × X2 and a measure µ, called the product measure on A1 ⊗ A2 with the following properties: (1) A1 ⊗ A2 ⊇ {A × B : A ∈ A1 , B ∈ A2 }. (2) µ(A × B) = µ1 (A)µ2 (B) ∀ A ∈ A1 , B ∈ A2 . Let f be a non-negative extended real valued function on X1 × X2 . For each x ∈ X1 and y ∈ X2 , let fx : X2 → [0, ∞] and f y : X1 → [0, ∞] be defined by fx (u) = f (x, u), f y (v) = f (v, y),

u ∈ X2 , v ∈ X1 ,

respectively. We would like to show that, if f is measurable, then (3) fx and f y are measurable with respect to A2 and A1 , respectively, for each x ∈ X1 and y ∈ X2 , (4) the functions g : X1 → [0, ∞] and h : X2 → [0, ∞] defined by Z g(x) = fx (y)dµ2 (y), x ∈ X1 , X2

161

162

Measure and Integration Z h(y) = f y (x)dµ1 (x), y ∈ X2 , X1

respectively, are measurable, and R R R (5) X1 ×X2 f dµ = X1 gdµ1 = X2 hdµ2 . The equalities in (5) are written, sometimes, as Z Z Z  f (x, y)dµ(x, y) = f (x, y)dµ2 (y) dµ1 (x) X1 ×X2 ZX1  ZX2  = f (x, y)dµ1 (x) dµ2 (y). X2

X1

We would also like to have results in (3), (4), (5) above for a complex valued measurable function f , whenever f is integrable with respect to the product measure. We shall show that the existence of a product measure with the required properties will be guaranteed whenever µ1 and µ2 are σ-finite measures.

6.2

Product σ-algebra and Product Measure

Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be measure spaces. Definition 6.2.1 Sets of the form A × B with A ∈ A1 and B ∈ A2 are called measurable rectangles, and the σ-algebra generated by the family of all measurable rectangles is called the product σ-algebra. We denote the product σ-algebra by A1 ⊗ A2 . ♦ Definition 6.2.2 For any set E ⊆ X1 × X2 and (x, y) ∈ X1 × X2 , let Ex Ey

:= {v ∈ X2 : (x, v) ∈ E}, := {u ∈ X1 : (u, y) ∈ E}.

The sets Ex and E y are called x-section and y-section, respectively, of the set E. ♦ It can be easily seen that E⊆F

⇒

Ex ⊆ Fx

and E y ⊆ F y .

In the following lemma, we state some easily verifiable facts. Lemma 6.2.3 Let E be a measurable rectangle, say E = A × B with A ∈ A1 and B ∈ A2 . Then,   B, x ∈ A, A, y ∈ B, Ex = Ey = ∅, x 6∈ A, ∅, y 6∈ B.

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Further, the following are true. (i) Ex ∈ A2 and E y ∈ A1 for all (x, y) ∈ X1 × X2 . (ii) The functions x 7→ µ2 (Ex ) and y 7→ µ1 (E y ) are measurable. R R (iii) X1 µ2 (Ex )dµ1 = µ1 (A)µ2 (B) = X2 µ1 (E y )dµ2 . The results in the above lemma prompt us to ask whether the following statements are true for every E ∈ A1 ⊗ A2 : (a) Ex ∈ A2 , E y ∈ A1 for every (x, y) ∈ X1 × X2 , (b) x 7→ µ2 (Ex ) and y 7→ µ1 (E y ) are measurable, R R (c) X1 µ2 (Ex )dµ1 (x) = X2 µ1 (E y )dµ2 (y), and R (d) E 7→ X1 µ2 (Ex )dµ1 (x) is a measure on A1 ⊗ A2 . Our first attempt in this chapter is to prove that the above results are true provided µ1 and µ2 are σ-finite. For this, we shall make use of another easily verifiable proposition. Proposition 6.2.4 For every E ⊆ X1 × X2 , (E c )x = Exc ,

(E c )y = (E y )c

and for En ⊆ X1 × X2 , n ∈ N, (∪En )x = ∪(En )x ,

(∪En )y = ∪(En )y .

Further, if {En : n ∈ N} is a disjoint family of subsets of X1 × X2 , then {(En )x : n ∈ N} and {Eny : n ∈ N} are disjoint families. Notation: For n ∈ N, if En ∈ X1 × X2 , then we shall use the notations En,x and Eny for (En )x and (En )y , respectively. Theorem 6.2.5 Let E ∈ A1 ⊗ A2 . Then for every (x, y) ∈ X1 × X2 , we have Ex ∈ A2 and E y ∈ A1 . Proof. Let S be the family of all E ∈ A1 ⊗ A2 such that Ex ∈ A2 for all x ∈ X1 . We have to show that S = A1 ⊗ A2 . For this, it is enough to show that S is a σ-algebra containing all measurable rectangles. We have already observed that if E is a measurable rectangle, then Ex ∈ A1 and E y ∈ A2 for every (x, y) ∈ X1 × X2 so that S contains all measurable rectangles. The fact that S is a σ-algebra follows from Proposition 6.2.4. Similarly we see that the family of all E ∈ A1 ⊗ A2 such that E y ∈ A1 for all y ∈ X2 is A1 ⊗ A2 .

164

Measure and Integration

Theorem 6.2.6 Suppose (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) are σ-finite measure spaces. Then for every E ∈ A1 ⊗ A2 , the functions x 7→ µ2 (Ex ),

y 7→ µ1 (E y )

are measurable with respect to A1 and A2 , respectively, and Z Z µ2 (Ex )dµ1 (x) = µ1 (E y )dµ2 (y). X1

X2

For the proof of the above theorem we shall also make use of a lemma (Lemma 6.2.9) whose statement requires the following two definitions. Definition 6.2.7 A subset of X1 × X2 is called an elementary set if it is a disjoint union of a finite number of measurable rectangles. ♦ Definition 6.2.8 Let X be a set. A family S of subsets of a set X is called a monotone class if it has the following two properties: (1) If Ai ∈ S and Ai ⊆ Ai+1 for all i ∈ N, then ∪Ai ∈ S, (2) If Ai ∈ S and Ai ⊇ Ai+1 for all i ∈ N, then ∩Ai ∈ S.

♦

We observe the following: (a) Given any family F of subsets of X, there exists a smallest monotone class containing F, called the monotone class generated by F. (b) A1 ⊗ A2 is a monotone class containing all elementary sets. Notation: Given any family F of subsets of X, the monotone class generated by F is denoted by MF . Now, we state the required lemma; its proof is given at the end of this subsection. Lemma 6.2.9 The σ-algebra A1 ⊗ A2 is the smallest monotone class containing all elementary sets. Proof of Theorem 6.2.6. Let S be the class of all E ∈ A1 ⊗ A2 such that the conclusions in the theorem hold. We prove that S = A1 ⊗ A2 so that the proof will be complete. The following facts can be verified easily: (a) S contains all measurable rectangles. (b) S contains all elementary sets. (c) S contains finite disjoint unions of its members.

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We claim that S has also the following properties: (i) If En ∈ S with En ⊆ En+1 for all n ∈ N, then ∪En ∈ S. (ii) If {En } is a disjoint family in S, then ∪En ∈ S. (iii) If En ∈ S such that En ⊇ En+1 for all n ∈ N and there exists a measurable rectangle A × B satisfying E1 ⊆ A × B, µ1 (A) < ∞ and µ2 (B) < ∞, then ∩En ∈ S. Note that, if µ1 and µ2 are finite measures, then (i) and (iii) imply that S is a monotone class (containing all measurable rectangles), so that by Lemma 6.2.9, S = A1 ⊗ A2 , and hence the proof is complete in this case. Proof of (i): Let En ∈ S with En ⊆ En+1 for all n ∈ N. By the definition of S, for each n ∈ N, the functions y 7→ µ1 (Eny )

x 7→ µ2 (En,x ), are measurable functions and Z Z µ2 (En,x )dµ1 (x) = X1

µ1 (Eny )dµ2 (x).

X2

Also we have En,x ⊆ En+1,x ,

y Eny ⊆ En+1

∀n ∈ N

so that µ2 (En,x ) → µ2 (∪En,x ) = µ2 ((∪En )x ), µ1 (Eny ) → µ1 (∪Eny ) = µ1 ((∪En )y ) as n → ∞. Hence, the functions y 7→ µ1 ((∪En )y )

x 7→ µ2 ((∪En )x ),

are measurable, and by MCT (Theorem 4.2.15), we have Z Z lim µ2 (En,x )dµ1 (x) = µ2 ((∪En )x )dµ1 (x), n→∞

X1

Z lim

n→∞

X2

X1

µ1 (Eny )dµ2 (y) =

Z

µ1 ((∪En )y )dµ2 (y).

X2

Therefore, ∪En ∈ S. Proof of (ii): Let {En } be a disjoint family in S. Then ∪En = ∪Fn where Fn = ∪ni=1 Ei , n ∈ N. Since each Fn is a finite disjoint union of members of S, we have Fn ∈ S for every n ∈ N. Also, Fn ⊆ Fn+1 for every n ∈ N. Hence, by (i), ∪En = ∪Fn ∈ S.

166

Measure and Integration

Proof of (iii): As in (i), the functions y 7→ µ1 (Eny )

x 7→ µ2 (En,x ), are measurable functions and Z Z µ2 (En,x )dµ1 (x) = X1

µ1 (Eny )dµ2 (y).

X2

Since A × B ⊇ E1 , we have E1y ⊆ (A × B)y .

E1,x ⊆ (A × B)x , Hence, by Lemma 6.2.3,

µ2 ((A × B)x ) = µ2 (B)χA (x) < ∞, µ1 ((A × B)y ) = µ1 (A)χB (y) < ∞. Also, the condition En ⊇ En+1 for all n ∈ N, consequently, the facts En,x ⊇ En+1,x ,

y Eny ⊇ En+1

∀n ∈ N,

imply the convergence µ2 (En,x ) → µ2 (∩En,x ) = µ2 ((∩En )x ), µ1 (Eny ) → µ1 (∩Eny ) = µ1 ((∩En )y ) as n → ∞. Hence, the functions y 7→ µ1 ((∩En )y )

x 7→ µ2 ((∩En )x ), are measurable. Again since

µ2 (En,x ) ≤ µ2 ((A × B)x ) = µ2 (B)χA (x), µ1 (Eny ) ≤ µ1 ((A × B)y ) = µ1 (A)χB (y) with Z

Z µ2 (B)χA (x)dµ1 (x) = µ2 (B)µ1 (A) =

X1

µ1 (A)χB (y)dµ2 (y) X2

for all n ∈ N, by DCT (Theorem 5.1.13), Z Z lim µ2 (En,x )dµ1 (x) = n→∞

X1

Z lim

n→∞

X2

µ2 ((∩En )x )dµ1 (x),

X1

µ1 (Eny )dµ2 (y) =

Z X2

µ1 ((∩En )y )dµ2 (y).

