Measure and Integration 3037191597, 978-3-03719-159-0

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Dietmar A. Salamon

Dietmar A. Salamon

The book is intended as a companion to a one semester introductory lecture course on measure and integration. After an introduction to abstract measure theory it proceeds to the construction of the Lebesgue measure and of Borel measures on locally compact Hausdorff spaces, Lp spaces and their dual spaces and elementary Hilbert space theory. Special features include the formulation of the Riesz Representation Theorem in terms of both inner and outer regularity, the proofs of the Marcinkiewicz Interpolation Theorem and the Calderon–Zygmund inequality as applications of Fubini’s theorem and Lebesgue differentiation, the treatment of the generalized Radon–Nikodym theorem due to Fremlin, and the existence proof for Haar measures. Three appendices deal with Urysohn’s Lemma, product topologies, and the inverse function theorem. The book assumes familiarity with first year analysis and linear algebra. It is suitable for second year undergraduate students of mathematics or anyone desiring an introduction to the concepts of measure and integration.

ISBN 978-3-03719-159-0

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Measure and Integration

Measure and Integration

Textbooks in Mathematics

Dietmar A. Salamon

Measure and Integration

EMS Textbooks in Mathematics EMS Textbooks in Mathematics is a series of books aimed at students or professional mathematicians seeking an introduction into a particular field. The individual volumes are intended not only to provide relevant techniques, results, and applications, but also to afford insight into the motivations and ideas behind the theory. Suitably designed exercises help to master the subject and prepare the reader for the study of more advanced and specialized literature. Jørn Justesen and Tom Høholdt, A Course In Error-Correcting Codes Markus Stroppel, Locally Compact Groups Peter Kunkel and Volker Mehrmann, Differential-Algebraic Equations Dorothee D. Haroske and Hans Triebel, Distributions, Sobolev Spaces, Elliptic Equations Thomas Timmermann, An Invitation to Quantum Groups and Duality Oleg Bogopolski, Introduction to Group Theory Marek Jarnicki and Peter Pflug, First Steps in Several Complex Variables: Reinhardt Domains Tammo tom Dieck, Algebraic Topology Mauro C. Beltrametti et al., Lectures on Curves, Surfaces and Projective Varieties Wolfgang Woess, Denumerable Markov Chains Eduard Zehnder, Lectures on Dynamical Systems. Hamiltonian Vector Fields and Symplectic Capacities Andrzej Skowronski ´ and Kunio Yamagata, Frobenius Algebras I. Basic Representation Theory Piotr W. Nowak and Guoliang Yu, Large Scale Geometry Joaquim Bruna and Juliá Cufí, Complex Analysis Eduardo Casas-Alvero, Analytic Projective Geometry Fabrice Baudoin, Diffusion Processes and Stochastic Calculus Olivier Lablée, Spectral Theory in Riemannian Geometry

Dietmar A. Salamon

Measure and Integration

Author: Dietmar A. Salamon Departement Mathematik ETH Zürich Rämistrasse 101 8092 Zürich Switzerland E-mail: [email protected]

2010 Mathematics Subject Classification: Primary: 28-01; Secondary: 28A05, 28A10, 28A12, 28A15, 28A20, 28A25, 28A30, 28A33, 28A35, 28C05, 28C10, 28C15, 35J05, 43A05, 44A35, 46B22, 46C05, 46E27, 46E30 Key words: sigma-Algebra, Lebesgue monotone convergence, Caratheodory criterion, Lebesgue measure, Borel measure, Dieudonné’s measure, Riesz Representation Theorem, Alexandrov Double Arrow Space, Sorgenfrey Line, separability, Cauchy–Schwarz inequality, Jensen’s inequality, Egoroff’s theorem, Hardy’s inequality, absolutely continuous measure, truly continuous measure, singular measure, signed measure, Radon–Nikodym Theorem, Lebesgue Decomposition Theorem, Hahn Decomposition Theorem, Jordan Decomposition Theorem, Hardy–Littlewood maximal inequality, Vitali’s Covering Lemma, Lebesgue point, Lebesgue Differentiation Theorem, Banach-Zarecki Theorem, Vitali–Caratheodory Theorem, Cantor function, product sigma-algebra, Fubini’s Theorem, convolution, Young’s inequality, mollifier, Marcinciewicz interpolation, Poisson identity, Green’s formula, Calderon–Zygmund inequality, Haar measure, modular character.

ISBN 978-3-03719-159-0 The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.helveticat.ch. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2016 European Mathematical Society

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Typeset using the author’s TEX files: M. Zunino, Stuttgart, Germany Printing and binding: Beltz Bad Langensalza GmbH, Bad Langensalza, Germany ∞ Printed on acid free paper 987654321

Preface

This book is based on notes for the lecture course Measure and Integration held at ETH Zürich in the spring semester 2014. Prerequisites are the first-year courses on analysis and linear algebra, including the Riemann integral [9, 18, 19, 21], as well as some basic knowledge of metric and topological spaces. The course material is based in large parts on Chapters 1–8 of the textbook real and complex analysis by Walter Rudin [17]. In addition to Rudin’s book, the lecture notes by Urs Lang [10, 11], the five volumes on measure theory by David H. Fremlin [4], the paper by Heinz König [8] on the generalized Radon–Nikodým theorem, the lecture notes by C. E. Heil [7] on absolutely continuous functions, Dan Ma’s topology blog [12] on exotic examples of topological spaces, and the paper by Gert K. Pedersen [16] on the Haar measure were very helpful in preparing this manuscript. The text also contains some material that was not covered in the lecture course, namely some of the results in Sections 4.5 and 5.2 (concerning the dual space of Lp ./ in the non- -finite case), Section 5.4 on the generalized Radon–Nikodým theorem, Sections 7.6 and 7.7 on Marcinkiewicz interpolation and the Calderón– Zygmund inequality, and Chapter 8 on the Haar measure. I am grateful to many people who helped to improve this manuscript. Thanks to the students at ETH who pointed out typos or errors in earlier drafts. Thanks to Andreas Leiser for his careful proofreading. Thanks to Theo Buehler for many enlightening discussions and for pointing out the book by Fremlin, Dan Ma’s topology blog, and the paper by Pedersen. Thanks to Urs Lang for his insightful comments on the construction of the Haar measure.

ETH, Zürich 1 August 2015

Dietmar A. Salamon

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1

Abstract measure theory . . . . . . . . . 1.1 -algebras . . . . . . . . . . . . . 1.2 Measurable functions . . . . . . . 1.3 Integration of nonnegative functions 1.4 Integration of real valued functions 1.5 Sets of measure zero . . . . . . . . 1.6 Completion of a measure space . . 1.7 Exercises . . . . . . . . . . . . . .

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The Lebesgue measure . . . . . . 2.1 Outer measures . . . . . . 2.2 The Lebesgue outer measure 2.3 The transformation formula 2.4 Lebesgue equals Riemann . 2.5 Exercises . . . . . . . . . .

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Borel measures . . . . . . . . . . . . . 3.1 Regular Borel measures . . . . . 3.2 Borel outer measures . . . . . . 3.3 The Riesz representation theorem 3.4 Exercises . . . . . . . . . . . . .

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Lp spaces . . . . . . . . . . 4.1 Hölder and Minkowski . 4.2 The Banach space Lp ./ 4.3 Separability . . . . . . 4.4 Hilbert spaces . . . . . 4.5 The dual space of Lp ./ 4.6 Exercises . . . . . . . .

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Contents

viii 5

The Radon–Nikodým theorem . . . . . 5.1 Absolutely continuous measures . 5.2 The dual space of Lp ./ revisited 5.3 Signed measures . . . . . . . . . 5.4 Radon–Nikodým generalized . . 5.5 Exercises . . . . . . . . . . . . .

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Differentiation . . . . . . . . . . . . 6.1 Weakly integrable functions . . 6.2 Maximal functions . . . . . . . 6.3 Lebesgue points . . . . . . . . 6.4 Absolutely continuous functions 6.5 Exercises . . . . . . . . . . . .

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Product measures . . . . . . . . . . . . 7.1 The product -algebra . . . . . . . 7.2 The product measure . . . . . . . . 7.3 Fubini’s Theorem . . . . . . . . . 7.4 Fubini and Lebesgue . . . . . . . . 7.5 Convolution . . . . . . . . . . . . 7.6 Marcinkiewicz interpolation . . . . 7.7 The Calderón–Zygmund inequality 7.8 Exercises . . . . . . . . . . . . . .

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The Haar measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Topological groups . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Haar measures . . . . . . . . . . . . . . . . . . . . . . . . . . . .

309 309 312

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Appendices A Urysohn’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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B The product topology

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343

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

347

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

351

C The inverse function theorem

Introduction

We learn already in high school that integration plays a central role in mathematics and physics. One encounters integrals in the notions of area or volume, when solving a differential equation, in the fundamental theorem of calculus, in Stokes’ theorem, or in classical and quantum mechanics. The first year analysis course at ETH includes an introduction to the Riemann integral, which is satisfactory for many applications. However, it has certain drawbacks, in that some very basic functions are not Riemann integrable, that the pointwise limit of a sequence of Riemann integrable functions need not be Riemann integrable, and that the space of Riemann integrable functions is not complete with respect to the L1 -norm. One purpose of this book is to introduce the Lebesgue integral, which does not suffer from these drawbacks and agrees with the Riemann integral whenever the latter is defined. Chapter 1 introduces abstract integration theory for functions on measure spaces. It includes proofs of the Lebesgue monotone convergence theorem, the Fatou lemma, and the Lebesgue dominated convergence theorem. In Chapter 2 we move on to outer measures and introduce the Lebesgue measure on Euclidean space. Borel measures on locally compact Hausdorff spaces are the subject of Chapter 3. Here the central result is the Riesz representation theorem. In Chapter 4 we encounter Lp spaces and show that the compactly supported continuous functions form a dense subspace of Lp for a regular Borel measure on a locally compact Hausdorff space when p < 1. Chapter 5 is devoted to the proof of the Radon–Nikodým theorem about absolutely continuous measures and to the proof that Lq is naturally isomorphic to the dual space of Lp when 1=p C1=q D 1 and 1 < p < 1. Chapter 6 deals with differentiation. Chapter 7 introduces product measures and contains a proof of Fubini’s Theorem, an introduction to the convolution product on L1 .Rn /, and a proof of the Calderón–Zygmund inequality. Chapter 8 constructs Haar measures on locally compact Hausdorff groups. Despite the overlap with the book of Rudin [17], there are some differences in exposition and content. A small expository difference is that in Chapter 1 measurable functions are defined in terms of pre-images of (Borel) measurable sets rather than pre-images of open sets. The Lebesgue measure in Chapter 2 is introduced in terms of the Lebesgue outer measure instead of as a corollary of the Riesz repre-

2

Introduction

sentation theorem. The notion of a Radon measure on a locally compact Hausdorff space in Chapter 3 is defined in terms of inner regularity, rather than outer regularity together with inner regularity on open sets. This leads to a somewhat different formulation of the Riesz representation theorem (which includes the result as formulated by Rudin). In Chapters 4 and 5 it is shown that Lq ./ is isomorphic to the dual space of Lp ./ for all measure spaces (not just the -finite ones) whenever 1 < p < 1 and 1=p C 1=q D 1. It is also shown that L1 ./ is isomorphic to the dual space of L1 ./ if and only if the measure space is localizable. Chapter 5 includes a generalized version of the Radon–Nikodým theorem for signed measures, due to Fremlin [4], which does not require that the underlying measure  is  -finite. In the formulation of König [8] it asserts that a signed measure admits a -density if and only if it is both absolutely continuous and inner regular with respect to . In addition, the present book includes a self-contained proof of the Calderón–Zygmund inequality in Chapter 7 and an existence and uniqueness proof for (left and right) Haar measures on locally compact Hausdorff groups in Chapter 8. The book is intended as a companion for a foundational one-semester lecture course on measure and integration theory and there are many topics that it does not cover. For example, the subject of probability theory is only touched upon briefly at the end of Chapter 1 and the interested reader is referred to the book of Malliavin [13] which covers many additional topics including Fourier analysis, limit theorems in probability theory, Sobolev spaces, and the stochastic calculus of variations. Many other fields of mathematics require the basic notions of measure and integration. They include functional analysis and partial differential equations (see, e.g., Gilbarg–Trudinger [5]), geometric measure theory, geometric group theory, ergodic theory and dynamical systems, and differential topology and geometry. There are many other textbooks on measure theory that cover most or all of the material in the present book, as well as much more, perhaps from somewhat different view points. They include the book of Bogachev [2] which also contains many historical references, the book of Halmos [6], and the aforementioned books of Fremlin [4], Malliavin [13], and Rudin [17].

Chapter 1

Abstract measure theory

The purpose of this first chapter is to introduce integration on abstract measure spaces. The basic idea is to assign to a real valued function on a given domain a number that gives a reasonable meaning to the notion of area under the graph. For example, to the characteristic function of a subset of the domain one would want to assign the length or area or volume of that subset. To carry this out one needs a sensible notion of measuring the size of the subsets of a given domain. Formally this can take the form of a function which assigns a nonnegative real number, possibly also infinity, to each subset of our domain. This function should have the property that the measure of a disjoint union of subsets is the sum of the measures of the individual subsets. However, as is the case with many beautiful ideas, this naive approach does not work. Consider for example the notion of the length of an interval of real numbers. In this situation each single point has measure zero. With the additivity requirement it would then follow that every subset of the reals, when expressed as the disjoint union of all its elements, must also have measure zero, thus defeating the original purpose of defining the length of an arbitrary subset of the reals. This reasoning carries over to any dimension and makes it impossible to define the familiar notions of area or volume in the manner outlined above. To find a way around this, it helps to recall the basic observation that any uncountable sum of positive real numbers must be infinity. Namely, if we are given a collection of positive real numbers whose sum is finite, then only finitely many of these numbers can be bigger than 1=n for each natural number n, and so it can only be a countable collection. Thus it makes sense to demand additivity only for countable collections of disjoint sets. Even with the restricted concept of countable additivity it will not be possible to assign a measure to every subset of the reals and recover the notion of the length of an interval. For example, call two real numbers equivalent if their difference is rational, and let E be a subset of the half unit interval that contains precisely one element of each equivalence class. Since each equivalence class has a nonempty intersection with the half unit interval, such a set exists by the Axiom of Choice. Assume that all translates of E have the same measure. Then countable additivity would imply that the unit interval has measure zero or infinity.

4

1. Abstract measure theory

One way out of this dilemma is to give up on the idea of countable additivity and replace it by the weaker requirement of countable subadditivity. This leads to the notion of an outer measure which will be discussed in Chapter 2. Another way out is to retain the requirement of countable additivity, but give up on the idea of assigning a measure to every subset of a given domain. Instead, one assigns a measure only to some subsets which are then called measurable. This idea will be pursued in the present chapter. A subtlety of this approach is that in some important cases it is not possible to give an explicit description of those subsets of a given domain that one wants to measure, and instead one can only impose certain axioms that the collection of all measurable sets must satisfy. By contrast, in topology the open sets can often be described explicitly. For example, the open subsets of the real line are countable unions of open intervals, while there is no such explicit description for the Borel measurable subsets of the real line. The precise formulation of this approach leads to the notion of a  -algebra which is discussed in Section 1.1. Section 1.2 introduces measurable functions and examines their basic properties. Measures and the integrals of positive measurable functions are the subject of Section 1.3. Here the nontrivial part is to establish additivity of the integral and the proof is based on the Lebesgue monotone convergence theorem. An important inequality is the Fatou lemma. It is needed to prove the Lebesgue dominated convergence theorem in Section 1.4 for real valued integrable functions. Section 1.5 deals with sets of measure zero, which are negligible for many purposes. For example, it is often convenient to identify two measurable functions if they agree almost everywhere, i.e., on the complement of a set of measure zero. This defines an equivalence relation. The quotient of the space of integrable functions by this equivalence relation is a Banach space and is denoted by L1 . Section 1.6 discusses the completion of a measure space. Here the idea is to declare every subset of a set of measure zero to be measurable as well.

1.1  -algebras For any fixed set X denote by 2X the set of all subsets of X and, for any subset A  X , denote by Ac WD X n A its complement.

1.1. -algebras

5

Definition 1.1 (measurable space). Let X be a set. A collection A  2X of subsets of X is called a -algebra if it satisfies the following axioms: (a) X 2 A; (b) if A 2 A, then Ac 2 A; (c) every countable union of elements of A is again an element of A, i.e., if Ai 2 A S for i D 1; 2; 3; : : : , then 1 iD1 Ai 2 A. A measurable space is a pair .X; A/ consisting of a set X and a  -algebra A  2X . The elements of a  -algebra A are called measurable sets. Lemma 1.2. Every -algebra A  2X satisfies the following: (d) ; 2 A; (e) if n 2 N and A1 ; : : : ; An 2 A, then

Sn

iD1

Ai 2 A;

(f) every finite or countable intersection of elements of A is an element of A; (g) if A; B 2 A, then A n B 2 A. Proof. Condition (d) follows from (a), (b) because X c D ;, and (e) follows from (c), (d) by taking A WD ; for i > n. Condition (f) follows from (b), (c), (e) because
c T Si c c A D i i i Ai , and (g) follows from (b), (f) because A n B D A \ B . This proves Lemma 1.2.  Example 1.3. The sets A WD ¹;; Xº and A WD 2X are  -algebras. Example 1.4. Let X be an uncountable set. Then the collection A  2X of all subsets A  X such that either A or Ac is countable is a -algebra. (Here countable means finite or countably infinite.) Example 1.5. Let X be a set and let ¹Ai ºi2I be a partition of X, i.e., Ai is a S nonempty subset of X for each i 2 I , Ai \ Aj D ; for i ¤ j , and X D i 2I Ai . Then [ A WD ¹AJ WD Aj j J  I º j 2J

is a -algebra. Exercise 1.6. (i) Let X be a set and let A; B  X be subsets such that the four sets A n B; B n A; A \ B; X n .A [ B/ are nonempty. What is the cardinality of the smallest -algebra A  X containing A and B? (ii) How many  -algebras on X are there when #X D k for k D 0; 1; 2; 3; 4? (iii) Is there an infinite -algebra with countable cardinality?

6

1. Abstract measure theory

Exercise 1.7. Let X be any set and let I be any nonempty index set. Suppose that for every i 2 I a -algebra Ai  2X is given. Prove that the intersection T A WD i 2I Ai D ¹A  X j A 2 Ai for all i 2 I º is a -algebra. Lemma 1.8. Let X be a set and E  2X be any set of subsets of X. Then there is a unique smallest  -algebra A  2X containing E (i.e., A is a -algebra, E  A, and if B is any other  -algebra with E  B, then A  B). Proof. Uniqueness follows directly from the definition. Namely, if A and B are two smallest -algebras containing E, we have both B  A and A  B and hence X A D B. To prove existence, denote by S  22 the collection of all  -algebras B  2X that contain E and define ˇ ² ³ \ ˇ if B  2X is a -algebra ˇ A WD BD AX ˇ : such that E  B; then A 2 B B2S

Thus A is a -algebra by Exercise 1.7. Moreover, it follows directly from the definition of A that E  A and that every -algebra B that contains E also contains A. This proves Lemma 1.8.  Lemma 1.8 is a useful tool to construct nontrivial -algebras. Before doing that, let us first take a closer look at Definition 1.1. The letter “ ” stands for countable and the crucial observation is that axiom (c) allows for countable unions. On the one hand, this is a lot more general than only allowing for finite unions, which would be the subject of Boolean algebra. On the other hand, it is a lot more restrictive than allowing for arbitrary unions, which one encounters in the subject of topology. Topological spaces will play a central role in this book and we recall here the formal definition. Definition 1.9 (topological space). Let X be a set. A collection U  2X of subsets of X is called a topology on X if it satisfies the following axioms: (a) ;; X 2 U; (b) if n 2 N and U1 ; : : : ; Un 2 U, then

Tn

Ui 2 U; S (c) if I is any index set and Ui 2 U for i 2 I , then i 2I Ui 2 U. iD1

A topological space is a pair .X; U/ consisting of a set X and a topology U  2X . If .X; U/ is a topological space, the elements of U are called open sets, and a subset F  X is called closed if its complement is open, i.e., F c 2 U. Thus finite intersections of open sets are open and arbitrary unions of open sets are open. Likewise, finite unions of closed sets are closed and arbitrary intersections of closed sets are closed.

1.1. -algebras

7

Conditions (a) and (b) in Definition 1.9 are also properties of every  -algebra. However, condition (c) in Definition 1.9 is not shared by -algebras because it permits arbitrary unions. On the other hand, complements of open sets are typically not open. Many of the topologies used in this book arise from metric spaces and are familiar from first-year analysis courses. Here is a recollection of the definition. Definition 1.10 (metric space). A metric space is a pair .X; d / consisting of a set X and a function dW X  X ! R satisfying the following axioms: (a) d.x; y/  0 for all x; y 2 X, with equality if and only if x D y; (b) d.x; y/ D d.y; x/ for all x; y 2 X ; (c) d.x; z/  d.x; y/ C d.y; z/ for all x; y; z 2 X . A function d W X X ! R that satisfies these axioms is called a distance function and the inequality in (c) is called the triangle inequality. A subset U  X of a metric space .X; d / is called open (or d -open) if for every x 2 U there exists a constant " > 0 such that the open ball B" .x/ WD B" .x; d / WD ¹y 2 X j d.x; y/ < "º (centered at x and of radius ") is contained in U . The collection of d -open subsets of X will be denoted by U.X; d / WD ¹U  X j U is d -openº: It follows directly from the definitions that the collection U.X; d /  2X of d -open sets in a metric space .X; d / satisfies the axioms of a topology in Definition 1.9. A subset F of a metric space .X; d / is closed if and only if the limit point of every convergent sequence in F is itself contained in F . Example 1.11. A normed vector space is a pair .X; k
k/ consisting of a real vector space X and a function X ! RW x 7 ! kxk satisfying the following: (a) kxk  0 for all x 2 X , with equality if and only if x D 0; (b) kxk D jj kxk for all x 2 X and  2 R; (c) kx C yk  kxk C kyk for all x; y 2 X.

8

1. Abstract measure theory

Let .X; k
k/ be a normed vector space. Then the formula d.x; y/ WD kx

yk

defines a distance function on X. X is called a Banach space if the metric space .X; d / is complete, i.e., if every Cauchy sequence in X converges. Example 1.12. The set X D R of real numbers is a metric space with the standard distance function d.x; y/ WD jx yj: The topology on R induced by this distance function is called the standard topology on R. The open sets in the standard topology are unions of open intervals. Exercise. Every union of open intervals is a countable union of open intervals. Exercise 1.13. Consider the set x WD Œ 1; 1 WD R [ ¹ 1; 1º: R For a; b 2 R define .a; 1 WD .a; 1/ [ ¹1º;

Œ 1; b/ WD . 1; b/ [ ¹ 1º:

x open if it is a countable union of open intervals in R and sets Call a subset U  R of the form .a; 1 or Œ 1; b/ for a; b 2 R. x satisfies the axioms of a topology. This (i) Show that the set of open subsets of R x is called the standard topology on R. x is induced by the distance function (ii) Prove that the standard topology on R x R x !R dW R defined by the following formulas for x; y 2 R: d.x; y/ WD

e xCy

2je x y e y x j C ex y C ey x C e

d.x; 1/ WD d.1; x/ WD

2e x ex C e

d.x; 1/ WD d. 1; x/ WD

x

2e x ex C e

d. 1; 1/ WD d.1; 1/ WD 2:

;

x

;

x y

;

1.1. -algebras

(iii) Prove that the map

9

x ! Œ 1; 1 f WR

defined by f .x/ WD tanh.x/ WD

ex e ex C e

x x

x 2 R;

;

f .˙1/ WD ˙1; is a homeomorphism. Prove that it is an isometry with respect to the metric x and the standard metric on the interval Œ 1; 1. Deduce that .R x; d/ in (ii) on R is a compact metric space. x by Exercise 1.14. Extend the total ordering of R to R Extend addition by 1 C a WD 1

x. 1  a  1 for a 2 R

for 1 < a  1,

and by 1 C a WD

1 for 1  a < 1.

(The sum a C b is undefined when ¹a; bº D ¹ 1; 1º.) Let a1 ; a2 ; a3 ; : : : and x. b1 ; b2 ; b3 ; : : : be sequences in R (i) Define lim sup an and lim inf an and show that they always exist. n!1

n!1

(ii) Show that lim sup. an / D

lim inf an : n!1

n!1

(iii) Assume ¹an ; bn º ¤ ¹ 1; 1º so the sum an C bn is defined for n 2 N. Prove the inequality lim sup.an C bn /  lim sup an C lim sup bn ; n!1

n!1

n!1

whenever the right-hand side exists. Find an example where the inequality is strict. (iv) If an  bn for all n 2 N, show that lim inf an  lim inf bn : n!1

n!1

Definition 1.15. Let .X; U/ be a topological space and let B  2X be the smallest -algebra containing U. Then B is called the Borel -algebra of .X; U/ and the elements of B are called Borel (measurable) sets.

10

1. Abstract measure theory

Lemma 1.16. Let .X; U/ be a topological space. Then the following holds. (i) Every closed subset F  X is a Borel set. S (ii) Every countable union 1 i D1 Fi of closed subsets Fi  X is a Borel set. (These are sometimes called F -sets.) T (iii) Every countable intersection 1 iD1 Ui of open subsets Ui  X is a Borel set. (These are sometimes called Gı -sets.) Proof. Part (i) follows from the definition of Borel sets and condition (b) in Definition 1.1, part (ii) follows from (i) and (c), and part (iii) follows from (ii) and (b), because the complement of an F -set is a Gı -set.  Consider for example the Borel  -algebra on the real axis R with its standard topology. In view of Lemma 1.16 it is a legitimate question whether there is any subset of R at all that is not a Borel set. The answer to this question is positive, which may not be surprising; however, the proof of the existence of subsets that are not Borel sets is surprisingly nontrivial. It will only appear much later in this book, after we have introduced the Lebesgue measure (see Lemma 2.15). For now it is useful to note that, roughly speaking, every set that one can construct in terms of some explicit formula, will be a Borel set, and one can only prove with the Axiom of Choice that subsets of R must exist that are not Borel sets.

Recollections about point set topology We close this section with a digression into some basic notions in topology that, at least for metric spaces, are familiar from first-year analysis and will be used throughout this book. The two concepts we recall here are compactness and continuity. A subset K  X of a metric space .X; d / is called compact if every sequence in K has a subsequence that converges to some element of K. Thus, in particular, every compact subset is closed. The notion of compactness carries over to general topological spaces as follows. Let .X; U/ be a topological space and let K  X. An open cover of K is a collection of open sets ¹Ui ºi2I , indexed by a set I , such that [ K Ui : i 2I

The set K is called compact if every open cover of K has a finite subcover, i.e., if for every open cover ¹Ui ºi2I of K there exist finitely many indices i1 ; : : : ; in 2 I such that K  Ui1 [


[ Uin :

1.2. Measurable functions

11

When .X; d / is a metric space and U D U.X; d / is the topology induced by the distance function (Definition 1.10), the two notions of compactness agree. Thus, for every subset K  X, every sequence in K has a subsequence converging to an element of K if and only if every open cover of K has a finite subcover. For a proof see, for example, Munkres [14] or [20, Appendix C.1]. We emphasize that when K is a compact subset of a general topological space .X; U/ it does not follow that K is closed. For example, a finite subset of X is always compact, but need not be closed or, if U D ¹;; Xº, then every subset of X is compact, but only the empty set and X itself are closed subsets of X. If, however, .X; U/ is a Hausdorff space (i.e., for any two distinct points x; y 2 X there exist open sets U; V 2 U such that x 2 U , y 2 V , and U \ V D ;), then every compact subset of X is closed (Lemma A.2). Next, recall that a map f W X ! Y between two metric spaces .X; dX / and .Y; dY / is continuous (i.e., for every x 2 X and every " > 0 there is a ı > 0 such that f .Bı .x; dX //  B" .f .x/; dY /) if and only if the pre-image f

1

.V / WD ¹x 2 X j f .x/ 2 V º

of every open subset of Y is an open subset of X . This second notion carries over to general topological spaces, i.e., a map f W X ! Y between topological spaces .X; UX / and .Y; UY / is called continuous if V 2 UY H) f

1

.V / 2 UX :

It follows directly from the definition that topological spaces form a category, in that the composition g ı f W X ! Z of two continuous maps f W X ! Y and gW Y ! Z between topological spaces is again continuous. Another basic observation is that if f W X ! Y is a continuous map between topological spaces and K is a compact subset of X, then its image f .K/ is a compact subset of Y .

1.2 Measurable functions In analogy to continuous maps between topological spaces, one can define measurable maps between measurable spaces as those maps under which pre-images of measurable sets are again measurable. A slightly different approach is taken by Rudin [17], who defines a measurable map from a measurable space to a topological space as one under which pre-images of open sets are measurable. Both definitions agree whenever the target space is equipped with its Borel  -algebra. As a warmup we begin with some recollections about pre-images of sets that are also relevant for the above discussion of continuity. For any map f W X ! Y between two sets X and Y and any subset B  Y , the pre-image f

1

.B/ WD ¹x 2 X j f .x/ 2 Bº

12

1. Abstract measure theory

of B under f is a well-defined subset of X, whether or not the map f is bijective, i.e., even if there does not exist any map f 1 W Y ! X . The pre-image defines a map from 2Y to 2X . It satisfies f

1

.Y / D X;

1

f

.;/ D ;;

(1.1)

and preserves union, intersection, and complement. Thus 1

f for every subset B  Y , and [  [ Bi D f f 1 i2I

.Y n B/ D X n f

1

.Bi /;

f

1

\

1

.B/

 \ Bi D f

i 2I

i 2I

(1.2)

1

.Bi /

(1.3)

i2I

for every collection of subsets Bi  Y , indexed by a set I . Definition 1.17 (measurable function). (i) Let .X; AX / and .Y; AY / be measurable spaces. A map f W X ! Y is called measurable if the pre-image of every measurable subset of Y under f is a measurable subset of X, i.e., B 2 AY H) f

1

.B/ 2 AX :

x is called mea(ii) Let .X; AX / be a measurable space. A function f W X ! R x associated to surable if it is measurable with respect to the Borel -algebra on R the standard topology in Exercise 1.13 (see Definition 1.15). (iii) Let .X; UX / and .Y; UY / be topological spaces. A map f W X ! Y is called Borel measurable if the pre-image of every Borel measurable subset of Y under f is a Borel measurable subset of X . Example 1.18. Let X be a set. The characteristic function of a subset A  X is the function A W X ! R defined by A .x/ WD

´ 1

if x 2 A;

0

if x … A:

(1.4)

Now assume .X; A/ is a measurable space, consider the Borel -algebra on R, and let A  X be any subset. Then A is a measurable function if and only if A is a measurable set.

1.2. Measurable functions

13

Part (iii) in Definition 1.17 is the special case of part (i), where AX  2X and AY  2Y are the -algebras of Borel sets (see Definition 1.15). Theorem 1.20 below shows that every continuous function between topological spaces is Borel measurable. It also shows that a function from a measurable space to a topological space is measurable with respect to the Borel  -algebra on the target space if and only if the pre-image of every open set is measurable. Since the collection of Borel sets is in general much larger than the collection of open sets, the collection of measurable functions is then also much larger than the collection of continuous functions. Theorem 1.19 (measurable maps). Let .X; AX /, .Y; AY /, and .Z; AZ / be measurable spaces. (i) The identity map idX W X ! X is measurable. (ii) If f W X ! Y and gW Y ! Z are measurable maps, then so is the composition g ı f W X ! Z. (iii) Let f W X ! Y be any map. Then the set f AX WD ¹B  Y j f

1

.B/ 2 AX º

(1.5)

is a -algebra on Y , called the pushforward of AX under f . (iv) A map f W X ! Y is measurable if and only if AY  f AX . Proof. Parts (i) and (ii) follow directly from the definitions. That f AX  2Y defined by (1.5) is a -algebra follows from (1.1) (for axiom (a)), (1.2) (for axiom (b)), and (1.3) (for axiom (c)). This proves part (iii). Moreover by Definition 1.17, f is measurable if and only if f 1 .B/ 2 AX for every B 2 AY , and this means that AY  f AX . This establishes (iv) and completes the proof of Theorem 1.19.  Theorem 1.20 (measurable and continuous maps). Let .X; AX / and .Y; AY / be measurable spaces. Assume UY  2Y is a topology on Y such that AY is the Borel -algebra of .Y; UY /. (i) A map f W X ! Y is measurable if an only if the pre-image of every open subset V  Y under f is measurable, i.e., V 2 UY H) f

1

.V / 2 AX :

(ii) Assume UX  2X is a topology on X such that AX is the Borel -algebra of .X; UX /. Then every continuous map f W X ! Y is (Borel) measurable.

14

1. Abstract measure theory

Proof. By part (iv) of Theorem 1.19, a map f W X ! Y is measurable if and only if AY  f AX . Since f AX is a -algebra on Y by part (iii) of Theorem 1.19, and the Borel  -algebra AY is the smallest -algebra on Y containing the collection of open sets UY by Definition 1.15, it follows that AY  f AX if and only if UY  f AX . By the definition of f AX in (1.5), this translates into the condition V 2 UY H) f

1

.V / 2 AX :

This proves part (i). If, in addition, AX is the Borel -algebra of a topology UX on X and f W .X; UX / ! .Y; UY / is a continuous map, then the pre-image of every open subset V  Y under f is an open subset of X and hence is a Borel subset of X ; thus it follows from part (i) that f is Borel measurable. This proves part (ii) and Theorem 1.20.  Theorem 1.21 (characterization of measurable functions). Let .X; A/ be a measurx be any function. Then the following are equivalent: able space and let f W X ! R (i) f is measurable; (ii) f

1

..a; 1/ is a measurable subset of X for every a 2 R;

(iii) f

1

.Œa; 1/ is a measurable subset of X for every a 2 R;

(iv) f

1

.Π1; b// is a measurable subset of X for every b 2 R;

(v) f

1

.Œ 1; b/ is a measurable subset of X for every b 2 R.

Proof. That (i) implies (ii), (iii), (iv), and (v) follows directly from the definitions. x be such that f 1 ..a; 1/ 2 AX We prove that (ii) implies (i). Thus let f W X ! R for every a 2 R and define x jf B WD f AX D ¹B  R

1

x

.B/ 2 AX º  2R :

x by part (iii) of Theorem 1.19 and .a; 1 2 B for every Then B is a -algebra on R x n .b; 1 2 B for every b 2 R by a 2 R by assumption. Therefore Œ 1; b D R axiom (b) and hence [ Œ 1; b/ D Œ 1; b n1  2 B n2N

by axiom (c) in Definition 1.1. Hence it follows from (f) in Lemma 1.2 that .a; b/ D Œ 1; b/ \ .a; 1 2 B

1.2. Measurable functions

15

x is a countfor every pair of real numbers a < b. Since every open subset of R able union of sets of the form .a; b/, .a; 1, Œ 1; b/, it follows from axiom (c) in x is an element of B. Hence it follows from Definition 1.1 that every open subset of R Theorem 1.20 that f is measurable. This shows that (ii) implies (i). That either of the conditions (iii), (iv), and (v) also implies (i) is shown by a similar argument which is left as an exercise for the reader. This proves Theorem 1.21.  Our next goal is to show that sums, products, and limits of measurable functions are again measurable. The next two results are useful for the proofs of these fundamental facts. Theorem 1.22 (vector valued measurable functions). Let .X; A/ be a measurable space and let f D .f1 ; : : : ; fn /W X ! Rn be a function. Then f is measurable if and only if fi W X ! R is measurable for each i . Proof. For i D 1; : : : ; n define the projection i W Rn ! R by i .x/ WD xi for x D .x1 ; : : : ; xn / 2 R. Since i is continuous, it follows from Theorems 1.19 and 1.20 that if f is measurable, so is fi D i ı f for all i. Conversely, suppose that fi is measurable for i D 1; : : : ; n. Let ai < bi for i D 1; : : : ; n and define Q.a; b/ WD ¹x 2 Rn j ai < xi < bi , for all i º D .a1 ; b1 / 


 .an ; bn /: Then f

1

.Q.a; b// D

n \

fi

1

..ai ; bi // 2 A

iD1

by property (f) in Lemma 1.2. Now every open subset of Rn can be expressed as a countable union of sets of the form Q.a; b/. (Prove this!) Hence it follows from axiom (c) in Definition 1.1 that f 1 .U / 2 A for every open set U  Rn , and so f is measurable. This proves Theorem 1.22.  Lemma 1.23. Let .X; A/ be a measurable space and let u; vW X ! R be measurable functions. If W R2 ! R is continuous, then the function hW X ! R defined by h.x/ WD .u.x/; v.x// for x 2 X is measurable. Proof. The map f WD .u; v/W X ! R2 is measurable (with respect to the Borel -algebra on R2 ) by Theorem 1.22 and the map W R2 ! R is Borel measurable by Theorem 1.20. Hence the composition h D  ı f W X ! R is measurable by Theorem 1.19. This proves Lemma 1.23. 

16

1. Abstract measure theory

Theorem 1.24 (properties of measurable functions). Let .X; A/ be a measurable space. (i) If f; gW X ! R are measurable functions, then so are the functions f C g;

fg;

max¹f; gº;

min¹f; gº;

jf j:

x , k D 1; 2; 3; : : : , be a sequence of measurable functions. (ii) Let fk W X ! R x are measurable: Then the following functions from X to R inf fk ; k

sup fk ; k

lim sup fk ; k!1

lim inf fk : k!1

Proof. (i) The functions W R2 ! R defined by .s; t/ WD s C t, .s; t/ WD st , .s; t/ WD max¹s; tº, .s; t/ WD min¹s; tº, or .s; t/ WD jsj are all continuous. Hence assertion (i) follows from Lemma 1.23. (ii) Define

x g WD sup fk W X ! R k

and let a 2 R. Then the set g

1

..a; 1/ D ¹x 2 X j sup fk .x/ > aº k

D ¹x 2 X j there exists k 2 N such that fk .x/ > aº [ D ¹x 2 X j fk .x/ > aº k2N

D

[

fk 1 ..a; 1/

k2N

is measurable. Hence it follows from Theorem 1.21 that g is measurable. It also follows from part (i) (already proved) that fk is measurable, hence so is supk . fk / by what we have just proved, and then so is the function inf k fk D supk . fk /. With this understood, we conclude that the functions lim sup fk D inf sup fk ; k!1

`2N k`

lim inf fk D sup inf fk k!1

are also measurable. This proves Theorem 1.24.

`2N k`



It follows from Theorem 1.24 that the pointwise limit of a sequence of measurable functions, if it exists, is again measurable. This is in sharp contrast to Riemann integrable functions.

1.2. Measurable functions

17

Step functions We close this section with a brief discussion of measurable step functions. Such functions will play a central role throughout this book. In particular, they are used in the definition of the Lebesgue integral. Definition 1.25 (step function). Let X be a set. A function sW X ! R is called a step function (or simple function) if it takes on only finitely many values, i.e., the image s.X/ is a finite subset of R. Let sW X ! R be a step function, write s.X/ D ¹˛1 ; : : : ; ˛` º

with ˛i ¤ ˛j for i ¤ j ,

and define Ai WD s

1

.˛i / D ¹x 2 X j s.x/ D ˛i º

for i D 1; : : : ; `. Then the sets A1 ; : : : ; A` form a partition of X , i.e., XD

` [

Ai ;

Ai \ Aj D ; for i ¤ j:

(1.6)

iD1

(See Example 1.5.) Moreover, sD

` X

˛i A ;

(1.7)

i

i D1

where A W X ! R is the characteristic function of the set Ai for i D 1; : : : ; ` i (see (1.4)). In this situation s is measurable if and only if the set Ai  X is measurable for each i. For later reference we prove the following. Theorem 1.26 (approximation). Let .X; A/ be a measurable space and let f W X ! Œ0; 1 be a function. Then f is measurable if and only if there exists a sequence of measurable step functions sn W X ! Œ0; 1/ such that 0  s1 .x/  s2 .x/ 


 f .x/;

f .x/ D lim sn .x/ for all x 2 X: n!1

18

1. Abstract measure theory

Proof. If f can be approximated by a sequence of measurable step functions, then f is measurable by Theorem 1.24. Conversely, suppose that f is measurable. For n 2 N define n W Œ0; 1 ! R by n .t/ WD

´ k2 n

n

if k2

n

 t < .k C 1/2

n

; k D 0; 1; : : : ; n2n

1;

if t  n:

(1.8)

These functions are Borel measurable and satisfy n .0/ D 0 and n .1/ D n for all n, as well as t 2 n  n .t/  nC1 .t/  t whenever n  t > 0. Thus lim n .t/ D t

n!1

for all t 2 Œ0; 1:

Hence the functions sn WD n ı f satisfy the requirements of the theorem.



1.3 Integration of nonnegative functions Our next goal is to define the integral of a measurable step function and then the integral of a general nonnegative measurable function via approximation. This requires the notion of volume or measure of a measurable set. The definitions of measure and integral will require some arithmetic on the space Œ0; 1. Addition to 1 and multiplication by 1 are defined by a C 1 WD 1 C a WD 1;

a
1 WD 1
a WD

´ 1

if a ¤ 0;

0

if a D 0:

With this convention, addition and multiplication are commutative, associative, and distributive. Moreover, if ai and bi are nondecreasing sequences in Œ0; 1, then the limits a WD lim ai i!1

and

b WD lim bi i!1

exists in Œ0; 1 and satisfy the familiar rules a C b D lim .ai C bi / and i !1

ab D lim .ai bi /: i!1

1.3. Integration of nonnegative functions

19

These rules must be treated with caution. The product rule does not hold when the sequences are not nondecreasing. For example ai WD i converges to a D 1, bi WD 1=i converges to b D 0, but ai bi D 1 does not converge to ab D 0. (Exercise. Show that the sum of two convergent sequences in Œ0; 1 always converges to the sum of the limits.) Also, for all a; b; c 2 Œ0; 1, a C b D a C c;

a < 1 H) b D c;

ab D ac;

0 < a < 1 H) b D c:

Neither of these assertions extend to the case a D 1. Definition 1.27 (measure). Let .X; A/ be a measurable space. A measure on .X; A/ is a function W A ! Œ0; 1 satisfying the following axioms. (a)  is  -additive, i.e., if Ai 2 A, i D 1; 2; 3; : : : , is a sequence of pairwise disjoint measurable sets, then 1 1 [  X  Ai D .Ai /: i D1

i D1

(b) There exists a measurable set A 2 A such that .A/ < 1. A measure space is a triple .X; A; / consisting of a set X , a  -algebra A  2X , and a measure W A ! Œ0; 1. The basic properties of measures are summarized in the next theorem. Theorem 1.28 (properties of measures). Let .X; A; / be a measure space. Then the following holds. (i) .;/ D 0. (ii) If n 2 N and A1 ; : : : ; An 2 A are such that Ai \ Aj D ; for i ¤ j , then .A1 [


[ An / D .A1 / C


C .An /: (iii) If A; B 2 A are such that A  B, then .A/  .B/:

20

1. Abstract measure theory

(iv) Let Ai 2 A be a sequence such that Ai  AiC1 for all i. Then 1  [  Ai D lim .Ai /: i !1

iD1

(v) Let Ai 2 A be a sequence such that Ai  AiC1 for all i. Then 1 \  .A1 / < 1 H)  Ai D lim .Ai /: i !1

i D1

Proof. We prove (i). Choose A1 2 A such that .A1 / < 1 and define Ai WD ; for i > 1. Then it follows from -additivity that X .A1 / D .A1 / C .;/; i>1

and hence .;/ D 0. This proves part (i). Part (ii) follows from (i) and -additivity by choosing Ai WD ; for i > n. We prove (iii). If A; B 2 A are such that A  B, then B nA 2 A by property (g) in Lemma 1.2 and hence .B/ D .A/C.B nA/  .A/ by part (ii). This proves part (iii). We prove (iv). Assume Ai  AiC1 for all i and set ´ A1 for i D 1, Bi WD Ai n Ai 1 for i > 1. Then Bi is measurable for all i and, for n 2 N, An D

n [

Bi ;

A WD

iD1

1 [

Ai D

iD1

1 [

Bi :

iD1

Since Bi \ Bj D ; for i ¤ j , it follows from  -additivity that .A/ D

1 X

.Bi / D lim

iD1

n!1

n X

.Bi / D lim .An /: n!1

i D1

Here the last equality follows from part (ii). This proves part (iv). We prove (v). Assume Ai  Ai C1 for all i and define Ci WD Ai n AiC1 . Then Ci is measurable for all i and, for n 2 N, An D A [

1 [ iDn

Ci ;

A WD

1 \ i D1

Ai :

1.3. Integration of nonnegative functions

21

Since Ci \ Cj D ; for i ¤ j ,  -additivity yields .An / D .A/ C

1 X

.Ci /

iDn

for all n 2 N. Since .A1 / < 1. It follows that lim .An / D .A/ C lim

n!1

n!1

P1

1 X

iD1

.Ci / < 1 and hence

.Ci / D .A/:

iDn



This proves part (v) and Theorem 1.28.

Exercise 1.29. Let .X; A; / be a measure
Pspace and let Ai 2 A be a sequence of S measurable sets. Prove that  A  i .Ai /. i i Example 1.30. Let .X; A/ be a measurable space. The counting measure W A ! Œ0; 1 is defined by .A/ WD #A

for A 2 A.

As an example, consider the counting measure W 2N ! Œ0; 1 on the natural numbers. Then the sets An WD ¹n; nC1;


º all have infinite measure and their intersection is the empty set and hence has measure zero. Thus the hypothesis .A1 / < 1 cannot be removed in part (v) of Theorem 1.28. Example 1.31. Let .X; A/ be a measurable space and fix an element x0 2 X. The Dirac measure at x0 is the measure ıx0 W A ! Œ0; 1 defined by ıx0 .A/ WD

´ 1

if x0 2 A;

0

if x0 … A;

for A 2 A:

Example 1.32. Let X be an uncountable set and let A be the -algebra of all subsets of X that are either countable or have countable complements (Example 1.4). Then the function W A ! Œ0; 1 defined by .A/ WD 0 when A is countable and by .A/ WD 1 when Ac is countable is a measure. S Example 1.33. Let X D i2I Ai be a partition and let A  2X be the  -algebra in Example 1.5. Then any function I ! Œ0; 1W i 7! ˛i determines a measure P S W A ! Œ0; 1 via .AJ / WD j 2J ˛j for J  I and AJ D j 2J Aj 2 A.

22

1. Abstract measure theory

With these preparations in place we are now ready to introduce the Lebesgue integral of a nonnegative measurable function Definition 1.34 (Lebesgue integral). Let .X; A; / be a measure space and let E 2 A be a measurable set. (i) Let sW X ! Œ0; 1/ be a measurable step function of the form sD

n X

˛i A

i

(1.9)

iD1

with ˛i 2 Œ0; 1/ and RAi 2 A for i D 1; : : : ; n. The (Lebesgue) integral of s over E is the number E s d 2 Œ0; 1 defined by Z n X s d WD ˛i .E \ Ai /: (1.10) E

i D1

(ii) Let f W X ! Œ0; 1 beR a measurable function. The (Lebesgue) integral of f over E is the number E f d 2 Œ0; 1 defined by Z Z f d WD sup s d; E

sf

E

where the supremum is taken over all measurable step function sW X ! Œ0; 1/ that satisfy s.x/  f .x/ for all x 2 X . The reader may verify that the right-hand side of (1.10) depends only on s and not on the choice of ˛i and Ai . The same definition can be used if f is only defined on the measurable set E  X. Then AE WD ¹A 2 A j A  Eº is a -algebra on E Rand E WD jAE is a measure. So .E; AE ; E / is a measure space and the integral E f dE is well defined. It agrees with the integral of the extended function on X defined by f .x/ WD 0 for x 2 X n E. Theorem 1.35 (basic properties of the Lebesgue integral). Let .X; A; / be a measure space, f; gW X ! Œ0; 1 be measurable functions, and E 2 A. Then the following holds: R R (i) if f  g on E, then E f d  E g d; R R (ii) E f d D X f E d; R (iii) if f .x/ D 0 for all x 2 E, then E f d D 0; R (iv) if .E/ D 0, then E f d D 0; R R (v) if A 2 A and E  A, then E f d  A f d; R R (vi) if c 2 Œ0; 1/, then E cf d D c E f d.

1.3. Integration of nonnegative functions

23

Proof. To prove (i), assume f  g on E. If sW X ! Œ0; 1/ is a measurable step function such that s  f , then sE  g, so Z

Z

Z

s d D E

E

sE d 

g d E

by the definition of the integral of g. Now take the supremum over all measurable step functions s  f to obtain Z Z f d  g d: E

E

This proves (i). We prove (ii). It follows from the definitions that Z Z f d D sup s d E

E

sf

Z D sup sf

X

sE d Z

D sup t f E

t d X

Z D X

f E d:

Here the supremum is over all measurable step functions sW X ! Œ0; 1/, respectively tW X ! Œ0; 1/, that satisfy s  f , respectively t  f E . The second equality Rfollows fromR the fact that every measurable step function sW X ! Œ0; 1/ satisfies E s d D X sE d by the definition of the integral. The third equation follows from the fact that a measurable step function t W X ! Œ0; 1/ satisfies t  f E if and only if it has the form t D sE for some measurable step function sW X ! Œ0; 1/ such that s  f . R Part (iii) is a consequence of (i) with g D 0 Rand the fact that E f d  0 by definition. Part (iv) follows from the fact that E s d D 0 for every measurable step function s when .E/ D 0. Part (v) follows from parts (i) and (ii) and the R fact that fRE  f A whenever E  A. Part (vi) follows from the fact that E cs d D c E s d for every c 2 Œ0; 1/ and every measurable step function s, by the commutativity, associativity, and distributivity rules for calculations with numbers in Œ0; 1. This proves Theorem 1.35. 

24

1. Abstract measure theory

Notably absent from the statements of Theorem 1.35 is the assertion that the integral of a sum is the sum of the integrals. This is a fundamental property that any integral should have. The proof that the integral in Definition 1.34 indeed satisfies this crucial condition requires some preparation. First one verifies this property for integrals of step functions and then one uses the Lebesgue monotone convergence Theorem 1.37. Lemma 1.36 (additivity for step functions). Let .X; A; / be a measure space and let s; tW X ! Œ0; 1/ be measurable step functions. (i) For every measurable set E 2 A Z Z Z .s C t/ d D s d C t d: E

E

E

(ii) If E1 ; E2 ; E3 ; : : : is a sequence of pairwise disjoint measurable sets, then Z 1 Z X [ s d D s d; E WD Ek : E

Ek

kD1

k2N

Proof. Write the functions s and t in the form sD

m X

tD

˛i A ; i

i D1

n X

ˇj B

j

j D1

where ˛i ; ˇj 2 Œ0; 1/ and Ai ; Bj 2 A are such that Ai \ Ai 0 D ; for i ¤ i 0 , S Sn Bj \ Bj 0 D ; for j ¤ j 0 , and X D m iD1 Ai D j D1 Bj . Then sCt D

m X n X

.˛i C ˇj /A

i \Bj

;

iD1 j D1

and hence Z m X n X .s C t/ d D .˛i C ˇj /.Ai \ Bj \ E/ E

iD1 j D1

D

m X iD1

D

˛i

n X

.Ai \ Bj \ E/ C

j D1

m X

j D1

˛i .Ai \ E/ C

iD1

Z

D

s d C E

n X j D1

Z

n X

t d: E

ˇj

m X i D1

ˇj .Bj \ E/

.Ai \ Bj \ E/

1.3. Integration of nonnegative functions

25

To prove (ii), let E1 ; E2 ; E3 ; : : : be a sequence of pairwise disjoint measurable sets S and define E WD 1 kD1 Ek : Then Z s d D E

m X

˛i .E \ Ai /

iD1

D

m X

˛i

iD1

D

m X

n X

˛i lim

D lim

n!1

m X

n!1

n!1

n X

.Ek \ Ai /

kD1

n X m X

˛i .Ek \ Ai /

kD1 iD1 n Z X

D lim

n!1

.Ek \ Ai /

kD1

˛i

iD1

D lim

kD1

1 Z X kD1

.Ek \ Ai /

kD1

iD1

D

1 X

s d Ek

s d: Ek



This proves Lemma 1.36.

Theorem 1.37 (Lebesgue monotone convergence theorem). Let .X; A; / be a measure space and let fn W X ! Œ0; 1 be a sequence of measurable functions such that fn .x/  fnC1 .x/ for all x 2 X and all n 2 N: Define f W X ! Œ0; 1 by f .x/ WD lim fn .x/ for x 2 X: n!1

Then f is measurable and Z lim

n!1 X

Z fn d D

f d: X

26

1. Abstract measure theory

Proof. By part (i) of Theorem 1.35, we have Z Z fn d  fnC1 d X

X

for all n 2 N and hence the limit Z ˛ WD lim

n!1 X

fn d

(1.11)

exists in Œ0; 1. Moreover, f D supn fn is a measurable function on X , by part (ii) of Theorem 1.24, and satisfies fn  f for all n 2 N. Thus it follows from part (i) of Theorem 1.35 that Z Z fn d  f d for all n 2 N; X

X

and hence

Z ˛

f d:

(1.12)

X

Now fix a measurable step function sW X ! Œ0; 1/ such that s  f . Define s W A ! Œ0; 1 by

Z

s d for E 2 A:

s .E/ WD

(1.13)

E

This function is a measure by part (ii) of Lemma 1.36 (which asserts that s is -additive) and by part (iv) of Theorem 1.35 (which asserts that s .;/ D 0). Now fix a constant 0 < c < 1 and define En WD ¹x 2 X j cs.x/  fn .x/º

for n 2 N:

Then En 2 A is a measurable set and En  EnC1 for all n 2 N. Moreover, 1 [

En D X:

(1.14)

nD1

(To spell it out, choose an element x 2 X . If f .x/ D 1, then fn .x/ ! 1 and hence cs.x/  s.x/  fn .x/ for some n 2 N, which means that x belongs to one of the sets En . If f .x/ < 1, then fn .x/ converges to f .x/ > cf .x/, hence fn .x/ > cf .x/  cs.x/ for some n 2 N, and for this n we have x 2 En .) Since cs  fn on En , it follows from parts (i) and (vi) of Theorem 1.35 that Z Z Z Z cs .En / D c s d D cs d  fn d  fn d  ˛: En

En

En

X

1.3. Integration of nonnegative functions

27

Here the last inequality follows from the definition of ˛ in (1.11). Hence s .En / 

˛ c

for all n 2 N:

(1.15)

Since s W A ! Œ0; 1 is a measure, by part (i) of Theorem 1.35, it follows from (1.14) and part (iv) of Theorem 1.28 that Z ˛ s d D s .X/ D lim s .En /  : (1.16) n!1 c X Here the last inequality R follows from (1.15). Since (1.16) holds for every constant 0 < c < 1, we have X s d  ˛ for every measurable step function sW X ! Œ0; 1/ such that s  f . Take the supremum over all such s to obtain Z Z f d D sup s d  ˛: X

sf

X

R

Combining this with (1.12) we obtain X f d D ˛ and hence the assertion of Theorem 1.37 follows from the definition of ˛ in (1.11).  Theorem 1.38 ( -additivity of the Lebesgue integral). Let .X; A; / be a measure space. (i) If f; gW X ! Œ0; 1 are measurable and E 2 A, then Z Z Z .f C g/ d D f d C g d: E

E

(1.17)

E

(ii) Let fn W X ! Œ0; 1 be a sequence of measurable functions and define f .x/ WD

1 X

fn .x/ for x 2 X:

nD1

Then f W X ! Œ0; 1 is measurable and, for every E 2 A, Z f d D E

1 Z X

fn d:

(1.18)

nD1 E

(iii) If f W X ! Œ0; 1 is measurable and E1 ; E2 ; E3 ; : : : is a sequence of pairwise disjoint measurable sets, then Z f d D E

1 Z X kD1

f d; Ek

E WD

[ k2N

Ek :

(1.19)

28

1. Abstract measure theory

Proof. (i) By Theorem 1.26, there exist sequences of measurable step functions sn ; tn W X ! Œ0; 1/ such that sn  snC1

and

tn  tnC1

for all n 2 N and f .x/ D lim sn .x/ and n!1

g.x/ D lim tn .x/ n!1

for all x 2 X . Then sn C tn is a monotonically nondecreasing sequence of measurable step functions converging pointwise to f C g. Hence, Z Z .f C g/ d D lim .sn C tn / d n!1 X

X

Z D lim

Z sn d C

n!1

X

X

Z D lim

n!1 X

Z sn d C lim

n!1 X

Z D

 tn d tn d

Z f d C

g d:

X

X

Here the first and last equalities follow from Theorem 1.37, while the second equality follows from part (i) of Lemma 1.36. This proves (i) for E D X. To prove it in general, replace f; g by f E ; gE and use part (ii) of Theorem 1.35. (ii) Define gn W X ! Œ0; 1 by gn WD

n X

fk :

kD1

This is a nondecreasing sequence of measurable functions, by part (i) of Theorem 1.24, and it converges pointwise to f by definition. Hence it follows from part (ii) of Theorem 1.24 that f is measurable and how the Lebesgue monotone convergence theorem (Theorem 1.37) shows that Z Z f d D lim gn d n!1 X

X

D D D

lim

Z X n

n!1 X

lim

n!1

n Z X kD1

1 Z X nD1 X

fk d

kD1

fk d

X

fn d:

1.3. Integration of nonnegative functions

29

Here the second equality follows from the definition of gn while the third one follows from part (i) of the present theorem (already proved). This proves (ii) for E D X. To prove it in general, replace f; fn by f E ; fn E and use part (ii) of Theorem 1.35. (iii) Let f W X ! Œ0; 1 be a measurable function and let Ek 2 A be a sequence of pairwise disjoint measurable sets. Define E WD

1 [

fn WD

Ek ;

kD1

n X

f E : k

kD1

Then it follows from part (i) of the present theorem (already proved) and part (ii) of Theorem 1.35 that Z fn d D

Z X n

X

X

D

k

n Z X X

kD1

D

f E d

kD1

f E d k

n Z X

f d: Ek

kD1

Now fn W X ! Œ0; 1 is a nondecreasing sequence of measurable functions converging pointwise to f E . Hence it follows from the Lebesgue monotone convergence theorem (Theorem 1.37) that Z Z f d D f E d E

X

Z D lim

n!1 X

n Z X

D lim

n!1

D This proves Theorem 1.38.

kD1

1 Z X kD1

fn d f d Ek

f d: Ek



30

1. Abstract measure theory

Exercise 1.39. Let W 2N ! Œ0; 1 be the counting measure on the natural numbers. Show that in this case equation (1.18) in part (ii) of Theorem 1.38 is equivalent to the formula 1 1 X 1 1 X   X X (1.20) aij aij D j D1

j D1

iD1

iD1

for every map N  N ! Œ0; 1W .i; j / 7! aij . The next theorem shows that every measurable function f W X ! Œ0; 1 induces another measure f on .X; A/. Theorem 1.40. Let .X; A; / be a measure space and let f W X ! Œ0; 1 be a measurable function. Then the function f W A ! Œ0; 1; defined by Z

f d for E 2 A

f .E/ WD

(1.21)

E

is a measure, and Z

Z g df D

fg d

E

(1.22)

E

for every measurable function gW X ! Œ0; 1 and every E 2 A. Proof. f is -additive by part (iii) of Theorem 1.38 and f .;/ D 0 by part (iv) of Theorem 1.35. Hence f is a measure (see Definition 1.27). Now let g WD A be the characteristic function of a measurable set A 2 A. Then Z Z Z A df D f .A/ D f d D f A d: X

A

X

Here the first equality follows from the definition of the integral for measurable step functions in Definition 1.34, the second equality from the definition of f , and the last equality from part (ii) of Theorem 1.35. Thus equation (1.22) (with E D X ) holds for characteristic functions of measurable sets. Taking finite sums and using part (vi) of Theorem 1.35 and part (i) of Theorem 1.38 we find that (1.22) (with E D X) continues to hold for all measurable step functions g D sW X ! Œ0; 1/. Now approximate an arbitrary measurable function gW X ! Œ0; 1 by a sequence of measurable step functions via Theorem 1.26 and use the Lebesgue monotone convergence theorem (Theorem 1.37) to deduce that (1.22) holds with E D X for all measurable functions gW X ! Œ0; 1. Now replace g by gE and use part (ii) of Theorem 1.35 to obtain equation (1.22) in general. This proves Theorem 1.40. 

1.3. Integration of nonnegative functions

31

It is one of the central questions in measure theory under which conditions a measure W A ! Œ0; 1 can be expressed in the form f for some measurable function f W X ! Œ0; 1. We return to this question in Chapter 5. The final result in this section is an inequality which will be used in the proof of the Lebesgue dominated convergence theorem (Theorem 1.45). Theorem 1.41 (Fatou’s lemma). Let .X; A; / be a measure space. Consider a sequence of measurable functions fn W X ! Œ0; 1. Then Z Z lim inf fn d  lim inf fn d: X n!1

n!1

X

Proof. For n 2 N define gn W X ! Œ0; 1 by gn .x/ WD inf fi .x/ i n

for x 2 X. Then gn is measurable, by Theorem 1.24, and g1 .x/  g2 .x/  g3 .x/ 


;

lim gn .x/ D lim inf fn .x/ DW f .x/

n!1

n!1

for all x 2 X . Moreover, gn  fi for all i  n. By part (i) of Theorem 1.35 this implies Z Z gn d  fi d X

X

for all i  n, and hence Z

Z gn d  inf

X

i n X

fi d

for all n 2 N. Thus, by the Lebesgue monotone convergence theorem (Theorem 1.37), Z Z Z Z f d D lim gn d  lim inf fi d D lim inf fn d: n!1 X

X

n!1 in X

n!1

X



This proves Theorem 1.41.

Example 1.42. Let .X; A; / be a measure space and E 2 A be a measurable set such that 0 < .E/ < .X/. Define fn WD E when n is even and fn WD 1 E when n is odd. Then lim inf fn D 0 and so n!1

Z

Z lim inf fn d D 0 < min¹.E/; .X n E/º D lim inf

X

n!1

Thus the inequality in Theorem 1.41 can be strict.

n!1

fn d: X

32

1. Abstract measure theory

1.4 Integration of real valued functions The integral of a real valued measurable function is defined as the difference of the integrals of its positive and negative parts. This definition makes sense whenever at least one of these numbers is not equal to infinity. It leads naturally to the concepts of integrability and the Lebesgue integral. The basic properties of the Lebesgue integral are summarized in Theorem 1.44 below. The main result of this section is the Lebesgue dominated convergence theorem (Theorem 1.45). Definition 1.43 (Lebesgue integrable functions). Let .X; A; / be a measure space. A function fR W X ! R is called (Lebesgue) integrable or -integrable if f is measurable and X jf j d < 1: Denote the set of -integrable functions by L1 ./ WD L1 .X; A; / WD ¹f W X ! R j f is -integrableº: The Lebesgue integral of f 2 L1 ./ over a set E 2 A is the real number Z Z Z C f d WD f d f d; E

E

(1.23)

E

where the functions f ˙ W X ! Œ0; 1/ are defined by f C .x/ WD max¹f .x/; 0º;

f .x/ WD max¹ f .x/; 0º

(1.24)

The functions f ˙ are measurable by Theorem 1.24 and 0  f ˙  jf j. Hence their integrals over E are finite by part (i) of Theorem 1.35. Theorem 1.44 (properties of the Lebesgue integral). Let .X; A; / be a measure space. Then the following holds. (i) The set L1 ./ is a real R vector space and, for every E 21 A, the function 1 L ./ ! RW f 7! E f d is linear, i.e., if f; g 2 L ./ and c 2 R, then f C g; cf 2 L1 ./, and Z Z Z Z Z .f C g/ d D f d C g d; cf d D c f d: (1.25) E

E

E

E

E

(ii) For all f; g 2 L1 ./ and all E 2 A, Z

Z

f  g on E H)

f d  E

g d:

(iii) If f 2 L1 ./, then jf j 2 L1 ./ and, for all E 2 A, ˇZ ˇ Z ˇ ˇ ˇ f dˇ  jf j d: ˇ ˇ E

(1.26)

E

E

(1.27)

1.4. Integration of real valued functions

33

(iv) If f 2 L1 ./ and E1 ; E2 ; E3 ; : : : is a sequence of pairwise disjoint measurable sets, then Z 1 Z X [ f d; E WD Ek : (1.28) f d D E

Ek

kD1

k2N

(v) For all E 2 A and all f 2 L1 ./, Z Z f d D f E d: E

(1.29)

X

(vi) Let E 2 A and f 2 L1 ./. If .E/ D 0 or f jE D 0, then Z f d D 0: E 1 Proof. (i) Let f; g 2 L1 ./ and R let c 2 R. Then f C g 2 L ./ because jf C gj  jf j C jgj and hence X jf C gj d < 1 by part R(i) of Theorem 1.38. Likewise, cf 2 L1 ./ because jcf j D jcjjf j and hence X jcf j d < 1 by part (vi) of Theorem 1.35. To prove the second equation in (1.25) assume first that c  0. Then .cf /˙ D cf ˙ and hence Z Z Z cf d D cf C d cf d E

E

E

Z Dc

f

C

Z d

c

f

E

d

E

Z Dc

f d: E

Here the second equation follows from part (vi) of Theorem 1.35. If c < 0, then .cf /C D . c/f and .cf / D . c/f C and hence, again using part (iv) of Theorem 1.35, we obtain Z Z Z cf d D . c/f d . c/f C d E

E

E

Z

Z

D . c/

f

d

f C d

. c/

E

E

Z Dc

f d: E

Now let h WD f C g. Then hC hC C f

h DfC

f

C gC

C g D h C f C C gC :

g and hence

34

1. Abstract measure theory

Hence it follows from part (i) of Theorem 1.38 that Z

Z

C

Z

h d C E

Z

d C

f E

g d D

h d C

E

Therefore, Z

Z

E

h d

Z

g C d:

d C

E

E

h d

E

E

Z D

f

C

Z

C

h d D E

Z

f

C

Z

Z

C

d C

g d

E

f

E

Z

Z d

g d

E

E

Z

D

f d C E

g d E

and this proves (i). C C C (ii) Assume fR D f C f  g D R g C g on E. Then f C g  g C f C on E and hence E .f C g / d  E .g C f / d by part (i) of Theorem 1.35. Now use the additivity of the integral in part (i) of Theorem 1.38 to obtain

Z f

C

Z

Z

d C

E

C

g d  E

Z

g d C E

f

d:

E

This implies (1.26). (iii) Since jf j  f  jf j it follows from (1.25) and (1.26) that Z

Z jf j d D E

Z

Z

. jf j/ d  E

f d  E

jf j d E

and this implies (1.27). (iv) Equation (1.28) holds for f ˙ by part (iii) of Theorem 1.38 and hence holds for f by Definition 1.43. R R (v) The formula E f d D X f E d in (1.29) follows from part (ii) of Theorem 1.35 since f ˙ E D .f E /˙ . R ˙ (vi) If f vanishes on E, then f ˙ also vanish on E R and˙ hence E f d D 0 by part (iii) of Theorem 1.35. If .E/ D 0, then E f d D 0 by part (iv) of Theorem 1.35. In R either case it follows from the definition of the integral in Definition 1.43 that E f d D 0. This proves Theorem 1.44. 

1.4. Integration of real valued functions

35

Theorem 1.45 (Lebesgue dominated convergence theorem). Let .X; A; / be a measure space, gW X ! Œ0; 1/ be an integrable function, let fn W X ! R be a sequence of integrable functions satisfying jfn .x/j  g.x/ for all x 2 X and n 2 N;

(1.30)

and converging pointwise to f W X ! R, i.e., f .x/ D lim fn .x/ for all x 2 X: n!1

Then f is integrable and, for every E 2 A, Z Z f d D lim fn d:

(1.32)

n!1 E

E

(1.31)

Proof. f is measurable by part (ii) of Theorem 1.24 and jf .x/j  g.x/ for all x 2 X by (1.30) and (1.31). Hence it follows from part (i) of Theorem 1.35 that Z Z jf j d  g d < 1; X

X

and so f is integrable. Moreover jfn

f j  jfn j C jf j  2g:

Therefore, by Fatou’s lemma (Theorem 1.41), that Z Z 2g d D lim inf .2g jfn f j/ d X n!1

X

Z  lim inf

.2g

n!1

jfn

f j/ d

X

Z D lim inf

Z jfn

2g d

n!1

X

 f j d

X

Z

Z

D

2g d

jfn

lim sup n!1

X

f j d:

X

Here the penultimate step follows from part (i) of Theorem 1.44. This implies Z lim sup jfn f j d  0: n!1

X

Hence Z lim

n!1 X

jfn

f j d D 0:

36

1. Abstract measure theory

Since ˇZ ˇ ˇ fn d ˇ E

Z E

ˇ Z ˇ f dˇˇ  jfn

Z f j d 

E

part (iii) of Theorem 1.44 yields ˇZ ˇ lim ˇˇ fn d n!1 E

jfn

f j d;

X

Z E

ˇ ˇ f dˇˇ D 0;

which is equivalent to (1.32). This completes the proof of Theorem 1.45.



1.5 Sets of measure zero Assume throughout this section that .X; A; / is a measure space. A set of measure zero (or null set) is a measurable set N 2 A such that .N / D 0. Let P stand for some property that a point x 2 X may have, or not have, depending on x. For example, if f W X ! Œ0; 1 is a measurable function on X, then P could stand for the condition f .x/ > 0, or for the condition f .x/ D 0, or for the condition f .x/ D 1. Or if fn W X ! R is a sequence of measurable functions, P could stand for the statement “the sequence fn .x/ converges.” In such a situation we say that P holds almost everywhere if there exists a set N  X of measure zero such that every element x 2 X n N has the property P. It is not required that the set of all elements x 2 X that have the property P is measurable, although that may often be the case. Example 1.46. Let fn W X ! R be any sequence of measurable functions. Then the set E WD ¹x 2 X j .fn .x//1 nD1 is a Cauchy sequenceº \ [ \ D ¹x 2 X j jfn .x/ fm .x/j < 2

k

º

k2N n0 2N n;mn0

is measurable. If N WD X n E is a set of measure zero, then fn converges almost everywhere to a function f W X ! R. This function can be chosen measurable by defining f .x/ WD lim fn .x/ for x 2 E n!1

and f .x/ WD 0

for x 2 N .

This is the pointwise limit of the sequence of measurable functions gn WD fn E and hence is measurable by part (ii) of Theorem 1.24. The first observation is that every nonnegative function with finite integral is almost everywhere finite.

1.5. Sets of measure zero

Lemma 1.47. Let f W X ! Œ0; 1 be a measurable function. If then f < 1 almost everywhere.

37

R X

f d < 1,

Proof. Define R N WD ¹xR 2 X j f .x/ D 1º and h WD 1N : Then h  f and so 1.N / D X h d  X f d < 1 by Theorem 1.35. Hence .N / D 0.  The second observation is that if two integrable, or nonnegative measurable, functions agree almost everywhere, then their integrals agree over every measurable set. Lemma 1.48. Assume either that f; gW X ! Œ0; 1 are measurable functions that agree almost everywhere, or that f; gW X ! R are -integrable functions that agree almost everywhere. Then Z Z f d D g d for all A 2 A: (1.33) A

A

Proof. Fix a measurable set A 2 A and define N WD ¹x 2 X j f .x/ ¤ g.x/º. Then N is measurable and .N / D 0 by assumption. Hence .A \ N / D 0 by part (iii) of Theorem 1.28. This implies Z Z Z f d D f d C f d A

AnN

A\N

Z D

f d AnN

Z D X

f AnN d:

Here the first equality follows from part (iii) of Theorem 1.38 in the nonnegative case and from part (iv) of Theorem 1.44 in the integrable case. The second equality follows from part (iv) of Theorem 1.35 in the nonnegative case and from part (vi) of Theorem 1.44 in the integrable case. The third equality follows from part (ii) of Theorem 1.35 in the nonnegative case and from part (v) of Theorem 1.44 in the integrable case. Since f AnN D gAnN , it follows that the integrals of f and g over A agree. This proves Lemma 1.48.  The converse of Lemma 1.48 fails for nonnegative measurable functions. For example, if X is a singleton and .X/ D 1, then the integrals of any two positive functions agree over every measurable set. However, the converse of Lemma 1.48 does hold for integrable functions. Since the difference of two integrable functions is again integrable, it suffices to assume g D 0, and in this case the converse also holds for nonnegative measurable functions. This is the content of the next lemma.