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167

Therefore, ∩En ∈ S. Now, since µ1 and µ2 are σ-finite measures, there exist disjoint families (n) (n) {X1 : n ∈ N} and {X2 : n ∈ N} of measurable subsets of X1 and X2 , respectively, such that ∞ [

X1 =

(n) X1 ,

X2 =

n=1 (n)

∞ [

(m)

X2

m=1 (m)

with µ1 (X1 ) < ∞ and µ2 (X2 ) < ∞ for all n, m ∈ N. For E ∈ A1 ⊗ A2 , let (n) (m) En,m := E ∩ (X1 × X2 ), n, m ∈ N. Clearly, E is a disjoint union of {En,m : n, m ∈ N}. Let A := {E ∈ A1 ⊗ A2 : En,m ∈ S ∀ n, m ∈ N}. By (i), (ii), and (iii), it can be seen (verify) that A is a monotone class containing all elementary sets. Hence, by Lemma 6.2.9, A = A1 ⊗ A2 . Thus, E ∈ A1 ⊗ A2 implies En,m ∈ S for all n, m ∈ N. Again by (ii) above, E = ∪En,m ∈ S. That is, for every E ∈ A1 ⊗ A2 , the conclusions of the theorem hold. Thus, we have proved that S = A1 ⊗ A2 , which completes the proof. The following corollary is immediate from Theorem 6.2.6. Corollary 6.2.10 Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be σ-finite measure spaces and E ∈ A1 ⊗ A2 . If µ2 (Ex ) = 0 for a.a. x ∈ X1 , then µ1 (Ey ) = 0 for a.a. y ∈ X2 . The following theorem leads to the definition of the product measure. Theorem 6.2.11 Suppose that (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) are σ-finite measure spaces. For E ∈ A1 ⊗ A2 , let Z Z µ(E) := µ2 (Ex )dµ1 (x) = µ1 (E y )dµ2 (y). X1

X2

Then µ is a measure on A1 ⊗ A2 . Proof. Clearly, µ(∅) = 0. Let {En : n ∈ N} be a disjoint family in A1 ⊗A2 . Then, by Proposition 6.2.4, we have Z Z µ(∪En ) = µ2 ((∪En )x )dµ1 (x) = µ2 (∪En,x )dµ1 (x). X1

X1

Now, using the fact that {En,x : n ∈ N} is a disjoint family in A2 and the monotone convergence theorem, we have Z X ∞ ∞ Z ∞ X X µ(∪En ) = µ2 (En,x )dµ1 (x) = µ2 (En,x )dµ1 (x) = µ(En ). X1 n=1

This completes the proof.

n=1

X1

n=1

168

Measure and Integration

Definition 6.2.12 The measure µ in Theorem 6.2.11 is called the product measure on A1 ⊗ A2 , and it is denoted by µ1 × µ2 . ♦ Proof of Lemma 6.2.9. The proof involves two main steps: Step (i): Let F be an algebra on a set X, MF be the monotone class generated by F and AF be the σ-algebra generated by F. Then MF is a σ-algebra and MF = AF . Step (ii): The family E of all elementary sets in A1 ⊗ A2 is an algebra. Since AE , the σ-algebra generated by E, is A1 ⊗ A2 , results in Step 1 and Step 2 will imply the required result ME = A1 ⊗ A2 . Proof of Step (i): Since MF is a monotone class, for showing that MF is a σ-algebra, it is enough to show that it is an algebra, because, in that case, for any (An ) in MF , ∞ [

An =

n=1

as

n [

Ak ∈ MF

∞  [ n [ n=1

and

k=1

 Ak ∈ MF ,

k=1 n [ k=1

Ak ⊆

n+1 [

Ak

∀ n ∈ N.

k=1

f := {A : Ac ∈ MF }. Then it can be seen that M f is a monotone class Let M f containing F. Hence, MF ⊆ M. Thus, A ∈ MF

⇒

Ac ∈ MF .

Next, we show that MF is closed under finite unions, that is, to show that A, B ∈ MF implies A ∪ B ∈ MF ; equivalently, to show that for every A ∈ b = MF , where MF , A b := {B ∈ MF : A ∪ B ∈ MF }. A b is a monotone class. Therefore, for showing So, let A ∈ MF . Note that A b = MF , it is enough to show that F ⊆ A. b So, let C ∈ F. Note that that A b because, D ∈ F implies D ∪ C ∈ F ⊆ MF . Thus, C b is a monotone F ⊆ C, class containing F. Since MF is the smallest monotone class containing F, b = MF . Thus, A ∈ MF = C b so that C ∈ A, b proving that F ⊆ A. b we obtain C Thus, the proof of Step (i) is completed. Proof of Step (ii): It can be easily seen that E is closed under finite disjoint unions and finite intersections. Also, for any measurable rectangle A1 × A2 , (A1 × A2 )c = (Ac1 × A2 ) ∪ (A1 × Ac2 )

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169

so that (A1 × A2 )c is a disjoint union of two measurable rectangles. Hence, (A1 × A2 )c ∈ E. Since each member A of E is a finite disjoint union of measurable rectangles, say A = ∪ni=1TRi , where {Ri : i = 1, . . . , n} is a disjoint family n of rectangles, we have Ac = i=1 Ric ∈ E. Now, if A, B ∈ E, then A ∪ B = (A \ B) ∪ B = (A ∩ B c ) ∪ B, which is a finite disjoint union of members of E. Thus, E is closed under finite unions as well.

6.3

Fubini’s Theorem

Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be measure spaces. Definition 6.3.1 Let f be a function defined on X1 × X2 taking values in another set Y . For each x ∈ X1 , the function fx : X2 → Y defined by fx (y) = f (x, y),

y ∈ X2 ,

is called the x-section of f , and for each y ∈ X2 , the function f y : X1 → Y defined by f y (x) = f (x, y), x ∈ X1 , is called the y-section of f .

♦

Proposition 6.3.2 Let f be a measurable function on X1 × X2 taking values in a topological space Y . Then for each (x, y) ∈ X1 × X2 , fx and f y are measurable with respect to A2 and A1 , respectively. Proof. Let G be an open set in the topological space in which f takes values. Then we see that, for each (x, y) ∈ X1 × X2 , {u ∈ X1 : f y (u) ∈ G} = {u ∈ X1 : f (u, y) ∈ G} = [f −1 (G)]y , {v ∈ X2 : fx (v) ∈ G} = {u ∈ X1 : f (x, v) ∈ G} = [f −1 (G)]x . Since f is measurable, by Proposition 6.2.4, both [f −1 (G)]y and [f −1 (G)]x are measurable sets. Hence the result. By the above proposition (Proposition 6.3.2), if f is an extended real valued and non-negative measurable function on X1 × X2 , then its sections fx and f y are R measurable for each R x ∈ X1 and y ∈ X2 , respectively. Thus, the integrals f (x, y)dµ (y) and f (x, y)dµ1 (x) are well-defined. 2 X2 X1 Now, we prove the Fubini’s theorem for non-negative measurable functions, which is also known as Tonelli’s theorem.

170

Measure and Integration

Theorem 6.3.3 (Fubini’s theorem - I) Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be σ-finite measure spaces and let f be an extended real valued and nonnegative measurable function on X1 × X2 . Then the functions Z Z x 7→ fx dµ2 y 7→ f y dµ1 X2

X1

are measurable with respect to A1 and A2 , respectively, and  Z Z Z f d(µ1 × µ2 ) = f (x, y)dµ2 (y) dµ1 (x) X1 ×X2 X X  Z 1 Z 2 = f (x, y)dµ1 (x) dµ2 (y). X2

Proof. For x ∈ X1 and y ∈ X2 , let Z g(x) = fx dµ2 ,

X1

Z

f y dµ1 .

h(y) =

X2

X1

Let us consider first the case f = χE for some E ∈ A1 ⊗ A2 . Then we have Z Z g(x) = χEx dµ2 = µ2 (Ex ), h(y) = χEy dµ1 = µ1 (E y ). X2

X1

Hence, for f = χE , the result is a consequence of Theorem 6.2.6. Next, let f be a non-negative simple measurable function. In this case, the result follows by using the linearity of integrals. Now, let f be any non-negative measurable function. Then, consider an increasing sequence (ϕn ) of simple measurable functions which converges to f pointwise. If we take Z Z ϕyn dµ1 , ϕn,x dµ2 , hn (y) = gn (x) = X2

X1

then, by MCT (Theorem 4.2.15), gn → g and hn → h pointwise. Again, applying MCT, we have the convergence Z Z Z Z gn dµ1 → gdµ1 , hn dµ2 → hdµ2 . X1

X1

X2

X2

Since Z

Z gn dµ1 =

X1

Z fn d(µ1 × µ2 ) ∀ n ∈ N,

hn dµ2 = X1 ×X2

X2

by taking limit, we obtain Z Z gdµ1 = X1

This completes the proof.