38

1. Abstract measure theory

Lemma 1.49. Assume either that f W X ! Œ0; 1 is measurable, or that f W X ! R is -integrable. Then the following are equivalent: (i) f D 0 almost everywhere; R (ii) A f d D 0 for all A 2 A; R (iii) X jf j d D 0. Proof. That (i) implies (ii) is the content of Lemma 1.48. That (ii) implies (iii) is obvious in the nonnegative case. In the integrable case, define AC WD ¹x 2 X j f .x/  0º;

A WD ¹x 2 X j f .x/ < 0º:

Then f C D f AC and f D f A by (1.24). Hence Z Z Z Z C jf j d D f d C f d D f d X

X

AC

X

Z f d D 0 A

by Theorem 1.44 and (ii). It remains to prove that (iii) implies (i). Let f W X ! Œ0; 1 be a measurable R function such that X f D 0 and define the measurable sets An WD ¹x 2 X j f .x/ > 2 Then 2

n

Z .An / D

2 X

n

n

º

for n 2 N: Z

An d 

f d D 0 X

for all n 2 N by Theorem 1.35. Hence .An / D 0 for all n 2 N, and so N WD ¹x 2 X j f .x/ ¤ 0º D

1 [

An

nD1

is a set of measure zero. In the integrable case apply this argument to the function jf jW X ! Œ0; 1/. This proves Lemma 1.49.  Lemma 1.50. Let f 2 L1 ./. Then ˇZ ˇ Z ˇ ˇ ˇ f dˇ D jf j d ˇ ˇ X

(1.34)

X

if and only if f D jf j almost everywhere or f D jf j almost everywhere. R R R R Proof. Assume R(1.34). Then X f d D X jf j d or X f d D X jf j d. In the first case X .jf j f / dRD 0, and so jf j f D 0 almost everywhere by Lemma 1.49. In the second case X .jf j C f / d D 0, and so jf j C f D 0 almost everywhere. This proves Lemma 1.50. 

1.5. Sets of measure zero

39

Definition 1.51 (the Banach space L1 ./). Define an equivalence relation on the real vector space of all measurable function from X to R by 

def

f  g () the set ¹x 2 X j f .x/ ¤ g.x/º has measure zero:

(1.35)

Thus two functions are equivalent if and only if they agree almost everywhere. (Verify that this is an equivalence relation!) By Lemma 1.48 the subspace L1 ./ is invariant under this equivalence relation, i.e., if f; gW X ! R are measurable,   f 2 L1 ./, and f  g, then g 2 L1 ./. Moreover, the set ¹f 2 L1 ./ j f  0º is a linear subspace of L1 ./ and hence the quotient space 

L1 ./ WD L1 ./= is again a real vector space. It is the space of all equivalence classes in L1 ./ under the equivalence relation (1.35). Thus an element of L1 ./ is not a function on X, but a set of functions on X . By Lemma 1.48, the map Z 1 L ./ ! RW f 7 ! jf j d DW kf kL1 X

takes on the same value on all the elements in a given equivalence class and so descends to the quotient space L1 ./. By Lemma 1.49 it defines a norm on L1 ./ and Theorem 1.53 below shows that L1 ./ is a Banach space with this norm (i.e., a complete normed vector space). Theorem 1.52 (convergent series of integrable functions). Let .X; A; / be a measure space and let fn W X ! R be a sequence of -integrable functions such that 1 Z X jfn j d < 1: (1.36) nD1 X

Then there are a set N of measure zero and a function f 2 L1 ./ such that 1 X

jfn .x/j < 1 and

f .x/ D

nD1

1 X

fn .x/ for all x 2 X n N;

(1.37)

nD1

Z f d D A

1 Z X

fn d for all A 2 A;

(1.38)

nD1 A

and Z ˇ ˇ ˇf lim ˇ

n!1 X

n X kD1

ˇ ˇ fk ˇˇ d D 0:

(1.39)

40

1. Abstract measure theory

Proof. Define 1 X

.x/ WD

jfk .x/j

kD1

for x 2 X . This function is measurable by part (ii) of Theorem 1.24. Moreover, it follows from the Lebesgue monotone convergence theorem (Theorem 1.37) and from part (i) of Theorem 1.38 that Z  d D lim

Z X n

n!1 X

X

n!1

D

kD1

1 Z X kD1

kD1

n X

D lim

jfk j d

Z jfk j d X

jfk j d

X

< 1: Hence the set N WD ¹x 2 X j .x/ D 1º has measure zero by Lemma 1.47 and 1 X

jfk .x/j < 1 for all x 2 X n N .

kD1

Define the function f W X ! R by 8 ˆ 0 for x 2 N , ˆ < 1 f .x/ WD X ˆ fk .x/ for x 2 X n N: ˆ : kD1

Then f satisfies (1.37). Define the functions gW X ! R and gn W X ! R by g WD XnN ;

gn WD

n X kD1

fk X nN

for n 2 N:

1.5. Sets of measure zero

41

These functions are measurable by part (i) of Theorem 1.24. Moreover, Z Z g d D  d < 1 X

X

by Lemma 1.48. Since jgn .x/j  g.x/ for all n 2 N and gn converges pointwise to f , it follows from the Lebesgue dominated convergence theorem (Theorem 1.45) that f 2 L1 ./ and, for all A 2 A, Z

Z f d D lim

A

n!1 A

gn d D lim

Z X n

n!1 A

fk d D

1 Z X

fn d:

nD1 A

kD1

P Here the second step follows from Lemma 1.48 because gn D nkD1 fk almost everywhere. The last step follows by interchanging sum and integral, using part (i) of Theorem 1.44. This proves (1.38). To prove (1.39) note that f

n X

fk D f

gn

kD1

almost everywhere, that f .x/ gn .x/ converges to zero for all x 2 X , and that jf gn j  jf j C g, where jf j C g is integrable. Hence, by Lemma 1.48 and the Lebesgue dominated convergence theorem (Theorem 1.45), Z ˇ ˇ ˇf lim ˇ

n!1 X

n X kD1

ˇ Z ˇ fk ˇˇ d D lim jf n!1

gn j d D 0;

X



This proves (1.39) and Theorem 1.52.

Theorem 1.53 (completeness of L1 ). Let .X; A; / be a measure space and let fn 2 L1 ./ be a sequence of integrable functions. Assume fn is a Cauchy sequence with respect to the L1 -norm, i.e., for every " > 0 there is an n0 2 N such that, for all m; n 2 N, Z n; m  n0 H)

jfn

fm j d < ":

(1.40)

X

Then there exists a function f 2 L1 ./ such that Z lim jfn f j d D 0: n!1 X

(1.41)

Moreover, there is a subsequence fni that converges almost everywhere to f .

42

1. Abstract measure theory

Proof. By assumption, there is a sequence ni 2 N such that Z jfniC1 fni j d < 2 i ; ni < niC1 ; for all i 2 N: X

Then the sequence gi WD fniC1

fni 2 L1 ./

satisfies (1.36). Hence, by Theorem 1.52, there exists a function g 2 L1 ./ such that 1 1 X X gD gi D .fniC1 fni / i D1

iD1

almost everywhere and Z ˇ kX1 ˇ ˇ gi ˇ k!1

0 D lim

X

i D1

ˇ Z ˇ g ˇˇ d D lim jfnk k!1 X

fn1

gj d:

(1.42)

Define f WD fn1 C g: Then fni D fn1 C

i 1 X

gj

j D1

converges almost everywhere to f . We prove (1.41). Let " > 0. By (1.42), there is an ` 2 N such that Z jfnk f j d < "=2 for all k  `. X

By (1.40), the integer ` can be chosen such that Z jfn fm j d < "=2 for all n; m  n` . X

Then Z

Z jfn

X

f j d 

Z jfn

X

fn` j d C

X

for all n  n` . This proves (1.41) and Theorem 1.53.

jfn`

f j d < " 

1.6. Completion of a measure space

43

1.6 Completion of a measure space The discussion in Section 1.5 shows that sets of measure zero are negligible in the sense that the integral of a measurable function remains the same if the function is modified on a set of measure zero. Thus also subsets of sets of measure zero can be considered negligible. However, such subsets need not be elements of our  -algebra A. It is sometimes convenient to form a new -algebra by including all subsets of sets of measure zero. This leads to the notion of a completion of a measure space .X; A; /. Definition 1.54. A measure space .X; A; / is called complete if N 2 A;

.N / D 0;

E  N H) E 2 A:

Theorem 1.55. Let .X; A; / be a measure space and define ˇ ² ³ ˇ there exist measurable sets A; B 2 A such that  ˇ A WD E  X ˇ : A  E  B and .B n A/ D 0 Then the following holds. (i) A is a -algebra and A  A . (ii) There exists a unique measure  W A ! Œ0; 1 such that  jA D : (iii) The triple .X; A ;  / is a complete measure space. It is called the completion of .X; A; /. (iv) If f W X ! R is -integrable, then f is  -integrable and, for E 2 A, Z Z  f d D f d (1.43) E

E

This continues to hold for all A-measurable functions f W X ! Œ0; 1. x is A -measurable, then there exists an A-measurable function (v) If f  W X ! R x f W X ! R such that the set N  WD ¹x 2 X j f .x/ ¤ f  .x/º 2 A has measure zero, i.e.,  .N  / D 0.

44

1. Abstract measure theory

Proof. (i) First, X 2 A because A  A . Second, let E 2 A and choose A; B 2 A such that A  E  B and .B n A/ D 0. Then B c  E c  Ac and Ac n B c D Ac \ B D B n A. Hence .Ac n B c / D 0 and so E c 2 A . Third, let Ei 2 A for i 2 N and choose Ai ; Bi 2 A such that Ai  Ei  Bi and .Bi n Ai / D 0. Define A WD

[

Ai ;

E WD

[

i

Then A  E  B and B n A D

B WD

Ei ;

[

i

S

i .Bi

.B n A/ 

n A/ 

X

Bi :

i

S

i .Bi

n Ai /. Hence

.Bi n Ai / D 0

i

and this implies E 2 A . Thus we have proved (i). (ii) For E 2 A define  .E/ WD .A/

where A; B 2 A;

(1.44)

A  E  B; .B n A/ D 0: This is the only possibility for defining a measure  W A ! Œ0; 1 that agrees with  on A because .A/ D .B/ whenever A; B 2 A such that A  B and .B n A/ D 0. To prove that  is well defined, let E 2 A and A; B 2 A as in (1.44). If A0 ; B 0 2 A is another pair such that A0  E  B 0 and .B 0 n A0 / D 0, then A n A0  E n A0  B 0 n A0 and hence .A n A0 / D 0. This implies .A/ D .A \ A0 / D .A0 /, where the last equality follows by interchanging the roles of the pairs .A; B/ and .A0 ; B 0 /. Thus the map  W A ! Œ0; 1 in (1.44) is well defined. We prove that  is a measure. Let Ei 2 A be a sequence of pairwise disjoint sets and choose sequences Ai ; Bi 2 A such that Ai  Ei  Bi for all i . Then the S Ai are pairwise disjoint and  .Ei / D .Ai / for all i . Moreover, A WD i Ai 2 A, S B WD i Bi 2 A, A  E  B, and .B n A/ D 0 as we have seen in the proof of P P part (i). Hence  .E/ D .A/ D i .Ai / D i  .Ei /. This proves (ii). (iii) Let E 2 A such that  .E/ D 0 and let E 0  E. Choose A; B 2 A such that A  E  B and .B n A/ D 0. Then .A/ D  .E/ D 0 and hence .B/ D .A/ C .B n A/ D 0. Since E 0  E  B, this implies that E 0 2 A (by choosing B 0 WD B and A0 WD ;). This shows that .X; A ;  / is a complete measure space.

1.6. Completion of a measure space

45

(iv) Assume f W X ! Œ0; 1 is A-measurable. By Theorem 1.26, there exists a sequence of A-measurable step functions sn W X ! R such that 0  s1  s2 


 f and f .x/ D lim sn .x/ for all x 2 X . n!1

Since  jA D  we have Z

Z sn d D

X

sn d

X

for all n and hence it follows from the Lebesgue monotone convergence theorem (Theorem 1.37) for both  and  that Z Z Z Z f d D lim sn d D lim sn d D f d : X

n!1 X

n!1 X

X

This proves (1.43) for E D X and all A-measurable functions f W X ! Œ0; 1. To prove it for all E replace f by f E and use part (ii) of Theorem 1.35. This proves equation (1.43) for all A-measurable functions f W X ! Œ0; 1. That it continues to hold for all f 2 L1 ./ follows directly from Definition 1.43. This proves (iv). (v) If f  D E for E 2 A , choose A; B 2 A such that A  E  B;

.B n A/ D 0;

and define f WD A . Then N  D ¹x 2 X j f  .x/ ¤ f .x/º D E n A  B n A: Hence  .N  /   .B n A/ D .B n A/ D 0. This proves (v) for characteristic functions of A -measurable sets. For A -measurable step functions the assertion follows by multiplication with real numbers and taking finite sums. Now let f  W X ! Œ0; 1 be an arbitrary A -measurable function. By Theorem 1.26, there exists a sequence of A -measurable step functions si W X ! Œ0; 1/ such that si converges pointwise to f  . For each i 2 N choose an A-measurable step function si W X ! Œ0; 1/ and a set Ni 2 A such that si D si on X n Ni and  .Ni / D 0. Then there is a sequence of sets Ni 2 A such that Ni  Ni and .Ni / D 0 for all i . Define f W X ! Œ0; 1 by ´ [ f  .x/ if x … N; f .x/ WD N WD Ni : 0 if x 2 N; i

46

1. Abstract measure theory

Then N 2 A, .N / D 0, and the sequence of A-measurable functions si XnN converges pointwise to f as i tends to infinity. Hence f is A-measurable by part (ii) of Theorem 1.24 and, agrees with f  on X n N by definition. Now let x be A -measurable. Then so are .f  /˙ WD max¹˙f  ; 0º. Construct f W X ! R ˙ f W X ! Œ0; 1 as above. Then f .x/ D 0 whenever f C .x/ > 0 and vice versa. Thus f WD f C f is well defined, A-measurable, and agrees with f  on the complement of a -null set. This proves Theorem 1.55.  Corollary 1.56. Let .X; A; / be a measure space and let .X; A ;  / be its completion. Denote the equivalence class of a -integrable function f 2 L1 ./ under the equivalence relation (1.35) in Definition 1.51 by Œf  WD ¹g 2 L1 ./ j .¹x 2 X j f .x/ ¤ g.x/º/ D 0º: Then the map L1 ./ ! L1 . /W Œf  7 ! Œf 

(1.45)

is a Banach space isometry. Proof. The map (1.45) is linear and injective by definition. It preserves the L1 -norm by part (iv) of Theorem 1.55 and is surjective by part (v) of Theorem 1.55.  As noted in Section 1.5, sets of measure zero can be neglected when integrating functions. Hence it is sometimes convenient to enlarge the notion of integrability. It is not even necessary that the function be defined on all of X , as long as it is defined on the complement of a set of measure zero. Thus, let .X; A; / be a measure space and call a function f W E ! R, defined on a measurable subset E  X, measurable if .X nE/ D 0 and the set f 1 .B/  E is measurable for every Borel set B  R. Call f integrable if the function on all of X , obtained by setting f jX nE D 0, is integrable. If .X; A; / is complete our integrable function f W E ! R can be extended in any manner whatsoever to all of X , and the extended function on X is then integrable in the original sense, regardless of the choice of the extension. Moreover, its integral over any measurable set A 2 A is unaffected by the choice of the extension (see Lemma 1.48). With this extended notion of integrability we see that the Lebesgue dominated convergence theorem (Theorem 1.45) continues to hold if (1.31) is replaced by the weaker assumption that fn only converges to f almost everywhere. That such an extended terminology might be useful can also be seen in TheoP rem 1.52, where the series 1 nD1 fn only converges on the complement of a set N of measure zero, and the function f can only be naturally defined on E WD X n N . Our choice in the proof of Theorem 1.52 was to define f jN WD 0, but this choice

1.7. Exercises

47

does not affect any of the statements of the theorem. Moreover, when working with  the quotient space L1 ./ D L1 ./=  we are only interested in the equivalence class of f under the equivalence relation (1.35) rather than a specific choice of an element of this equivalence class.

1.7 Exercises Exercise 1.57. Let X be an uncountable set and let A  2X be the set of all subsets A  X such either A or Ac is countable. Define ´ 0 if A is countable; .A/ WD 1 if Ac is countable; for A 2 A. Show that .X; A; / is a measure space. Describe the measurable functions and their integrals. (See Examples 1.4 and 1.32.) Exercise 1.58. Let .X; A; / be a measure space such that .X/ < 1 and let fn W X ! Œ0; 1/ be a sequence of bounded measurable functions that converges uniformly to f W X ! Œ0; 1/. Prove that Z Z f d D lim fn d: (1.46) n!1 X

X

Find an example of a measure space .X; A; / with .X/ D 1 and a sequence of bounded measurable functions fn W X ! Œ0; 1/ converging uniformly to f such that (1.46) does not hold. Exercise 1.59. (i) Let fn W Œ0; 1 ! Œ 1; 1 be a sequence of continuous functions that converges uniformly to zero. Show that Z 1 lim fn .x/ dx D 0: n!1 0

(ii) Let fn W Œ0; 1 ! Œ 1; 1 be a sequence of continuous functions such that lim fn .x/ D 0

n!1

Prove that

Z lim

n!1 0

for all x 2 Œ0; 1:

1

fn .x/ dx D 0;

without using Theorem 1.45. A good reference is Eberlein [3].

48

1. Abstract measure theory

(iii) Construct a sequence of continuous functions fn W Œ0; 1 ! Œ 1; 1 that converges pointwise, but not uniformly, to zero. R 1 (iv) Construct a sequence of continuous functions fn W Œ0; 1 ! Œ 1; 1 such that 0 fn .x/ dx D 0 for all n and fn .x/ does not converge for any x 2 Œ0; 1. Exercise 1.60. Let .X; A; / be a measure space and f W X ! Œ0; 1 be a measurable function such that Z 0 < c WD f d < 1: X

Prove that Z lim

n!1 X



n log 1 C

˛

f n˛

8 ˆ 1;

for 0 < ˛ < 1:

Hint. The integrand can be estimated by ˛f when ˛  1. Exercise 1.61. Let X WD N and A WD 2N and let W 2N ! Œ0; 1 be the counting measure (Example 1.30). Prove that a function f W N ! R is -integrable if and only if the sequence .f .n//n2N of real numbers is absolutely summable and that in this case Z 1 X f d D f .n/: N

nD1

Exercise 1.62. Let .X; A/ be a measurable space and let n W A ! Œ0; 1 be a sequence of measures. Show that the formula .A/ WD

1 X

n .A/

nD1

for A 2 A defines a measure W A ! Œ0; 1. Let f W X ! R be a measurable function. Show that f is -integrable if and only if 1 Z X

jf j dn < 1:

nD1 X

If f is -integrable, prove that Z f d D X

1 Z X nD1 X

f dn :

1.7. Exercises

49

Exercise 1.63. Let .X; A; / be a measure space such that .X/ < 1 and let f W X ! R be a measurable function. Show that f is integrable if and only if 1 X

j.¹x 2 X j jf .x/j > nº/j < 1:

nD1

Exercise 1.64. Let .X; A; / be a measure space and let f W X ! R be a -integrable function. (i) Prove that for every " > 0 there exists a ı > 0 such that, for any A 2 A, ˇZ ˇ ˇ ˇ .A/ < ı H) ˇˇ f dˇˇ < ": A

Hint. Argue indirectly. See Lemma 5.21. (ii) Prove that for every " > 0 there exists a measurable set A 2 A such that, for any B 2 A, ˇ ˇZ Z ˇ ˇ ˇ B  A H) ˇ f d f dˇˇ < ": X

B

Exercise 1.65. Let .X; A/ be a measurable space and define .A/ WD

´ 0 1

if A D ;; if A 2 A and A ¤ ;:

Determine the completion .X; A ;  / and the space L1 ./. Exercise 1.66. Let .X; A; / be a measure space such that  D ıx0 is the Dirac measure at some point x0 2 X (Example 1.31). Determine the completion .X; A ;  / and the space L1 ./. Exercise 1.67. Let .X; A; / be a complete measure space. Prove that .X; A; / is equal to its own completion. Exercise 1.68. Let .X; A; / and .X; A0 ; 0 / be two measure spaces with A  A0 and 0 jA D . Prove that L1 ./  L1 .0 / and Z Z f d D f d0 X

for every f 2 L1 ./.

X

50

1. Abstract measure theory

Hint. Prove the following. (i) Let f W X ! Œ0; 1 be A-measurable and define 8 ˆ 0 if f .x/  ı; ˆ < fı .x/ WD f .x/ if ı < f .x/  ı ˆ ˆ : 1 ı if f .x/ > ı 1 :

1

;

Then fı is A-measurable for every ı > 0 and Z Z lim fı d D f d: ı!0 X

X

(ii) Let 0 < c < 1, let f W X ! Œ0; c be A-measurable, and assume that .¹x 2 X j f .x/ > 0º/ < 1: Then

Z

Z f d D

X

(Consider also the function c

f d0 :

X

f .)

Exercise 1.69 (pushforward of a measure). Let .X; A; / be a measure space, let Y be a set, and let W X ! Y be a map. The pushforward of A is the -algebra  A WD ¹B  Y j 

1

.B/ 2 Aº  2Y :

(1.47)

The pushforward of  is the function  W  A ! Œ0; 1 defined by . /.B/ WD .

1

.B//;

for B 2  A:

(1.48)

(i) Prove that .Y;  A;  / is a measure space. (ii) Let .X; A ;  / be the completion of .X; A; / and let .Y; . A/ ; . / / be the completion of .Y;  A;  /. Prove that . / .E/ D  .

1

.E//

for all E 2 . A/   A :

(1.49)

Deduce that .Y;  A;  / is complete whenever .X; A; / is complete. Find an example where . A/ ¨  A .

1.7. Exercises

51

(iii) Fix a function f W Y ! Œ0; 1. Prove that f is  A-measurable if and only if f ı  is A-measurable. If f is  A-measurable, prove that Z Z f d. / D .f ı / d: (1.50) Y

X

(iv) Determine the pushforward of .X; A; / under a constant map. The following extended remark contains a brief introduction to some of the basic concepts and terminology in probability theory. It will not be used elsewhere in this book and can be skipped at first reading. Remark 1.70 (probability theory). A probability space is a measure space .; F;P / such that P ./ D 1: The underlying set  is called the sample space, the -algebra F  2 is called the set of events, and the measure P W F ! Œ0; 1 is called a probability measure. Examples of finite sample spaces are • the set  D ¹h; tº for tossing a coin, • the set  D ¹1; 2; 3; 4; 5; 6º for rolling a dice, • the set  D ¹00; 0; 1; : : : ; 36º for spinning a roulette wheel, and • the set  D ¹2; : : : ; 10; j; q; k; aº  ¹}; ~; ; |º for drawing a card from a deck. Examples of infinite sample spaces are the set  D N [ ¹1º for repeatedly tossing a coin until the first tail shows up, a compact interval of real numbers for random arrival times, and a disc in the plane for throwing a dart. A random variable is an integrable function XW  ! R: Its expectation E.X/ and variance V.X/ are defined by Z E.X/ WD X dP; 

52

1. Abstract measure theory

and Z V.X/ WD

E.X//2 dP D E.X 2 /

.X

E.X/2 ;



respectively. Given a random variable X W  ! R one is interested in the value of the probability measure on the set X 1 .B/ for a Borel set B  R. This value is the probability of the event that the random variable X takes its value in the set B and is denoted by P .X 2 B/ WD P .X 1 .B// D .X P /.B/: Here X P denotes the pushforward of the probability measure P to the Borel  -algebra B  2R (Exercise 1.69). By (1.50) the expectation and variance of X are given by Z E.X/ D x d.X P /.x/ R

and Z V.X/ D

.x R

E.X//2 d.X P /.x/;

respectively. The (cumulative) distribution function of a random variable X is the function FX W R ! Œ0; 1 defined by FX .x/ WD P .X  x/ D P .¹! 2  j X.!/  xº/ D .X P /.. 1; x/: It is nondecreasing and right continuous, satisfies lim FX .x/ D 0;

x! 1

lim FX .x/ D 1;

x!1

and the integral of a continuous function on R with respect to the pushforward measure X P agrees with the Riemann–Stieltjes integral (Exercise 6.20) with respect to FX . Moreover, FX .x/ lim FX .t/ D P .X 1 .x// t !x

by Theorem 1.28. Thus FX is continuous at x if and only if P .X 1 .x// D 0. This leads to the following notions of convergence. Let X W  ! R be a random variable.

1.7. Exercises

53

A sequence .Xi /i2N of random variables is said to • converge in probability to X if X j  "/ D 0

lim P .jXi

i!1

for all " > 0,

• converge in distribution to X if FX .x/ D lim FXi .x/ i!1

for every x 2 R such that FX is continuous at x. We prove that that convergence almost everywhere implies convergence in probability. Let " > 0 and define Ai WD ¹! 2  j jXi .!/

X.!/j  "º

Let E   be the set of all ! 2  such that the sequence Xi .!/ does not converge to X.!/. This set is measurable by Example 1.46 and has measure zero by convergence almost everywhere. Moreover, \ [ Aj  E i2N j i

and so lim P

[

i!1

 Aj D P .E/ D 0

j i

by Theorem 1.28. Thus lim P .Ai / D 0:

i !1

We prove that convergence in probability implies convergence in distribution. Let x 2 R such that FX is continuous at x. Let " > 0 and choose ı > 0 such that FX .x/

" < FX .x 2

" ı/  FX .x C ı/ < FX .x/ C : 2

Now choose i0 2 N such that P .jXi

X j  ı/
0. Define [ U WD ¹x 2 X j there exists a 2 A such that d.a; x/ < "º D B" .a/: a2A

Then U is open, A  U , and U \ B D ;. Hence U 2 B  A./ by assumption and hence .A [ B/ D ..A [ B/ \ U / C ..A [ B/ n U / D .A/ C .B/: Thus the outer measure  satisfies (ii).

62

2. The Lebesgue measure

We prove that (ii) implies (i). Thus assume that  satisfies (ii). We show that every closed set A  X is -measurable, i.e., .D/ D .D \ A/ C .D n A/ for all D  X . Since .D/  .D \ A/ C .D n A/, by definition of an outer measure, it suffices to prove the following. Claim 1. Fix a closed set A  X and a set D  X such that .D/ < 1. Then .D/  .D \ A/ C .D n A/: To see this, replace the set D n A by the smaller set D n Uk , where Uk WD ¹x 2 X j there exists a 2 A such that d.a; x/ < 1=kº D

[

B1=k .a/:

a2A

For each k 2 N, the set Uk is open and d.x; y/  1=k for all x 2 D \ A and all y 2 D n Uk . Hence 1 d.D \ A; D n Uk /  : k By (ii) and axiom (b) this implies .D \ A/ C .D n Uk / D ..D \ A/ [ .D n Uk //  .D/

(2.6)

for every subset D  X and every k 2 N. We will prove the following. Claim 2. lim .D n Uk / D .D n A/. k!1

Claim 1 follows directly from Claim 2 and (2.6). To prove Claim 2 note that AD

1 \

Ui ;

iD1

because A is closed. (If x 2 Ui for all i 2 N, then there exists a sequence ai 2 A such that d.ai ; x/ < 1=i and hence x D lim ai 2 A.) This implies i !1

Uk n A D

1 [

.Uk n Ui / D

i D1

1 [

Ui n Ui C1




iDk

and hence D n A D .D n Uk / [ .D \ .Uk n A// D .D n Uk / [

1 [

.D \ .Ui n UiC1 //:

i Dk

Thus D n A D .D n Uk / [

1 [ i Dk

Ei ;

Ei WD .D \ Ui / n Ui C1 :

(2.7)

2.1. Outer measures

63

Claim 3. The outer measures of the Ei satisfy 1 X

.Ei / < 1:

iD1

Claim 3 H) Claim 2. It follows from Claim 3 that the sequence "k WD

1 X

.Ei /

iDk

converges to zero. Moreover, it follows from equation (2.7) and axiom (c) in Definition 2.3 that .D n A/  .D n Uk / C

1 X

.Ei / D .D n Uk / C "k :

i Dk

Hence it follows from axiom (b) in Definition 2.3 that .D n A/

"k  .D n Uk /  .D n A/

for every k 2 N. Since "k converges to zero, this implies Claim 2. The proof of Claim 3 relies on the next assertion. Claim 4. d.Ei ; Ej / > 0 for i  j C 2. Claim 4 H) Claim 3. It follows from Claim 4, axiom (b), and (ii) that n X

.E2i / D 

i D1

n [

 E2i  .D/

iD1

and n X

.E2i

i D1

1/

D

n [

E2i

 1

 .D/

iD1

for every n 2 N. Hence 1 X

.Ei /  2.D/ < 1

i D1

and this shows that Claim 4 implies Claim 3.

64

2. The Lebesgue measure

Proof of Claim 4. We show that d.Ei ; Ej / 

1 .i C 1/.i C 2/

for j  i C 2:

To see this, fix indices i; j with j  i C 2. Let x 2 Ei and y 2 X such that d.x; y/


1 i C1

d.x; y/ 1 .i C 1/.i C 2/

1 i C2 1  j D

for all a 2 A. Hence y … Uj and consequently y … Ej , because Ej  Uj . This proves Claim 4 and Theorem 2.5. 

2.2 The Lebesgue outer measure The purpose of this section is to introduce the Lebesgue outer measure  on Rn , construct the Lebesgue measure as the restriction of  to the  -algebra of all -measurable subsets of Rn , and prove Theorem 2.1. Definition 2.6. A closed cuboid in Rn is a set of the form Q WD Q.a; b/ WD Œa1 ; b1   Œa2 ; b2  


 Œan ; bn 

(2.8)

n

D ¹x D .x1 ; : : : ; xn / 2 R j aj  xj  bj for j D 1; : : : ; nº for a1 ; : : : ; an ; b1 ; : : : ; bn 2 R with aj < bj for all j . The (n-dimensional) volume of the cuboid Q.a; b/ is defined by Vol.Q.a; b// WD Voln .Q.a; b// WD

n Y j D1

.bj

aj /:

(2.9)

2.2. The Lebesgue outer measure

65

The volume of the open cuboid n Y

U WD int.Q/ D

.ai ; bi /

iD1

is defined by Vol.U / WD Vol.Q/: The set of all closed cuboids in Rn will be denoted by ˇ ² ³ ˇ a ; : : : ; an ; b1 ; : : : ; bn 2 R; Qn WD Q.a; b/ ˇˇ 1 : aj < bj for j D 1; : : : ; n Definition 2.7. A subset A  Rn is called a Jordan null set if, for every " > 0, there exist finitely many closed cuboids Q1 ; : : : ; Q` 2 Qn such that A

` [

` X

Qi ;

iD1

Vol.Qi / < ":

iD1

Definition 2.8. A subset A  Rn is called a Lebesgue null set if, for every " > 0, there is a sequence of closed cuboids Qi 2 Qn , i 2 N, such that A

1 [

1 X

Qi ;

iD1

Vol.Qi / < ":

iD1

Definition 2.9. The Lebesgue outer measure on Rn is the function n

 D n W 2 R

! Œ0; 1

defined by .A/ WD inf

1 °X

1 ˇ ± [ ˇ Voln .Qi / ˇ Qi 2 Qn ; A  Qi

i D1

for A  Rn :

(2.10)

i D1 n

Theorem 2.10 (the Lebesgue outer measure). Let W 2R ! Œ0; 1 be the function defined by (2.10). Then the following holds. (i)  is an outer measure; (ii)  is translation-invariant, i.e., for all A  Rn and all x 2 Rn .A C x/ D .A/; (iii) if A; B  Rn are such that d.A; B/ > 0, then .A [ B/ D .A/ C .B/; (iv) .int.Q// D .Q/ D Vol.Q/ for all Q 2 Qn .

66

2. The Lebesgue measure

Proof. We prove (i). The empty set is contained in every cuboid Q 2 Qn . Since there are cuboids with arbitrarily small volume, it follows that .;/ D 0. If A  B  Rn it follows directly from Definition 2.9 that .A/  .B/. Now let Ai  Rn for i 2 N, define 1 [ Ai ; A WD iD1

and fix a constant " > 0. Then it follows from Definition 2.9 that, for i 2 N, there exists a sequence of cuboids Qij 2 Qn , j 2 N, such that Ai 

1 [

Qij ;

j D1

1 X

Vol.Qij /
0 and choose a sequence of open cuboids Ui  Rn such that " Qi  Ui ; Vol.Ui / < Vol.Qi / C i : 2 Since Q is compact and the Ui form an open cover of Q, there exists a constant k 2 N such that k [ Q Ui : iD1

This implies k 1 n X 1  1 n 1  # Q \ # U \ Z  Z i Nn N Nn N i D1

k X 1 n 1 x # U \ Z : i Nn N



i D1

Take the limit N ! 1 and use (2.13) to obtain Vol.Q/ 

k X

Vol.Ui /

i D1



1 X

Vol.Ui /

i D1



1   X " C Vol.Q / i 2i i D1

D"C

1 X iD1

Vol.Qi /:

2.2. The Lebesgue outer measure

69

Since " > 0 can be chosen arbitrarily small, this proves (2.12) and (2.11). Thus we have proved that .Q/  Vol.Q/  .Q/, and so .Q/ D Vol.Q/. To prove that .int.Q// D Vol.Q/, fix a constant " > 0 and choose a closed cuboid P 2 Qn such that P  int.Q/; Vol.Q/ " < Vol.P /: Then Vol.Q/ Thus Vol.Q/

" < Vol.P / D .P /  .int.Q//:

" < .int.Q// for all " > 0. Therefore, by axiom (b), Vol.Q/  .int.Q//  .Q/ D Vol.Q/;

and hence .int.Q// D Vol.Q/. This proves part (iv) and Theorem 2.10.



Rn

Definition 2.11. Let W 2 ! Œ0; 1 be the Lebesgue outer measure. A subset A  Rn is called Lebesgue measurable if A is -measurable, i.e., .D/ D .D \ A/ C .D n A/

for all D  Rn :

The set of all Lebesgue measurable subsets of Rn will be denoted by A WD ¹A  Rn j A is Lebesgue measurableº: The function

m WD jA W A ! Œ0; 1

is called the Lebesgue measure on Rn . A function f W Rn ! R is called Lebesgue measurable if it is measurable with respect to the Lebesgue  -algebra A on Rn (and the Borel -algebra on the target space R). Corollary 2.12. (i) .Rn ; A; m/ is a complete measure space. (ii) m is translation-invariant, i.e., if A 2 A and x 2 Rn , then A C x 2 A and

m.A C x/ D m.A/:

(iii) Every Borel set in Rn is Lebesgue measurable. (iv) If Q 2 Qn , then Q; int.Q/ 2 A;

and

m.int.Q// D m.Q/ D Vol.Q/:

Proof. Assertion (i) follows from Theorem 2.4 and part (i) of Theorem 2.10. Assertion (ii) follows from the definitions and part (ii) of Theorem 2.10. Assertion (iii) follows from Theorem 2.5 and part (iii) of Theorem 2.10. Finally, assertion (iv) follows from (iii) and part (iv) of Theorem 2.10. 

70

2. The Lebesgue measure

The restriction of the measure m in Corollary 2.12 to the Borel -algebra of Rn satisfies the requirements of Theorem 2.1 (translation invariance and normalization) and hence settles the existence problem. The uniqueness proof relies on certain regularity properties of the measure m which are established in the next theorem along with continuity from below for the Lebesgue outer measure . Theorem 2.14 shows that m is the completion of its restriction to the Borel -algebra of Rn and, with that at hand, we can then prove uniqueness in Theorem 2.1. Theorem 2.13 (regularity of the Lebesgue outer measure). The Lebesgue outer n measure W 2R ! Œ0; 1 satisfies the following. (i) For every subset A  Rn , .A/ D inf¹.U / j A  U  Rn and U is openº: (ii) If A  Rn is Lebesgue measurable, then .A/ D sup¹.K/ j K  A and K is compactº: (iii) If Ai is a sequence of subsets of Rn such that Ai  Ai C1 for all i 2 N, then their union 1 [ A WD Ai i D1

has Lebesgue outer measure .A/ D lim .Ai /: i!1

Proof. (i) Fix a subset A  Rn and a constant " > 0. The assertion is obvious when .A/ D 1. Hence assume .A/ < 1 and choose a sequence of closed cuboids Qi 2 Qn such that A

1 [

Qi ;

iD1

1 X iD1

" Vol.Qi / < .A/ C : 2

Now choose a sequence of open cuboids Ui  Rn such that Q i  Ui ; Then U WD

S1

.U / 

i D1

1 X iD1

Vol.Ui / < Vol.Qi / C

" 2i C1

:

Ui is an open subset of Rn containing A and

.Ui / D

1 X iD1

Vol.Ui /
0 so large that A  Br WD ¹x 2 Rn j jxj < rº: Fix a constant " > 0. By (i), there exists an open set U  Rn such that Bxr n A  U and .U /  .Bxr n A/ C ". Hence K WD Bxr n U is a compact subset of A and .K/ D .Bxr /

.U /  .Bxr /

.Bxr n A/

" D .A/

":

Here the first equality uses the fact that K and U are disjoint Lebesgue measurable sets with union Bxr and the last equality uses the fact that A and Bxr n A are disjoint Lebesgue measurable sets with union Bxr . This proves (ii) for bounded Lebesgue measurable sets. If A 2 A is unbounded, then .A/ D sup .A \ Bxr / r>0

D sup sup¹.K/ j K  .A \ Bxr / and K is compactº r>0

D sup¹.K/ j K  A and K is compactº: (iii) If .Ai / D 1 for some i , then the assertion is obvious. Hence assume .Ai / < 1 for all i and fix a constant " > 0. By (i), there is a sequence of open sets Ui  Rn such that .Ui / < .Ai / C "2 i for all i. Since Ai  Ui \ UiC1 , this yields .UiC1 n Ui / D .UiC1 / < .AiC1 /

.Ui \ Ui C1 / .Ai / C "2

i 1

for all i 2 N. It follows that k k [  X1  Ui D .U1 / C .UiC1 n Ui / < .Ak / C ": i D1

iD1

Take the limit k ! 1 to obtain 1 [   Ui  lim .Ak / C ": iD1

k!1

Thus .A/  lim .Ak / C "; k!1

for all " > 0,

and so .A/  lim .Ak /: k!1

The converse inequality is obvious. This proves Theorem 2.13.



72

2. The Lebesgue measure n

Theorem 2.14 (the Lebesgue measure as a completion). Let W 2R ! Œ0; 1 be the Lebesgue outer measure in Definition 2.9, let m D jA W A ! Œ0; 1 be the Lebesgue measure, let B  A be the Borel -algebra of Rn , and define  WD jB W B ! Œ0; 1: Then .Rn ; A; m/ is the completion of .Rn ; B; /. Proof. Let .Rn ; B ;  / denote the completion of .Rn ; B; /. Claim. Let A  Rn . Then the following are equivalent. (I) A 2 A, i.e., .D/ D .D \ A/ C .D n A/ for all D  Rn . (II) A 2 B , i.e., there exist Borel measurable sets B0 ; B1 2 B such that B0  A  B1

and

.B1 n B0 / D 0:

If the set A satisfies both (I) and (II), then .A/  .B1 / D .B0 / C .B1 n B0 / D .B0 /  .A/; and hence

m.A/ D .A/ D .B0 / D  .A/:

This shows that A D B and m D  . Thus it remains to prove the claim. Fix a subset A  Rn . We show that (II) implies (I). Thus assume that A 2 B and choose Borel measurable sets B0 ; B1 2 B such that B0  A  B1 ;

.B1 n B0 / D 0:

Then .A n B0 /  .B1 n B0 / D 0 and hence .A n B0 / D 0. Since  is an outer measure, by part (i) of Theorem 2.10, it follows from Theorem 2.4 that A n B0 2 A and hence A D B0 [ .A n B0 / 2 A. We now show that (I) implies (II). Thus assume that A 2 A. Suppose first that .A/ < 1. By Theorem 2.13, there exist a sequence of compact sets Ki  Rn and a sequence of open sets Ui  Rn such that Ki  A  Ui ;

.A/

1 1  .Ki /  .Ui /  .A/ C : i i

2.2. The Lebesgue outer measure

Define

1 [

B0 WD

Ki ;

1 \

B1 WD

73

Ui :

i D1

i D1

These are Borel sets satisfying B0  A  B1 and .A/

1 1  .Ki /  .B0 /  .B1 /  .Ui /  .A/ C : i i

Take the limit i ! 1 to obtain .A/  .B0 /  .B1 /  .A/; hence .B0 / D .B1 / D .A/ < 1; and hence .B1 n B0 / D .B1 /

.B0 / D 0:

This shows that A 2 B for every A 2 A with .A/ < 1. Now suppose that our set A 2 A satisfies .A/ D 1 and define Ak WD ¹x 2 A j jxi j  k for i D 1; : : : ; nº for k 2 N: Then Ak 2 A and .Ak /  .2k/n for all k. Hence Ak 2 B for all k and so there exist sequences of Borel sets Bk ; Bk0 2 B such that Bk  Ak  Bk0

and

.Bk0 n Bk / D 0:

and

B 0 WD

Define B WD

1 [

Bk

kD1

1 [

Bk0 :

kD1

Then B; B 0 2 B, B  A  B 0 , and 0

.B n B/ 

1 X kD1

.Bk0

n B/ 

1 X

.Bk0 n Bk / D 0:

kD1

This shows that A 2 B for every A 2 A. Thus we have proved that (I) implies (II) and this completes the proof of Theorem 2.14. 

74

2. The Lebesgue measure

Proof of Theorem 2.1. The existence of a translation-invariant normalized Borel measure on Rn follows from Corollary 2.12. We prove uniqueness. Thus assume that 0 W B ! Œ0; 1 is a translation-invariant measure such that 0 .Œ0; 1n / D 1. Define  WD 0 .Œ0; 1/n /. Then 0    1. We show in five steps that  D 1 and 0 D . Step 1. For x D .x1 ; : : : ; xn / and k 2 N0 WD N [ ¹0º define R.x; k/ WD Œx1 ; x1 C 2 Then 0 .R.x; k// D 2

nk

k

/ 


 Œxn ; xn C 2

k

/:

D .R.x; k//.

Fix an integer k 2 N0 . Since R.x; k/ D R.0; k/ C x for every x 2 Rn , it follows from the translation invariance of 0 that there is a constant ck  0 such that for all x 2 Rn :

0 .R.x; k// D ck

Since R.x; 0/ can be expressed as the disjoint union [ R.x; 0/ D R.x C 2 `2Zn ; 0`j 2k

k

`; k/;

1

this implies  D 0 .R.x; 0// D

X

0 .R.x C 2

k

`; k// D 2nk ck :

`2Zn ; 0`j 2k 1

Hence ck D 2 nk D .R.x; k//. Here the last equation follows from the fact that .0; 2 k /n  R.0; k/  Œ0; 2 k n and hence .R.x; k// D .R.0; k// D 2 k by part (iv) of Corollary 2.12. Step 2. 0 .U / D .U / for every open set U  Rn . Let U  Rn be open. We prove that U can be expressed as a countable union of sets Ri D R.xi ; ki / as in Step 1. To see this, define R0 WD ¹R.x; 0/ j x 2 Zn ; R.x; 0/  U º; ˇ ² ³ ˇ x 2 2 1 Zn ; R.x; 1/  U; ˇ R1 WD R.x; 1/ ˇ ; R.x; 1/ 6 R, for all R 2 R0 ˇ ² ˇ x 2 2 k Zn ; R.x; k/  U; Rk WD R.x; k/ ˇˇ R.x; k/ 6 R, for all R 2 R0 [ R1 [


[ Rk and denote R WD

1 [ kD0

Rk :

³ 1

for k  2,

2.2. The Lebesgue outer measure

Then U can be expressed as the disjoint union U D for all R 2 R by Step 1. Hence X

0 .U / D

R2R

0 .R/ D

X

S

R2R

75

R and 0 .R/ D .R/

.R/ D .U /:

R2R

Step 3. 0 .K/ D .K/ for every compact set K  Rn . Let K  Rn be compact. Choose r > 0 so large that K  U WD . r; r/n : Then U and U n K are open. Hence, by Step 2, 0 .K/ D 0 .U /

0 .U n K/ D .U /

.U n K/ D .K/:

Step 4. 0 .B/ D .B/ for every Borel set B 2 B. Let B 2 B. It follows from Step 2, Step 3, and Theorem 2.13 that 0 .B/  inf¹0 .U / j B  U  Rn and U is openº D inf¹.U / j B  U  Rn and U is openº D .B/ D sup¹.K/ j K  B and K is compactº D sup¹0 .K/ j K  B and K is compactº  0 .B/: Step 5.  D 1 and 0 D . By Step 4, we have  D .Œ0; 1n / D 0 .Œ0; 1n / D 1 and hence 0 D  D . This proves Step 5 and Theorem 2.1.  We have given two definitions of the Lebesgue measure mW A ! Œ0; 1: The first one, in Definition 2.2, uses the existence and uniqueness of a normalized translation-invariant Borel measure W B ! Œ0; 1; established in Theorem 2.1, and then defines .Rn ; A; m/ as the completion of that measure. The second one, n in Definition 2.11, uses the Lebesgue outer measure W 2R ! Œ0; 1 of Definition 2.9 and Theorem 2.10, and defines the Lebesgue measure as the restriction of  to the -algebra of -measurable subsets of Rn (see Theorem 2.4). Theorem 2.14 asserts that the two definitions agree. The next lemma uses the Axiom of Choice to establish the existence of subsets of Rn that are not Lebesgue measurable.

76

2. The Lebesgue measure

Lemma 2.15. Let A  R be a Lebesgue measurable set such that m.A/ > 0. Then there exists a set B  A that is not Lebesgue measurable. Proof. Consider the equivalence relation on R defined by def

x  y () x

y2Q

for x; y 2 R. By the Axiom of Choice, there exists a subset E  R which contains precisely one element of each equivalence class. This means that E satisfies the following two conditions: (I) for every x 2 R there exists a rational number q 2 Q such that x (II) if x; y 2 E and x ¤ y, then x

q 2 E;

y … Q.

For q 2 Q define the set Bq WD A \ .E C q/ D ¹x 2 A j x q 2 Eº: S Then it follows from (I) that A D q2Q Bq : Fix a rational number q 2 Q. We prove that if Bq is Lebesgue measurable, then m.Bq / D 0. Indeed, let n 2 N, and define Bq;q 0 ;n WD .Bq \ Œ n; n/ C q 0 D ¹x C q 0 j x 2 Bq ; jxj  nº

for q 0 2 Q:

This set is Lebesgue measurable, its Lebesgue measure is independent of q 0 , and Bq;q 0 ;n \ Bq;q 00 ;n D ; for all q 0 ; q 00 2 Q with q 0 ¤ q 00 by condition (II). Since P Bq;q 0 ;n  Œ n; nC1 for q 0 2 Œ0; 1\Q, we have q 0 2Œ0;1\Q m.Bq;q 0 ;n /  2n C 1: This sum is infinite and all summands agree, so m.Bq \ Œ n; n/ D 0. This holds for all n 2 N, and hence m.Bq / D 0, as claimed. S If Bq is Lebesgue measurable for all q 2 Q, it follows that A D q2Q Bq is a Lebesgue null set, a contradiction. Thus one of the sets Bq is not Lebesgue measurable and this proves Lemma 2.15.  Remark 2.16. (i) Using Lemma 2.15 one can construct a continuous function f W R ! R and a Lebesgue measurable function gW R ! R such that the composition g ı f is not Lebesgue measurable (see Example 6.24). (ii) Let E  R be the set constructed in the proof of Lemma 2.15. Then the set E  R  R2 is not Lebesgue measurable. This follows from a similar argument as in Lemma 2.15 using the sets ..E \ Œ n; n/ C q/  Œ0; 1. On the other hand, the set E  ¹0º  R2 is Lebesgue measurable and has Lebesgue measure zero. However, it is not a Borel set, because its pre-image in R under the continuous map R ! R2 W x 7! .x; 0/ is the original set E and hence is not a Borel set.

2.3. The transformation formula

77

2.3 The transformation formula The transformation formula describes how the integral of a Lebesgue measurable function transforms under composition with a C 1 diffeomorphism. Fix a positive integer n 2 N and denote by .Rn ; A; m/ the Lebesgue measure space. For any Lebesgue measurable set X  Rn denote by AX WD ¹A 2 A j A  X º the restricted Lebesgue -algebra and by mX WD mjAX W AX ! Œ0; 1 the restriction of the Lebesgue measure to AX . Theorem 2.17 (transformation formula). Suppose W U ! V is a C 1 diffeomorphism between open subsets of Rn . (i) If f W V ! Œ0; 1 is Lebesgue measurable, then f ı W U ! Œ0; 1 is Lebesgue measurable and Z Z .f ı /jdet.d/j d m D f d m: (2.14) U

V

(ii) If E 2 AU and f 2 L1 .mV /, then .E/ 2 AV , .f ı /jdet.d/j 2 L1 .mU /, and Z Z .f ı /jdet.d/j d m D f d m: (2.15) E

.E /



Proof. See page 82.

The proof of Theorem 2.17 relies on the next two lemmas. The first lemma is the special case where  is linear and f is the characteristic function of a Lebesgue measurable set. The second lemma is a basic estimate that follows from the linear case and implies formula (2.14) for the characteristic functions of open sets. Lemma 2.18. Let ˆW Rn ! Rn be a linear transformation and let A  Rn be a Lebesgue measurable set. Then ˆ.A/ is a Lebesgue measurable set and m.ˆ.A// D jdet.ˆ/jm.A/:

(2.16)

78

2. The Lebesgue measure

Proof. If det.ˆ/ D 0, then ˆ.A/ is contained in a proper linear subspace of Rn and hence is a Lebesgue null set for every A 2 A. In this case both sides of equation (2.16) vanish. Hence it suffices to assume that ˆ is a vector space isomorphism. For vector space isomorphisms we prove the assertion in six steps. Denote n by B  2R the Borel  -algebra and by  WD mjB the restriction of the Lebesgue measure to the Borel -algebra. Thus  is the unique translation-invariant Borel measure on Rn that satisfies the normalization condition .Œ0; 1n / D 1 (Theorem 2.1) and .Rn ; A; m/ is the completion of .Rn ; B; / (Theorem 2.14). Step 1. There exists a unique map W GL.n; R/ ! .0; 1/ such that .ˆ.B// D .ˆ/.B/

(2.17)

for every ˆ 2 GL.n; R/ and every Borel set B 2 B. Fix a vector space isomorphism ˆW Rn ! Rn . Since ˆ is a homeomorphism of Rn with its standard topology, it follows that ˆ.B/ 2 B for every B 2 B. Define the number .ˆ/ 2 Œ0; 1 by .ˆ/ WD .ˆ.Œ0; 1/n //:

(2.18)

Since ˆ.Œ0; 1/n / has nonempty interior, it follows that .ˆ/ > 0 and since ˆ.Œ0; 1/n / is contained in the compact set ˆ.Œ0; 1n /, it follows that .ˆ/ < 1. Now define the map ˆ W B ! Œ0; 1 by ˆ .B/ WD

.ˆ.B// .ˆ/

for B 2 B:

Then ˆ is a normalized translation-invariant Borel measure. The -additivity follows directly from the -additivity of , the formula ˆ .;/ D 0 is obvious from the definition, that compact sets have finite measure follows from the fact that ˆ.K/ is compact if and only if K  Rn is compact, the translation invariance follows immediately from the translation invariance of  and the fact that ˆ.B C x/ D ˆ.B/ C ˆ.x/ for all B 2 B and all x 2 Rn , and the normalization condition ˆ .Œ0; 1/n / D 1 follows directly from the definition of ˆ . Hence ˆ D  by Theorem 2.1.

2.3. The transformation formula

79

Step 2. Let  be as in Step 1 and let A 2 A and ˆ 2 GL.n; R/. Then ˆ.A/ 2 A and m.ˆ.A// D .ˆ/m.A/. By Theorem 2.14, there exist Borel sets B0 ; B1 2 B such that B0  A  B1 and .B1 n B0 / D 0. Then ˆ.B0 /  ˆ.A/  ˆ.B1 / and, by Step 1, .ˆ.B1 / n ˆ.B0 // D .ˆ.B1 n B0 // D .ˆ/.B1 n B0 / D 0: Hence ˆ.A/ is a Lebesgue measurable set and m.ˆ.A// D .ˆ.B0 // D .ˆ/.B0 / D .ˆ/m.A/ by Theorem 2.14 and Step 1. Step 3. Let  be as in Step 1 and let ˆ D diag.1 ; : : : ; n / be a diagonal matrix with nonzero diagonal entries i 2 R n ¹0º. Then .ˆ/ D j1 j


jn j. Define I WD Œ 1; 1 and Ii WD Œ ji j; ji j for i D 1; : : : ; n. Then Q WD I n has Lebesgue measure m.Q/ D 2n and the cuboid ˆ.Q/ D I1 


 In has Lebesgue measure m.ˆ.Q// D 2n j1 j


jn j by part (iv) of Corollary 2.12. Hence Step 3 follows from Step 2. Step 4. The map W GL.n; R/ ! .0; 1/ in Step 1 is a group homomorphism from the general linear group of automorphisms of Rn to the multiplicative group of positive real numbers. Let ˆ; ‰ 2 GL.n; R/. Then it follows from (2.17) with B WD ‰.Œ0; 1/n / and from the definition of .‰/ in (2.18) that .ˆ‰/ D .ˆ‰.Œ0; 1/n // D .ˆ/.‰.Œ0; 1/n // D .ˆ/.‰/: Thus  is a group homomorphism as claimed and this proves Step 4. Step 5. The map W GL.n; R/ ! .0; 1/ in Step 1 is continuous with respect to the standard topologies on GL.n; R/ and .0; 1/. It suffices to prove continuity at the identity. Define the norms kxk1 WD max jxi j ; iD1;:::;n

kˆk1 WD

sup

0¤x2Rn

kˆxk1 kxk1

(2.19)

for x 2 Rn and a linear map ˆW Rn ! Rn . Denote the closed unit ball in Rn by Q WD ¹x 2 Rn j kxk1  1º D Œ 1; 1n . Fix a constant 0 < ı < 1 and a linear map ˆW Rn ! Rn such that kˆ 1k1 < ı. Then ˆ 2 GL.n; R/ and ˆ

1

D

1 X kD0

.1

ˆ/k ;

kˆ

1

k1
0 by Steps 2 and 3, this shows that .1 ı/n  .ˆ/  .1 C ı/n . Given " > 0, choose a constant 0 < ı < 1 such that 1 " < .1 ı/n < .1 C ı/n < 1 C ". Then kˆ

1k1 < ı H) j.ˆ/

1j1 < "

for all ˆ 2 GL.n; R/. Step 6. .ˆ/ D jdet.ˆ/j for all ˆ 2 GL.n; R/. If ˆ 2 GL.n; R/ is diagonalizable with real eigenvalues, then .ˆ/ D jdet.ˆ/j by Step 3 and Step 4. If ˆ 2 GL.n; R/ has only real eigenvalues, then it can be approximated by a sequence of diagonalizable automorphisms with real eigenvalues and hence it follows from Step 5 that .ˆ/ D jdet.ˆ/j. Since every automorphism of Rn is a finite composition of automorphisms with real eigenvalues (elementary matrices), this proves Step 6. Lemma 2.18 follows immediately from Step 2 and Step 6.  Define the metric d1 W Rn  Rn ! Œ0; 1/ by d1 .x; y/ WD kx

yk1

for x; y 2 Rn ,

where k
k1 is as in (2.19). The open ball of radius r > 0 about a point a D .a1 ; : : : ; an / 2 Rn with respect to this metric is the open cube Br .a/ WD .a1 and its closure is Bxr .a/ D Œa1

r; a1 C r/ 


 .an

r; an C r/

r; a1 C r 


 Œan

r; an C r:

Lemma 2.19. Let U  Rn be an open set and let K  U be a compact subset. Let W U ! Rn be a continuously differentiable map such that det.d.x// ¤ 0 for all x 2 K. For every " > 0 there exists a constant ı > 0 such that the following holds: if 0 < r < ı, a 2 Rn , and R  Rn satisfy Br .a/  R  Bxr .a/  K, then jm..R//

jdet.d.a//j m.R/j < " m.R/:

(2.20)

2.3. The transformation formula

81

Proof. The maps K ! RW x 7! kd.x/ 1 k1 and K ! RW x 7! jdet.d.x//j are continuous by assumption. Since K is compact, these maps are bounded. Hence there is a constant c > 0 such that

d.x/

1 1

 c;

jdet.d.x//j  c

for all x 2 K:

(2.21)

Let " > 0 and choose a constant 0 < ˛ < 1 so small that 1

" ˛/n < .1 C ˛/n < 1 C : c

" < .1 c

(2.22)

Choose ı > 0 so small that, for all x; y 2 Rn , x; y 2 K; kx

yk1 < ı H) kd.x/

d.y/k1
0, Step 1 follows. Step 2. K  U is compact, then Z m..K// D

jdet.d/j d m: K

S  U . Then Choose an open set W  K with compact closure W m..K// D m..W // m..W n K// Z Z D jdet.d/j d m jdet.d/j d m W

W nK

Z D

jdet.d/j d m: K

Here the second equality follows from Step 1.

(2.27)

84

2. The Lebesgue measure

Step 3. If B 2 B has compact closure Bx  U , then .B/ 2 B and Z m..B// D jdet.d/j d m: B

That .B/ is a Borel set follows from the fact that it is the pre-image of the Borel set B under the continuous map  1 W V ! U (Theorem 1.20). Abbreviate b WD m..B//: Assume first that b < 1 and fix a constant " > 0. Then it follows from Theorem 2.13 that there exist an open set W 0  Rn with compact closure W 0  V such that .B/  W 0 and m.W 0 / < b C " and a compact set K 0  B such that .K 0 / > b ". Define K WD  1 .K 0 / and W WD  1 .W 0 /. Then K is compact, S  U is compact, and W is open, W K  B  W;

b

" < m..K//  m..W // < b C ":

Hence it follows from Step 1 and Step 2 that Z Z Z b " < jdet.d/j d m  jdet.d/j d m  jdet.d/j d m < b C ": K

B

Thus

W

Z b

jdet.d/j d m < b C "

"
0; then d .A [ B/ D d .A/ C d .B/: Hence, by Theorems 2.4 and 2.5, the set Ad WD ¹A  X j A is d -measurableº is a -algebra containing the Borel sets and d WD d jAd W Ad ! Œ0; 1 is a measure. It is called the d -dimensional Hausdorff measure on X . Hausdorff measures play a central role in geometric measure theory. (iii) If d D 0, then A0 D 2X and 0 D 0 is the counting measure. (iv) The n-dimensional Hausdorff measure on Rn agrees with the Lebesgue measure up to a factor (the Lebesgue measure of the ball of radius 1=2).

2.5. Exercises

93

(v) Let A  X be nonempty. The Hausdorff dimension of A is the number dim.A/ WD sup¹r  0 j r .A/ D 1º D inf¹s  0 j s .A/ D 0º:

(2.37)

The second equality follows from the fact that d .A/ > 0 implies r .A/ D 1 for 0  r < d , and d .A/ < 1 implies s .A/ D 0 for s > d . (vi) The Hausdorff dimension of a smooth embedded curve €  Rn is d D 1 and its 1-dimensional Hausdorff measure 1 .€/ is the length of the curve. (vii) The Hausdorff dimension of the Cantor set is d D log.2/= log.3/.

Chapter 3

Borel measures

The regularity properties established for the Lebesgue (outer) measure in Theorem 2.13 play an important role in much greater generality. The present chapter is devoted to the study of Borel measures on locally compact Hausdorff spaces that satisfy similar regularity properties. The main result is the Riesz representation theorem (Theorem 3.15). We begin with some further recollections about topological spaces. Let .X; U/ be a topological space (see Definition 1.9). A neighborhood of a point x 2 X is a subset A  X that contains x in its interior, i.e., x 2 U  A for some open set U . X is called a Hausdorff space if any two distinct points in X have disjoint neighborhoods, i.e., for all x; y 2 X with x ¤ y there exist open sets U; V  X such that x 2 U , y 2 V , and U \ V D ;. X is called locally compact if every point in X has a compact neighborhood. It is called -compact if there exists a sequence of compact sets Ki  X, i 2 N, such that Ki  Ki C1 for all i and S XD 1 iD1 Ki .

3.1 Regular Borel measures Assume throughout that .X; U/ is a locally compact Hausdorff space and denote by B  2X the Borel -algebra. Thus B is the smallest -algebra on X that contains all open sets. In the context of this chapter it is convenient to include local finiteness (compact sets have finite measure) in the definition of a Borel measure. There are other geometric settings, such as the study of Hausdorff measures (Exercise 2.35), where one allows for compact sets to have infinite measure, but these are not discussed here. Definition 3.1. A measure W B ! Œ0; 1 is called a Borel measure if .K/ < 1 for every compact set K  X. A measure W B ! Œ0; 1 is called outer regular if .B/ D inf¹.U / j B  U  X and U is openº

(3.1)

96

3. Borel measures

for every Borel set B 2 B, is called inner regular if .B/ D sup¹.K/ j K  B and K is compactº

(3.2)

for every Borel set B 2 B, and is called regular if it is both outer and inner regular. A Radon measure is an inner regular Borel measure. Example 3.2. The restriction of the Lebesgue measure on X D Rn to the Borel -algebra is a regular Borel measure by Theorem 2.13. Example 3.3. The counting measure on X D N with the discrete topology U D B D 2N is a regular Borel measure. Example 3.4. Let .X; U/ be any locally compact Hausdorff space and fix a point x0 2 X . Then the Dirac measure  D ıx0 at x0 is a regular Borel measure (Example 1.31). Example 3.5. Let X be an uncountable set equipped with the discrete topology U D B D 2X . Define W B ! Œ0; 1 by .B/ WD

´ 0 1

if B is countable; if B is uncountable:

This is a Borel measure. Moreover, a subset K  X is compact if and only if it is finite. Hence .X/ D 1 and .K/ D 0 for every compact set K  X . Thus  is not a Radon measure. The next example occupies four pages and illustrates the subtlety of the subject (see also Exercise 18 in Rudin [17, page 59]). It constructs a compact Hausdorff space .X; U/ and a Borel measure  on X that is not a Radon measure. More precisely, there is a point  2 X such that the open set U WD X n ¹º is not -compact and satisfies .U / D 1 and .K/ D 0 for every compact subset K  U . This example is a kind of refinement of Example 3.5. It is due to Dieoudonné. Example 3.6 (Dieudonné’s measure). (i) Let .X; 4/ be an uncountable well-ordered set with a maximal element  2 X such that every element x 2 X n ¹º has only countably many predecessors. Here a set is called countable if it is finite or countably infinite. (Think of this as the uncountable Mount Everest: no sequence reaches the mountain peak .) Thus the relation 4 on X satisfies the following axioms.

3.1. Regular Borel measures

97

(a) If x; y; z 2 X satisfy x 4 y and y 4 z, then x 4 z. (b) If x; y 2 X satisfy x 4 y and y 4 x, then x D y. (c) If x; y 2 X , then x 4 y or y 4 x. (d) If ; ¤ A  X , then there is an a 2 A such that a 4 x for all x 2 A. (e) If x 2 X n ¹º, then x 4  and the set ¹y 2 X j y 4 xº is countable. Define the relation  on X by x  y if x 4 y and x ¤ y. For ; ¤ A  X denote by min.A/ 2 A the unique element of A that satisfies min.A/ 4 x for all x 2 A. (See conditions (b) and (d).) For x 2 X define Sx WD ¹y 2 X j x  yº;

Px WD ¹y 2 X j y  xº:

Thus Px is the set of predecessors of x and Sx is the set of successors of x. If x 2 X n ¹º, then Px is countable and Sx is uncountable. Define the map sW X n ¹º ! X n ¹º;

s.x/ WD min.Sx /:

Then X n Sx D Ps.x/ D Px [ ¹xº for all x 2 X . Let U  2X be the smallest topology that contains the sets Px and Sx for all x 2 X . A set U  X is open in this topology if it is a union of sets of the form Pb , Sa and Sa \ Pb . (ii) We prove that .X; U/ is a Hausdorff space. Let x; y 2 X such that x ¤ y and suppose without loss of generality that x  y. Then Ps.x/ and Sx are disjoint open sets such that x 2 Ps.x/ and y 2 Sx . (iii) We prove that every nonempty compact set K  X contains a largest element max.K/ 2 K such that K \ Smax.K/ D ;. This is obvious when  2 K because S D ;. Thus assume  … K and define V WD ¹x 2 X j K  Px º: Since  2 V , this set is nonempty and min.X/  min.V / DW v because K ¤ ;. Since X n K is open and v 2 X n K, there exist elements a; b 2 X such that a  v  b and Sa \ Pb \ K D ;. This implies K  Pv n .Sa \ Pb /  Pb n .Sa \ Pb /  X n Sa D Ps.a/ : Hence K n ¹aº  Ps.a/ n ¹aº D Pa and K 6 Pa because a  v and so a … V . This implies a 2 K  Ps.a/ , and hence K \ Sa D K n Ps.a/ D ;.

98

3. Borel measures

(iv) We prove that .X; U/ is compact. Let ¹Ui ºi2I be an open cover of X. We prove by induction that there exist finite sequences x1 ; : : : ; x ` 2 X

and

i1 ; : : : ; i` 2 I

such that xk 2 Uik n Uik

1

and Sxk  Ui1 [


[ Uik for k  2, and X D

S`

j D1

1

Uij . Define x1 WD 

and choose i1 2 I such that  2 Ui1 . If Ui1 D X the assertion holds with ` D 1. Now suppose, by induction, that x1 ; : : : ; xk and i1 ; : : : ; ik have been constructed such that xj 2 Uij for j D 1; : : : ; k and Sxk  Ui1 [


[ Uik 1 . If Ui1 [


[ Uik D X; then we are done with ` D k. Otherwise Ck WD X n Ui1 [


[ Uik is a nonempty compact set and we define xkC1 WD max.Ck / by part (iii). Then xkC1 2 Ck and Ck \SxkC1 D ;. Hence SxkC1  Ui1 [


[Uik . Choose ikC1 2 I such that xkC1 2 UikC1 . This completes the induction argument. The induction must stop because xkC1  xk for all k and every strictly decreasing sequence in X is finite by the well ordering axiom (d). This shows that .X; U/ is compact. (v) Let Ki  X, i 2 N, be a sequence of uncountable compact sets. We prove that the compact set \ K WD Ki i2N

is uncountable. To see this, we first prove that K n ¹º ¤ ;:

(3.3)

3.1. Regular Borel measures

99

Choose a sequence xn 2 X n¹º such that xn  xnC1 for all n 2 N and x2k Ci 2 Ki for 1  i  2k 1 and k 2 N. That such a sequence exists follows by induction from the fact that the set X n Sxn D Ps.xn / is countable for each n, while the sets Ki are uncountable for all i. Now the set [ P WD Pxn n2N

is countable and hence the set S WD X n P [ DXn Pxn n2N

DXn

[

Ps.xn /

n2N

D

\

.X n Ps.xn / /

n2N

D

\

Sxn

n2N

is uncountable. Hence x WD min.S/  : We prove that x 2 Ki for all i 2 N. Assume by contradiction that x … Ki for some i . Then there are elements a; b 2 X such that a  x  b and U WD Pb \ Sa  X n Ki : If xn 4 a for all n 2 N, then P  Pa and so a 2 X n P D S , which is impossible because a  x D min.S/. Thus there must be an integer n0 2 N such that a  xn0 . This implies a  xn  x  b and hence xn 2 U  X n Ki for all n  n0 , contradicting the fact that x2k Ci 2 Ki for all k 2 N. This contradiction shows that our assumption that x … Ki for some i 2 N must have been wrong. Thus x 2 K and this proves (3.3). We prove that K is uncountable. Assume by contradiction that K is countable and choose a sequence xi 2 K such that K n ¹º D ¹xi j i 2 Nº. Then s.xi /   and Ki0 WD Ki \ Sxi D Ki n Ps.xi / is an uncountable compact set for every i 2 N. Moreover, \ K 0 WD Ki0  K n ¹xi j i 2 Nº D ¹º; i2N

contradicting the fact that K 0 n ¹º ¤ ; by (3.3). This contradiction shows that K is uncountable as claimed.

100

3. Borel measures

(vi) Define A  2X by ˇ ² ³ ˇ A [ ¹º contains an uncountable compact set; ˇ A WD A  X ˇ : or Ac [ ¹º contains an uncountable compact set: We prove that this is a -algebra. To see this, note first that X 2 A and that A 2 A implies Ac 2 A by definition. Now choose a sequence Ai 2 A and denote [ A WD Ai : i2N

If one of the sets Ai [ ¹º contains an uncountable compact set, then so does the set A [ ¹º. If none of the sets Ai [ ¹º contains an uncountable compact set, then the set Aci [ ¹º contains an uncountable compact set for all i 2 N and hence so T does the set i2N .Aci [ ¹º/ D Ac [ ¹º by part (v). In both cases it follows that A 2 A. (vii) Define the map

W A ! Œ0; 1

by .A/ WD

´ 1 0

if A [ ¹º contains an uncountable compact set, if Ac [ ¹º contains an uncountable compact set.

This map is well defined because the sets A [ ¹º and Ac [ ¹º cannot both contain uncountable compact sets by part (v). It satisfies .;/ D 0. Moreover, if Ai 2 A is a sequence of pairwise disjoint measurable sets, then at most one of the sets Ai [¹º S P can contain an uncountable compact set and hence . i 2N Ai / D i2N .Ai /. Therefore,  is a measure. (viii) The  -algebra B  2X of all Borel sets in X is contained in A. To see this, let U  X be open. If U c is uncountable, then U c [ ¹º is an uncountable compact set and hence U 2 A. If U c is countable choose a sequence xi 2 U c such that U c n ¹º D ¹xi j i 2 Nº and define \ S WD Sxi : i2N

Then X n S D

S

i2N .X

n Sxi / D

S

i2N

Ps.xi / is a countable set and hence

s WD min.S/  : Since xi  s for all i 2 N, it follows that U c n ¹º  Ps . Hence X n Ps is an uncountable compact subset of U [ ¹º and so U 2 A.