X2

Z f d(µ1 × µ2 ).

hdµ2 = X1 ×X2

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171

Now, we state and prove the Fubini’s theorem for a complex measurable function f on X1 × X2 . Before stating the theorem, let us recall from Proposition 6.3.2 that, if f is a complex measurable function on X1 × X2 , then its sections fx and f y are measurable for each x ∈ X1 and y ∈ X2 , respectively. Thus, the integrals Z Z |f (x, y)|dµ2 (y), |f (x, y)|dµ1 (x) X2

X1

are well-defined. Also, we know from Theorem 6.3.3 that the functions Z Z x 7→ |fx |dµ2 y 7→ |f y |dµ1 X2

X1

are measurable with respect to A1 and A2 , respectively. Theorem 6.3.4 (Fubini’s theorem - II) Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be σ-finite measure spaces and let f be a complex measurable function on the product measure space (X1 × X2 , A1 ⊗ A2 , µ1 × µ2 ). Suppose that at least one of the integrals Z Z Z Z Z   |fx |dµ2 dµ1 , |f y |dµ1 dµ2 , |f |d(µ1 × µ2 ) X1

X2

X1 ×X2

X1

X2

is finite. Then they are equal and the following results hold. (i) fx ∈ L1 (µ2 ) for a.a. x ∈ X1 , f y ∈ L1 (µ1 ) for a.a. y ∈ X2 , and f ∈ L1 (µ1 × µ2 ), R R (ii) the functions x 7→ X2 fx dµ2 and y 7→ X1 f y dµ1 belong to L1 (µ1 ) and L1 (µ2 ), respectively. (iii) The integrals Z Z  fx dµ2 dµ1 , X1

Z

X2

Z

 f y dµ1 dµ2 ,

X1

X2

Z f d(µ1 × µ2 ) X1 ×X2

are equal. Proof. Since |fx | = |f |x and |f y | = |f |y , by Theorem 6.3.3, the integrals Z Z Z Z Z   y |fx |dµ2 dµ1 , |f |dµ1 dµ2 , |f |d(µ1 × µ2 ) X1

X2

X2

X1 ×X2

X1

are equal. Hence, if one of these integrals is finite, then all of them are finite. In particular, results in (i) hold. To prove (ii) and (iii), first we assume that f ∈ L1 (µ1 × µ2 ) is real valued. Note that Z Z f + d(µ1 × µ2 ) ≤ |f |d(µ1 × µ2 ) < ∞. X1 ×X2

X1 ×X2

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Measure and Integration

Hence, the integrals Z Z  (f + )x dµ2 dµ1 ,

Z

X1

X2

Z

X2

Z

 (f + )y dµ1 dµ2 ,

f + d(µ1 × µ2 )

X1 ×X2

X1

are equal and finite. In particular, the functions Z Z x 7→ f + (x, y)dµ2 (y) y 7→ f + (x, y)dµ1 (x) X2

X1

belong to L1 (µ1 ) and L1 (µ2 ), respectively. Hence, (ii) and (iii) hold with f + in place of f . Similarly, we have the conclusions in (ii) and (iii) with f − in place of f . Therefore, we have (ii) and (iii) for f as well. The case for complex valued f follows by writing f as f = Re(f ) + iIm(f ) and applying the results for the real valued functions Re(f ) and Im(f ), and observing the facts that |Re(f )| ≤ |f |, |Im(f )| ≤ |f |, and the linearity of taking integrals.

6.4

Counter Examples

6.4.1

σ-finiteness condition cannot be dropped

Let X1 = [0, 1] with Lebesgue measure µ1 and X2 = [0, 1] with counting measure µ2 . Let D := {(x, y) ∈ X1 × X2 : x = y}, the diagonal set. Since D = ∩Dn , where    n  [ j−1 j j−1 j Dn := , × , , n ∈ N, n n n n j=1 it follows that D is a measurable subset of X1 × X2 . Note that for x ∈ [0, 1], Dx = {x},

Dy = {y},

µ2 (Dx ) = 1,

µ1 (Dy ) = 0,

so that Z

Z µ2 (Dx )dµ1 = µ1 (X1 ) = 1,

X1

µ1 (Dy )dµ2 = 0.

X2

Hence, the integrals involved in the definition of product measure are not equal for the measurable set D. Also, taking f = χD , the characteristic function of D, we have Z Z Z Z Z   fx dµ2 dµ1 = χDx dµ2 dµ1 = µ2 (Dx )dµ1 = 1, X1

Z X2

X2

Z X1

X1



f y dµ1 dµ2 =

Z X2

X2

Z X1

X1



Z

χDy dµ1 dµ2 =

µ1 (Dy )dµ2 = 0.

X2

Thus, the iterated integrals in Fubini’s theorem are not equal for f = χD . Note that µ2 is not σ-finite.

Integration on Product Spaces

6.4.2

173

Product of complete measures need not be complete

Suppose (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) are complete σ-finite measure spaces such that there exists A ∈ A1 with µ1 (A) = 0 and there exists B ⊆ X2 such that B 6∈ A2 . Then we have A × B ⊆ A × X2 with (µ1 × µ2 )(A × X2 ) = µ1 (A)µ2 (X2 ) = 0. But, A × B 6∈ A1 ⊗ A2 , since for every x ∈ A, (A × B)x = B 6∈ A2 . Thus, µ1 × µ2 is not complete. As an example, consider (X1 , A1 , µ1 ) = (X2 , A2 , µ2 ) = (R, M, m). We know that Q ∈ A1 with m(Q) = 0 and there exists B ⊆ R such that B 6∈ M. Thus, (R, M, m) is complete, whereas (R × R, M ⊗ M, m × m) is not complete. It can be shown that the completion of (R × R, M ⊗ M, m × m) is the Lebesgue measure space (R2 , M2 , m2 ).

6.5

Problems

1. Prove Lemma 6.2.3. 2. Prove Proposition 6.2.4. 3. Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be σ-finite measure spaces and let S be the family of all E ∈ A1 ⊗ A2 such that the functions x 7→ µ2 (Ex ) and y y with respect to A1 and A2 , respectively, and Z 7→ µ1 (E ) are measurable Z µ1 (E y )dµ2 . Show that

µ2 (Ex )dµ1 = X1

X2

(a) S contains all measurable rectangles. (b) S contains all elementary sets. (c) S contains finite disjoint unions of its members. 4. Supply details of the proof of Corollary 6.2.6. 5. Prove that if A1 = Bm and A2 = Bn , then Bm ⊗ Bn = Bm+n . [Hint: Observe: Every open set in Rm+n is a countable union of sets of the form A × B where A and B are rectangles in Rm and Rn , respectively, and prove: Bm m + n contains sets of the form A × Rn and Rm × B where A ∈ Bm and B ∈ Bn .] 6. Let (X1 , A1 , µ1 ) and (X2 , A2 , µ2 ) be σ-finite measure spaces. Given measurable functions f1 : X1 → R and f2 : X2 → R, let f (x, y) = f1 (x)f2 (y), Prove the following:

(x, y) ∈ X1 × X2 .

174

Measure and Integration (a) f is measurable on the product space (X1 × X2 , A1 × A2 ). (b) If f1 ∈ L1 (µ1 ) and f2 ∈ L1 (µ2 ), then f ∈ L1 (µ1 × µ2 ) and Z   Z Z f2 dµ2 . f1 dµ1 f d(µ1 × µ2 ) = X1 ×X2

X2

X1

xy for (x, y) ∈ I × I \ {(0, 0)} and 7. Let I = [−1, 1] and f (x, y) = (x2 + y 2 )2 f (0, 0) = 0. Show that the integrals Z Z Z Z   f (x, y)dm(y) dm(x) f (x, y)dm(x) dm(y), I

I

I

I

Z f (x, y)d(m × m)(x, y) does not exist.

exist and are equal, but I×I

8. Let I = [0, 1], S = I × I \ {(0, 0)} and f : S → R be defined by ( x2 −y 2 , (x, y) ∈ I × I \ {(0, 0)}, (x2 +y 2 )2 f (x, y) = 0, (x, y) = (00). Show that the integrals Z Z  f (x, y)dm(x) dm(y), I

Z Z

I

I

 f (x, y)dm(y) dm(x) I

exist and are unequal. 9. Let f be a non-negative measurable function on a σ-finite measure space (X, A, µ) and S = {(x, y) ∈ X × R : 0 ≤ y ≤ f (x)}. Show that S ∈ A × M and Z Z f (x)dµ(x) = χS (x, y)d(µ × m). X

X×R

Z τ sin x π 10. Using Fubini’s theorem prove that lim dx = . τ →∞ 0 x 2 R∞ [Hint: Use the relation x1 = 0 e−xy dy.] 11. Let X1 = [0, 1] = X2 with Lebesgue measure and f : X × X → R be defined by  xy if (x, y) 6= (0, 0), x2 +y 2 f (x, y) = 0 if (x, y) = (0, 0). Show that (a) f is not integrable,   R R R R (b) X f (x, y)dm(x) dm(y) and X f (x, y)dm(y) d(x) exist and X X are equal. 12. Let (X, A, µ) be a complete measure space and f : X → [0, ∞) be an integrable function. Let g(t) := µ({x ∈ X : f (x) ≥ t}), t ≥ 0. Show that Z Z ∞ f dµ = g(t)dt. X

0

[Hint: Use the function h(t, x) = χEt (x), where Et := {x ∈ X : f (x) ≥ t} for (t, x) ∈ [0, ∞) × X.]

Integration on Product Spaces

175

13. For f, g ∈ L1 ((0, 2π]), extending them as 2π-periodic functions on R and using the notation introduced in the beginning of Section 5.3.1, define the convolution of f and g by Z 2π 1 (f ∗ g)(x) := f (x − y)g(y)dm(y), x ∈ R. 2π 0 \ Using Fubini’s theorem, prove that (f ∗ g)(n) = fˆ(n)ˆ g (n) for all n ∈ Z, where, for f ∈ L1 ((0, 2π]), fˆ(n) denotes the n-th Fourier coefficient defined as in Problem 19 in Section 5.5.

Chapter 7 Fourier Transform

In this chapter we introduce the concept of Fourier transform which is part and parcel of the theoretical study of partial differential equations. The purpose of including this chapter in the book is to show some applications of some of the basic theorems in the subject of measure and integration, such as dominated convergence theorem and Fubini’s theorem, to another branch of analysis.

7.1 7.1.1

Fourier Transform on L1 (R) Definition and some basic properties Z

We shall use the notation

Z f (x)dx for

R

f dm whenever f is integrable R

with respect to the Lebesgue measure Rm. Thus, f ∈ L1 (R) if and only if f : R → C is Lebesgue measurable and R |f (x)|dx < ∞. We shall also use the notation C(R) for the set of all continuous complex valued functions defined on R, and by C0 (R), we mean the set of all f ∈ C(R) such that for every ε > 0, there exists a compact set K ⊆ R with the property that |f (x)| < ε for every x ∈ R \ K. Thus, f ∈ C0 (R) ⇐⇒ |f (x)| → 0 as |x| → ∞. We may observe that every f ∈ C0 (R) is uniformly continuous. Definition 7.1.1 Let f ∈ L1 (R). The Fourier transform of f is the function fˆ : R → C defined by Z 1 ˆ f (t) = √ f (x)e−itx dx, t ∈ R. 2π R The map f 7→ fˆ is also called the Fourier transform on L1 (R).

♦

Note that the assumption f ∈ L1 (R) ensures the existence of its Fourier transform fˆ.

177

178

Measure and Integration  1, |x| ≤ 1, Example 7.1.2 For x ∈ R, let f (x) = that is, f = χ[−1,1] . 0, |x| > 1, Then, for t 6= 0, we have r Z 1 r Z 1 2 2 sin t 1 e−itx dx = cos(tx)dx = , fˆ(t) = √ π 0 π t 2π −1 q and fˆ(0) = π2 . We observe that, though the function f is discontinuous, fˆ is continuous, since limt→0 sint t = 1, so that fˆ(t) → fˆ(0) as t → 0. Note also that fˆ 6∈ L1 (R). ♦  1 − |x|, |x| ≤ 1, Example 7.1.3 For x ∈ R, let f (x) = Then we have 0, |x| > 1. Z 1 1 = √ (1 − |x|)e−itx dx 2π −1 r Z 1 2 = (1 − x) cos(tx)dx π 0 r  2  2 sin (t/2) = . π (t/2)2

fˆ(t)

Thus both f and fˆ belong to L1 (R). 2

Example 7.1.4 Let f (x) = e−x

/2

♦

, x ∈ R. We show that

2 fˆ(t) = e−t /2 ,

t ∈ R.

We note that 1 fˆ(t) = √ 2π

∞

Z

e−(

x2 2

+ixt)

−∞

2 1 dx = √ e−t /2 2π

Z

∞

2

e−(x+it)

/2

dx.

−∞

Now, Z

∞

e

−(x+it)2 /2

Z R→∞

−∞

where

R

Z

e

R

−(x+it)2 /2

/2

dx,

−R

Z

e−z

dx =

−R

2

e−(x+it)

dx := lim

2

/2

dz,

ΓR

with ΓR := {x + it : −t ≤ x ≤ R}. Considering the rectangle with vertices (−R, 0), (−R, t), (R, t), (R, 0), by Cauchy’s theorem, we see that Z

R

−R

e−x

2

/2

Z dx = Γ1,t,R

e−z

2

/2

Z dz + ΓR

e−z

2

/2

Z dz − Γ2,t,R

e−z

2

/2

dz,

Fourier Transform

179

where Γ1,t,R is the line segment path joining (−R, 0) and (−R, t), and Γ2,t,R is the line segment path joining (R, 0) and (R, t), that is, Γ1,t,R := {−R + iy : 0 ≤ y ≤ t},

Γ2,t,R := {R + iy : 0 ≤ y ≤ t}.

We observe that, Z Z t 2 −z 2 /2 e−(−R+iy) /2 dy → 0, e dz = Γ1,t,R 0 Z Z t −(R+iy)2 /2 −z 2 /2 e dy → 0 e dz = Γ2,t,R 0 as R → ∞. Hence, we have Z Z R 2 2 e−x /2 dx − e−z /2 dz → 0 −R ΓR

as R → ∞.

Consequently, Z

∞

e

−(x+it)2 /2

Z dx = lim

−∞

Thus, using the fact that 2

e−t /2 fˆ(t) = √ 2π

Z

R

−R

R∞

e−x

−∞

e

−(x+it)2 /2

∞

Z

e−x

dx =

2

/2

dx.

−∞ 2

/2

dx = 1, we obtain 2

∞

−∞

e

R→∞

√1 2π

−(x+it)2 /2

e−t /2 dx = √ 2π

Z

∞

e−x

2

/2

2

dx = e−t

/2

.

−∞

In this case both f and fˆ belong to L1 (R).

♦

In the above examples, not only is fˆ continuous, but we also have fˆ ∈ C0 (R). This is true, in fact, for any f ∈ L1 (R). First, let us prove the continuity of fˆ. Theorem 7.1.5 For f ∈ L1 (R), the function fˆ is continuous, bounded and kf k1 kfˆk∞ := sup |fˆ(t)| ≤ √ . 2π t∈R Proof. Let f ∈ L1 (R). Then, for every t ∈ R, we have Z Z 1 1 kf k1 −itx ˆ √ √ |f (t)| = f (x)e dx ≤ |f (x)|dx = √ . 2π R 2π R 2π √ Thus, fˆ is a bounded function and kfˆk∞ ≤ kf k1 / 2π. To see that fˆ is continuous, let t ∈ R and (tn ) be a sequence in R such

180

Measure and Integration

that tn → t as n → ∞. We have to show that fˆ(tn ) → fˆ(t) as n → ∞. Note that Z 1 ˆ ˆ f (t) − f (tn ) = √ f (x)[e−itx − e−itn x ]dx. 2π R Taking gn (x) := f (x)[e−itx − e−itn x ] for n ∈ N, x ∈ R, we have |gn (x)| ≤ 2|f (x)| a.e., and gn (x) → 0 a.e. Hence, by DCT (Theorem 5.1.13), fˆ(t) − fˆ(tn ) → 0 as n → ∞. This completes the proof. As in the case of real valued functions defined on R, we say that a function (i) f : R → C is an odd function if f (−x) = −f (x) for every x ∈ R, and (ii) f : R → C is an even function if f (−x) = f (x) for every x ∈ R. With these definitions, we observe the following. Proposition 7.1.6 Let f ∈ L1 (R). If f is an odd (respectively, even) function a.e., then fˆ is an odd (respectively, even) function. Proof. Recall that, e−itx = cos(tx) − i sin(tx) for every t, x ∈ R. Therefore, Z Z i 1 ˆ √ √ f (x) cos(tx)dx − f (x) sin(tx)dx f (t) = 2π R 2π R and 1 fˆ(−t) = √ 2π

Z

i f (x) cos(tx)dx + √ 2π R

Z f (x) sin(tx)dx. R

Now, the results will follow by using the facts that f ∈ L1 (R), and that the functions y 7→ cos(y) and y 7→ sin(y) are even and odd, respectively. In order to study further properties of the Fourier transform, we shall make use of the following two lemmas, wherein we use the following notation: For f : R → C and for τ ∈ R, let fτ (x) := f (x − τ ),

x ∈ R.

The following lemma is a consequence of the translation invariance of the Lebesgue measure, and the details of its proof are left as an exercise (see Problem 7). Lemma 7.1.7 Let 1 ≤ p < ∞. Then, for any f ∈ Lp (R), fτ ∈ Lp (R) and for any f, g ∈ Lp (R), kf − gkp = kfτ − gτ kp ∀ τ ∈ R.

Fourier Transform

181

Lemma 7.1.8 Let 1 ≤ p < ∞ and f ∈ Lp (R). Then Z lim |f (x − τ ) − f (x)|p dx = 0, τ →0

R

that is, kf − fτ kp → 0

as

τ → 0.

Further, the map τ 7→ fτ from R to Lp (R) is continuous. Proof. Let f ∈ Lp (R) and let ε > 0 be given. By Theorem 5.2.21, there exists g ∈ Cc (R) such that kf − gkp < ε. Then, by Lemma 7.1.7, we also have kfτ − gτ kp < ε, and hence kf − fτ kp

≤ kf − gkp + kg − gτ kp + kgτ − fτ kp < 2ε + kg − gτ kp .

Now, by the uniform continuity of g, there exists δ > 0 such that |g(x) − g(x − τ )| < ε for all x ∈ R and τ ∈ R with |τ | < δ. Since g ∈ Cc (R), there exists a closed interval [a, b] ⊂ R with b−a > 2δ such that g(x) = 0 for every x 6∈ [a+δ, b−δ]. Hence, Z b p kg − gτ kp = |g(x) − g(x − τ )|p dx ≤ (b − a)εp a

for every τ with |τ | < δ. Thus, kf − fτ kp < 2ε + (b − a)1/p ε so that lim kf − fτ kp = 0. Again, by Lemma 7.1.7, for τ, τ0 ∈ R, τ →0

kfτ − fτ0 kp = kfτ −τ0 − f kp → 0

as τ → τ0 .

Hence, τ 7→ fτ is continuous on R. Theorem 7.1.9 Let f ∈ L1 (R). Then fˆ ∈ C0 (R). In particular, fˆ is uniformly continuous. Proof. Let f ∈ L1 (R). Note that Z Z 1 1 −it(x+π/t) ˆ f (t) = − √ f (x)e dx = − √ f (x − π/t)e−itx dx. 2π R 2π R Hence, Z 1 2fˆ(t) = √ [f (x) − f (x − π/t)]e−itx dx. 2π R Therefore, by Lemma 7.1.8, |fˆ(t)| → 0 as |t| → ∞. The particular case follows from the fact that every function in C0 (R) is uniformly continuous.

182

Measure and Integration

Theorem 7.1.10 Suppose f, g ∈ L1 (R). Then the integrals R f (t)ˆ g (t)dt exist and R Z Z ˆ f (t)g(t)dt = f (t)ˆ g (t)dt. R

R R

fˆ(t)g(t)dt and

R

Proof. By Theorem 7.1.5, we know that fˆ is a bounded function. Hence, Z Z |fˆ(t)g(t)| dt ≤ kfˆk∞ |g(t)|dt = kfˆk∞ kgk1 . R

R

R Thus, the integral R fˆ(t)g(t)dt exists. Similarly, R |f (t)ˆ g (t)| dt exists. Since Z Z Z Z  |f (x)e−itx g(t)|dxdt = |f (x)|dx |g(t)|dt = kf k1 kgk1 , R

R

R

R

R

using Fubini’s theorem (Theorem 6.3.4), we obtain  Z Z Z 1 f (x)e−itx dx g(t)dt fˆ(t)g(t)dt = √ 2π R R R Z  Z 1 = √ f (x) g(t)e−itx dt dx 2π R R Z = f (x)ˆ g (x)dx. R

This completes the proof. We shall make use of the following proposition, which follows from Proposition 5.2.17 by taking the measure space as (R, M, m). Proposition 7.1.11 Let 1 ≤ p < ∞, f ∈ Lp (R), and (αn ) be a sequence of positive real numbers such that αn → ∞ as n → ∞. Let fn := χ[−αn ,αn ] f for n ∈ N. Then, for every r ∈ [1, p], fn ∈ Lr (R) for all n ∈ N and Z lim |f − fn |r dm = 0. n→∞

R

1

In particular, if f ∈ L (R), then Z Z f dm = lim R

n→∞

αn

f dm.

−αn

Theorem 7.1.12 Let f ∈ L1 (R) and let (αn ) and (fn ) be as in Proposition 7.1.11. Then (fˆn ) converges to fˆ uniformly on R. Proof. Clearly, fn ∈ L1 (R) for all n ∈ N. As in Theorem 7.1.5, Z 1 |f (x) − fn (x)|dx. sup |fˆ(t) − fˆn (t)| ≤ √ 2π R t∈R R By Proposition 7.1.11, R |f (x) − fn (x)|dx → 0 as n → ∞. Hence, (fˆn ) converges to fˆ uniformly on R.