3.1. Regular Borel measures

101

(ix) The set U WD X n ¹º is uncountable and every compact subset of U is countable by part (v). Hence .K/ D 0 for every compact subset K  U and .U / D 1 because U [ ¹º D X is an uncountable compact set. Thus jB W B ! Œ0; 1 is a Borel measure, but not a Radon measure. The next lemma and theorem are included here in preparation for the Riesz representation theorem (Theorem 3.15). They explain the relation between the various regularity properties of Borel measures. Lemma 3.7. Let W B ! Œ0; 1 be an outer regular Borel measure that is inner regular on open sets, i.e., .U / D sup¹.K/ j K  U and K is compactº

(3.4)

for every open set U  X. Then the following holds: (i) every Borel set B  X with .B/ < 1 satisfies (3.2); (ii) if X is -compact, then  is regular. Proof. (i) Fix a Borel set B  X with .B/ < 1 and a constant " > 0. Since  is outer regular, there exists an open set U  X such that B  U;

" .U / < .B/ C : 2

Thus U n B is a Borel set and .U n B/ D .U / .B/ < "=2. Use the outer regularity of  again to obtain an open set V  X such that U n B  V;

.V /
0 and choose a sequence of compact sets Ki  Bi such that 1 .Ki / > 0 .Bi / 2 i " for all i. Then, for every n 2 N, the union K1 [


[ Kn is a compact subset of B and 0 .B/  1 .K1 [


[ Kn / D

n X

1 .Ki / >

iD1

n X

0 .Bi /

":

iD1

Now take the limit n ! 1 to obtain 0 .B/ 

1 X

0 .Bi /

":

iD1

P Since this holds for all " > 0, it follows that 0 .B/  1 i D1 0 .Bi / and hence P1 0 .B/ D i D1 0 .Bi / by (3.7). This shows that 0 is a measure. Moreover, it follows directly from the definition of 0 that 0 .K/ D 1 .K/ for every compact set K  X . Since 1 is inner regular on open sets it follows that 0 .U / D 1 .U / for every open set U  X. Since 0 .K/ D 1 .K/ for every compact set K  X , the definition of 0 in (3.5) shows that 0 is inner regular and hence is a Radon measure. The inequality 0 .B/  1 .B/ for B 2 B follows directly from the definition of 0 . This proves part (i). Part (ii) follows directly from the definition of 0 and part (ii) of Lemma 3.7.

104

3. Borel measures

We prove part (iii). Assume first that sW X ! R is a Borel measurable step function with compact support. Then sD

` X

˛i B ; i

iD1

where ˛i 2 R and Bi 2 B with 1 .Bi / < 1. Hence 0 .Bi / D 1 .Bi / by part (i) of Lemma 3.7 and hence Z s d0 D X

` X

Z ˛i 0 .Bi / D

s d1 : X

i D1

Now let f W X ! Œ0; 1 be a Borel measurable function with compact support. By Theorem 1.26, there exists a sequence of Borel measurable step functions sn W X ! Œ0; 1/ such that 0  s1 .x/  s2 .x/ 




and f .x/ D lim sn .x/ n!1

for all x 2 X. Thus sn has compact support for each n. By the Lebesgue monotone convergence theorem (Theorem 1.37) this implies Z Z Z Z f d0 D lim sn d0 D lim sn d1 D f d1 : X

n!1 X

n!1 X

X

If f W X ! R is a 1 -integrable function with compact support, then by what we have just proved, Z Z f ˙ d0 D

X

f ˙ d1 < 1;

X

so f is 0 -integrable and satisfies (3.6). This proves part (iii). We prove part (iv) in four steps. Step 1. Let W B ! Œ0; 1 be a Borel measure such that Z Z f d D f d1 X

X

for every compactly supported continuous function f W X ! R. Then .K/  1 .K/;

1 .U /  .U /

for every compact set K  X and every open set U  X .

(3.8)

3.1. Regular Borel measures

105

Fix an open set U  X and a compact set K  U . Then Urysohn’s lemma (Theorem A.1) asserts that there exists a compactly supported continuous function f W X ! R such that f jK  1;

supp.f /  U;

0  f  1:

Hence it follows from (3.8) that Z Z .K/  f d D f d1  1 .U / X

and likewise

X

Z 1 .K/ 

Z f d1 D

X

f d  .U /: X

Since .K/  1 .U / for every open set U  X containing K and 1 is outer regular, we obtain .K/  inf¹1 .U / j K  U  X and U is openº D 1 .K/: Since 1 .K/  .U / for every compact set K  U and 1 is inner regular on open sets, we obtain 1 .U / D sup¹1 .K/ j K  U and K is compactº  .U /: Step 2. Let  be as in Step 1 and assume in addition that  is inner regular on open sets. Then .K/ D 1 .K/ for every compact set K  X and .U / D 1 .U / for every open set U  X. If U  X is an open set, then .U / D sup¹.K/ j K  U and K is compactº  sup¹1 .K/ j K  U and K is compactº D 1 .U /  .U /: Here the two inequalities follow from Step 1. It follows that .U / D 1 .U /. Now let K be a compact set. Then 1 .K/ < 1. Since 1 is outer regular, there exists an open set U  X such that K  U and 1 .U / < 1. Since  and 1 agree on open sets it follows that .K/ D .U /

.U n K/ D 1 .U /

1 .U n K/ D 1 .K/:

106

3. Borel measures

Step 3. Let  be as in Step 2. Then 0 .B/  .B/  1 .B/ for all B 2 B:

(3.9)

Fix a Borel set B 2 B. Then, by Step 2, 0 .B/ D sup¹1 .K/ j K  B and K is compactº D sup¹.K/ j K  B and K is compactº  .B/  inf¹.U / j B  U  X and U is openº D inf¹1 .U / j B  U  X and U is openº D 1 .B/: Step 4. Let W B ! Œ0; 1 be a Borel measure that satisfies (3.9). Then Z Z Z f d D f d0 D f d1 X

X

X

for every continuous function f W X ! R with compact support. By the definition of the integral and part (iii), Z Z Z Z f d0  f d  f d1 D f d0 X

X

X

X

for every compactly supported continuous function f W X ! Œ0; 1/. Hence Z Z Z f d D f d0 D f d1 X

X

X

for every compactly supported continuous function f W X ! Œ0; 1/, and hence also for every compactly supported continuous function f W X ! R. This proves Step 4 and Theorem 3.8.  Example 3.9. Let .X; U/ be the compact Hausdorff space in Example 3.6 and let W B ! Œ0; 1 be Dieudonné’s measure. (i) Take 1 WD  and define the function 0 W B ! Œ0; 1 by (3.5). Then 0 .X/ D 1;

0 .¹º/ D 0;

0 .X n ¹º/ D 0;

and so 0 is not a measure. Hence the assumptions on 1 cannot be removed in part (i) of Theorem 3.8.

3.2. Borel outer measures

107

(ii) Take 1 WD ı to be the Dirac measure at the point  2 X. This is a regular Borel measure and so the measure 0 in (3.5) agrees with 1 . It is an easy exercise to show that the integral of a continuous function f W X ! R with respect to the Dieudonné measure  is given by Z Z Z f d D f ./ D f d0 D f d1 : X

X

X

Moreover, the compact set K D ¹º satisfies .K/ D 0 < 1 D 1 .K/ and the open set U WD X n ¹º satisfies 1 .U / D 0 < 1 D .U /. This shows that the inequalities in Step 1 in the proof of Theorem 3.8 can be strict and that the hypothesis that  is inner regular on open sets cannot be removed in part (iv) of Theorem 3.8. Remark 3.10. As Example 3.6 shows, it may sometimes be convenient to define a Borel measure first on a -algebra that contains the -algebra of all Borel measurable sets and then restrict it to B. Thus let A  2X be a -algebra containing B and let W A ! Œ0; 1 be a measure. Call  outer regular if it satisfies (3.1) for all B 2 A, call it inner regular if it satisfies (3.2) for all B 2 A, and call it regular if it is both outer and inner regular. If  is regular and .X; B ;  / denotes the completion of .X; B; jB /, it turns out that the completion is also regular (exercise). If in addition .X; A; / is -finite (see Definition 4.29 below), then A  B ;

 D  jA :

(3.10)

To see this, let A 2 A such that .A/ < 1. Choose a sequence of compact sets Ki  X and a sequence of open sets Ui  X such that Ki  A  Ui and S .A/ 2 i  .Ki /  .Ui /  .A/ C 2 i for all i 2 N. Then B0 WD 1 iD1 Ki T1 and B1 WD i D1 Ui are Borel sets such that B0  A  B1 and .B1 n B0 / D 0. Thus every set A 2 A with .A/ < 1 belongs to B and  .A/ D .A/. This proves (3.10) because every A-measurable set is a countable union of A-measurable sets with finite measure. Note that if X is  -compact and .K/ < 1 for every compact set K  X, then .X; A; / is -finite.

3.2 Borel outer measures This section is of preparatory nature. It discusses outer measures on a locally compact Hausdorff space that satisfy suitable regularity properties and shows that the resulting measures on the Borel  -algebra are outer/inner regular. The result will play a central role in the proof of the Riesz representation theorem. As in Section 3.1, we assume that .X; U/ is a locally compact Hausdorff space and denote by B the Borel -algebra of .X; U/.

108

3. Borel measures

Definition 3.11. A Borel outer measure on X is an outer measure W 2X ! Œ0; 1 that satisfies the following axioms: (a) if K  X is compact, then .K/ < 1; (b) if K0 ; K1  X are disjoint compact sets, then .K0 [ K1 / D .K0 / C .K1 /; (c) .A/ D inf¹.U / j A  U  X; U is openº for every subset A  X ; (d) .U / D sup¹.K/ j K  U; K is compactº for every open set U  X . Theorem 3.12. Let W 2X ! Œ0; 1 be a Borel outer measure. Then jB is an outer regular Borel measure and is inner regular on open sets. One can deduce Theorem 3.12 from Carathéodory’s theorem (Theorem 2.4) and use axioms (a) and (b) (instead of the Carathéodory criterion in Theorem 2.5) to show that the  -algebra of -measurable sets contains the Borel  -algebra. That the resulting Borel measure has the required regularity properties is then obvious from axioms (c) and (d). We choose a different route, following Rudin [17], and give a direct proof of Theorem 3.12 which does not rely on Theorem 2.4. The former approach is left to the reader as well as the verification that both proofs give rise to the the same -algebra, i.e., the -algebra A in (3.11) agrees with the  -algebra of -measurable subsets of X . Proof of Theorem 3.12. Define Ae WD ¹E  X j .E/ D sup¹.K/ j K  E; K is compactº < 1º

(3.11a)

and A WD ¹A  X j A \ K 2 Ae for every compact set K  X º:

(3.11b)

Here the subscript “e” stands for “endlich” and indicates that the elements of Ae are sets of finite measure. We prove in seven steps that A is a -algebra containing B, that  WD jA W A ! Œ0; 1 is a measure, and that .X; A; / is a complete measure space. That  is outer regular and is inner regular on open sets follows immediately from conditions (c) and (d) in Definition 3.11.

3.2. Borel outer measures

109

Step 1. Let E1 ; E2 ; E3 ; : : : be a sequence of pairwise disjoint sets in Ae and define E WD

1 [

Ei :

iD1

Then the following holds: P (i) .E/ D 1 i D1 .Ei /; (ii) if .E/ < 1, then E 2 Ae . P The assertions are obvious when .E/ D 1 because .E/  1 iD1 .Ei /. Hence assume .E/ < 1. We argue as in the proof of Theorem 3.8. Fix a constant " > 0. Since Ei 2 Ae for all i , there is a sequence of compact sets Ki  Ei such that .Ki / > .Ei / 2 i " for all i . Then for all n 2 N .E/  .K1 [


[ Kn / D .K1 / C


C .Kn /  .E1 / C


C .En /

(3.12) ":

Here the equality follows from condition (b) in Definition 3.11. Take the limit n ! 1 to obtain 1 X .Ei /  .E/ C ": i D1

Since this holds for all " > 0, it follows that 1 X

.Ei /  .E/ 

i D1

and hence

1 X

.Ei /

iD1 1 X

.Ei / D .E/:

(3.13)

i D1

Now (3.12) and (3.13) yield .E/  .K1 [


[ Kn / 

n X

.Ei /

"

i D1

D .E/

1 X iDnC1

for all n 2 N.

.Ei /

"

110

3. Borel measures

By (3.13), there exists an n" 2 N such that 1 X

.Ei / < ":

iDn" C1

Hence the compact set K" WD K1 [


[ Kn"  E satisfies .E/  .K" /  .E/

2":

Since this holds for all " > 0, we obtain .E/ D sup¹.K/ j K  E; K is compactº and hence E 2 Ae . This proves Step 1. Step 2. If E0 ; E1 2 Ae , then E0 [ E1 2 A e ;

E0 \ E1 2 Ae ;

E0 n E1 2 Ae :

We first prove that E0 n E1 2 Ae . Fix a constant " > 0. Since E0 ; E1 2 Ae , and by condition (c) in Definition 3.11, there exist compact sets K0 ; K1  X and open sets U0 ; U1  X such that Ki  Ei  Ui ;

.Ei /

" < .Ki /  .Ui / < .Ei / C ";

i D 0; 1:

Moreover, every compact set with finite outer measure is an element of Ae , by definition, and every open set with finite outer measure is an element of Ae by condition (d) in Definition 3.11. Hence Ki ; Ui ; Ui n Ki 2 Ae for i D 0; 1 and it follows from Step 1 that .Ei n Ki /  .Ui n Ki / D .Ui /

.Ki /  2";

(3.14a)

.Ui n Ei /  .Ui n Ki / D .Ui /

.Ki /  2";

(3.14b)

for i D 0; 1. Define K WD K0 n U1  E0 n E1 :

(3.15)

3.2. Borel outer measures

111

Then K is a compact set and E0 n E1  .E0 n K0 / [ .K0 n U1 / [ .U1 n E1 /: By the definition of an outer measure, this implies .E0 n E1 /  .E0 n K0 / C .K0 n U1 / C .U1 n E1 /  .K/ C 4": Here the last inequality follows from the definition of K in (3.15) and the inequalities in (3.14). Since " > 0 was chosen arbitrarily, it follows that .E0 n E1 / D sup¹.K/ j K  E0 n E1 ; K is compactº and hence E0 n E1 2 Ae . With this understood, it follows from Step 1 that E0 [ E1 D .E0 n E1 / [ E1 2 Ae ;

E0 \ E1 D E0 n .E0 n E1 / 2 Ae :

This proves Step 2. Step 3. A is a -algebra. First, X 2 A because K 2 Ae for every compact set K  X . Second, assume A 2 A and let K  X be a compact set. Then by definition A \ K 2 Ae . Moreover K 2 Ae and hence, by Step 2, Ac \ K D K n .A \ K/ 2 Ae : Since this holds for every compact set K  X , we have Ac 2 Ae . Third, let Ai 2 A for i 2 N and denote A WD

1 [

Ai :

iD1

Fix a compact set K  X . Then Ai \ K 2 Ae for all i by the definition of A. Hence, by Step 2, Bi WD Ai \ K 2 Ae for all i and then, again by Step 2, Ei WD Bi n .B1 [


[ Bi for all i .

1/

2 Ae

112

3. Borel measures

The sets Ei are pairwise disjoint and 1 [ i D1

Ei D

1 [

Bi D A \ K:

iD1

Since .A \ K/  .K/ < 1 by condition (a) in Definition 3.11, it follows from Step 1 that A\K 2 Ae . This holds for every compact set K  X and hence A 2 A. This proves Step 3. Step 4. B  A. Let F  X be closed. If K  X is compact, then F \ K is a closed subset of a compact set and hence is compact (see Lemma A.2). Thus F \ K 2 Ae for every compact subset K  X and so F 2 A. Therefore we have proved that A contains all closed subsets of X . Since A is a -algebra by Step 3, it also contains all open subsets of X and thus B  A. This proves Step 4. Step 5. Let A  X. Then A 2 Ae if and only if A 2 A and .A/ < 1. If A 2 Ae , then A \ K 2 Ae for every compact set K  X by Step 2 and hence A 2 A. Conversely, let A 2 A such that .A/ < 1. Fix a constant " > 0. By condition (c) in Definition 3.11, there exists an open set U  X such that A  U and .U / < 1. By condition (d) in Definition 3.11, there exists a compact set K  X such that K  U; .K/ > .U / ": Since K; U 2 Ae and U D .U n K/ [ K, it follows from Step 1 that .U n K/ D .U /

.K/ < ":

Moreover, A \ K 2 Ae because A 2 A. Hence it follows from the definition of Ae that there exists a compact set H  A \ K such that .H /  .A \ K/

"

D .A n .A n K//

"

 .A/

.A n K/

"

 .A/

.U n K/

"

 .A/

2":

Since " > 0 was chosen arbitrarily, we conclude that .A/ D sup¹.K/ j K  A; K is compactº and hence A 2 Ae . This proves Step 5.

3.3. The Riesz representation theorem

113

Step 6.  WD jA is an outer regular extended Borel measure and  is inner regular on open sets. We prove that  is a measure. By definition, .;/ D 0. Now let Ai 2 A be a sequence of pairwise disjoint measurable sets and define A WD

1 [

Ai :

iD1

P If .Ai / < 1 for all i , then Ai 2 Ae by Step 5 and hence .A/ D 1 i D1 .Ai / by Step 1. If .Ai / D 1 for some i, then .A/  .Ai / and so .A/ D 1. Thus  is a measure. Moreover, B  A by Step 4, .K/ < 1 for every compact set K  X by condition (a) in Definition 3.11,  is outer regular by condition (c) in Definition 3.11, and  is inner regular on open sets by condition (d) in Definition 3.11. This proves Step 6. Step 7. .X; A; / is a complete measure space. If E  X satisfies .E/ D 0, then E 2 Ae by the definition of Ae , and hence E 2 A by Step 5. This proves Step 7 and Theorem 3.12. 

3.3 The Riesz representation theorem Let .X; U/ be a locally compact Hausdorff space and B be its Borel -algebra. A function f W X ! R is called compactly supported if its support supp.f / WD ¹x 2 X j f .x/ ¤ 0º is a compact subset of X. The set of compactly supported continuous functions on X will be denoted by ˇ ² ³ ˇ f is continuous and ˇ Cc .X/ WD f W X ! R ˇ : supp.f / is a compact subset of X Thus a continuous function f W X ! R belongs to Cc .X/ if and only if there exists a compact set K  X such that f .x/ D 0 for all x 2 X n K. The set Cc .X/ is a real vector space. Definition 3.13. A linear functional ƒW Cc .X/ ! R is called positive if f  0 H) ƒ.f /  0 for all f 2 Cc .X/.

114

3. Borel measures

The next lemma shows that every positive linear functional on Cc .X/ is continuous with respect to the topology of uniform convergence when restricted to the subspace of functions with support contained in a fixed compact subset of X. Lemma 3.14. Let ƒW Cc .X/ ! R be a positive linear functional and let fi 2 Cc .X/ be a sequence of compactly supported continuous functions that converges uniformly to f 2 Cc .X/. If there exists a compact set K  X such that supp.fi /  K

for all i 2 N,

then ƒ.f / D lim ƒ.fi /: i !1

Proof. Since fi converges uniformly to f , the sequence "i WD sup jfi .x/

f .x/j

x2X

converges to zero. By Urysohn’s lemma (Theorem A.1), there exists a compactly supported continuous function W X ! Œ0; 1 such that .x/ D 1 for all x 2 K. This function satisfies "i   fi

f  "i 

for all i .

Hence "i ƒ./  ƒ.fi /

ƒ.f /  "i ƒ./;

because ƒ is positive, and hence jƒ.fi / ƒ.f /j  "i ƒ./ for all i . Since "i converges to zero, so does jƒ.fi / ƒ.f /j and this proves Lemma 3.14.  Let W B ! Œ0; 1 be a Borel measure. Then every continuous function f W X ! R with compact support is integrable with respect to . Define the map ƒ W Cc .X/ ! R by Z ƒ .f / WD

f d:

(3.16)

X

Then ƒ is a positive linear functional. The Riesz representation theorem asserts that every positive linear functional on Cc .X/ has this form. It also asserts uniqueness under certain regularity hypotheses on the Borel measure. The following theorem includes two versions of the uniqueness statement.

3.3. The Riesz representation theorem

115

Theorem 3.15 (Riesz representation theorem). Let ƒW Cc .X/ ! R be a positive linear functional. Then the following holds. (i) There exists a unique Radon measure 0 W B ! Œ0; 1 such that ƒ0 D ƒ: (ii) There exists a unique outer regular Borel measure 1 W B ! Œ0; 1 such that 1 is inner regular on open sets and ƒ1 D ƒ: (iii) The Borel measures 0 and 1 in (i) and (ii) agree on all compact sets and on all open sets. Moreover, 0 .B/  1 .B/ for all B 2 B. (iv) Let W B ! Œ0; 1 be a Borel measure that is inner regular on open sets. Then ƒ D ƒ () 0 .B/  .B/  1 .B/ for all B 2 B. Proof. The proof has nine steps. Step 1 defines a function W 2X ! Œ0; 1, Step 2 shows that it is an outer measure, and Steps 3, 4, and 5 show that it satisfies the axioms of Definition 3.11. Step 6 defines 1 and Step 7 shows that ƒ1 D ƒ. Step 8 defines 0 and Step 9 proves uniqueness. Step 1. Define the function U W U ! Œ0; 1 by U .U / WD sup¹ƒ.f / j f 2 Cc .X/; 0  f  1; supp.f /  U º

(3.17)

for every open set U  X and define W 2X ! Œ0; 1 .A/ WD inf¹U .U / j A  U  X; U is openº for every subset A  X . Then .U / D U .U / for every open set U  X.

(3.18)

116

3. Borel measures

If U; V  X are open sets such that U  V , then U .U /  U .V / by definition. Hence .U / D inf¹U .V / j U  V  X; V is openº D U .U / for every open set U  X and this proves Step 1. Step 2. The function W 2X ! Œ0; 1 in Step 1 is an outer measure. By definition, .;/ D U .;/ D 0. Since U .U /  U .V / for all open sets U; V  X with U  V , it follows also from the definition that .A/  .B/ whenever A  B  X . Next we prove that for all open sets U; V  X U .U [ V /  U .U / C U .V /:

(3.19)

To see this, let f 2 Cc .X/ such that 0  f  1 and K WD supp.f /  U [ V: 2 Cc .X/ such that

By Theorem A.4, there exist functions ; supp./  U;

supp. /  V;

 0;

;

C

 1;

. C

/jK  1:

Hence f D f C f and so ƒ.f / D ƒ.f C f / D ƒ.f / C ƒ. f /  U .U / C U .V /: This proves (3.19). Now choose a sequence of subsets Ai  X and define 1 [

A WD

Ai :

iD1

We must prove that .A/ 

1 X

.Ai /:

(3.20)

iD1

If there exists an i 2 N such that .Ai / D 1, then .A/ D 1 because Ai  A P and hence 1 iD1 .Ai / D 1 D .A/. Hence assume .Ai / < 1 for all i . Fix a constant " > 0. By the definition of  in (3.18), there exists a sequence of open sets Ui  X such that Ai  Ui ;

U .Ui / < .Ai / C 2 i ":

Define U WD

1 [ iD1

Ui :

3.3. The Riesz representation theorem

117

Let f 2 Cc .X/ such that 0  f  1 and supp.f /  U . Since f has compact support, there exists an integer k 2 N such that k [

supp.f / 

Ui :

i D1

By the definition of U and (3.19), this implies ƒ.f /  U .U1 [


[ Uk /  U .U1 / C


C U .Uk / < .A1 / C


C .Ak / C ": Hence ƒ.f / 

1 X

.Ai / C "

iD1

for every f 2 Cc .X/ such that 0  f  1 and supp.f /  U . Consequently .A/  U .U / 

1 X

.Ai / C "

i D1

by the definition of U .U / in (3.17). Thus .A/ 

1 X

.Ai / C " for every " > 0,

iD1

and hence .A/ 

1 X

.Ai /:

iD1

This proves (3.20) and Step 2. Step 3. Let U  X be an open set. Then U .U / D sup¹.K/ j K  U; K is compactº:

(3.21)

Let f 2 Cc .X/ such that 0  f  1;

K WD supp.f /  U:

Then it follows from the definition of U in (3.17) that ƒ.f /  U .V / for every open set V  X with K  V . Hence it follows from the definition of  in (3.18) that ƒ.f /  .K/:

118

3. Borel measures

Hence U .U / D sup¹ƒ.f / j f 2 Cc .X/; 0  f  1; supp.f /  U º  sup¹.K/ j K  U; K is compactº  .U / D U .U /: Hence U .U / D sup¹.K/ j K  U; K is compactº and this proves Step 3. Step 4. Let K  X be an compact set. Then .K/ D inf¹ƒ.f / j f 2 Cc .X/; f  0; f jK  1º:

(3.22)

In particular, .K/ < 1. Define a WD inf¹ƒ.f / j f 2 Cc .X/; f  0; f jK  1º: We prove that a  .K/. Let U  X be any open set containing K. By Urysohn’s lemma (Theorem A.1), there exists a function f 2 Cc .X/ such that 0  f  1;

supp.f /  U;

f jK  1:

Hence, a  ƒ.f /  U .U /: This shows that a  U .U / for every open set U  X containing K. Take the infimum over all open sets containing K and use the definition of  in equation (3.18) to obtain a  .K/. We prove that .K/  a. Choose a function f 2 Cc .X/ such that f  0 and f .x/ D 1 for all x 2 K. Fix a constant 0 < ˛ < 1 and define U˛ WD ¹x 2 X j f .x/ > ˛º: Then U˛ is open and K  U˛ . Hence .K/  U .U˛ /: Moreover, every function g 2 Cc .X/ with 0  g  1 and supp.g/  U˛ satisfies ˛g.x/  ˛  f .x/ for x 2 U˛ , hence ˛g  f , and so ˛ƒ.g/  ƒ.f /. Take the supremum over all such g to obtain ˛U .U˛ /  ƒ.f / and hence .K/  U .U˛ / 

1 ƒ.f /: ˛

3.3. The Riesz representation theorem

119

This shows that .K/  ˛1 ƒ.f / for all ˛ 2 .0; 1/, and hence .K/  ƒ.f /: Since this holds for every function f 2 Cc .X/ with f  0 and f jK  1, it follows that .K/  a. This proves Step 4. Step 5. Let K0 ; K1  X be compact sets such that K0 \ K1 D ;. Then .K0 [ K1 / D .K0 / C .K1 /: The inequality .K0 [ K1 /  .K0 / C .K1 / holds because  is an outer measure by Step 2. To prove the converse inequality, choose f 2 Cc .X/ such that 0  f  1;

f jK0  0;

f jK1  1:

That such a function exists follows from Urysohn’s lemma (Theorem A.1) with K WD K1

and

U WD X n K0 :

Now fix a constant " > 0. Then it follows from Step 4 that there exists a function g 2 Cc .X/ such that g  0;

gjK0 [K1  1;

ƒ.g/ < .K0 [ K1 / C ":

It follows also from Step 4 that .K0 / C .K1 /  ƒ..1

f /g/ C ƒ.fg/ D ƒ.g/ < .K0 C K1 / C ":

Hence .K0 / C .K1 / < .K0 C K1 / C "

for every " > 0,

and therefore .K0 / C .K1 /  .K0 C K1 /: This proves Step 5. Step 6. The function 1 WD jB W B ! Œ0; 1 is an outer regular Borel measure that is inner regular on open sets.

120

3. Borel measures

The function  is an outer measure by Step 2. It satisfies condition (a) in Definition 3.11 by Step 4, it satisfies condition (b) by Step 5, it satisfies condition (c) by Step 1, and it satisfies condition (d) by Step 3. Hence  is a Borel outer measure. Hence Step 6 follows from Theorem 3.12. Step 7. Let 1 be as in Step 6. Then ƒ1 D ƒ: We will prove that

Z ƒ.f / 

f d1

(3.23)

X

for all f 2 Cc .X/. Once this is understood, it follows that Z Z ƒ.f / D ƒ. f /  . f / d1 D f d1 X

X

and hence Z X

f d1  ƒ.f / for all f 2 Cc .X/.

Thus

Z ƒ.f / D

for all f 2 Cc .X/,

f d1 X

and this proves Step 7. Thus it remains to prove the inequality (3.23). Consider a continuous function f W X ! R with compact support and denote K WD supp.f /;

a WD inf f .x/; x2X

b WD sup f .x/: x2X

Fix a constant " > 0 and choose real numbers y0 < a < y1 < y2 <


< yn

1

< yn D b

such that yi

yi

1

< ";

i D 1; : : : ; n:

For i D 1; : : : ; n define Ei WD ¹x 2 K j yi

1

< f .x/  yi º:

Then Ei is the intersection of the open set f 1 ..yi 1 ; 1// with the closed set f 1 .. 1; yi / and hence is a Borel set. Moreover, Ei \ Ej D ; for i ¤ j and n [ KD Ei : i D1

3.3. The Riesz representation theorem

121

Since 1 is outer regular, there exist open sets U1 ; : : : ; Un  X such that " Ei  Ui ; 1 .Ui / < 1 .Ei / C ; sup f < yi C " n Ui

(3.24)

for all i . (For each i , choose first an open set that satisfies the first two conditions in (3.24) and then intersect it with the open set f 1 .. 1; yi C "//.) By Theorem A.4, there exist functions 1 ; : : : ; n 2 Cc .X/ such that i  0;

supp.i /  Ui ;

n X

i  1;

iD1

n X

i jK  1:

(3.25)

iD1

It follows from (3.24), (3.25), and Step 4 that f D

n X

i f;

i f  .yi C "/i ;

ƒ.i /;

ƒ.i /  1 .Ui / < 1 .Ei / C

iD1

and 1 .K/ 

n X iD1

" : n

Hence, ƒ.f / D  D

n X

ƒ.i f /

i D1 n X

.yi C "/ƒ.i /

i D1 n X

.yi C jaj C "/ƒ.i /

jaj

i D1

 D

n X

jaj1 .K/

n

.yi C "/1 .Ei / C

i D1



ƒ.i /

iD1

 " .yi C jaj C "/ 1 .Ei / C n

i D1 n X

n X

n X

.yi

"X .yi C jaj C "/ n iD1

"/1 .Ei / C ".21 .K/ C b C jaj C "/

ZiD1  f d1 C ".21 .K/ C b C jaj C "/: X

Here we have used the inequality yi C Rjaj C "  0. Since " > 0 can be chosen arbitrarily small it follows that ƒ.f /  X f d1 . This proves (3.23).

122

3. Borel measures

Step 8. Define 0 W B ! Œ0; 1 by 0 .B/ WD sup¹.K/ j K  B; K is compactº Then 0 is a Radon measure, ƒ0 D ƒ, and 0 and 1 satisfy (iii) and (iv). It follows from Step 6 and part (i) of Theorem 3.8 that 0 is a Radon measure and it follows from Step 7 and part (iii) of Theorem 3.8 that ƒ0 D ƒ1 D ƒ. That the measures 0 and 1 satisfy assertions (iii) and (iv) follows from parts (i) and (iv) of Theorem 3.8. Step 9. We prove uniqueness in (i) and (ii). By definition, 0 .K/ D .K/ D 1 .K/ for every compact set K  X. Second, it follows from and Steps 1 and 3 that 0 .U / D U .U / D .U / D 1 .U / for every open set U  X. Third, Steps 7 and 8 assert that ƒ0 D ƒ1 D ƒ. Hence it follows from part (iv) of Theorem 3.8 that every Borel measure W B ! Œ0; 1 that is inner regular on open sets and satisfies ƒ D ƒ agrees with  on all compact sets and on all open sets. Hence, every Radon measure W B ! Œ0; 1 that satisfies ƒ D ƒ is given by .B/ D sup¹.K/ j K  B; K is compactº D 0 .B/ for every B 2 B. Likewise, every outer regular Borel measure W B ! Œ0; 1 that is inner regular on open sets and satisfies ƒ D ƒ is given by .B/ D inf¹.U / j B  U  X; U is openº D .B/ D 1 .B/ for every B 2 B. This proves Step 9 and Theorem 3.15.



The following corollary is the converse of Theorem 3.8. Corollary 3.16. Let 0 W B ! Œ0; 1 be a Radon measure and define 1 .B/ WD inf¹0 .U / j B  U  X; U is openº

for all B 2 B:

(3.26)

Then 1 is an outer regular Borel measure, is inner regular on open sets, and 0 .B/ D sup¹1 .K/ j K  B; K is compactº

for all B 2 B:

(3.27)

Proof. Let 1 be the unique outer regular Borel measure on X that is inner regular on open sets and satisfies ƒ1 D ƒ0 . Then Theorem 3.15 asserts that 0 and 1 agree on all compact sets and all open sets. Since 1 is outer regular, it follows that 1 is given by (3.26). Since 0 is inner regular it follows that 0 satisfies (3.27). This proves Corollary 3.16. 

3.3. The Riesz representation theorem

123

Corollary 3.17. Every Radon measure is outer regular on compact sets. Proof. Equation (3.26) with B D K compact and 0 .K/ D 1 .K/.



The next theorem formulates a condition on a locally compact Hausdorff space which guarantees that all Borel measures are regular. The condition (every open subset is  -compact) is shown below to be strictly weaker than second countability. Theorem 3.18. Let X be a locally compact Hausdorff space. (i) Assume X is -compact. Then every Borel measure on X that is inner regular on open sets is regular. (ii) Assume every open subset of X is -compact. Then every Borel measure on X is regular. Proof. (i) Let W B ! Œ0; 1 be a Borel measure that is inner regular on open sets and let 0 ; 1 W B ! Œ0; 1 be the Borel measures associated to ƒ WD ƒ in parts (i) and (ii) of the Riesz representation theorem (Theorem 3.15). Since  is inner regular on open sets, it follows from part (iii) of Theorem 3.15 that 0 .B/  .B/  1 .B/ for all B 2 B. Since X is -compact, it follows from part (ii) of Theorem 3.8 that 0 D  D 1 . Hence  is regular. (ii) Let W B ! Œ0; 1 be a Borel measure. We prove that  is inner regular on open sets. Fix an open set U  X . Since U is  -compact, there exists a sequence S of compact sets Ki  U such that Ki  KiC1 for all i 2 N and U D 1 iD1 Ki . Hence .U / D lim .Ki / i!1

by Theorem 1.28, so .U / D sup¹.K/ j K  U and K is compactº: This shows that  is inner regular on open sets and hence it follows from (i) that  is regular. This proves Theorem 3.18.  Example 3.9 shows that the assumption that every open set is  -compact cannot be removed in part (ii) of Theorem 3.18 even if X is compact. Note also that Theorem 3.18 provides another proof of regularity for the Lebesgue measure, which was established in Theorem 2.13. Corollary 3.19. Let X be a locally compact Hausdorff space such that every open subset of X is  -compact. Then for every positive linear functional ƒW Cc .X/ ! R there exists a unique Borel measure  such that ƒ D ƒ:

124

3. Borel measures

Proof. This follows from Theorem 3.15 and part (ii) of Theorem 3.18.



Remark 3.20. Let X be a compact Hausdorff space and let C.X/ D Cc .X/ be the space of continuous real valued functions on X. From a functional analytic viewpoint it is interesting to understand the dual space of C.X/, i.e., the space of all bounded linear functionals on C.X/ (Definition 4.23). Exercise 5.35 below shows that every bounded linear functional on C.X/ is the difference of two positive linear functionals. If every open subset of X is -compact, it then follows from Corollary 3.19 that every bounded linear functional on C.X/ can be represented uniquely by a signed Borel measure. (See Definition 5.10 in Section 5.3 below.) An important class of locally compact Hausdorff spaces that satisfy the hypotheses of Theorem 3.18 and Corollary 3.19 are the second countable ones. Here are the definitions. A basis of a topological space .X; U/ is a collection V  U of open sets such that every open set U  X is a union of elements of V. A topological space .X; U/ is called second countable if it admits a countable basis. It is called first countable if, for every x 2 X, there is a sequence of open sets Wi , i 2 N, such that x 2 Wi for all i and every open set that contains x contains one of the sets Wi . Lemma 3.21. Let X be a locally compact Hausdorff space: (i) if X is second countable, then every open subset of X is -compact; (ii) if every open subset of X is -compact, then X is first countable. Proof. (i) Let V be a countable basis of the topology and let U  X be an open set. Denote by V.U / the collection of all sets V 2 V such that Vx  U and Vx is compact. Let x 2 U . By Lemma A.3 there is an open set W  X with compact S  U . Since V is a basis of the topology, there is an closure such that x 2 W  W element V 2 V such that x 2 V  W . Hence Vx is a closed subset of the compact S and so is compact by Lemma A.2. Thus V 2 V.U / and x 2 V . This shows set W that [ U D V: V 2V.U /

Since V is countable, so is V.U /. Choose a bijection N ! V.U /W i 7! Vi and define Ki WD Vx1 [


[ Vxi S for i 2 N. Then Ki  Ki C1 for all i and U D 1 iD1 Ki . Hence U is  -compact. (ii) Fix an element x 2 X. Since X is a Hausdorff space, the set X n ¹xº is open and hence is -compact by assumption. Choose a sequence of compact sets Ki  X n ¹xº

3.3. The Riesz representation theorem

125

such that Ki  KiC1 and

1 [

for all i 2 N

Ki D X n ¹xº:

iD1

Then each set Ui WD X n Ki is open and contains x. By Lemma A.3, there exists a sequence of open sets Vi  X with compact closure such that x 2 Vi  Vxi  Ui D X n Ki

for all i 2 N.