Fourier Transform

183

Corollary 7.1.13 Let f ∈ L1 (R). Then, for each t ∈ R, Z α 1 f (x)e−itx dx. fˆ(t) = lim √ α→∞ 2π −α Proof. Let f ∈ L1 (R) and for α > 0, let Z α 1 gα (t) := √ f (x)e−itx dx for t ∈ R. 2π −α Now, let (αn ) be any sequence of positive real numbers such that αn → ∞ as n → ∞. Then, by Theorem 7.1.12, (gαn ) converges uniformly to fˆ. In particular, for each t ∈ R, (gαn (t)) converges to fˆ(t), which proves the result. Remark 7.1.14 Suppose f ∈ L2 (R) and for n ∈ N, let fn := χ[−n,n] f . By Proposition 7.1.11, we know that fn ∈ L1 (R)∩L2 (R). Hence, fˆn is well-defined for every n ∈ N. We shall prove in Section 7.2 that, fˆn ∈ L2 (R) for all n ∈ N and the sequence (fˆn ) converges in L2 (R), so that Φ(f ) := lim fˆn n→∞

2

can be defined in the sense of L -convergence. We shall call the above Φ as the Fourier-Plancherel transform of f . ♦ Next we prove some results involving the derivatives of functions. Theorem 7.1.15 Let f ∈ L1 (R) be such that it is differentiable on R and f 0 ∈ L1 (R). Then fb0 (t) = itfˆ(t) ∀ t ∈ R. For its proof we shall use the following lemma. Lemma 7.1.16 Let f ∈ L1 (R) be such that it is differentiable on R and f 0 ∈ L1 (R). Then f ∈ C0 (R). Proof. Since f 0 ∈ L1 (R), by fundamental theorem of Lebesgue integration (Theorem 5.3.18), for any x > 0, we have Z x f (x) = f (0) + f 0 (y)dy (∗) 0

and f is continuous, in fact, absolutely continuous. We first show that Rx lim f (x) exists. For this, by (∗), it is enough to show that lim 0 f 0 (y)dy x→∞ x→∞ exists. Let (xn ) be any sequence of positive real numbers such that xn → ∞. Since f 0 ∈ L1 (R) and χ[0,xn ] → 1 pointwise, by Proposition 5.2.17, we have Z xn Z ∞ f 0 (y)dy → f 0 (y)dy as n → ∞. 0

0

184

Measure and Integration Rx x→∞ 0

Thus, lim

f 0 (y)dy exists. Consequently, lim f (x) exists. x→∞

Let ` = lim f (x). We claim that ` = 0. Assume for a moment that ` 6= 0. x→∞

Then there exists α > 0 such that |f (x)| > |`|/2 for all x ≥ α. Hence, Z y ` |f (x)|dx ≥ (y − α) ∀ y ≥ α. 2 α Thus, we arrive at a contradiction, since L1 (R). Thus, lim f (x) = 0. Similar x→∞

analysis will lead to lim f (x) = 0. x→−∞

Proof of Theorem 7.1.15. Since f 0 ∈ L1 (R), applying Corollary 7.1.13 for f 0 , we have Z a Z 0 −itx f (x)e dx = lim f 0 (x)e−itx dx. a→∞

R

−a

By integration by parts, for a > 0, we have Z a Z  −itx a 0 −itx f (x)e dx = e f (x) −a + (it) −a

a

f (x)e−itx dx.

−a

 a By Lemma 7.1.16, lim e−itx f (x) −a = 0. Hence, a→∞

Z

a

lim

a→∞

f 0 (x)e−itx dx = (it)

−a

Z

f (x)e−itx dx.

R

Thus, we have proved that fb0 (t) = it fˆ(t). Corollary 7.1.17 Let k ∈ N and f ∈ L1 (R) be such that its k-th derivative f (k) exists a.e. and f (k) ∈ L1 (R). Then (k) (t) = (it)k fˆ(t) (i) fd

∀ t ∈ R;

(ii) |tk fˆ(t)| → 0 as |t| → ∞. Proof. (i) This is obtained by repeated application of Theorem 7.1.15. (ii) By Theorem 7.1.9 and the result in (i), (k) (t)| → 0 |tk fˆ(t)| = |(it)k fˆ(t)| = |fd

as |t| → ∞.

Thus, the proof is complete. The result in Corollary 7.1.17 (ii) is usually written as  1  fˆ(t) = o as |t| → ∞. |t|k The following theorem gives a sufficient condition for the differentiability of the Fourier transform of a function f ∈ L1 (R) and also gives a relation between the derivative of fˆ and f .

Fourier Transform

185

Theorem 7.1.18 Let f ∈ L1 (R) and g(x) = −ixf (x) for almost all x ∈ R. If g ∈ L1 (R), then fˆ is differentiable and fˆ 0 (t) = gˆ(t). Proof. For t, h ∈ R, we have fˆ(t + h) − fˆ(t)

= =

Z 1 √ f (x)[e−ix(t+h) − e−ixt ]dx 2π R Z 1 √ f (x)e−ixt [e−ixh − 1]dx. 2π R

Hence, Z 1 fˆ(t + h) − fˆ(t) xf (x)e−ixt ψh (x)dx, =√ ih 2π R where  ψh (x) =

e−ixh −1 ixh

if x 6= 0, if x = 0.

0

Note that, |ψh (x)| ≤ 1 and ψh (x) → −1 as |h| → 0. Hence, |xf (x)e−ixt ψh (x)| ≤ |xf (x)| and xf (x)e−ixt ψh (x) → −xf (x)e−ixt

as

|h| → 0.

Therefore, by DCT (Theorem 5.1.13), Z Z xf (x)e−ixt ψh (x)dx → (−x)f (x)e−ixt dx R

R

as |h| → 0. Thus, Z fˆ(t + h) − fˆ(t) 1 (−ix)f (x)e−ixt dx, =√ h→0 h 2π R lim

that is, fˆ is differentiable and its derivative at t ∈ R is gˆ(t).

7.1.2

Fourier transform as a linear operator

Recall that C0 (R) denotes the set of all continuous functions f : R → C such that |f (x)| → 0 as |x| → ∞. It can be shown that C0 (R) is a complex vector space and f 7→ kf k∞ := sup |f (x)|,

f ∈ C0 (R),

x∈R

defines a norm on C0 (R). In fact, C0 (R) is a Banach space, that is, complete, with respect to the above norm (see, e.g., [9]).

186

Measure and Integration

Recall from Theorem 7.1.9 √ that if f ∈ L1 (R), then fˆ ∈ C0 (R), and by ˆ Theorem 7.1.5, kf k∞ ≤ kf k1 / 2π. We also see from the definition of the Fourier transform that, for f, g ∈ L1 (R) and α ∈ C, (f\ + g)(t) = fˆ(t) + gˆ(t),

[)(t) = αfˆ(t). (αf

Thus, the map F : L1 (R) → C0 (R) defined by (Ff )(t) = fˆ(t),

f ∈ L1 (R), t ∈ R,

is a linear operator. Further, from the above properties, we have kfˆ − gˆk∞ ≤

kf − gk1 √ 2π

∀ f, g ∈ L1 (R)

so that F is continuous as well. A natural question is whether it is surjective. We answer this question in the negative by displaying an example of a function g ∈ C0 (R) which is not the Fourier transform of any function in L1 (R). This example is taken from [3]. First let us observe the following result, which is a modified form of a result from [3]. Lemma 7.1.19 Let f ∈ L1 (R) be an odd function. Then, there exists M > 0 such that Z R fˆ(t) dt ≤ M r t for all r, R with 0 < r < R < ∞. Proof. Let f ∈ L1 (R) be an odd function a.e. Then, by Proposition 7.1.6, fˆ is an odd function, and it is given by Z ∞ 2i fˆ(t) = − √ f (x) sin(tx)dx. 2π 0 Now, let R > r > 0. Then, using Fubini’s theorem, we have  Z R Z ∞ Z R ˆ 2i 1 f (t) √ − dt = f (x) sin(tx)dx dt t 2π r t r 0 ! Z ∞ Z R sin(tx) 2i dt dx f (x) = √ t 2π 0 r ! Z ∞ Z Rx 2i sin(s) = √ f (x) ds dx. s 2π 0 rx R b We know from calculus that there exists M0 > 0 such that a sinx x dx ≤ M0 for all (a, b) ⊆ R. Thus, Z Z Z ∞ Rx sin(s) R fˆ(t) 2 2M0 kf k1 dt ≤ √ |f (x)| ds dx ≤ √ . r rx t s 2π 0 2π This completes the proof.

Fourier Transform

187

Theorem 7.1.20 The map f 7→ fˆ from L1 (R) to C0 (R) is not surjective. Proof. Assume for a moment that the map f 7→ fˆ from L1 (R) to C0 (R) is surjective. In particular, for every odd function g ∈ C0 (R), there exists f ∈ L1 (R) such that fˆ = g. Let ϕ ∈ L1 (R) be an odd extension of f on [0, ∞). That is, ϕ(x) = f (x) for x ≥ 0 and ϕ(x) = −f (−x) for x < 0. Then ϕˆ = g on [0, ∞), and by Lemma 7.1.19, there exists M > 0 such that Z R g(t) Z R ϕ(t) ˆ dt = dt ≤ M r r t t for all r, R > 0 with 0 < r < R < ∞. Taking g as the odd extension of the function ψ : [0, ∞) → R defined by  t/e, 0 ≤ t ≤ e, ψ(t) := 1/ ln(t), t > e, we see that Z e

R

g(t) dt = t

Z e

R

ψ(t) dt = ln(ln(R)) → ∞ as t

R → ∞.

Thus, we arrive at a contradiction to the conclusion in the last paragraph. Consequently, the map f 7→ fˆ from L1 (R) to C0 (R) is not surjective.

7.1.3

Fourier inversion theorem

Another question one may ask is whether f can be recovered from fˆ. We are going to answer this question affirmatively if fˆ also belongs to L1 (R). In fact, we prove the following theorem, called the Fourier inversion theorem. Theorem 7.1.21 (Fourier inversion theorem) Let f ∈ L1 (R) be such that fˆ ∈ L1 (R). Then Z 1 √ f (x) = fˆ(t)eitx dt 2π R for almost all x ∈ R. First we consider a function φ ∈ L1 (R) satisfying the following properties: (a) 0 < φ(x) ≤ 1 for all x ∈ R. (b) φ(λx) → 1 as λ → 0+ for each x ∈ R. (c) The function ψ : R → C defined by Z 1 ψ(x) := φ(t)eitx dt, 2π R R is non-negative and R ψ(x)dx = 1.

x ∈ R,

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Measure and Integration

There are functions φ satisfying the conditions (a), (b), (c) listed above. For example, one may take φ(x) = e−|x| , x ∈ R. Clearly, this function satisfies properties (a) and (b). To see (c), we note that Z

e−|t| eitx dt =

Z

0

et eitx dt +

Z

−∞

R

so that ψ(x) :=

1 2π

∞

e−t eitx dt =

0

Z

φ(t)eitx dt =

R

2 1 + x2

1 . π(1 + x2 )

dx R 1+x2

R

Since = π, it follows that (c) is also satisfied. The following two lemmas are crucial for proving Theorem 7.1.21. Lemma 7.1.22 Let f ∈ L1 (R) and let φ ∈ L1 (R) R be with properties in (a), (b), and (c) listed above. Then the integrals R f (x − λs)ψ(s) ds and R φ(λt)fˆ(t)eixt dt are well defined for all x ∈ R and λ > 0, and R Z Z 1 f (x − λs)ψ(s) ds = fλ (x) := √ φ(λt)fˆ(t)eixt dt. 2π R R R Proof. Since φ ∈ L1 (R) and fˆ is a bounded function, the integrals R f (x− R λs)ψ(s) ds and R φ(λt)fˆ(t)eixt dt are well defined for all x ∈ R and λ > 0. We also observe that Z Z 1 u du f (x − λs)ψ(s) ds = f (x − u) ψ λ λ R R  Z Z 1 i(x−y)t = f (y)φ(λt)e dt dy. 2π R R Since Z Z

i(x−y)t

|φ(λt)f (y)e R

R

Z

Z |dtdy =

|f (y)|dy dt =

φ(λt) R



R

1 kf k1 kφk1 , λ

Fubini’s theorem (Theorem 6.3.4) can be applied to obtain Z Z Z  1 1 i(x−y)t f (y)φ(λt)e dt dy = √ φ(λt)fˆ(t)eixt dt. 2π R 2π R R Thus, Z

1 f (x − λs)ψ(s) ds = √ 2π R

for all x ∈ R and λ > 0.

Z R

φ(λt)fˆ(t)eixt dt = fλ (x)

Fourier Transform

189

Lemma 7.1.23 Let f ∈ L1 (R) and for λ > 0, let fλ be as in Lemma 7.1.22. Then, fλ ∈ L1 (R) and for every x ∈ R, Z 1 fλ (x) → √ fˆ(t)eixt dt as λ → 0, 2π R and kfλ − f k1 → 0 as λ → 0. Proof. Let f ∈ L1 (R) and x ∈ R. Since φ(λt) → 1 as λ → 0+ for each t ∈ R, we have φ(λt)fˆ(t)eitx → fˆ(t)eitx as λ → 0 for each t ∈ R. Also, |φ(λt)fˆ(t)eitx | ≤ |fˆ(t)| for all t ∈ R and for all λ > 0. Hence, by DCT (Theorem 5.1.13), we have Z Z 1 1 ixt ˆ φ(λt)f (t)e dt → √ fˆ(t)eixt dt as λ → 0. fλ (x) = √ 2π R 2π R Next, let λ > 0. By Lemma 7.1.22, Z fλ (x) − f (x) = [f (x − λs) − f (x)]ψ(s) ds. R

R Hence, by Fubini’s theorem and using the fact that ψ ≥ 0 and R ψ(s) ds = 1, we have  Z Z Z |fλ (x) − f (x)|dx ≤ |f (x − λs) − f (x)|ψ(s) ds dx R R R  Z Z = |f (x − λs) − f (x)| dx ψ(s)ds R ZR = gλ (s)ψ(s)ds, R

Z |f (x−λs)−f (x)| dx. Recall from Lemma 7.1.8 that gλ (s) →

where gλ (s) := R

0 as λ → 0 for all s ∈ R. Also, we have Z gλ (s) = |f (x − λs) − f (x)| dx ≤ 2kf k1 R

so that

Z

Z |gλ (s)ψ(s)|ds =

R

gλ (s)ψ(s)ds ≤ 2kf k1 . R

R Hence, by DCT (Theorem 5.1.13), R |fλ (x) − f (x)|dx → 0 as λ → 0. In particular, fλ ∈ L1 (R) and kfλ − f k1 → 0 as λ → 0. Proof of Theorem 7.1.21. Let φ and fλ be as in Lemma 7.1.22. By Lemma 7.1.23, kfλ − f k1 → 0 as λ → 0. Hence, by Proposition 5.2.13, there

190

Measure and Integration

exists a sequence (λn ) of positive reals such that λn → 0 and fλn → f a.e. Again by Lemma 7.1.23, Z Z 1 1 fλn (x) = √ φ(λn t)fˆ(t)eixt dt → √ fˆ(t)eixt dt 2π R 2π R R as n → ∞ for every x ∈ R. Therefore, f (x) = √12π R fˆ(t)eixt dt for almost all x ∈ R. An immediate corollary to the inversion theorem (Theorem 7.1.21) is the following. Corollary 7.1.24 The map f 7→ fˆ from L1 (R) to C0 (R) is injective. Proof. If f ∈ L1 (R) is such that fˆ = 0, then f satisfies the assumption that fˆ ∈ L1 (R) so that Theorem 7.1.21 can be applied to see that f = 0 a.e. Remark 7.1.25 Recall that, in Example 7.1.2, we had a function f ∈ L1 (R) such that fˆ 6∈ L1 (R). Thus, the condition fˆ ∈ L1 (R) in Theorem 7.1.21 is not redundant. ♦ We close this section by another consequence of the inversion theorem. Theorem 7.1.26 Let f ∈ L1 (R) be such that fˆ ∈ L1 (R). Then f ∈ L2 (R) if and only if fˆ ∈ L2 (R), and in that case kfˆk2 = kf k2 . In particular, if f ∈ L1 (R) ∩ L2 (R) such that fˆ ∈ L1 (R), then fˆ ∈ L2 (R) and kfˆk2 = kf k2 . Proof. By Fourier inversion theorem (Theorem 7.1.21), Z Z 2 |f (x)| dx = f (x)f (x)dx R R Z  Z 1 itx ˆ = √ f (x) f (t)e dt dx 2π R R  Z Z 1 −itx ˆ f (x) f (t)e dt dx. = √ 2π R R  R R Since R R |f (x)fˆ(t)e−itx |dt dx ≤ kf k1 kfˆk1 , by Fubini’s theorem 6.3.4, Z  Z 1 −itx ˆ √ |f (x)| dx = f (t) f (x)e dx dt 2π R R R Z Z = fˆ(t)fˆ(t)dt = |fˆ(t)|2 dt.

Z

2

R

R

Thus f ∈ L2 (R) if and only if fˆ ∈ L2 (R), and in that case kfˆk2 = kf k2 . The particular case is obvious.

Fourier Transform

191

In view of Theorem 7.1.21 and Theorem 7.1.26, one may look for some sufficient conditions on f such that fˆ ∈ L1 (R). Here is one. Theorem 7.1.27 Let f ∈ L1 (R) ∩ L∞ (R) and fˆ ≥ 0 a.e. Then fˆ ∈ L1 (R). 2

Proof. Let ϕ(x) := e−x /2 , x ∈ R. Then we have (see Example 7.1.4) ϕˆ = ϕ. For ε > 0, let ϕε (x) = ϕ(εx), x ∈ R. Then, it can be seen that Z Z √ 1 t , ϕˆε (t) = ϕˆ ϕˆε (t)dt = ϕ(τ ˆ )dτ = 2π. ε ε R R By Theorem 7.1.10, Z

fˆ(t)ϕε (t)dt =

R

Z f (t)ϕˆε (t)dt. R

Hence, using the hypothesis that fˆ ≥ 0, we obtain, Z Z Z √ fˆ(t)ϕε (t)dt = f (t)ϕˆε (t)dt ≤ |f (t)|ϕˆε (t)dt ≤ 2πkf k∞ . R

R

R

Since ϕε (t) increases to 1 as ε → 0, fˆ(t)ϕε (t) increases to fˆ(t) as ε → 0. Hence, by MCT (Theorem 4.2.15). Z Z Z √ |fˆ(t)|dt = fˆ(t)dt = lim fˆ(t)ϕε (t)dt ≤ 2πkf k∞ . R

R

ε→0

R

Thus, fˆ ∈ L1 (R).

7.2

Fourier-Plancherel Transform

As promised in Remark 7.1.14, now we plan to define an analogue of the Fourier transform for any f ∈ L2 (R). For this we shall make use of the concept of convolution of L1 -functions and some of their properties. Theorem 7.2.1 Let f and g belong to L1 (R). Then the integral Z 1 f (x − y)g(y) dy (f ∗ g)(x) := √ 2π R is well-defined for almost all x ∈ R, f ∗ g ∈ L1 (R) and f[ ∗ g = fˆgˆ.

192

Measure and Integration

Proof. First let us assume that f and g are Borel measurable. Note that the functions (x, y) → x−y and (x, y) → y are continuous from R2 to R. Therefore, the functions (x, y) 7→ f (x−y) and (x, y) 7→ g(y) are Borel measurable. Hence, (x, y) 7→ f (x − y)g(y) is Borel measurable. Now, using the fact that f and g belong to L1 (R) and Fubini’s theorem, we have   Z Z Z Z |f (x − y)g(y)|dy dx = |f (x − y)g(y)|dx dy R R R R Z  Z = |g(y)| |f (x − y)|dx dy R

≤

R

kf k1 kgk1 .

Z |f (x − y)g(y)|dy < ∞ for almost all x ∈ R, f ∗ g is well defined

Thus, R

for almost all x ∈ R and f ∗ g ∈ L1 (R). Next, let us assume that f and g are Lebesgue measurable. Then we know that there exist Borel measurable functions f0 and g0 such that f = f0 and g = g0 a.e. Hence, we obtain the conclusion of the theorem by applying the first part to f0 and g0 . Now, we prove the last relation. We note that Z 1 [ (f ∗ g)(x)e−itx dx f ∗ g(t) = √ 2π R  Z Z 1 = f (x − y)g(y) dy e−itx dx. 2π R R  R R Since R R |f (x − y)g(y)|dy dx ≤ kf k1 kgk1 , applying Fubini’s theorem, Z Z   f (x − y)e−itx dx g(y) dy f (x − y)g(y) dy e−itx dx = R ZR  ZR  f (x − y)e−it(x−y) dx g(y)e−ity dy. =

Z Z R

R

R

Thus, f[ ∗ g(t) = fˆ(t)ˆ g (t) for all t ∈ R. Definition 7.2.2 For f, g in L1 (R), the convolution of f and g is the function f ∗ g defined in Theorem 7.2.1, that is, Z 1 (f ∗ g)(x) = √ f (x − y)g(y) dy 2π R for almost all x ∈ R.