Define Wi WD V1 \


\ Vi

for i 2 N.

Then Si  W

i \

.X n Kj / D X n Ki

j D1

and hence

1 \

Si D ¹xº: W

iD1

This implies that each open set U  X that contains x also contains one of the Si . Namely, if x 2 U and U is open, then W S1 n U is a compact set contained sets W in 1 [ Si /; X n ¹xº D .X n W iD1

hence there exists a j 2 N such that S1 n U  W

j [

Si / D X n W Sj ; .X n W

iD1

Sj  U . Thus the sets Wj form a countable neighborhood basis of x and and so W this proves Lemma 3.21. 

126

3. Borel measures

Example 3.22. The Alexandrov double arrow space is an example of a compact Hausdorff space in which every open subset is -compact and which is not second countable. It is defined as the ordered space .X; /, where X WD Œ0; 1  ¹0; 1º and  denotes the lexicographic ordering ´ s < t or .s; i/  .t; j / () s D t and i D 0 and j D 1: The topology U  2X is defined as the smallest topology containing the sets Sa WD ¹x 2 X j a  xº;

Pb WD ¹x 2 X j x  bº;

a; b 2 X:

It has a basis consisting of the sets Sa , Pb , Sa \ Pb for all a; b 2 X. This topological space .X; U/ is a compact Hausdorff space and is perfectly normal, i.e., for any two disjoint closed subsets F0 ; F1  X there exists a continuous function f W X ! Œ0; 1 such that F0 D f

1

.0/;

F1 D f

1

.1/:

(For a proof see Dan Ma’s topology blog [12].) This implies that every open subset of X is -compact. Moreover, the subsets Y0 WD .0; 1/  ¹0º;

Y1 WD .0; 1/  ¹1º

are both homeomorphic to the Sorgenfrey line, defined as the real axis with the (nonstandard) topology in which the open sets are the unions of half open intervals Œa; b/. Since the Sorgenfrey line is not second countable, neither is the double arrow space .X; U/. (The Sorgenfrey line is Hausdorff and perfectly normal, but is not locally compact because every compact subset of it is countable.)

3.4 Exercises Exercise 3.23. This exercise shows that the measures 0 ; 1 in Theorem 3.15 need not agree. Let .X; d / be the metric space given by X WD R2 and ´ 0 if x1 D x2 ; d..x1 ; y1 /; .x2 ; y2 // WD jy1 y2 j C 1 if x1 ¤ x2 : Let B  2X be the Borel -algebra of .X; d /.

3.4. Exercises

127

(i) Show that .X; d / is locally compact. (ii) Show that for every compactly supported continuous function f W X ! R there exists a finite set Sf  R such that supp.f /  Sf  R. (iii) Define the positive linear functional ƒW Cc .X/ ! R by X Z 1 ƒ.f / WD f .x; y/ dy: x2Sf

1

(Here the integrals on the right are understood as the Riemann integrals or, equivalently by Theorem 2.24, as the Lebesgue integrals.) Let W B ! Œ0; 1 be a Borel measure such that Z f d D ƒ.f / for all f 2 Cc .X/: X

Prove that every one-element subset of X has measure zero. (iv) Let  be as in (iii) and let E WD R  ¹0º. This set is closed. If  is inner regular, prove that .E/ D 0. If  is outer regular, prove that .E/ D 1. Exercise 3.24. This exercise shows that the Borel assumption cannot be removed in Theorem 3.18. (The measure  in part (ii) is not a Borel measure.) Let .X; U/ be the topological space defined by X WD N [ ¹1º and U WD ¹U  X j U  N or #U c < 1º: Thus .X; U/ is the (Alexandrov) one-point compactification of the set N of natural numbers with the discrete topology. (If 1 2 U , then the condition #U c < 1 is equivalent to the assertion that U c is compact.) (i) Prove that .X; U/ is a compact Hausdorff space and that every subset of X is  -compact. Prove that the Borel -algebra of X is B D 2X . (ii) Let W 2X ! Œ0; 1 be the counting measure. Prove that  is inner regular, but not outer regular. Exercise 3.25. Let .X; UX / and .Y; UY / be locally compact Hausdorff spaces and denote their Borel -algebras by BX  2X and BY  2Y . Let W X ! Y be a continuous map and let X W BX ! Œ0; 1 be a measure. (i) Prove that BY   BX (see Exercise 1.69). (ii) If X is inner regular, show that  X jBY is inner regular.

128

3. Borel measures

(iii) Find an example where X is outer regular and  X jBY is not outer regular. Hint. Consider the inclusion of N into its one-point compactification and use Exercise 3.24. (In this example X is a Borel measure, however,  X is not a Borel measure.) Exercise 3.26. Let .X; d / be a metric space. Prove that .X; d / is perfectly normal, i.e., if F0 ; F1  X are disjoint closed subsets, then there is a continuous function f W X ! Œ0; 1 such that F0 D f 1 .0/ and F1 D f 1 .1/. Compare this with Urysohn’s lemma (Theorem A.1). Hint. An explicit formula for f is given by f .x/ WD

d.x; F0 / ; d.x; F0 / C d.x; F1 /

where d.x; F / WD inf d.x; y/ y2F

for x 2 X and F  X. Exercise 3.27. Recall that the Sorgenfrey line is the topological space .R; U/, where U  2R is the smallest topology that contains all half open intervals Œa; b/ with a < b. Prove that the Borel  -algebra of .R; U/ agrees with the Borel  -algebra of the standard topology on R. Exercise 3.28. Recall from Example 3.22 that the double arrow space is X WD Œ0; 1  ¹0; 1º with the topology induced by the lexicographic ordering. Prove that B  X is a Borel set for this topology if and only if there exist a Borel set E  Œ0; 1 and two countable sets F; G  X such that B D ..E  ¹0; 1º/ [ F / n G:

(3.28)

Hint 1. Show that the projection f W X ! Œ0; 1 onto the first factor is continuous with respect to the standard topology on the unit interval. Hint 2. Denote by B  2X the set of all sets of the form (3.28) with E  Œ0; 1 a Borel set and F; G  X countable. Prove that B is a -algebra. Exercise 3.29 (the Baire  -algebra). Let .X; U/ be a locally compact Hausdorff space and define ˇ ² ³ ˇ K is compact and there is a sequence of open sets ˇ T Ka WD K  X ˇ : Ui such that UiC1  Ui for all i and K D 1 iD1 Ui

3.4. Exercises

129

Let Ba  2X be the smallest -algebra that contains Ka . It is contained in the Borel  -algebra B  2X and is called the Baire  -algebra of .X; U/. The elements of Ba are called Baire sets. A function f W X ! R is called Baire measurable if f 1 .U / 2 Ba for every open set U  R. A Baire measure is a measure W Ba ! Œ0; 1 such that .K/ < 1 for all K 2 Ka . (i) Let f W X ! R be a continuous function with compact support. Prove that f 1 .c/ 2 Ka for every nonzero real number c. (ii) Prove that Ba is the smallest -algebra such that every continuous function f W X ! R with compact support is Ba -measurable. (iii) If every open subset of X is  -compact, prove that Ba D B. Hint. Show first that every compact set belongs to Ka and then that every open set belongs to Ba . Exercise 3.30. (i) Let X be an uncountable set and let U WD 2X be the discrete topology. Prove that B  X is a Baire set if and only if B is countable or has a countable complement. Define W Ba ! Œ0; 1 by .B/ WD

´ 0 1

if B is countable; if B c is countable:

R Show that X f d D 0 for every f 2 Cc .X/. Thus positive linear functionals ƒW Cc .X/ ! R need not be uniquely represented by Baire measures. (ii) Let X be the compact Hausdorff space of Example 3.6. Prove that the Baire sets in X are the countable subsets of X n ¹º and their complements. ˇ (iii) Let X be the Stone–Cech compactification of N in Example 4.60 below. Prove that the Baire sets in X are the subsets of N and their complements. (iv) Let X D R2 be the locally compact Hausdorff space in Exercise 3.23 (with a nonstandard topology). Show that B  X is a Baire set if and only if the set Bx WD ¹y 2 R j .x; y/ 2 Bº is a Borel set in R for every x 2 R and one of the sets S0 WD ¹x 2 R j Bx ¤ ;º and S1 WD ¹x 2 R j Bx ¤ Rº is countable.

130

3. Borel measures

Exercise 3.31. Let .X; U/ be a locally compact Hausdorff space and let Ba  B  2X be the Baire and Borel -algebras. Let F.X/ denote the real vector space of all functions f W X ! R. For F  F.X/ consider the following conditions: (a) Cc .X/  F; (b) If fi 2 F is a sequence converging pointwise to f 2 F.X/, then f 2 F. Let Fa  F.X/ be the intersection of all subsets F  F.X/ that satisfy conditions (a) and (b). Prove the following. (i) Fa satisfies (a) and (b). (ii) Every element of Fa is Baire measurable. Hint. The set of Baire measurable functions on X satisfies (a) and (b). (iii) If f 2 Fa and g 2 Cc .X/, then f C g 2 Fa . Hint. Let g 2 Cc .X/. Then the set Fa g satisfy (a) and (b) and hence contains Fa . (iv) If f; g 2 Fa , then f Cg 2 Fa . Hint. Let g 2 Fa . Then the set Fa g satisfy (a) and (b) and hence contains Fa . (v) If f 2 Fa and c 2 R, then cf 2 Fa . Hint. Fix a real number c ¤ 0. Then the set c 1 Fa satisfy (a) and (b) and hence contains Fa . (vi) If f 2 Fa and g 2 Cc .X/, then fg 2 Fa . Hint. Fix a real number c such that c C g.x/ > 0 for all x 2 R. Then the set .c C g/ 1 Fa satisfy (a) and (b) and hence contains Fa . Now use (iv) and (v). (vii) If A  X such that A 2 Fa and f 2 Fa , then f A 2 Fa . Hint. The set .1 C A / 1 Fa satisfy (a) and (b) and hence contains Fa . (viii) The set A WD ¹A  X j A 2 Fa or X nA 2 Fa º is a  -algebra. Hint. If A ; B 2 FA , then A[B D A C B A B 2 Fa . If XnA ; XnB 2 FA , then X n.A[B/ D XnA X nB 2 Fa . If A ; XnB 2 FA , then X n.A[B/ D .X nA/\.X nB/ D XnB A X nB 2 Fa . Thus A; B 2 A H) A [ B 2 A:

3.4. Exercises

131

(ix) A D Ba . Hint. Let K 2 Ka . Use Urysohn’s lemma (Theorem A.1) to construct a sequence gi 2 Cc .X/ that converges pointwise to K . (x) For every f 2 Fa there exists a sequence of compact sets Ki 2 Ka such that S Ki  KiC1 for all i and supp.f /  i2N Ki . Hint. The set of functions f W X ! R with this property satisfies conditions (a) and (b). Exercise 3.32. Show that, for every locally compact Hausdorff space X and any two Borel measures 0 ; 1 as in Theorem 3.8, there is a Baire set N  X such that 0 .N / D 0 and 0 .B/ D 1 .B/ for every Baire set B  X n N . Hint 1. Show first that 0 .B/ D sup¹0 .K/W K 2 Ka ; K  Bº;

(3.29)

where Ka is as in Exercise 3.29. To see this, prove that the right-hand side of equation (3.29) defines a Borel measure  on X that is inner regular on open sets and satisfies   0 and ƒ D ƒ0 . Hint 2. Suppose there exists a Baire set N  X such that 0 .N / < 1 .N /. Show that 1 .N / D 1 and that N can be chosen such that 0 .N / D 0. Next show that XnN 2 Fa , where Fa is as in Exercise 3.31, and deduce that X n N is contained in a countable union of compact sets. ˇ Example 3.33. Let X be the Stone–Cech compactification of N discussed in Example 4.60 below and denote by Ba  B  2X the Baire and Borel  -algebras. Thus B  X is a Baire set if and only if either B  N or X n B  N. (See part (iii) of Exercise 3.30.) For a Borel set B  X, define 0 .B/ WD

X1 ; n

n2B

ˇ ² ³ ˇ B  U  X; ˇ 1 .B/ WD inf 0 .U / ˇ : U is open As in Example 4.60, denote by X0  X the union of all open sets U  X with 0 .U / < 1. Then the restriction of 0 to X0 is a Radon measure, the restriction of 1 to X0 is outer regular and is inner regular on open sets, and 0 is given by (3.5) as in Theorem 3.8. Moreover, X0 n N is a Baire set in X0 and 0 .X0 n N/ D 0 while 1 .X0 n N/ D 1. Thus we can choose N WD X0 n N in Exercise 3.32 and 0 and 1 do not agree on the Baire  -algebra.

132

3. Borel measures

Example 3.34. Let X D R2 be the locally compact Hausdorff space in Example 3.23 and let 0 ; 1 be the Borel measures of Theorem 3.15 associated to the linear functional ƒW Cc .X/ ! R in that example. Then it follows from part (iv) of Exercise 3.30 that 0 .B/ D 1 .B/ for every Baire set B  X. Thus we can choose N D ; in Exercise 3.32. However, there does not exist any Borel set N  X such that 0 .N / D 0 and 0 agrees with 1 on all Borel subsets of X n N . (A set N  X is a Borel set with 0 .N / D 0 if and only if Nx WD ¹y 2 R j .x; y/ 2 N º is a Borel set and m.Nx / D 0 for all x 2 R.) Exercise 3.35. Let Z be the disjoint union of the locally compact Hausdorff spaces X0 in Example 3.33 and X D R2 in Example 3.34. Find Baire sets B0  X0 and B  X whose (disjoint) union is not a Baire set in Z.

Chapter 4

Lp spaces

This chapter discusses the Banach space Lp ./ associated to a measure space .X; A; / and a number 1  p  1. Section 4.1 introduces the inequalities of Hölder and Minkowski and Section 4.2 shows that Lp ./ is complete. In Section 4.3 we prove that, when X is a locally compact Hausdorff space,  is a Radon measure, and 1  p < 1, the subspace of continuous functions with compact support is dense in Lp ./. If in addition X is second countable, it follows that Lp ./ is separable. When 1 < p < 1 (or p D 1 and the measure space .X; A; / is localizable) the dual space of Lp ./ is isomorphic to Lq ./, where 1=pC1=q D 1. For p D 2 this follows from elementary Hilbert space theory and is proved in Section 4.4. For general p the proof requires the Radon–Nikodým theorem and is deferred to Chapter 5. Some preparatory results are proved in Section 4.5.

4.1 Hölder and Minkowski Assume throughout that .X; A; / is a measure space and that p; q are real numbers such that 1 1 C D 1; 1 < p < 1; 1 < q < 1: (4.1) p q Then any two nonnegative real numbers a and b satisfy Young’s inequality ab 

1 p 1 q a C b ; p q

(4.2)

and equality holds in (4.2) if and only if ap D b q . Exercise. Prove this by examining the critical points of the function .0; 1/ ! RW x 7 !

1 p x p

xb:

4. Lp spaces

134

Theorem 4.1. Let f; gW X ! Œ0; 1 be measurable functions. Then f and g satisfy the Hölder inequality Z 1=p Z 1=q Z p q (4.3) fg d  f d g d X

X

X

and the Minkowski inequality Z 1=p Z 1=p Z 1=p p p p  C : .f C g/ d f d g d X

X

(4.4)

X

Proof. Define Z A WD

1=p f d ; p

Z B WD

X

1=q g d : q

X

If A D 0, then f D 0 almost everywhere by Lemma 1.49, hence fg D 0 almost R everywhere, and hence X fg d D 0 by Lemma 1.48. This proves the Hölder inequality (4.3) in the case A D 0. If A D 1 and B > 0, then AB D 1 and so (4.3) holds trivially. Interchanging A and B if necessary, we find that (4.3) holds whenever one of the numbers A; B is zero or infinity. Hence assume 0 < A < 1 and 0 < B < 1. Then it follows from (4.2) that R Z f g X fg d D d AB X AB Z   p 1 f 1  g q   C d q B X p A R R 1 X g q d 1 X f p d C D p Ap q Bq D

1 1 C p q

D 1: This proves the Hölder inequality. To prove the Minkowski inequality, put Z 1=p a WD f p d ; X

Z b WD

1=p g d ; p

X

Z c WD X

1=p .f C g/p d :

4.1. Hölder and Minkowski

135

We must prove that c  a C b. This is obvious when a D 1 or b D 1. Hence assume a; b < 1. We first show that c < 1. This holds because f  .f p C g p /1=p

g  .f p C g p /1=p ;

and

hence f C g  2.f p C g p /1=p ; therefore .f C g/p  2p .f p C g p /; and integrating this inequality and raising the integral to the power 1=p we obtain c  2.ap C b p /1=p < 1: With this understood, it follows from the Hölder inequality that Z

p

p 1

c D

f .f C g/

Z d C

X

Z 

g.f C g/p

1

d

X

1=p Z f p d .f C g/.p

X

1/q

1=q d

X

1=p Z g p d .f C g/.p

Z C X

1/q

1=q d

X

Z D .a C b/

1 .f C g/ d

1=p

p

X

D .a C b/c

p 1

:

Here we have used the identity pq proves Theorem 4.1.

q D p. It follows that c  a C b and this 

R R Exercise 4.2. (i) Assume 0 < X f p d < 1 and 0 < X g q d < 1. Prove that equality holds in (4.3) if and only if there exists a constant ˛ > 0 such that g q D ˛f p almost everywhere. Hint. Use the proof of the Hölder inequality and the fact that equality holds in (4.2) if and only ap D b q . R R (ii) Assume 0 < X f p d < 1 and 0 < X g p d < 1. Prove that equality holds in (4.4) if and only if there is a real number  > 0 such that g D f almost everywhere. Hint. Use part (i) and the proof of the Minkowski inequality.

4. Lp spaces

136

4.2 The Banach space Lp ./ Definition 4.3. Let .X; A; / be a measure space and let 1  p < 1. Consider a x . The Lp -norm of f is the number measurable function f W X ! R Z kf kp WD

1=p : jf jp d

(4.5)

X

A function f W X ! R is called p-integrable or an Lp -function if it is measurable and kf kp < 1. The space of Lp -functions is denoted by Lp ./ WD ¹f W X ! R j f is A-measurable and kf kp < 1º:

(4.6)

It follows from the Minkowski inequality (4.4) that the sum of two Lp -functions is again an Lp -function, and hence Lp ./ is a real vector space. Moreover, the function Lp ./ ! Œ0; 1/W f 7 ! kf kp satisfies the triangle inequality kf C gkp  kf kp C kgkp for all f; g 2 Lp ./ by (4.4), and kf kp D jj kf kp ; for all  2 R and f 2 Lp ./ by definition. However, in general k
kp is not a norm on Lp ./ because kf kp D 0 if and only if f D 0 almost everywhere by Lemma 1.49. We can turn the space Lp ./ into a normed vector space by identifying two functions f; g 2 Lp ./ whenever they agree almost everywhere. Thus we introduce the equivalence relation 

f  g () f D g

-almost everywhere:

(4.7)

Denote the equivalence class of a function f 2 Lp ./ under this equivalence relation by Œf  and the quotient space by 

Lp ./ WD Lp ./= :

(4.8)

4.2. The Banach space Lp ./

137

This is again a real vector space. (For p D 1 see Definition 1.51.) The Lp -norm in (4.5) depends only on the equivalence class of f and so the map Lp ./ ! Œ0; 1/W Œf  7 ! kf kp is well defined. It is a norm on Lp ./ by Lemma 1.49. Thus we have defined the normed vector space Lp ./ for 1  p < 1. It is often convenient to abuse notation and write f 2 Lp ./ instead of Œf  2 Lp ./, always bearing in mind that then f denotes an equivalence class of p-integrable functions. If .X; A ;  / denotes the completion of .X; A; /, it follows as in Corollary 1.56 that Lp ./ is naturally isomorphic to Lp . /. Remark 4.4. Assume 1 < p < 1 and let f; g 2 Lp ./ such that kf C gkp D kf kp C kgkp ;

kf kp ¤ 0:

Then it follows from part (ii) of Exercise 4.2 that there exists a real number   0 such that g D f almost everywhere. Example 4.5. If .Rn ; A; m/ is the Lebesgue measure space we write Lp .Rn / WD Lp .m/: (See Definition 2.2 and Definition 2.11.) Example 4.6. If W 2N ! Œ0; 1 is the counting measure we write `p WD Lp ./: Thus the elements of `p are sequences .xn /n2N of real numbers such that k.xn /kp WD

1 X

jxn jp

1=p

< 1:

pD1

If we define f WN ! R by f .n/ WD xn then Z

jf jp d D N

for n 2 N, 1 X

jxn jp :

pD1

For p D 1 there is a similar normed vector space L1 ./, defined next.

4. Lp spaces

138

Definition 4.7. Let .X; A; / be a measure space and let f W X ! Œ0; 1 be a measurable function. The essential supremum of f is the number ess sup f 2 Œ0; 1 defined by ess sup f WD inf¹c 2 Œ0; 1 j f  c almost everywhereº

(4.9)

A function f W X ! R is called an L1 -function if it is measurable and kf k1 WD ess sup jf j < 1

(4.10)

The set of L1 -functions on X will be denoted by L1 ./ WD ¹f W X ! R j f is measurable and ess supjf j < 1º and the quotient space by the equivalence relation (4.7) by 

L1 ./ WD L1 ./= :

(4.11)

This is a normed vector space with the norm defined by (4.10), which depends only on the equivalence class of f . Lemma 4.8. For every f 2 L1 ./ there exists a measurable set E 2 A such that .E/ D 0 and

sup jf j D kf k1 :

X nE

Proof. The set En WD ¹x 2 X j jf .x/j > kf k1 C 1=nº has measure zero for all n. Hence E WD

[

En

n2N

is also a set of measure zero and jf .x/j  kf k1 for all x 2 X n E. Therefore supX nE jf j D kf k1 . This proves Lemma 4.8.  Theorem 4.9. Lp ./ is a Banach space for 1  p  1. Proof. Assume first that 1  p < 1. In this case the argument is a refinement of the proof of Theorem 1.52 and Theorem 1.53 for the case p D 1. Let fn 2 Lp ./ be a Cauchy sequence with respect to the norm (4.5). Choose a sequence of positive integers n1 < n2 < n3 <


such that kfni for all i 2 N.

fni C1 kp < 2

i

4.2. The Banach space Lp ./

139

Define gk WD

k X

fni j;

jfniC1

g WD

1 X

jfniC1

fni j D lim gk : k!1

iD1

iD1

Then it follows from Minkowski’s inequality (4.4) that kgk kp 

k X

kfni

fniC1 kp
0 .B/ 2 i for all i . Define [ A WD Ki  B: i2N

Then 1 .A/ D 0 .A/ D lim 0 .Ki / D 0 .B/: i !1

This implies A 2 Lp .1 / and ŒB 0 D ŒA 0 2 X. By Lemma 4.12, it follows that X D Lp .0 / and this proves Lemma 4.17.  Proof of Theorem 4.13. Let V  U be a countable basis for the topology. Assume without loss of generality that Vx is compact for all V 2 V. (If W  U is any countable basis for the topology, then the set V WD ¹V 2 W j Vx is compactº is also a countable basis for the topology by Lemma A.3.) Choose a bijection N ! VWi 7 ! Vi

and let I WD ¹I  N j #I < 1º be the set of finite subsets of N. Then the map I ! NWI 7 !

X

2i

1

i2I

is a bijection, so the set I is countable. For I 2 I define [ VI WD Vi : i 2I

Define the set V  Lp ./ by ` ° X V WD s D ˛j V j D1

Ij

ˇ ± ˇ ˇ ` 2 N and ˛j 2 Q; Ij 2 I for j D 1; : : : ; ` :

4.3. Separability

145

This set is contained in Lp ./ because Vx is compact for all V 2 V. It is countable and its closure x X WD V in Lp ./ is a closed linear subspace. By Lemma 4.12, it suffices to prove that ŒB  2 X for every B 2 B with .B/ < 1. To see this, fix a Borel set B 2 B with .B/ < 1 and a constant " > 0. Since X is second countable, every open subset of X is -compact (Lemma 3.21). Hence  is regular by Theorem 3.18. Therefore, there exist a compact set K  X and an open set U  X such that K  B  U;

.U n K/ < "p :

Define I WD ¹i 2 N j Vi  U º: S Since V is a basis of the topology, we have K  U D i 2I Vi . Since K is compact, there is a finite set I  I such that K  VI  U: Next, since B

V vanishes on X n .U n K/ and jB I

kB

V j  1, I

V kp  .U n K/1=p < ": I

x Since V 2 V and " > 0 was chosen arbitrary, it follows that ŒB  2 X D V. I This proves Theorem 4.13.  Proof of Theorem 4.15. By Corollary 3.16 and Lemma 4.17, it suffices to consider the case where  is outer regular and is inner regular on open sets. Define S WD ¹Œf  2 Lp ./ j for all " > 0, there exists s 2 Sc .X/ such that kf

skp < "º;

and C WD ¹Œf  2 Lp .1 / j for all " > 0, there exists g 2 Cc .X/ such that kf

gkp < "º:

We must prove that Lp ./ D S D C. Since S and C are closed linear subspaces of Lp ./, it suffices to prove that ŒB  2 S \ C for every Borel set B 2 B with .B/ < 1 by Lemma 4.12. Let B 2 B with .B/ < 1 and let " > 0. By Lemma 3.7, there is a compact set K  X and an open set U  X such that K  B  U and .U n K/ < "p . By Urysohn’s lemma (Theorem A.1), there

4. Lp spaces

146

is a function f 2 Cc .X/ such that 0  f  1, f jK  1, and supp.f /  U . This implies 0  f K  U nK and 0  B K  U nK : Hence, kB and likewise kf

K kp  kU nK kp D .U n K/1=p < "

K kp < ". By Minkowski’s inequality (4.4), this implies kB

f kp  kB

K kp C kK

f kp < 2":

This shows that ŒB  2 S \ C. This proves Theorem 4.15.



Remark 4.18. The reader may wonder whether Theorem 4.15 continues to hold for all Borel measures W B ! Œ0; 1 that are inner regular on open sets. To answer this question one can try to proceed as follows. Let 0 ; 1 be the Borel measures on X in Theorem 3.15 that satisfy ƒ0 D ƒ1 D ƒ . Then 0 is a Radon measure, 1 is outer regular and is inner regular on open sets, and 0 .B/  .B/  1 .B/ for all B 2 B. Thus Lp .1 /  Lp ./  Lp .0 / and one can consider the maps Lp .1 / ! Lp ./ ! Lp .0 /: Their composition is a Banach space isometry by Lemma 4.17. The question is now whether or not the first map Lp .1 / ! Lp ./ is surjective or, equivalently, whether the second map Lp ./ ! Lp .0 / is injective. If this holds, then the  subspace Cc .X/= is dense in Lp ./, otherwise it is not. The proof of Lemma 4.17 shows that the answer is affirmative if and only if every Borel set B  X with 0 .B/ < .B/ satisfies .B/ D 1. Thus the quest for a counterexample can be rephrased as follows. Question. Do there exist a locally compact Hausdorff space .X; U/ and Borel measures 0 ; 1 ; W B ! Œ0; 1 on its Borel -algebra B  2X such that all three measures are inner regular on open sets, 1 is outer regular, 0 is given by (3.5), 0 .B/  .B/  1 .B/ for all Borel sets B 2 B, and 0 D 0 .B/ < .B/ < 1 .B/ D 1 for some Borel set B 2 B?

4.4. Hilbert spaces

147

This leads to deep problems in set theory. A probability measure on a measurable space .X; A/ is a measure W A ! Œ0; 1 such that .X/ D 1. A measure W A ! Œ0; 1 is called nonatomic if countable sets have measure zero. Now consider the measure on X D R2 in Exercise 3.23 with 0 .R  ¹0º/ D 0 and 1 .R  ¹0º/ D 1, and define W R ! R2 by .x/ WD .x; 0/. If there is a nonatomic probability measure W 2R ! Œ0; 1, then the measure 0 C   provides a positive answer to the above question, and thus Theorem 4.15 would not extend to all Borel measures that are inner regular on open sets. The question of the existence of a nonatomic probability measure is related to the continuum hypothesis. The generalized continuum hypothesis asserts that, if X is any set, then each subset of 2X whose cardinality is strictly larger than that of X admits a bijection to 2X . It is independent of the other axioms of set theory and implies that nonatomic probability measures W 2X ! Œ0; 1 do not exist on any set X . This is closely related to the theory of measure-free cardinals. (See Fremlin [4, Section 4.3.7].)

4.4 Hilbert spaces This section introduces some elementary Hilbert space theory. It serves two purposes. First, it shows that the Hilbert space L2 ./ is isomorphic to its own dual space. Second, this result in turn will be used in the proof of the Radon–Nikodým theorem for -finite measure spaces in the next chapter. Definition 4.19. Let H be a real vector space. A bilinear map H  H ! RW .x; y/ 7 ! hx; yi

(4.16)

is called an inner product if it is symmetric, i.e., hx; yi D hy; xi for all x; y 2 H and positive definite, i.e., hx; xi > 0 for all x 2 H n ¹0º. The norm associated to an inner product (4.16) is the function p (4.17) H ! RW x 7 ! kxk WD hx; xi: Lemma 4.20. Let H be a real vector space equipped with an inner product (4.16) and the associated norm (4.17). The inner product and norm satisfy the Cauchy– Schwarz inequality jhx; yij  kxk kyk (4.18) and the triangle inequality kx C yk  kxk C kyk for all x; y 2 H . Thus (4.17) is a norm on H .

(4.19)

4. Lp spaces

148

Proof. The Cauchy–Schwarz inequality is obvious when x D 0 or y D 0. Hence assume x ¤ 0 and y ¤ 0, and put  WD kxk 1 x and  WD kyk 1 y. Then kk D kk D 1. Hence 0  k

h; ik2 D h; 

h; ii D 1

h; i2 :

This implies jh; ij  1 and hence jhx; yij  kxk kyk. In turn it follows from the Cauchy–Schwarz inequality that kx C yk2 D kxk2 C 2hx; yi C kyk2  kxk2 C 2 kxk kyk C kyk2 D .kxk C kyk/2 : 

This proves the triangle inequality (4.19) and Lemma 4.20.

Definition 4.21. An inner product space .H; h
;
i/ is called a Hilbert space if the norm (4.17) is complete, i.e., every Cauchy sequence in H converges. Example 4.22. Let .X; A; / be a measure space. Then H WD L2 ./ is a Hilbert space. The inner product is induced by the bilinear map Z L2 ./  L2 ./ ! RW .f; g/ 7 ! hf; gi WD fg d: (4.20) X

It is well defined because the product of two L2 -functions f; gW X ! R is integrable by (4.3) with p D q D 2. That it is bilinear follows from Theorem 1.44, and that it is symmetric is obvious. In general, it is not positive definite. However, it descends to a symmetric bilinear form Z 2 2 L ./  L ./ ! RW .Œf  ; Œg / 7 ! hf; gi D fg d: (4.21) X

by Lemma 1.48 which is positive definite by Lemma 1.49. Hence (4.21) is an inner product on L2 ./. It is called the L2 inner product. The norm associated to this inner product is 2

L ./ ! RW Œf  7 ! kf k2 D

Z

1=2 p f d D hf; f i: 2

(4.22)

X

This is the L2 -norm in (4.5) with p D 2. By Theorem 4.9, L2 ./ is complete with the norm (4.22) and hence is a Hilbert space.

4.4. Hilbert spaces

149

Definition 4.23. Let .V; k
k/ be a normed vector space. A linear functional ƒW V ! R is called bounded if there exists a constant c  0 such that jƒ.x/j  c kxk

for all x 2 V:

The norm of a bounded linear functional ƒW V ! R is the smallest such constant c and will be denoted by kƒk WD sup 0¤x2V

jƒ.x/j : kxk

(4.23)

The set of bounded linear functionals on V is denoted by V  and is called the dual space of V . Exercise 4.24. Prove that a linear functional on a normed vector space is bounded if and only if it is continuous. Exercise 4.25. Let .V; k
k/ be a normed vector space. Prove that the dual space V  with the norm (4.23) is a Banach space. (See Example 1.11.) Theorem 4.26 (Riesz). Let H be a Hilbert space and let ƒW H ! R be a bounded linear functional. Then there exists a unique element y 2 H such that ƒ.x/ D hy; xi

for all x 2 H:

(4.24)

jhy; xij D kƒk : kxk

(4.25)

This element y 2 H satisfies kyk D sup 0¤x2H

Thus the map H ! H  W y 7! hy;
i is an isometry of normed vector spaces. Theorem 4.27. Let H be a Hilbert space and let E  H be a nonempty closed convex subset. Then there exists a unique element x0 2 E such that kx0 k  kxk for all x 2 E. 

Proof. See page 151.

Theorem 4.27 implies Theorem 4.26. We prove existence. If ƒ D 0, then y D 0 satisfies (4.24). Hence assume ƒ ¤ 0 and define E WD ¹x 2 H j ƒ.x/ D 1º: Then E ¤ ; because there exists an element  2 H such that ƒ./ ¤ 0 and hence x WD ƒ./ 1  2 E. The set E is a closed because ƒW H ! R is continuous,

4. Lp spaces

150

and it is convex because ƒ is linear. Hence Theorem 4.27 yields an element x0 2 E such that kx0 k  kxk for all x 2 E: We prove that x 2 H;

ƒ.x/ D 0 H) hx0 ; xi D 0:

(4.26)

To see this, fix an element x 2 H such that ƒ.x/ D 0. Then x0 C tx 2 E for all t 2 R. This implies kx0 k2  kx0 C txk2 D kx0 k2 C 2thx0 ; xi C t 2 kxk2

for all t 2 R:

Thus the differentiable function t 7! kx0 C txk2 attains its minimum at t D 0, and so its derivative vanishes at t D 0. Hence, ˇ d ˇˇ 0D kx0 C txk2 D 2hx0 ; xi; dt ˇ tD0 and this proves (4.26). Now put y WD

x0 : kx0 k2

Fix an element x 2 H and put  WD ƒ.x/. Then ƒ.x Hence it follows from (4.26) that 0 D hx0 ; x This implies hy; xi D

x0 i D hx0 ; xi

x0 / D ƒ.x/

 D 0.

kx0 k2 :

hx0 ; xi D  D ƒ.x/: kx0 k2

Thus y satisfies (4.24). We prove (4.25). Assume y 2 H satisfies (4.24). If y D 0, then ƒ D 0, and so kyk D 0 D kƒk. Hence assume y ¤ 0. Then kyk D

ƒ.y/ jƒ.x/j kyk2 D  sup D kƒk : kyk kyk 0¤x2H kxk

Conversely, it follows from the Cauchy–Schwarz inequality that jƒ.x/j D jhy; xij  kykkxk for all x 2 H and hence kƒk  kyk. This proves (4.25).

4.4. Hilbert spaces

151

We prove uniqueness. Assume y; z 2 H satisfy hy; xi D hz; xi D ƒ.x/ for all x 2 H . Then hy

z; xi D 0 for all x 2 H . Take x WD y ky

and hence y

zk2 D hy

z; y

z to obtain

zi D 0;

z D 0. This proves Theorem 4.26, assuming Theorem 4.27.



Proof of Theorem 4.27. Define ı WD inf¹kxk j x 2 Eº: We prove uniqueness. Let x0 ; x1 2 E such that kx0 k D kx1 k D ı: Then 21 .x0 C x1 / 2 E because E is convex, and so kx0 C x1 k  2ı. Thus kx0

x1 k2 D 2 kx0 k2 C 2 kx1 k2

kx0 C x1 k2 D 4ı 2

kx0 C x1 k2  0

and therefore x0 D x1 . We prove existence. Choose a sequence xi 2 E such that lim kxi k D ı:

i!1

We claim that xi is a Cauchy sequence. Indeed, fix a constant " > 0. Then there exists an integer i0 2 N such that " i 2 N; i  i0 H) kxi k2 < ı 2 C : 4 Let i; j 2 N such that i  i0 and j  i0 . Then 12 .xi C xj / 2 E because E is convex, and hence kxi C xj k  2ı. This implies kxi

xj k2 D 2kxi k2 C 2kxj k2  " < 4 ı2 C 4ı 2 4

kxi C xj k2

D ": Thus xi is a Cauchy sequence. Since H is complete, the limit x0 WD lim xi i !1

exists. Moreover, x0 2 E because E is closed and kx0 k D ı because the norm function (4.17) is continuous. This proves Theorem 4.27. 