♦

Recall from Theorem 7.1.26 that if f ∈ L1 (R)∩L2 (R) and fˆ ∈ L1 (R), then ˆ f ∈ L2 (R) and kfˆk2 = kf k2 . The concept of convolution helps us in dropping the condition fˆ ∈ L1 (R).

Fourier Transform

193

Theorem 7.2.3 Let f ∈ L1 (R) ∩ L2 (R). Then fˆ ∈ L2 (R) and kfˆk2 = kf k2 . Proof. We note that Z Z 1 1 ˆ fˆ(t) = √ f (x)eitx dx = √ f (−x)e−itx dx = f˜(t), 2π R 2π R where f˜(s) = f (−s), s ∈ R. Therefore, by Theorem 7.2.1, ˆ ˜ ∗ f (t). |fˆ(t)|2 = fˆ(t)fˆ(t) = f˜(t)fˆ(t) = f[ Note that f˜ ∗ f ∈ L1 (R) with |(f˜ ∗ f )(x)| ≤ kf k22

˜ ∗ f = |fˆ|2 ≥ 0. and f[

˜ ∗ f ∈ L1 (R). Thus, by the relation Hence, by Theorem 7.1.27, we have f[ ˜ ∗ f (t), we obtain fˆ ∈ L2 (R) so that, by Fourier inversion theorem |fˆ(t)|2 = f[ (Theorem 7.1.21), Z Z 1 1 [ itx ˜ ˜ (f ∗ f )(x) = √ f ∗ f (t)e dt = √ |fˆ(t)|2 eitx dt. 2π R 2π R This also shows, as in Theorem 7.1.5, that f˜ ∗ f is continuous on R. In particular, (f ∗ f˜)(0) is well-defined and Z Z Z Z 2 ˜ ˜ f (x)f (−x)dx = (f ∗ f )(0) = |fˆ(t)|2 dt. |f (x)| dx = f (x)f (x)dx = R

R

R

R

This completes the proof. Theorem 7.2.4 Let f ∈ L2 (R) and fn := χ[−n,n] f for n ∈ N. Then the following are true. (i) fn ∈ L1 (R) ∩ L2 (R), fˆn ∈ L2 (R) and kfˆn k2 = kfn k2 for every n ∈ N. (ii) (fˆn ) converges in L2 (R). (iii) The map Φ : L2 (R) → L2 (R) defined by Φ(f ) := lim fˆn , n→∞

f ∈ L2 (R),

is a linear operator, that is, Φ(f + g) = Φ(f ) + Φ(g) for all f, g ∈ L2 (R) and for all α ∈ C.

and

Φ(αf ) = αΦ(f )

194

Measure and Integration

(iv) The map Φ in (iii) satisfies Φ(f ) = fˆ ∀ f ∈ L1 (R) ∩ L2 (R) and ∀ f ∈ L2 (R).

kΦ(f )k2 = kf k2

Proof. (i) We have seen in Proposition 7.1.11 that fn ∈ L1 (R) ∩ L2 (R) so that by Theorem 7.2.3, fˆn ∈ L2 (R) and kfˆn k2 = kfn k2 for every n ∈ N. (ii) By (i), kfˆn − fˆm k2 = kfn\ − fm k2 = kfn − fm k2 .

(∗)

Further, by Proposition 7.1.11, kfn − f k2 → 0 as n → ∞. In particular, (fn ) is a Cauchy sequence in L2 (R). Hence, by (∗) above, (fˆn ) is also a Cauchy sequence in L2 (R). Since L2 (R) is complete, (fˆn ) converges in L2 (R). (iii) We note that for f, g ∈ L2 (R) and α ∈ C, ˆ Φ(f + g) = lim (f\ ˆn ) = Φ(f ) + Φ(g), n + gn ) = lim (fn + g n→∞

n→∞

\ ˆ Φ(αf ) = lim (αf n ) = α lim fn = αΦ(f ). n→∞

n→∞

Thus, Φ is a linear operator. (iv) From the definition of Φ, it is clear that Φ(f ) = fˆ for all f ∈ L1 (R) ∩ 2 L (R). Also, by Theorem 7.2.4, kΦ(f )k2 = lim kfˆn k2 = lim kfn k2 = kf k2 . n→∞

n→∞

This completes the proof. Definition 7.2.5 The linear operator Φ : L2 (R) → L2 (R) defined as in Theorem 7.2.4 is called the Fourier-Plancherel transform on L2 (R), and for f ∈ L2 (R), Φ(f ) is called the Fourier-Plancherel transform of f . ♦ We know that the Fourier-Plancherel transform is a linear isometry. Now, we prove that it is surjective as well. For this, we shall make use of the following lemma. Lemma 7.2.6 Let D be a subspace of L2 (R). If D is not dense in L2 (R), then there exists a non-zero f ∈ L2 (R) such that Z f (x)g(x)dx = 0 ∀ g ∈ D. R

The above lemma is an immediate consequence of Theorem 7.2.7 below, which is a reformulation of projection theorem in functional analysis (see e.g.,

Fourier Transform

195

Nair [9]). For the convenience of the reader, proof of Theorem 7.2.7 is also given. One may recall that a Hilbert space H is an inner product space such that the norm k · kH induced by the inner product, h·, ·iH , namely, 1/2

kukH := hu, uiH ,

u ∈ H,

is complete. Theorem 7.2.7 Let H be a Hilbert space and H0 be a closed subspace of H. Then, for every f ∈ H, there exists a unique g ∈ H0 such that hf − g, hiH = 0

∀ h ∈ H0 .

Proof. Let f ∈ H \ H0 . Then d := inf{kf − gk : g ∈ H0 } > 0. First we show that there exists g ∈ H0 such that kf − gkH = d, then we show that this g satisfies the requirements. Let (gn ) be a sequence in H0 such that kf − gn kH → d as n → ∞. Then (gn ) is a Cauchy sequence in H0 . To see this, first we note that, for every n, m ∈ N, kgn − gm kH = k(gn − f ) − (gm − f )kH . But, by parallelogram law in an inner product space, we have k(gn − f ) − (gm − f )k2H + k(gn − f ) + (gm − f )k2H = 2(kgn − f k2H + kgm − f k2H ). Since k(gn − f ) + (gm − f )kH = 2k(gn + gm )/2 − f kH ≥ 2d, we obtain k(gn − f ) − (gm − f )k2H

≤ 2(kgn − f k2H + kgm − f k2H ) − 4d2 .

Note that the right-hand side of the above inequality tends to 0 as n, m → ∞. Therefore, we obtain kfn − fm kH → 0 as n, m → ∞. Since H0 is complete, there exists g ∈ H0 such that kgn − gkH → 0 as n → ∞. Thus, kf − gkH = d. Next, let h ∈ H0 be such that khkH = 1. We note that hf − g, hiH h and (f − g) − hf − g, hiH h are orthogonal to each other. Hence, by Pythagoras’ theorem, kf − gk2H = khf − g, hiH hk2H + k(f − g) − hf − g, hiH hk2H . Since g + hf − g, hiH h ∈ H0 and since kf − gk2H = d2 , it follows that hf − g, hiH = 0. From this we have hf − g, hiH = 0 for all h ∈ H0 . To see the uniqueness, suppose g1 , g2 are in H0 such that |hf − g1 , hi| = 0 = |hf − g2 , hi|

∀ h ∈ H0 .

Then we have |hg2 − g1 , hi| = 0 for all h ∈ H0 . Taking h = g2 − g1 , we obtain |hg2 − g1 , g2 − g1 i| = 0, so that g2 = g1 .

196

Measure and Integration

Recall that Z

f (x)g(x)dx for f, g ∈ L2 (R),

hf, gi := R

defines an inner product on L2 (R) which makes it a Hilbert space. We shall use the above notation in the following. Now, from Theorem 7.1.10, we derive the following result. Proposition 7.2.8 For f, g ∈ L2 (R), hΦ(f ), g¯i = hΦ(g), f¯i. Proof. Let f, g ∈ L2 (R), and for n ∈ N, let fn := χ[−n,n] f,

gn := χ[−n,n] g.

Then fn , gn ∈ L2 (R) ∩ L1 (R) for every n ∈ N, and the sequences (fn ) and (gn ) converge to f and g, respectively, in L2 (R). Also, since Φ : L2 (R) → L2 (R) is an isometry, the sequences (Φ(fn )) and (Φ(gn )) converge to Φ(f ) and Φ(g), respectively, in L2 (R). Now, by Theorem 7.1.10, we obtain hΦ(fn ), g¯n i = hfˆn , g¯n i = hˆ gn , f¯n i = hΦ(gn ), f¯n i for each n ∈ N. Hence, taking limits as n → ∞, we obtain, hΦ(f ), g¯i = hΦ(g), f¯i. This completes the proof. Theorem 7.2.9 (Fourier-Plancherel theorem) The Fourier-Plancherel transform Φ : L2 (R) → L2 (R) is a surjective linear isometry. Proof. We have already seen in Theorem 7.2.4 that the Fourier-Plancherel transform Φ : L2 (R) → L2 (R) is a linear isometry. Now, we show that it is surjective as well. Since kΦ(f )k2 = kf k2 for all f ∈ L2 (R), it follows that R(Φ), the range of Φ, is closed in L2 (R). Hence, it is enough to prove that R(Φ) is dense in L2 (R). For this we shall use Lemma 7.2.6, that is, we prove that if g ∈ L2 (R) and hΦ(f ), gi = 0 for all f ∈ L2 (R), then g = 0. Let g ∈ L2 (R) be such that hΦ(f ), gi = 0 for all f ∈ L2 (R). Then, by Proposition 7.2.8, hΦ(¯ g ), f¯i = 0 for all f ∈ L2 (R), so that we also obtain hΦ(¯ g ), f i = 0 ∀ f ∈ L2 (R). Consequently, Φ(¯ g ) = 0. Therefore, g¯ = 0 and hence g = 0. This completes the proof.