4. Lp spaces

152

Corollary 4.28. Let .X; A; / be a measure space and let ƒW L2 ./ ! R be a bounded linear functional. Then there exists a function g 2 L2 ./, unique up to equality almost everywhere, such that Z ƒ.Œf  / D fg d for all f 2 L2 ./: X

Moreover kƒk D kgk2 . Thus L2 ./ is isomorphic to L2 ./. Proof. This follows immediately from Theorem 4.26 and Example 4.22.



4.5 The dual space of Lp ./ We wish to extend Corollary 4.28 to the Lp -spaces in Definition 4.3 and equation (4.8) (for 1  p < 1) and in Definition 4.7 (for p D 1). When 1 < p < 1 it turns out that the dual space of Lp ./ is always isomorphic to Lq ./, where 1=p C 1=q D 1. For p D 1 the natural homomorphism L1 ./ ! L1 ./ is an isometric embedding, however, in most cases the dual space of L1 ./ is much larger than L1 ./. For p D 1 the situation is more subtle. The natural homomorphism L1 ./ ! L1 ./ need not be injective or surjective. However, it is bijective for a large class of measure spaces and one can characterize those measure spaces for which it is injective, respectively bijective. This requires the following definition. Definition 4.29. A measure space .X; A; / is called -finite if there exists a sequence of measurable subsets Xi 2 A such that XD

1 [

Xi ;

Xi  Xi C1 ;

.Xi / < 1 for all i 2 N:

(4.27)

iD1

It is called semi-finite if every measurable set A 2 A satisfies .A/ > 0 H)

there exists E 2 A such that E  A and 0 < .E/ < 1:

(4.28)

It is called localizable if it is semi-finite and, for every collection of measurable sets E  A, there is a set H 2 A satisfying the following two conditions: (L1) .E n H / D 0 for all E 2 E; (L2) if G 2 A satisfies .E n G/ D 0 for all E 2 E, then .H n G/ D 0. A measurable set H satisfying (L1) and (L2) is called an envelope of E.

4.5. The dual space of Lp ./

153

The geometric intuition behind the definition of localizable is as follows. The collection E  A will typically be uncountable, so one cannot expect its union to be measurable. The envelope H is a measurable set that replaces the union of the sets in E. It covers each set E 2 E up to a set of measure zero and, if any other measurable set G covers each set E 2 E up to a set of measure zero, it also covers H up to a set of measure zero. The next lemma clarifies the notion of semi-finiteness. Lemma 4.30. Let .X; A; / be a measure space: (i) .X; A; / is semi-finite if and only if .A/ D sup¹.E/ j E 2 A; E  A; .E/ < 1º

(4.29)

for every measurable set A 2 A; (ii) if .X; A; / is -finite, then it is semi-finite. Proof. (i) Assume .X; A; / is semi-finite, let A 2 A, and define a WD sup¹.E/ j E 2 A; E  A; .E/ < 1º: Then a  .A/ and we must prove that a D .A/. This is obvious when a D 1. Hence assume a < 1. Choose a sequence of measurable sets Ei  A such that .Ei / < 1 and .Ei / > a 2 i for all i . Define Bi WD E1 [


[ Ei ;

B WD

1 [ i D1

Bi D

1 [

Ei :

iD1

Then Bi 2 A, Ei  Bi  A, and .Bi / < 1. Hence .Ei /  .Bi /  a for all i 2 N, and hence .B/ D lim .Bi / D a < 1: i !1

If .A n B/ > 0, then since .X; A; / is semi-finite, there exists a measurable set F 2 A such that F  A n B and 0 < .F / < 1, and hence B [ F  A;

a < .B [ F / D .B/ C .F / < 1;

contradicting the definition of a. This shows that .A n B/ D 0 and hence .A/ D .B/ C .A n B/ D a, as claimed. Thus we have proved that every semi-finite measure space satisfies (4.29). The converse is obvious and this proves part (i). (ii) Assume that .X; A; / is  -finite and choose a sequence of measurable sets Xi 2 A that satisfies (4.27). If A 2 A, then it follows from Theorem 1.28 that .A/ D limi!1 .A \ Xi /. Since .A \ Xi / < 1 for all i, this shows that every measurable set A satisfies (4.29) and so .X; A; / is semi-finite. This proves Lemma 4.30. 

4. Lp spaces

154

It is also true that every -finite measure space is localizable. This can be derived as a consequence of Theorem 4.35 (see Corollary 5.9 below). A more direct proof is outlined in Exercise 4.58. Example 4.31. Define .X; A; / by X WD ¹a; bº; A WD 2X ; .¹aº/ WD 1;

.¹bº/ WD 1:

This measure space is not semi-finite. Thus the linear map L1 ./ ! L1 ./ in Theorem 4.33 below is not injective. In fact, L1 ./ has dimension two and L1 ./ has dimension one. Example 4.32. Let X be an uncountable set, let A  2X be the  -algebra of all subsets A  X such that A or Ac is countable, and let W A ! Œ0; 1 be the counting measure. Then .X; A; / is semi-finite, but it is not localizable. For example, let H  X be an uncountable set with an uncountable complement and let E be the collection of all finite subsets of H . Then the only possible envelope of E would be the set H itself, which is not measurable. Thus Theorem 4.33 below shows that the map L1 ./ ! L1 ./ is injective, while Theorem 4.35 below shows that it is not surjective. An example of a bounded linear functional ƒW L1 ./ ! R that cannot be represented by an L1 -function is given by X ƒ.f / WD f .x/ x2H

for f 2 L1 ./ D L1 ./. The next theorem is the first step towards understanding the dual space of Lp ./ and is a fairly easy consequence of the Hölder inequality. It asserts that for 1=p C 1=q D 1 every element of Lq ./ determines a bounded linear functional on Lp ./ and that the resulting map Lq ./ ! Lp ./ is an isometric embedding (for p D 1 under the semi-finite hypothesis). The key question is then whether every bounded linear functional on Lp ./ is of that form. That this is indeed the case for 1 < p < 1 (and for p D 1 under the localizable hypothesis) is the content of Theorem 4.35 below. This is a much deeper theorem whose proof for p ¤ 2 requires the Radon–Nikodým theorem and will be carried out in Chapter 5.

4.5. The dual space of Lp ./

155

Theorem 4.33. Let .X; A; / be a measure space and fix constants 1  p  1;

1 1 C D 1: p q

1  q  1;

(4.30)

Then the following holds. (i) Let g 2 Lq ./. Then the formula Z ƒg .Œf  / WD fg d for f 2 Lp ./

(4.31)

X

defines a bounded linear functional ƒg W Lp ./ ! R and kƒg k D

sup

ˇZ ˇ ˇ ˇ ˇ fg dˇ ˇ ˇ X

f 2Lp ./; kf kp ¤0

kf kp

 kgkq :

(4.32)

(ii) The map g 7! ƒg in (4.31) descends to a bounded linear operator Lq ./ ! Lp ./ W Œg 7 ! ƒg :

(4.33)

(iii) Assume 1 < p  1 Then kƒg k D kgkq

for all g 2 Lq ./.

(iv) Assume p D 1. Then the map L1 ./ ! L1 ./ in (4.33) is injective if and only if it is an isometric embedding if and only if .X; A; / is semi-finite. Proof. See page 158.



The heart of the proof is the next lemma. It is slightly stronger than what is required to prove Theorem 4.33 in that the hypothesis on g to be q-integrable is dropped in part (iii) and replaced by the assumption that the measure space is semifinite. In this form Lemma 4.34 is needed in the proof of Theorem 4.35 and will also be useful for proving the Minkowski and Calderón–Zygmund inequalities in Theorems 7.19 and 7.43 below.

4. Lp spaces

156

Lemma 4.34. Let .X; A; / be a measure space and let p; q be as in (4.30). Let gW X ! Œ0; 1 be a measurable function and suppose that there exists a constant c  0 such that Z p fg d  c kf kp : f 2 L ./; f  0 H) (4.34) X

Then the following holds: (i) if q D 1, then kgk1  c; (ii) if 1 < q < 1 and kgkq < 1, then kgkq  c; (iii) if 1 < q < 1 and .X; A; / is semi-finite, then kgkq  c; (iv) if q D 1 and .X; A; / is semi-finite, then kgk1  c. Proof. (i) If q D 1 take f  1 in (4.34) to obtain kgk1  c. (ii) Assume 1 < q < 1 and kgkq < 1. Then it follows from Lemma 1.47 that the set A WD ¹x 2 X j g.x/ D 1º has measure zero. Define the function hW X ! Œ0; 1/ by ´ h.x/ WD

g.x/ for x 2 X n A, for x 2 A.

0

Then h is measurable and Z khkq D kgkq < 1;

Z f h d D

X

X

fg d  c kf kp

for all f 2 Lp ./ with f  0 by Lemma 1.48. Define f W X ! Œ0; 1/ by f .x/ WD h.x/q

1

for x 2 X .

Then f p D hp.q

1/

D hq D f h

4.5. The dual space of Lp ./

157

and hence Z kf kp D

1 h d

1=q

q

khkqq 1

D

X

Z ; X

f h d D khkqq :

Thus f 2 Lp ./ and so khkqq D

Z X

f h d  c kf kp D c khkqq

1

:

Since khkq < 1, it follows that kgkq D khkq  c, and this proves part (ii). (iii) Assume .X; A; / is semi-finite and 1 < q < 1. Suppose, by contradiction, that kgkq > c. We will prove that there exists a measurable function hW X ! Œ0; 1/ such that 0  h  g;

c < khkq < 1:

(4.35)

By (4.34), this function h satisfies Z Z f h d  fg d  c kf kp X

X

for all f 2 L ./ with f  0. Since khkq < 1 it follows from part (ii) that khkq  c, which contradicts the inequality khkq > c in (4.35). It remains to prove the existence of h. Since kgkq > c, it follows from Definition 1.34 that Rthere exists a measurable step function sW X ! Œ0; 1/ such that 0  s  g and X s q d > c q . If kskq < 1, take h WD s. If kskq D 1, there exist a measurable set A  X and a constant ı > 0 such that .A/ D 1 and ıA  s  g. Since .X; A; / is semi-finite, Lemma 4.30 asserts that there exists a measurable set E 2 A such that E  A and c q < ı q .E/ < 1. Then the function h WD ıE W X ! Œ0; 1/ satisfies 0  h  g and khkq D ı.E/1=q > c, as required. This proves part (iii). p

(iv) Let q D 1 and assume .X; A; / is semi-finite. Suppose, by contradiction, that kgk1 > c. Then there exists a constant ı > 0 such that the set A WD ¹x 2 X j g.x/  c C ıº has positive measure. Since .X; A; / is semi-finite there exists a measurable 1 set R E  A such that 0 < .E/ < 1. Hence, f WD E 2 L ./ and we get X fg d  .c C ı/.E/ > c.E/ D c kf k1 , in contradiction to (4.34). This proves (iv) and Lemma 4.34. 

4. Lp spaces

158

Proof of Theorem 4.33. The proof has four steps. Step 1. Let f 2 Lp ./, g 2 Lq ./. Then fg 2 L1 ./ and kfgk1  kf kp kgkq . R If 1 < p < 1, then X jfgj d  kf kp kgkq by the Hölder inequality (4.3). If p D 1, then jfgj  jf j kgk1 almost everywhere by Lemma 4.8, so fg 2 L1 ./ and kfgk1  kf k1 kgk1 . If p D 1, interchange the pairs .f; p/ and .g; q/. Step 2. We prove (i) and (ii). By Step 1, the right hand side of (4.31) is well defined, and by Lemma 1.48 it depends only on the equivalence class of f under equality almost

everywhere. Hence ƒg is well defined. It is linear by Theorem 1.44 and ƒg  kgkq by Step 1. This proves (i). The bounded linear functional ƒg W Lp ./ ! R depends only on the equivalence class of g, again by Lemma 1.48. Hence the map (4.33) is well defined. By Theorem 1.44 and (4.32), it is a bounded linear operator of norm less than or equal to one. This proves (ii). Step 3. If 1 < p  1, then kƒg k D kgkq for all g 2 Lq ./. This continues to hold for p D 1 when .X; A; / is semi-finite. Let g 2 Lq ./. For t 2 R define sign.t/ 2 ¹ 1; 0; 1º by sign.t/ WD 1 for t > 0, sign.t/ WD 1 for t < 0, and sign.0/ D 0. If f 2 Lp ./ is nonnegative, then the function f sign.g/W X ! R is p-integrable and Z f jgj d D ƒg .f sign.g//  kƒg k kf sign.g/kp  kƒg k kf kp : X

Hence kgkq  kƒg k by Lemma 4.34 and so kƒg k D kgkq by Step 2. Step 4. If the map L1 ./ ! L1 ./ is injective, then .X; A; / is semi-finite. Let A 2 A such that .A/ > 0 and define g WD A . Then ƒg W L1 ./ ! R is nonzero by assumption. Hence there is an f 2 L1 ./ such that Z Z 0 < ƒg .f / D fg d D f d: (4.36) X

A

For i 2 N define Ei WD ¹x 2 A j f .x/ > 2 i º. Then Ei 2 A, Ei  A, and Z i .Ei /  2 f d  2i kf k1 < 1: Ei

S1

Moreover, E WD iD1 Ei D ¹x 2 A j f .x/ > 0º is not a null set by (4.36). Hence one of the sets Ei has positive measure. Thus .X; A; / is semi-finite. This proves Step 4 and Theorem 4.33. 

4.5. The dual space of Lp ./

159

The next theorem asserts that, for 1 < p < 1, every bounded linear functional on Lp ./ has the form (4.31) for some g 2 Lq ./. For p ¤ 2 this is a much deeper result than Corollary 4.28. The proof requires the Radon–Nikodým theorem and will be deferred to the next chapter. Theorem 4.35 (the dual space of Lp ). Let .X; A; / be a measure space and fix constants 1 1 C D 1: 1  p < 1; 1 < q  1; p q Then the following holds. (i) Assume 1 < p < 1. Then the map Lq ./ ! Lp ./ W Œg 7! ƒg ; defined by (4.31) is bijective and hence is a Banach space isometry. (ii) Assume p D 1. Then the map L1 ./ ! L1 ./ W Œg 7! ƒg defined by (4.31) is bijective if and only if .X; A; / is localizable. 

Proof. See page 195.

This next example shows that, in general, Theorem 4.35 does not extend to the case p D 1 (regardless of whether or not the measure space .X; A; / is -finite). By Theorem 4.33, the Banach space L1 ./ is equipped with an isometric inclusion L1 ./ ! L1 ./ , however, the dual space of L1 ./ is typically much larger than L1 ./. Example 4.36. Let W 2N ! Œ0; 1 be the counting measure on the positive integers. Then `1 WD L1 ./ D L1 ./ is the Banach space of bounded sequences x D .xn /n2N of real numbers, equipped with the supremum norm kxk1 WD sup jxn j: n2N

1

An interesting closed subspace of `

is

c WD ¹x D .xn /n2N 2 `1 j x is a Cauchy sequenceº: It is equipped with a bounded linear functional ƒ0 W c ! R, defined by ƒ0 .x/ WD lim xn n!1

for x D .xn /n2N 2 c:

4. Lp spaces

160

The Hahn–Banach theorem, one of the fundamental principles of functional analysis, asserts that every bounded linear functional on a linear subspace of a Banach space extends to a bounded linear functional on the entire Banach space (whose norm is no larger than the norm of the original bounded linear functional on the subspace). In the case at hand this means that there is a bounded linear functional ƒW `1 ! R such that ƒjc D ƒ0 . This linear functional cannot have the form (4.31) for any g 2 L1 ./. To see this, note that `1 WD L1 ./ D L1 ./ is the space of summable sequences of real numbers. Let y D .yn /n2N 2 `1 be a sequence of real P numbers such that 1 nD1 jyn j < 1, and define the linear functional ƒy W `1 ! R by ƒy .x/ WD

1 X

xn y n

for x D .xn /n2N 2 `1 :

nD1

P1

Choose N 2 N such that nDN jyn j DW ˛ < 1 and define x D .xn /n2N 2 c by xn WD 0 for n < N and xn WD 1 for n  N . Then ƒy .x/  ˛ < 1 D ƒ.x/ and hence ƒy ¤ ƒ. This shows that ƒ does not belong to the image of the isometric inclusion `1 ,! .`1 / . Exercise 4.37. Let ƒ0 W c ! R be the functional in Example 4.36 and denote c0 WD ker ƒ0 . Thus c0 is the set of all sequences of real numbers that converge to zero, i.e., c0 D ¹x D .xn /n2N 2 `1 j lim xn D 0º: n!1

Prove that c0 is a closed linear subspace of ` to the dual space of c0 . Thus `1 Š .c0 / ;

1

and that `1 is naturally isomorphic

c0 ¨ `1 Š .`1 / Š .c0 / ;

`1 ¨ .`1 / Š .`1 / :

In the language of functional analysis, this means that the Banach spaces c0 and `1 are not reflexive, and neither is `1 . We close this section with two results that will be needed in the proof of Theorem 4.35. The first asserts that every bounded linear functional on Lp ./ can be written as the difference of two positive bounded linear functionals (Theorem 4.39). The second asserts that every positive bounded linear functional on Lp ./ is supported on a  -finite subset of X (Theorem 4.40). When ƒW Lp ./ ! R is a bounded linear functional it will be convenient to abuse notation and write ƒ.f / WD ƒ.Œf  /

for f 2 Lp ./.

4.5. The dual space of Lp ./

161

Definition 4.38. Let .X; A; / be a measure space and let 1  p < 1. A bounded linear functional ƒW Lp ./ ! R is called positive if f  0 H) ƒ.f /  0 for all f 2 Lp ./. Theorem 4.39. Let .X; A; / be a measure space and let 1  p < 1. Consider a bounded linear functional ƒW Lp ./ ! R. Define ˙ W A ! Œ0; 1 by ˙ .A/ WD sup¹ƒ.˙E / j E 2 A; E  A; .E/ < 1º

(4.37)

Then the maps ˙ are measures, Lp ./  L1 .C / \ L1 . /, and the formulas Z ˙ ƒ .f / WD f d˙ for f 2 Lp ./ (4.38) X

define positive bounded linear functionals ƒ˙ W Lp ./ ! R such that ƒ D ƒC

ƒ ;

kƒC k C kƒ k D kƒk:

(4.39)

Proof. The proof is divided into four steps. Step 1. The maps ˙ W A ! Œ0; 1 in (4.37) are measures. It follows directly from the definition that ˙ .;/ D 0. We must prove that C is -additive. That  is then also -additive follows by reversing the sign of ƒ. Thus, let Ai 2 A be a sequence of pairwise disjoint measurable sets and set A WD

1 [

Ai :

iD1

Let E 2 A such that E  A and .E/ < 1. Then it follows from the definition of C that ƒ.E \A /  C .Ai / for all i 2 N: (4.40) i

Moreover, the sequence of measurable functions fn WD E

n X

E \A  0 i

iD1

4. Lp spaces

162

converges pointwise to zero and satisfies 0  fnp  E for all n. Since .E/ < 1, the function E is integrable, and so it follows from the Lebesgue dominated convergence theorem (Theorem 1.45) that Z lim fnp d D 0; n!1 X

i.e.,

lim E

n X

n!1

E \A D 0: i p

i D1

Then (4.40) yields that ƒ.E / D lim

n!1

n X

ƒ.E \A / D

1 X

i

iD1

ƒ.E \A / 

1 X

i

i D1

C .Ai /:

i D1

Take the supremum over all E 2 A with E  A and .E/ < 1 to obtain C

 .A/ 

1 X

C .Ai /:

iD1

To prove the converse inequality, assume first that C .Ai / D 1 for some i ; since P C Ai  A, this implies C .A/ D 1 D 1 i D1  .Ai /. Hence it suffices to assume C  .Ai / < 1 for all i . Fix a constant " > 0 and choose a sequence of measurable sets Ei 2 A such that Ei  Ai and ƒ.E / > C .Ai / 2 i " for all i . Since i E1 [


[ En  A, it follows from the definition of C that C .A/  ƒ.E

1

/D [


[En

n X

ƒ.E / >

n X

i

i D1

C .Ai /

iD1

Take the limit n ! 1 to obtain C

 .A/ 

1 X

C .Ai /

"

for all " > 0,

iD1

so C .A/ 

1 X

C .Ai /;

i D1

as claimed. Thus C is -additive and this proves Step 1.

":

4.5. The dual space of Lp ./

163

Step 2. Let c WD kƒk. Then every measurable function f W X ! R satisfies Z Z jf j dC C jf j d  c kf kp : (4.41) X

X

In particular, Lp ./  L1 .C / \ L1 . /. Assume first that f D sW X ! Œ0; 1/ is a measurable step function in Lp ./. Then there exist real numbers ˛i > 0 and measurable sets Ai 2 A for i D 1; : : : ; ` P such that Ai \ Aj D ; for i ¤ j , .Ai / < 1 for all i , and s D `iD1 ˛i A : i Now fix a real number " > 0 and choose "i > 0 such that ` X

˛i "i D

i D1

" : 2

For i D 1; : : : ; ` choose Ei˙ 2 A such that Ei˙  Ai ;

ƒ.E C /  C .Ai /

ƒ.E /   .Ai /

"i ;

i

i

"i :

Then Z

C

Z

s d C X

s d D X

` X

˛i .C .Ai / C  .Ai //

iD1



` X

˛i .ƒ.E C / i

iD1

` X Dƒ ˛i .E C

ƒ.E / C 2"i / i

i

 E / C "

`

X

 c ˛i .E C

E / C "

iD1

i

iD1

Dc

` X

i

i

p

1=p ˛ip ..EiC n Ei / C .Ei n EiC // C"

iD1

c

` X

1=p ˛ip .Ai / C"

iD1

D c kskp C ":

4. Lp spaces

164

Take the limit " ! 0 to obtain (4.41) for f D s. To prove (4.41) in general it suffices to assume that f 2 Lp ./ is nonnegative. By Theorem 1.26, there is a sequence of measurable step functions 0  s1  s2 


that converges pointwise to f . Then .f sn /p converges pointwise to zero and is bounded above by f p 2 L1 ./. Hence lim kf sn kp D 0 n!1

by the Lebesgue dominated convergence theorem (Theorem 1.45), and Z Z lim sn d˙ D f d˙ n!1 X

X

by the Lebesgue monotone convergence theorem (Theorem 1.37). This proves the inequality (4.41). It follows from (4.41) that Lp ./  L1 .C / \ L1 . / and this proves Step 2. Step 3. If A 2 A and .A/ < 1, then ˙ .A/ < 1;

ƒ.A / D C .A/

 .A/

(4.42)

It follows from the inequality (4.41) in Step 2 that Z Z C C  .A/ C  .A/ D A d C A d  ckA kp D c.A/1=p < 1: X

X

Now let " > 0 and choose E 2 A such that E  A and ƒ.E / > C .A/ Since ƒ.AnE /   .A/, this implies ƒ.A / D ƒ.E / C ƒ.AnE / > C .A/

 .A/

".

":

Since this holds for all " > 0, we obtain ƒ.A /  C .A/  .A/. Reversing the sign of ƒ we also obtain ƒ.A /   .A/ C .A/ and this proves Step 3. Step 4. If f 2 Lp ./, then Z ƒ.f / D

f dC

X

Z f d :

(4.43)

X

Let sW X ! R be a p-integrable step function. Then there are real numbers ˛i and measurable sets Ai 2 A for i D 1; : : : ; ` such that .Ai / < 1 for all i and P s D `iD1 ˛i A . Hence it follows from Step 3 that i

ƒ.s/ D

` X

˛i ƒ.A / D

` X

i

iD1

iD1

C

˛i . .Ai /

Z  .Ai // D

C

Z

s d X

s d : X

4.5. The dual space of Lp ./

165

This proves (4.43) for p-integrable step functions. Now let f 2 Lp ./ and assume f  0. By Theorem 1.26, there is a sequence of measurable step functions 0  s1  s2 


that converges pointwise to f . Then .f sn /p converges pointwise to zero and is bounded above by f p 2 L1 ./. Hence lim kf

n!1

sn kp D 0

by the Lebesgue dominated convergence theorem and so lim ƒ.sn / D ƒ.f /:

n!1

Moreover,

Z X

by Step 2, and

f d˙  c kf kp < 1 Z

lim

n!1 X

sn d˙ D

Z

f d˙

X

by the Lebesgue monotone convergence theorem. Thus every nonnegative Lp -function f W X ! Œ0; 1/ satisfies (4.43). If f 2 Lp ./, then f ˙ 2 Lp ./ satisfy (4.43) by what we have just proved, and hence so does f D f C f . This proves Step 4. It follows from Steps 2 and 4 that the linear functionals ƒ˙ W Lp ./ ! R in (4.38) are bounded and satisfy (4.39). This proves Theorem 4.39.  Theorem 4.40. Let .X; A; / be a measure space and let 1 < p < 1. Consider a positive bounded linear functional ƒW Lp ./ ! R. Define .A/ WD sup¹ƒ.E / j E 2 A; E  A; .E/ < 1º

(4.44)

for A 2 A. Then the map W A ! Œ0; 1 is a measure, Lp ./  L1 ./, and Z ƒ.f / D f d for all f 2 Lp ./: (4.45) X

Moreover, there are measurable sets N 2 A and Xn 2 A for n 2 N such that X nN D

1 [

Xn ;

.N / D 0;

.Xn / < 1;

Xn  XnC1

(4.46)

nD1

for all n 2 N. Proof. That  is a measure satisfying Lp ./  L1 ./ and (4.45) follows from Theorem 4.39 and the fact that C D  and  D 0 because ƒ is positive. Now define c WD kƒk. We prove in three steps that there exist measurable sets N 2 A and Xn 2 A for n 2 N satisfying (4.46).

4. Lp spaces

166

Step 1. For every " > 0 there exist a measurable set A 2 A and a measurable function f W X ! Œ0; 1/ such that f jX nA D 0;

kf kp D 1;

inf f > 0; A

ƒ.f / > c

":

(4.47)

In particular, p

.A/  .inf f / A

< 1:

Choose h 2 Lp ./ such that khkp D 1 and ƒ.h/ > c ". Assume without loss of generality that h  0. (Otherwise replace h by jhj.) Define Ai WD ¹x 2 X j h.x/ > 2 i º: Then .h hA /p converges pointwise to zero as i ! 1 and is bounded by the i integrable function hp . Hence it follows from the Lebesgue dominated convergence theorem (Theorem 1.45) that lim kh

hA kp D 0; i

i !1

and therefore lim ƒ.hA / D ƒ.h/ > c

":

i

i !1

Choose i 2 N such that ƒ.hA / > c i

and define A WD Ai ;

f WD

" hA

i

khA kp

:

i

Then A and f satisfy (4.47) and so .A/  .inf f /

p

Z

A

f p d D .inf f /

p

A

X

:

This proves Step 1. Step 2. Let "; A; f be as in Step 1 and let E 2 A. Then .E \ A D ;; .E/ < 1/ H)

ƒ.E / .E/1=p

< "1=q

where 1 < q < 1 is chosen such that 1=p C 1=q D 1.

c p

 C1 ;

(4.48)

4.5. The dual space of Lp ./

167

Define g WD f C



" 1=p E : .E/

Then Z kgkp D

1=p

p

f d C "

D .1 C "/1=p

X

and, by (4.47), ƒ.g/ D ƒ.f / C



" 1=p ƒ.E / > c .E/

" C "1=p

ƒ.E / .E/1=p

:

Since ƒ.g/  c kgkp , it follows that c

" C "1=p

Further since .1 C "/1=p "1=p Finally since "1

.E/1=p

< c.1 C "/1=p :

1  "=p for all "  0, this implies

ƒ.E / .E/1=p

1=p

ƒ.E /

< c..1 C "/1=p

c  1/ C "  " C1 : p

D "1=q , Step 2 is proved.

Step 3. There exist measurable sets N; X1 ; X2 ; X3 ; : : : satisfying (4.46). Choose An 2 A and fn 2 Lp ./ as in Step 1 with " D 1=n. For n 2 N, set Xn WD A1 [


[ An ;

N WD X n

1 [

An D X n

nD1

1 [

Xn :

nD1

By Step 2, every measurable set E  N with .E/ < 1 satisfies ƒ.E / .E/1=p


0, there exists K  X such that : ˇ ; ˇ K is compact and sup XnK jf j < "

Prove that X is a Banach space with respect to the sup-norm. Prove that Cc .X/ is dense in C0 .X/. Exercise 4.48. Let .X; A; / be a measure space such that .X/ D 1 and let f; gW X ! Œ0; 1 be measurable functions such that fg  1. Prove that kf k1 kgk1  1: Exercise 4.49. Let .X; A; / be a measure space such that .X/ D 1 and let f W X ! Œ0; 1 be a measurable function. Prove that q

1 C kf k21 

Z q X

1 C f 2 d  1 C kf k1 :

(4.55)

Find a geometric interpretation of this inequality when  is the restriction of the Lebesgue measure to the unit interval X D Œ0; 1 and f D F 0 is the derivative of a continuously differentiable function F W Œ0; 1 ! R. Under which conditions does equality hold in either of the two inequalities in (4.55)? Exercise 4.50. Let .X; A; / be R a measure space and let f W X ! R be a measurable function such that f > 0 and X f d D 1. Let E  X be a measurable set such that 0 < .E/ < 1. Prove that  1  .E/

(4.56)

for 0 < p < 1:

(4.57)

Z log.f / d  .E/ log E

and Z

f p d  .E/1

p

E

Exercise 4.51. Let f W Œ0; 1 ! .0; 1/ be Lebesgue measurable. Prove that 1

Z

Z

1

Z log.f .t// dt 

f .s/ ds 0

1

0

f .x/ log.f .x// dx: 0

(4.58)

4.6. Exercises

171

Exercise 4.52. Fix two constants 1 < p < 1 and a > 0. (i) Let f W .0; 1/ ! R be Lebesgue measurable and suppose that the function .0; 1/ ! RW x 7! x p 1 a jf .x/jp ; is integrable. Show that the restriction of f to each interval .0; x is integrable and prove Hardy’s inequality 1

Z

x

1 a

0

ˇZ ˇ ˇ ˇ

x 0

ˇp 1=p Z 1 ˇ p ˇ f .t/ dt ˇ dx  xp a 0

1 a

1=p jf .x/j dx : p

(4.59) Show that equality holds in (4.59) if and only if f D 0 almost everywhere. Hint. Assume first that f is nonnegative with compact support and define Z 1 x f .t/ dt for x > 0. F .x/ WD x 0 Use integration by parts to obtain Z 1 Z p 1 p p 1 a p x F .x/ dx D x a 0 0

1 a

F .x/p

1

f .x/ dx:

Use Hölder’s inequality. (ii) Show that the constant p=a in Hardy’s inequality is sharp. Hint. Choose  < 1 a=p and take f .x/ WD x  for x  1 and f .x/ WD 0 for x > 1. (iii) Prove that every sequence .an /n2N of positive real numbers satisfies N 1 1  X 1 X p  p p X p an  a : N nD1 p 1 nD1 n

(4.60)

N D1

Hint. If an is nonincreasing, then (4.60) follows from (4.59) with a D p for a suitable function f . Deduce the general case from the special case.

1

(iv) Let f W .0; 1/ ! R be Lebesgue measurable and suppose that the function .0; 1/ ! RW x 7! x p 1Ca jf .x/jp ; is integrable. Show that the restriction of f to each interval Œx; 1/ is integrable and prove the inequality 1

Z

x 0

a 1

ˇZ ˇ ˇ ˇ

1 x

ˇp 1=p Z 1 ˇ p ˇ xp f .t/ dt ˇ dx  a 0

1Ca

1=p jf .x/j dx : p

(4.61) Hint. Apply the inequality (4.59) to the function g.x/ WD x

2

f .x

1

/.

4. Lp spaces

172

Exercise 4.53. Let .X; U/ be a locally compact Hausdorff space and let W B ! Œ0; 1 be an outer regular Borel measure on X that is inner regular on open sets. Let g 2 L1 ./. Prove that the following are equivalent: (i) the function g vanishes -almost everywhere; R (ii) X fg d D 0 for all f 2 Cc .X/. Hint. Assume (ii). Let K  X be compact. Use Urysohn’s lemma (Theorem A.1) to show that there is a sequence fn 2 CcR.X/ such that 0  fn  1 and Rfn converges almost everywhere to K . Deduce R that K g d D 0. Then prove that U g d D 0 for every open set U  X and B g d D 0 for all B 2 B. Warning! The regularity hypotheses on  cannot be removed. Find an example of a Borel measure where (ii) does not imply (i). (See Example 4.16.) Exercise 4.54. Egoroff’s theorem. Let .X; A; / be a measure space such that .X/ < 1 and let fn W X ! R be a sequence of measurable functions that converges pointwise to f W X ! R. Fix a constant " > 0. Then there exists a measurable set E 2 A such that .X n E/ < " and fn jE converges uniformly to f jE . Hint. Define S.k; n/ WD ¹x 2 X j jfi .x/

fj .x/j < 1=k, for all i; j > nº

for k; n 2 N:

Prove that lim .S.k; n// D .X / for all k 2 N: T Deduce that there is a sequence nk 2 N such that E WD k2N S.k; nk / satisfies the required conditions. Show that Egoroff’s theorem does not extend to  -finite measure spaces. n!1

Exercise 4.55. Let .X; A; / be a measure space and let 1 < p < 1. Let f 2 Lp ./ and let fn 2 Lp ./ be a sequence such that lim kfn kp D kf kp

n!1

and fn converges to f almost everywhere. Prove that lim kf

n!1

fn kp D 0:

4.6. Exercises

173

Prove that the hypothesis lim kfn kp D kf kp

n!1

cannot be removed. Hint 1. Fix a constant " > 0. Use Egoroff’s R theorem to construct disjoint measurable sets A; B 2 A such that X D A [ B, A jf jp d < ", .B/ < 1, and fn converges R to fp uniformly on B. Use Fatou’s lemma (Theorem 1.41) to prove that lim sup A jfn j d < ". n!1

Hint 2. Let gn WD 2p 1 .jfn jp C jf jp / jf fn jp and use Fatou’s lemma (Theorem 1.41) as in the proof of the Lebesgue dominated convergence theorem (Theorem 1.45). Exercise 4.56. Let .X; A; / be a measure space, let fn W X ! R be a sequence of measurable functions, and let f W X ! R be a measurable function. The sequence fn is said to converge in measure to f if lim .¹x 2 X j jfn .x/

n!1

f .x/j > "º/ D 0

for all " > 0. (On page 53 this is called convergence in probability.) Assume .X/ < 1 and prove the following. (i) If fn converges to f almost everywhere, then fn converges to f in measure. Hint. See page 53. (ii) If fn converges to f in measure, then a subsequence of fn converges to f almost everywhere. (iii) If 1  p  1 and fn ; f 2 Lp ./ satisfy lim kfn

n!1

f kp D 0;

then fn converges to f in measure. Exercise 4.57. Let .X; U/ be a compact Hausdorff space and W B ! Œ0; 1 be a Borel measure. Let C.X/ D Cc .X/ be the space of continuous real-valued functions on X. Consider the following conditions: (a) every nonempty open subset of X has positive measure; (b) there exists a Borel set E  X and an element x0 2 X such that every open neighborhood U of x0 satisfies .U \ E/ > 0 and .U n E/ > 0; (c)  is outer regular and is inner regular on open sets.