Fourier Transform

7.3

197

Problems

2 1. For a > 0, let f (x) = e−ax for x ∈ R. Show that f ∈ L1 (R) and fˆ(t) = 2

t √1 e− 4a . 2a

q 2 1 2. Let f (x) = e|x| for x ∈ R. Show that f ∈ L1 (R) and fˆ(t) = . Show π 1+x2 R dx 1 also that fˆ ∈ L (R) and, using inversion formula, find R 1+x2 . 3. For b 6= 0 in R  and f ∈ L1 (R), let g(x) = f (bx) for x ∈ R. Show that f ∈ L1 (R) 1 ˆ t and gˆ(t) = f . b

b

4. For a ∈ R and f ∈ L1 (R), let g(x) = f (x − a) for x ∈ R. Show that gˆ(t) = e−iat fˆ(t), t ∈ R. ˆ(t) = f¯ ˆ(−t) for every t ∈ R. 5. For f ∈ L1 (R), show that f¯ 6. If f, g ∈ L1 (R) are bounded, then verify that R (a) the integral hf, gi := R f (t)g(t)dt is well-defined, (b) hfˆ, g¯i = hˆ g , f¯i. 7. Give details of the proof of Lemma 7.1.7. 8. Show that C0 (R) is a complex vector space and the map f 7→ kf k∞ := sup |f (x)|,

f ∈ C0 (R),

x∈R

defines a complete norm on C0 (R). 9. For τ ∈ R and f ∈ Lp (R) let fτ (x) := f (x−τ ), x ∈ R. Show that, if 1 ≤ p < ∞ then fτ ∈ Lp (R) and τ 7→ fτ is a uniformly continuous function from R to Lp (R). 10. For f ∈ L1 (R), show that Z Z f (x)e−it(x+π/t) dx = f (x − π/t)e−itx dx. R

R

1 11. Let f (x) = 1+|x| , x ∈ R. Show that f ∈ L2 (R) but f 6∈ L1 (R). Does Z n lim f (x)e−ixt dx exist for each t ∈ R? n→∞

−n

12. Give detailed proof of Theorem 7.1.11. 13. Justify: If f ∈ L1 (R) and fˆ ∈ L1 (R), then there exists g ∈ C0 (R) such that f = g a.e. 14. Justify: Suppose f ∈ L1 (R) is such that fˆ ∈ L1 (R). If f is continuous, then f ∈ C0 (R) and Z 1 f (x) = √ fˆ(t)eitx dt ∀ x ∈ R. 2π R 15. For f, g ∈ L1 (R) prove that f ∗ g is continuous on R and f ∗ g = g ∗ f .

198

Measure and Integration

16. Suppose that the heat equation ut (x, t) = c2 uxx (x, t), x ∈ R, t > 0, with initial condition u(x, 0) = f (x), x ∈ R, is uniquely solvable and the functions x 7→ u(x, t) for each t > 0 and f are in L1 (R). Then show that 2 2

u ˆ(ξ, t) = e−c

ξ t

fˆ(ξ),

ξ ∈ R, t > 0.

17. Suppose the wave equation utt (x, t) = c2 uxx (x, t), x ∈ R, t > 0, with initial conditions u(x, 0) = f (x), ut (x, 0) = g(x) x ∈ R, is uniquely solvable and the functions x 7→ u(x, t) for each t > 0 and f, g are in L1 (R). Then show that u ˆ(ξ, t) = fˆ(ξ) cos(cξt) + gˆ(ξ) 18. For f ∈ L2 (R) and r > 0, let Z r ϕr (t) := f (x)e−itx dx,

sin(cξt) , cξ

ξ ∈ R, t > 0.

Z

r

Φ(f )(t)eitx dt

ψr (x) :=

−r

−r

for all x, t ∈ R. Show that ϕr ∈ L2 (R) and ψr ∈ L2 (R). 19. For s ≥ 0, let Z n o H s (R) := f ∈ L2 (R) : (1 + |t|2 )s |fˆ(t)|2 dt < ∞ . R

Prove the following: (a) H s (R) is a complex vector space. R g (t)dt, f, g ∈ H s (R), defines an inner product (b) hf, gis := R (1 + |t|2 )s fˆ(t)ˆ s on H (R). [It can be shown that the norm Z kf ks := (1 + |t|2 )s |fˆ(t)|2 dt,

f ∈ H s (R),

R

induced by the above inner product is complete. The space H s (R) with the above inner product is called a Sobolev space.] 20. Deduce Lemma 7.2.6 from Theorem 7.2.7.

Bibliography

[1] C.D. Aliprantis and O. Burkinshaw, Principles of Real Analysis, Academic Press (An imprint of Elsevier), 1998. [2] R.G. Bartle and D.R. Sherbert, Introduction to Real Analysis, Third Edition, John Wiley & Sons, Inc. 2000. [3] K. Chandrasekharan, Classical Fourier Transforms, Springer–Verlag, 1989. [4] C. G. Denlinger, Elements of Real Analysis, Jones and Bartlett Learning, 2011. [5] G. de Barra, Measure Theory and Integration, Wiley Eastern, New Delhi, 1981. [6] G. B. Folland, Real Analysis: Modern Techniques and Their Applications, John Wiley & Sons, Inc., New York, 1999. [7] S.R. Ghorpade and B.V. Limaye, A Course in Calculus and Real Analysis, Springer, 2006. [8] W. A. J. Luxemburg, Arzela’s dominated convergence theorem for the Riemann integral, The American Mathematical Monthly, Vol. 78, No. 9 (Nov., 1971), pp. 970-979. [9] M.T. Nair, Functional Analysis: A First Course, PHI Learning, New Delhi, 2002 (Fourth Print: 2014). [10] M.T. Nair, Calculus of One Variable, Ane Books Pvt. Ltd., New Delhi, 2015. [11] M.H. Protter, Basic Elements of Real Analysis, Springer, 1998. [12] H.L. Royden, Real Analysis, 3rd Edition, Prentice-Hall of India, New Delhi, 1995. [13] W. Rudin, Principles of Mathematical Analysis (Third Edition), International Student Edition, McGraw-Hill Kogakusha Ltd., Tokyo, 1964. [14] W. Rudin, Real and Complex Analysis (Third Edition), McGraw-Hill Book Co., Singapore, 1987.

199

Index

C0 (R), 177 Cc (Ω), 138 Fδ -set, 19 Gδ -set, 19 Lp - Spaces, 127 Va,b (ϕ), 146 σ-algebra generated by, 43 σ-filed, 60 σ-finite measure space, 39 σ-algebra, 38 k-cells, 58 x-section, 162, 169 y-section, 162, 169 Lp (Ω), 129 Lp (Ω, MΩ , m), 129 L(X), 114 L(µ), 114 L∞ (X, A, µ), 128 L∞ (µ), 128 Lp (X), 127 Lp (X, A, µ), 127 Lp (Ω), 129 Lp (Ω, MΩ , m), 129 Lp (µ), 127

bounded intervals, 11 bounded variation, 146

absolutely continuous, 103, 144 algebra, 38 almost everywhere, 67 convergence, 66

Darboux-Riemann integral, 3 DCT, 121, 123 De-Morgan, 30 dense, 43 Dirac measure, 40 Dirichlet’s function, 2, 7, 10 disjoint family, 23 distribution function, 61, 104 cumulative, 61 dominated convergence theorem, 121, 123

Borel σ-algebra, 43, 76 Borel measure, 46 Borel sets, 43 bounded above, 12 below, 12

canonical representation, 71 Cantor set, 22 Chandrasekharan, 186 characteristic function, 10, 58 closed set, 43 complete measure, 48 completion, 50 complex measurable function, 56 conditional expectation, 105 continuous, 56 convergence in measure, 71 converges almost everywhere, 66 pointwise, 65 convolution, 175, 192 countable additivity, 30 countable family, 11 counter examples, 172 counting measure, 40

201

202 Egoroff’s theorem, 69 elementary set, 164 essentially bounded, 127 even function, 180 events, 60 Fatou’s lemma, 101, 155 finite measure, 39 finite measure space, 39 Fourier coefficient, 158 inversion theorem, 187 transform, 177 Fourier-Plancherel theorem, 196 transform, 183, 194 Fubini’s theorem, 169, 171 fundamental theorem of Lebesgue integration, 142, 146, 151 greatest lower bound, 12 H¨ older’s inequality, 131 improper integral, 137 indefinite integral, 141 infimum, 13, 64 integrability, 113, 114 integrable, 114 integral, 113, 114 integral of ϕ over E, 82 over X, 81 integral of f , 90 over E, 90 integration on product spaces, 161 inversion theorem, 187 Jensen’s inequality, 129 least upper bound, 12 Lebesgue integrable, 119 integral, 11, 115 measurable, 25

Index measurable sets, 16 measure, 30, 42 outer measure, 15, 42 Lebesgue integrable function, 114 length of the curve, 146 limit inferior, 64 superior, 64 locally compact, 138 lower bound, 12 Darboux integral, 3 Darboux sum, 3 MCT, 102 measurable, 59 Borel, 57 extended real valued, 57 function, 56 Lebesgue, 57 rectangles, 162 sets, 38 space, 38 measure, 38 induced by ϕ and µ, 88 induced by f and µ, 60, 100 positive, 39 space, 38 zero, 8 mesh, 5 Minkowski inequality, 132 modulus, 63 monotone class, 164 convergence theorem, 97 monotonicity, 19, 39 negative part, 63 non-Lebesgue measurable sets, 16 non-measurable sets, 26 norm, 119, 129 nowhere dense, 35 odd function, 180 outer measure, 50

Index pointwise convergence, 65, 66 limit, 65 positive part, 63 probability density function, 61, 104 distribution, 60 measure, 39 space, 60, 104 theory, 59, 104 product σ-algebra, 162 measure, 162, 168 projection theorem, 194 Radon-Nikodym derivative, 103 theorem, 103 random variable, 60, 104 real measurable function, 56 restriction, 45 restriction of the σ-algebra, 46 the measure space, 46 Riemann integrable, 95, 119, 120 integral, 1, 2, 95 sum, 5

203 separable, 43 simple function, 71 measurable function, 63, 72 Sobolev space, 198 step function, 72 subadditivity, 20 supremum, 13, 64 symmetric difference, 79 tag set, 5 Tonelli’s theorem, 169 topological space, 43 total variation, 146 translation invariance, 19 unbounded intervals, 11 upper bound, 12 Darboux integral, 3 Darboux sum, 3 Vitali cover, 151 covering lemma, 151 Young’s inequality, 129

sample space, 60 seminorm, 118

zero measure, 40