4. Lp spaces

174

Prove the following. (i) Assume (a). Then the map C.X/ ! L1 ./ in (b) is an isometric embedding and hence its image is a closed linear subspace of L1 ./. (ii) Assume (a) and (b). Then there is a nonzero bounded linear functional ƒW L1 ./ ! R that vanishes on the image of the inclusion C./ ! L1 ./. Hint. If f D E almost everywhere, then f is discontinuous at x0 . (iii) Assume (a), (b), and (c). Then the isometric embedding L1 ./ ! L1 ./ of Theorem 4.33 is not surjective. Hint. Use part (ii) and Exercise 4.53. (iv) The Lebesgue measure on Œ0; 1 satisfies (a), (b), and (c). Exercise 4.58. Prove that every -finite measure space .X; A; / is localizable. Hint. Assume first that .X/ < 1. Let E  A and define c WD sup¹.E1 [


[ En / j n 2 N; E1 ; : : : ; En 2 Eº: S Show that there is a sequence Ei 2 E such that . 1 iD1 Ei / D c. Prove that S1 H WD i D1 Ei is an envelope of E. Exercise 4.59. Let .X; A; / be a localizable measure space. Prove that it satisfies the following. (F) Let F be a collection of measurable functions f W Af ! R, each defined on a measurable set Af 2 A. Suppose that any two functions f1 ; f2 2 F agree almost everywhere on Af1 \ Af2 . Then there exists a measurable function gW X ! R such that gjAf D f almost everywhere for all f 2 F. We will see in the next chapter that condition (F) is equivalent to localizability for semi-finite measure spaces. Hint. Let F be a collection of measurable functions as in (F). For a 2 R and f 2 F define Afa WD ¹x 2 Af j f .x/ < aº: For q 2 Q, let H q 2 A be an envelope of the collection Eq WD ¹Afq j f 2 Fº: Define the measurable sets X a WD

[ q2Q q a and x … X r for all r < a:

(4.62)

Then g is well defined and measurable and g D f on Af n Ef for all f 2 F. Example 4.60. This example is closely related to Exercise 3.24, however, it requires a considerable knowledge of functional analysis and the details go much beyond ˇ the scope of the present manuscript. It introduces the Stone–Cech compactification X of the natural numbers. This is a compact Hausdorff space containing N and satisfying the universality property that every continuous map from N to another compact Hausdorff space Y extends uniquely to a continuous map from X to Y .

4. Lp spaces

176

The space C.X/ of continuous functions on X can be naturally identified with the space `1 . Hence the space of positive bounded linear functionals on `1 is isomorˇ phic to the space of Radon measures on X by Theorem 3.15. Thus the Stone–Cech compactification of N can be used to understand the dual space of `1 . Moreover, it gives rise to an interesting example of a Radon measure which is not outer regular (explained to me by Theo Buehler). Consider the inclusion N ! .`1 / W n 7 ! ƒn

which assigns to each natural number n 2 N the bounded linear functional ƒn W `1 ! R defined by ƒn ./ WD n

for  D .i /i 2N 2 `1 .

This functional has norm one. Now the space of all bounded linear functionals on `1 of norm at most one, i.e., the unit ball in .`1 / , is compact with respect to the weak- topology by the Banach–Alaoglu theorem. Define X to be the closure of the set ¹ƒn j n 2 Nº in .`1 / with respect to the weak- topology. Thus 8 ˇ 9 ˇ For all finite sequences c 1 ; : : : ; c ` 2 R ˆ > ˇ ˆ > ˆ ˆ ˇ and  1 D .i1 /i2N ; : : : ;  ` D .i` /i2N 2 `1> > < = ˇ X WD ƒ 2 .`1 / ˇˇ satisfying ƒ. j / < c j for j D 1; : : : ; `; : ˆ > ˆ ˇ > ˆ > ˆ ˇ there exists an n 2 N such that > : ; ˇ  j < c j for j D 1; : : : ; ` n The weak- topology U  2X is the smallest topology such that the map f W X ! R;

f .ƒ/ WD ƒ./;

is continuous for each  2 `1 . The topological space .X; U/ is a separable compact ˇ Hausdorff space, called the Stone–Cech compactification of N. It is not second countable and one can show that the complement of a point in X that is not equal to one of the ƒn is not -compact. The only continuous functions on X are those of the form f , so the map `1 ! C.X/W  7 ! f is a Banach space isometry. (Verify that kf k WD sup jf .ƒ/j D kk1 for all ƒ2X

 2 `1 .) Thus the dual space of `1 can be understood in terms of the Borel measures on X.

4.6. Exercises

177

By Theorem 3.18, every Radon measure on X is regular. However, the Borel -algebra B  2X does carry  -finite measures W B ! Œ0; 1 that are inner regular, but not outer regular (and must necessarily satisfy .X/ D 1). Here is an example pointed out to me by Theo Buehler. Define .B/ WD

X 1 n

n2N ƒn 2B

for every Borel set B  X. This measure is -finite and inner regular, but not outer regular. (The set U WD ¹ƒn j n 2 Nº is open, its complement K WD X n U is compact and has measure zero, and every open set containing K misses only a finite subset of U and hence has infinite measure.) Now let X0  X be the union of all open sets in X with finite measure. Then X0 is not  -compact and the restriction of  to the Borel -algebra of X0 is a Radon measure, but is not outer regular.

Chapter 5

The Radon–Nikodým theorem

Recall from Theorem 1.40 that every measurable function f W X ! Œ0; 1/ on a measure space .X; A; / determines a measure f W A ! Œ0; 1 defined by Z f .A/ WD f d for A 2 A. A

By Theorem 1.35 it satisfies f .A/ D 0 whenever .A/ D 0. A measure with this property is called absolutely continuous with respect to . The Radon–Nikodým theorem asserts that, when  is -finite, every -finite measure that is absolutely continuous with respect to  has the form f for some measurable function f W X ! Œ0; 1/. It was proved by Johann Radon in 1913 for the Lebesgue measure space and extended by Otton Nikodým in 1930 to general  -finite measure spaces. A proof is given in Section 5.1. Consequences of the Radon–Nikodým theorem include the proof of Theorem 4.35 about the dual space of Lp ./ (Section 5.2) and the decomposition theorems of Lebesgue, Hahn, and Jordan for signed measures (Section 5.3). An extension of the Radon–Nikodým theorem to general measure spaces is proved in Section 5.4.

5.1 Absolutely continuous measures Definition 5.1. Let .X; A; / be a measure space. A measure W A ! Œ0; 1/ is absolutely continuous with respect to  if .A/ D 0 H) .A/ D 0 for all A 2 A.  is singular with respect to  if there exists a measurable set A such that .A/ D 0 and

.Ac / D 0:

In this case we also say that  and  are mutually singular. We write  if  is absolutely continuous with respect to  and  ?  if  and  are mutually singular.

180

5. The Radon–Nikodým theorem

Lemma 5.2. Let .X; A/ be a measurable space and let ; ; 1 ; 2 be measures on A. Then the following holds: (i) if 1 ?  and 2 ? , then 1 C 2 ? ; (ii) if 1   and 2  , then 1 C 2  ; (iii) if 1   and 2 ? , then 1 ? 2 ; (iv) if    and  ? , then  D 0. Proof. (i) Suppose that 1 ?  and 2 ? . Then there exist measurable sets Ai 2 A such that i .Ai / D 0 and .Aci / D 0 for i D 1; 2. Define A WD A1 \ A2 . Then Ac D Ac1 [ Ac2 is a null set for  and A is a null set for both 1 and 2 , and hence also for 1 C 2 . Thus 1 C 2 ?  and this proves (i). (ii) Suppose that 1   and 2  . If A 2 A satisfies .A/ D 0, then 1 .A/ D 2 .A/ D 0 and so .1 C 2 /.A/ D 1 .A/ C 2 .A/ D 0. Thus 1 C 2   and this proves (ii). (iii) Suppose that 1   and 2 ? . Since 2 ? , there exists a measurable set A 2 A such that 2 .A/ D 0 and .Ac / D 0. Since 1   it follows that 1 .Ac / D 0 and hence 1 ? 2 .This proves (iii). (iv) Suppose that    and  ? . Since  ? , there exists a measurable set A 2 A such that .A/ D 0 and .Ac / D 0. Since   , it follows that .Ac / D 0 and hence .X/ D .A/ C .Ac / D 0. This proves (iv) and Lemma 5.2.  Theorem 5.3 (Lebesgue decomposition theorem). Let .X; A; / be a -finite measure space and let  be a -finite measure on A. Then there exist unique measures a ; s W A ! Œ0; 1 such that  D a C s ;

a  ;

s ? :

(5.1) 

Proof. See page 186.

Theorem 5.4 (Radon–Nikodým). Let .X; A; / be a -finite measure space and let W A ! Œ0; 1 be a measure. The following are equivalent: (i)  is  -finite and absolutely continuous with respect to ; (ii) there exists a measurable function f W X ! Œ0; 1/ such that Z .A/ D f d for all A 2 A:

(5.2)

A

If (i) holds, then equation (5.2) determines f uniquely up to equality -almost everywhere. Moreover, f 2 L1 ./ if and only if .X/ < 1.

5.1. Absolutely continuous measures

181

Proof. The last assertion follows by taking A D X in (5.2). We prove that (ii) H) (i). Thus assume that there exists a measurable function f W X ! Œ0; 1/ such that  is given by (5.2). Then  is absolutely continuous with respect to  by Theorem 1.35. Since  is -finite, there exists a sequence of S measurable sets X1  X2  X3 


such that .Xn / < 1 and X D 1 nD1 Xn . Define An WD ¹x 2 Xn j f .x/  nº: S Then An  AnC1 and .An /  n.Xn / < 1 for all n and X D 1 nD1 An . Thus  is -finite and this shows that (ii) H) (i). It remains to prove that (i) H) (ii) and that f is uniquely determined by (5.2) up to equality -almost everywhere. This is proved in three steps. The first step is uniqueness, the second step is existence under the assumption .X/ < 1 and .X/ < 1, and the last step establishes existence in general. Step 1. Let .X; A; / be a measure space, let W A ! Œ0; 1 be a  -finite measure, and let f; gW X ! Œ0; 1/ be two measurable functions such that Z Z .A/ D f d D g d for all A 2 A: (5.3) A

A

Then f and g agree -almost everywhere. Since .X; A; / is a -finite measure space, there exists a sequence of measurable S sets A1  A2  A3 


such that .An / < 1 for all n 2 N and X D 1 nD1 An . For n 2 N put An WD ¹E 2 A j E  An º; n WD jAn : Take A D An in (5.3) to obtain f; g 2 L1 .n / for all n. Thus Z 1 f g 2 L .n /; .f g/ dn D 0 for all E 2 An : E

Hence f

g vanishes n -almost everywhere by Lemma 1.49. Thus the set En WD ¹x 2 An j f .x/ ¤ g.x/º

satisfies .En / D n .En / D 0, and hence the set E WD ¹x 2 X j f .x/ ¤ g.x/º D

1 [ nD1

satisfies .E/ D 0. This proves Step 1.

En

182

5. The Radon–Nikodým theorem

Step 2. Let .X; A/ be a measurable space and let ; W A ! Œ0; 1 be measures such that .X/ < 1, .X/ < 1, and   . Then there exists a measurable function hW X ! Œ0; 1/ such that

Z .A/ D

h d for all A 2 A.

A

By assumption,  C W A ! Œ0; 1 is a finite measure defined by . C /.A/ WD .A/ C .A/

for A 2 A:

Since . C /.X/ < 1, it follows from the Cauchy–Schwarz inequality that H WD L2 . C /  L1 . C /: Namely, if f 2 L2 . C /, then sZ Z jf j2 d. C / < 1; jf j d. C /  c

c WD

p .X/ C .X/:

X

X

Define

ƒW L2 . C / ! R

by

Z ƒ.f / WD

f d: X

for f 2 L2 .C/. (Here we abuse notation and use the same letter f for a function in L2 . C / and its equivalence class in L2 . C /.) Then Z Z jƒ.f /j  jf j d  jf j d. C /  c kf kL2 .C/ X

X

2

for all f 2 L . C /. Thus ƒ is a bounded linear functional on L2 . C / and it follows from Corollary 4.28 that there exists an L2 -function g 2 L2 . C / such that Z Z f d D fg d. C / (5.4) X

X

for all f 2 L2 . C /. This implies Z Z f .1 g/ d. C / D f d. C / X

X

Z f d. C /

X

f d X

f d X

Z D for all f 2 L2 . C /.

fg d. C / X

Z D

Z

(5.5)

5.1. Absolutely continuous measures

183

We claim that the inequalities 0  g < 1 hold . C /-almost everywhere. To see this, consider the measurable sets E0 WD ¹x 2 X j g.x/ < 0º;

E1 WD ¹x 2 X j g.x/  1º:

Then it follows from (5.4) with f WD E that 0

Z

Z

0  .E0 / D

E d D 0

X

X

E g d. C /  0: 0

Hence Z X

E g d. C / D 0 0

and Lemma 1.49 shows that the function f WD E g vanishes . C /-almost 0 everywhere. Hence . C /.E0 / D 0. Likewise, (5.5) with f WD E yields that 1

Z

Z

.E1 / D

E d D 1

X

g/ d. C /  0:

.1 E1

Hence .E1 / D 0. Since  is absolutely continuous with respect to , it follows that .E1 / D 0 and hence . C /.E1 / D 0 as claimed. Assume from now on that 0  g.x/ < 1 for all x 2 X . (Namely, redefine g.x/ WD 0 for x 2 E0 [ E1 without changing the identities (5.4) and (5.5).) Apply identity (5.5) to the characteristic function f WD A 2 L2 . C / of a measurable set A to obtain the identity Z .A/ D

g/ d. C /

.1

for all A 2 A:

A

By Theorem 1.40 this implies that (5.5) continues to hold for every measurable function f W X ! Œ0; 1/, whether or not it belongs to L2 . C /. Now define the measurable function hW X ! Œ0; 1/ by h.x/ WD

g.x/ 1 g.x/

for x 2 X:

184

5. The Radon–Nikodým theorem

By (5.4) with f D A and (5.5) with f D A h it satisfies Z .A/ D X

A d

Z D X

A g d. C /

Z D X

g/ d. C /

A h.1

Z D X

A h d

Z D

h d A

for all A 2 A. This proves Step 2. Step 3. We prove that (i) H) (ii). Since  and  are -finite measures, there exist sequences of measurable sets An ; Bn 2 A such that An  AnC1 , .An / < 1, Bn  BnC1 , .Bn / < 1 S S1 for all n, and X D 1 nD1 An D nD1 Bn . Define Xn WD An \ Bn . Then Xn  XnC1 ;

.Xn / < 1;

.Xn / < 1

S for all n, and X D 1 nD1 Xn . Thus, by Step 2, there exists a sequence of measurable functions fn W Xn ! Œ0; 1/ such that Z .A/ D

fn d

for all n 2 N and all A 2 A such that A  Xn :

(5.6)

A

It follows from Step 1 that the restriction of fnC1 to Xn agrees with fn -almost everywhere. Thus, modifying fnC1 on a set of measure zero if necessary, we may assume without loss of generality that fnC1 jXn D fn for all n 2 N. With this understood, define f W X ! Œ0; 1/ by f jXn WD fn

for n 2 N:

5.1. Absolutely continuous measures

185

This function is measurable because f

1

.Œ0; c/ D

1 [

1

.Xn \ f

1 [

.Œ0; c// D

fn 1 .Œ0; c/ 2 A

nD1

nD1

for all c  0. Now let E 2 A and define En WD E \ Xn 2 A for n 2 N. Then E1  E2  E3 


;

ED

1 [

En :

nD1

Hence it follows from part (iv) of Theorem 1.28 that .E/ D lim .En / n!1 Z f d D lim n!1 E n

Z D lim

n!1 X

En f d

Z D X

E f d

Z D

f d: E

Here the last but one equality follows from the Lebesgue monotone convergence theorem (Theorem 1.37). This proves Step 3 and Theorem 5.4.  Example 5.5. Let X be a one-element set and let A WD 2X . Define the measure W 2X ! Œ0; 1 by .;/ WD 0 and .X/ WD 1. (i) Choose .;/ WD 0 and .X/ WD 1. Then   , R but there does not exist a (measurable) function f W X ! Œ0; 1 such that X f d D .X/. Thus the hypothesis that .X; A; / is -finite cannot be removed in Theorem 5.4. R (ii) Choose  WD . Then we get .A/ D A f d for every nonzero function f W X ! Œ0; 1/. Thus the hypothesis that .X; A; / is -finite cannot be removed in Step 1 in the proof of Theorem 5.4. Example 5.6. Let X be an uncountable set and denote by A  2X the set of all subsets A  X such that either A or Ac is countable. Choose an uncountable subset H  X with an uncountable complement and define ; ; W A ! Œ0; 1 by ´ 0 if A is countable; .A/ WD .A/ WD #.A \ H /; .A/ WD #A: 1 if Ac is countable;

186

5. The Radon–Nikodým theorem

Then      and  and  are not -finite. R There does not exist any measurable function f W X ! Œ0; 1 such that .X/R D X f d. Nor is there any measurable function hW X ! R such that .A/ D A h d for all A 2 A. (The only possible such function would be h WD H which is not measurable.) Proof of Theorem 5.3. We prove uniqueness. Let a ; s ; 0a ; 0s W A ! Œ0; 1 be measures such that  D a C s D 0a C 0s ;

a  ;

0a  ;

s ? ;

0s ? :

Then there exist measurable sets A; A0 2 A such that s .A/ D 0;

.X n A/ D 0;

0s .A0 / D 0;

.X n A0 / D 0:

Since X n .A \ A0 / D .X n A/ [ .X n A0 /, this implies .X n .A \ A0 // D 0: Let E 2 A. Then s .E \ A \ A0 / D 0 D 0s .E \ A \ A0 / and hence a .E \ A \ A0 / D .E \ A \ A0 / D 0a .E \ A \ A0 /: Moreover .E n .A \ A0 // D 0, hence a .E n .A \ A0 // D 0 D 0a .E n .A \ A0 // and so

s .E n .A \ A0 // D .E n .A \ A0 // D 0s .E n .A \ A0 //:

This implies a .E/ D a .E \ A \ A0 / D 0a .E \ A \ A0 / D 0a .E/ s .E/ D s .E n .A \ A0 // D 0s .E n .A \ A0 // D 0s .E/: This proves uniqueness. We prove existence. The measure  WD  C W A ! Œ0; 1 is -finite. Hence it follows from the Radon–Nikodým theorem (Theorem 5.4) that there exist measurable functions f; gW X ! Œ0; 1/ such that Z Z .E/ D f d; .E/ D g d for all E 2 A: (5.7) E

E

5.2. The dual space of Lp ./ revisited

187

Define A WD ¹x 2 X j g.x/ > 0º

(5.8)

and a .E/ WD .E \ A/;

s .E/ WD .E \ Ac / for E 2 A:

(5.9)

Then it follows directly from (5.9) that the maps a ; s W A ! Œ0; 1 are measures and satisfy a C s D . Moreover, by (5.9), s .A/ D .A \ Ac / D .;/ D 0 and from (5.8) that gjAc D 0, so by (5.7) Z g d D 0: .Ac / D Ac

This shows that s ? . It remains to prove that a is absolutely continuous with respect to . To see this, let E 2 A such that .E/ D 0. Then by (5.7) Z Z E g d D g d D .E/ D 0: X

E

Hence it follows from Lemma 1.49 that E g vanishes -almost everywhere. Thus E \A g D A E g vanishes -almost everywhere. Since g.x/ > 0 for all x 2 E \ A, this implies .E \ A/ D 0: Hence Z a .E/ D .E \ A/ D

f d D 0: E \A

This shows that a   and completes the proof of Theorem 5.3.



5.2 The dual space of Lp ./ revisited This section is devoted to the proof of Theorem 4.35. Assume throughout that .X; A; / is a measure space and fix two constants 1  p < 1;

1 < q  1;

1 1 C D 1: p q

(5.10)

188

5. The Radon–Nikodým theorem

As in Section 4.5 we abuse notation and write ƒ.f / WD ƒ.Œf  / for the value of a bounded linear functional ƒW Lp ./ ! R on the equivalence class of a function f 2 Lp ./. Recall from Theorem 4.33 that every g 2 Lq ./ determines a bounded linear functional ƒg W Lp ./ ! R via

Z

fg d for f 2 Lp ./:

ƒg .f / WD X

The next result proves Theorem 4.35 in -finite case. Theorem 5.7. Assume .X; A; / is -finite and let ƒW Lp ./ ! R be a bounded linear functional. Then there exists a function g 2 Lq ./ such that ƒg D ƒ: Proof. Assume first that ƒ is positive. We prove in six steps that there exists a function g 2 Lq ./ such that g  0 and ƒg D ƒ. Step 1. Define .A/ WD sup¹ƒ.E / j E 2 A; E  A; .E/ < 1º

(5.11)

for A 2 A. Then the map W A ! Œ0; 1 is a measure, Lp ./  L1 ./, and Z ƒ.f / D f d for all f 2 Lp ./. X

This follows directly from Theorem 4.40. Step 2. Let  be as in Step 1 and define c WD kƒk. Then .A/  c.A/1=p for all A 2 A. By assumption, ƒ.f /  c kf kp for all f 2 Lp ./. Take f WD E to obtain ƒ.E /  c.E/1=p  c.A/1=p for all E 2 A with E  A and .E/ < 1. Take the supremum over all such E to obtain .A/  c.A/1=p by (5.11). Step 3. Let  be as in Step 1. Then there exists a measurable function gW X ! Œ0; 1/ such that

Z .A/ D A

g d for all A 2 A.

5.2. The dual space of Lp ./ revisited

189

By Step 2,  is -finite and   . Hence Step 3 follows from the Radon–Nikodým theorem (Theorem 5.4) for -finite measure spaces. Step 4. Let  be as in Step 1 and g be as in Step 3. Then Z Z fg d D f d X

X

for every measurable function f W X ! Œ0; 1/. This follows immediately from Step 3 and Theorem 1.40. Step 5. Let c be as in Step 2 and g be as in Step 3. Then kgkq  c. Let  be as in Step 1 and let f 2 Lp ./ such that f  0. Then Z Z Step 4 Step 1 fg d D f d D ƒ.f /  c kf kp : X

(5.12)

X

Moreover, the measure space .X; A; / is semi-finite by Lemma 4.30. Hence it follows from parts (iii) and (iv) of Lemma 4.34 that kgkq  c. Step 6. Let g be as in Step 3. Then ƒ D ƒg : Since g 2 Lq ./ by Step 5, the function g determines a bounded linear functional R p ƒg W L ./ ! R via ƒg .f / WD X fg d for f 2 Lp ./. By (5.12), it satisfies ƒg .f / D ƒ.f / for all f 2 Lp ./ with f  0. Apply this identity to the functions f ˙ W X ! Œ0; 1/ for all f 2 Lp ./ to obtain ƒ D ƒg . This proves the assertion of Theorem 5.7 for every positive bounded linear functional ƒW Lp ./ ! R. Let ƒW Lp ./ ! R be any bounded linear functional. By Theorem 4.39 there exist positive bounded linear functionals ƒ˙ W Lp ./ ! R such that ƒ D ƒC ƒ . Hence, by what we have just proved, there exist functions g ˙ 2 Lq ./ such that g ˙  0 and ƒ˙ D ƒg ˙ . Define g WD g C

g :

Then g 2 Lq ./ and ƒg D ƒg C This proves Theorem 5.7.

ƒg D ƒC

ƒ D ƒ: 

190

5. The Radon–Nikodým theorem

The next result proves Theorem 4.35 in the case p D 1. Theorem 5.8. Assume p D 1. Then the following are equivalent: (i) the measure space .X; A; / is localizable; (ii) the measure space .X; A; / is semi-finite and satisfies condition (F) in Exercise 4.59, i.e., if F is a collection of measurable functions f W Af ! R, each defined on a measurable set Af 2 A, such that any two functions f1 ; f2 2 F agree almost everywhere on Af1 \ Af2 , then there exists a measurable function gW X ! R such that gjAf D f almost everywhere for all f 2 F; (iii) the linear map

L1 ./ ! L1 ./ W g 7 ! ƒg

(5.13)

is bijective. Proof. The proof that (i) H) (ii) is outlined in Exercise 4.59. We prove that (ii) H) (iii). Since .X; A; / is semi-finite, the linear map (5.13) is injective by Theorem 4.33. We must prove that it is surjective. Assume first that ƒW L1 ./ ! R is a positive bounded linear functional. Define E WD ¹E 2 A j .E/ < 1º and, for E 2 E, define AE WD ¹A 2 A j A  Eº;

E WD jAE :

(5.14)

Then there is an extension operator E W L1 .E / ! L1 ./ defined by E .f /.x/ WD

´ f .x/ 0

for x 2 E; for x 2 X n E:

(5.15)

It descends to a bounded linear operator from L1 .E / to L1 ./ which will still be denoted by E . Define ƒE D ƒ ı E W L1 .E / ! R: This is a positive bounded linear functional for every E 2 E. Hence it follows from Theorem 5.7 (and the Axiom of Choice) that there is a collection of bounded measurable functions gE W E ! Œ0; 1/; E 2 E;

5.2. The dual space of Lp ./ revisited

191

such that Z ƒE .f / D

fgE dE

for all E 2 E and all f 2 L1 .E /:

E

If E; F 2 E, then E \ F 2 E and the functions gE jE \F , gF jE \F , and gE \F all represent the same bounded linear functional ƒE \F W L1 .E \F / ! R. Hence they agree almost everywhere by Theorem 4.33. This shows that the collection F WD ¹gE j E 2 Eº satisfies the hypotheses of condition (F) on page 174. Thus it follows from (ii) that there exists a measurable function gW X ! R such that, for all E 2 E, the restriction gjE agrees with gE almost everywhere on E. We claim that g  0 almost everywhere. Suppose on the contrary that the set A WD ¹x 2 X j g.x/ < 0º has positive measure. Since .X; A; / is semi-finite, there exists a set E 2 E such that E  A and .E/ > 0. Since g.x/ < 0  gE .x/ for all x 2 E it follows that gjE does not agree with gE almost everywhere, a contradiction. This shows that g  0 almost everywhere. We prove that g  kƒk almost everywhere. Suppose otherwise that the set AC WD ¹x 2 X j g.x/ > kƒkº has positive measure. Since .X; A; / is semi-finite, there exists a set E 2 E such that E  AC and .E/ > 0. Since kgE k1 D kƒE k  kƒk, it follows from Lemma 4.8 that gE .x/  kƒk < g.x/ for almost every x 2 E. Hence gjE does not agree with gE almost everywhere, a contradiction. This shows that g  kƒk almost everywhere and we may assume without loss of generality that 0  g.x/  kƒk for all x 2 X. We prove that ƒg D ƒ. Fix a function f 2 L1 ./ such that f  0. Then there exists a sequence Ei 2 E such that E1  E2  E3 


and E f converges pointwise to f . Namely, by Theorem 1.26, there exists a i sequence of measurable step functions si W X ! Œ0; 1/

192

5. The Radon–Nikodým theorem

such that 0  s1  s2 


and si converges pointwise to f . Since Z Z si d  f d < 1 for all i , X

X

the sets Ei WD ¹x 2 X j si .x/ > 0º have finite measure and 0  si  E f  f

for all i .

i

Thus the Ei are as required. Since the sequence jf E f j converges pointi wise to zero and is bounded above by the integrable function f , it follows from the Lebesgue dominated convergence theorem (Theorem 1.45) that lim kf

E f k1 D 0: i

i!1

Hence ƒ.f / D lim ƒ.E f / i

i !1

D lim ƒEi .f jEi / i !1 Z D lim fgEi d i !1 E i

Z D lim

i !1 E i

fg d

Z D

fg d: X

Here the last step follows from the Lebesgue monotone convergence theorem (Theorem 1.37). This shows that ƒ.f / D ƒg .f / for every nonnegative integrable function f W X ! Œ0; 1/. Consequently, ƒ.f / D ƒ.f C /

ƒ.f / D ƒg .f C /

for all f 2 L1 ./. Thus ƒ D ƒg as claimed.

ƒg .f / D ƒg .f /

5.2. The dual space of Lp ./ revisited

193

This shows that every positive bounded linear functional on L1 ./ belongs to the image of the map (5.13). Since every bounded linear functional on L1 ./ is the difference of two positive bounded linear functionals by Theorem 4.39, it follows that the map (5.13) is surjective. Thus we have proved that (ii) H) (iii). We prove that (iii) H) (i). Assume that the map (5.13) is bijective. Then .X; A; / is semi-finite by part (iv) of Theorem 4.33. Now let E  A be any collection of measurable sets. Assume without loss of generality that E1 ; : : : ; E` 2 E H) E1 [


[ E` 2 E: (Otherwise, replace E by the collection E0 of all finite unions of elements of E; then every measurable envelope of E0 is also an envelope of E.) For E 2 E define AE and E by (5.14) and define the bounded linear functional ƒE W L1 .E / ! R by Z

for f 2 L1 .E /:

(5.16)

f  0 H) ƒE .f /  ƒF .f /:

(5.17)

ƒE .f / WD

f dE E

Then for all E; F 2 A and f 2 L1 ./ E  F; Define ƒW L1 ./ ! R by ƒ.f / WD sup ƒE .f C jE /

sup ƒE .f jE /:

E 2E

(5.18)

E 2E

We prove that this is a well defined bounded linear functional with kƒk  1. To see this, note that Z ƒE .f jE / 

f d X

for every nonnegative function f 2 L1 ./, and so Z Z jƒ.f /j  f C d C f d D kf k1 X

for all f 2 L1 ./.

X

Moreover, it follows directly from the definition that ƒ.cf / D cƒ.f / for all c  0 and ƒ. f / D ƒ.f /. Now let f; g 2 L1 ./ be nonnegative integrable functions.

194

5. The Radon–Nikodým theorem

Then ƒ.f C g/ D sup ƒE .f jE C gjE / E 2E

 sup ƒE .f jE / C sup ƒE .gjE / E 2E

E 2E

D ƒ.f / C ƒ.g/: To prove the converse inequality, let " > 0 and choose E; F 2 E such that ƒE .f jE / > ƒ.f /

";

ƒF .gjF / > ƒ.g/

":

Then E [ F 2 E and it follows from (5.17) that ƒE [F ..f C g/jE [F / D ƒE [F .f jE [F / C ƒE [F .gjE [F /  ƒE .f jE / C ƒF .gjF / > ƒ.f / C ƒ.g/

2":

Hence ƒ.f C g/ > ƒ.f / C ƒ.g/

2" for all " > 0

and so ƒ.f / C ƒ.g/  ƒ.f C g/  ƒ.f / C ƒ.g/: This shows that ƒ.f C g/ D ƒ.f / C ƒ.g/ for all f; g 2 L1 ./ such that f; g  0. If f; g 2 L1 ./, then .f C g/C C f

C g D .f C g/ C f C C g C ;

and hence ƒ..f C g/C / C ƒ.f / C ƒ.g / D ƒ..f C g/ / C ƒ.f C / C ƒ.g C / by what we have just proved. Since ƒ.f / D ƒ.f C / it follows that ƒ.f C g/ D ƒ.f / C ƒ.g/

ƒ.f / by definition,

for all f; g 2 L1 ./.

This shows that ƒW L1 ! R is a positive bounded linear functional of norm kƒk  1.

5.2. The dual space of Lp ./ revisited

195

With this understood, it follows from (iii) that there exists a function g 2 L1 ./ such that ƒ D ƒg . Define H WD ¹x 2 X j g.x/ > 0º: We prove that H is an envelope of E. Fix a set E 2 E and suppose, by contradiction, that .E n H / > 0. Then, since .X; A; / is semi-finite, there exists a measurable set A 2 A such that 0 < .A/ < 1 and A  E n H . Since g.x/  0 for all x 2 A, it follows that Z 0 < .A/ D ƒE .A jE / D

g d  0; A

a contradiction. This shows that our assumption .E n H / > 0 was wrong. Hence .E n H / D 0 for all E 2 E, as claimed. Now let G 2 A be any measurable set such that .E n G/ D 0 for all E 2 E. We must prove that .H n G/ D 0. Suppose, by contradiction, that .H n G/ > 0. Since .X; A; / is semi-finite there exists a measurable set A 2 A such that 0 < .A/ < 1 and A  H n G. Then Z g d D ƒ.A / D sup ƒE .A jE / D sup .E \ A/ D 0: A

E 2E

E 2E

Here the second equality follows from (5.18), the third follows from (5.16), and the last follows from the fact that E \ A  E n G for all E 2 E. Since g > 0 on A, it follows from Lemma 1.49 that .A/ D 0, a contradiction. This shows that our assumption that .H n G/ > 0 was wrong and so .H n G/ D 0 as claimed. Thus we have proved that every collection of measurable sets E  A has a measurable envelope, and this completes the proof of Theorem 5.8.  Now we are in a position to prove Theorem 4.35 in general. Proof of Theorem 4.35. For p D 1 the assertion of Theorem 4.35 follows from the equivalence of (i) and (iii) in Theorem 5.8. Hence assume p > 1. We must prove that the linear map Lq ./ ! Lp ./ W g 7! ƒg is surjective. Let ƒW Lp ./ ! R be a positive bounded linear functional and define .A/ WD sup¹ƒ.E / j E 2 A; E  A; .E/ < 1º

for A 2 A:

Then W A ! Œ0; 1 is a measure by Theorem 4.40 and Z p 1 L ./  L ./; ƒ.f / D f d for all f 2 Lp ./: X

196

5. The Radon–Nikodým theorem

Theorem 4.40 also asserts that there exists a measurable set N 2 A such that .N / D 0 and the restriction of  to X n N is -finite. Define X0 WD X n N;

A0 WD ¹A 2 A j A  X0 º;

0 WD jA0

as in (5.14), let 0 W Lp .0 / ! Lp ./ be the extension operator as in (5.15), and define ƒ0 WD ƒ ı 0 W Lp .0 / ! R: Then ƒ0 is a positive bounded linear functional on Lp .0 / and Z Z ƒ.f / D f d D f d D ƒ0 .f jX0 / for all f 2 Lp ./: X

X nN

Since .X0 ; A0 ; 0 / is -finite, Theorem 5.7 shows that there exists a function g0 2 Lq .0 / such that g0  0 and Z ƒ0 .f0 / D f0 g0 d0 for all f0 2 Lp .0 /: X0

Define gW X ! Œ0; 1/ by g.x/ WD

´ g0 .x/ for x 2 X0 D X n N , 0

for x 2 N .

Then, by Theorem 4.33, kgkLq ./ D kg0 kLq .0 / D kƒ0 k D kƒk and, for all f 2 Lp ./, Z ƒ.f / D ƒ0 .f jX0 / D

Z fg0 d0 D

X0

fg d: X

This proves the assertion for positive bounded linear functionals. Since every bounded linear functional ƒW Lp ./ ! R is the difference of two positive bounded linear functionals by Theorem 4.39, this proves Theorem 4.35.  Corollary 5.9. Every -finite measure space is localizable. Proof. Let .X; A; / be a  -finite measure space. Then .X; A; / is semi-finite by Lemma 4.30. Hence the map L1 ./ ! L1 ./ W g 7! ƒg in (4.31) is injective by Theorem 4.33 and is surjective by Theorem 5.7. Hence it follows from Theorem 5.8 that .X; A; / is localizable. 

5.3. Signed measures

197

5.3 Signed measures Throughout this section .X; A/ is a measurable space, i.e., X is a set and A  2X is a -algebra. The following definition extends the notion of a measure on .X; A/ to a signed measure which can have positive and negative values. As an example from physics one can think of electrical charge. Definition 5.10. A function W A ! R is called a signed measure if it is -additive, i.e., for every sequence Ei 2 A of pairwise disjoint measurable sets, it holds that 1 1 1 [  X X j.Ei /j < 1;  Ei D .Ei /: (5.19) iD1

i D1

iD1

Lemma 5.11. Every signed measure W A ! R satisfies the following: (1) .;/ D 0; (2) if E1 ; : : : ; E` 2 A are pairwise disjoint, then ` ` [  X  Ei D .Ei /: i D1

i D1

Proof. To prove (i) take Ei WD ; in (5.19). To prove (ii) take Ei WD ; for all i > `.  Given a signed measure W A ! R, it is a natural question to ask whether it can be written as the difference of two measures ˙ W A ! Œ0; 1/. Closely related to this is the question whether there exists a measure W A ! Œ0; 1/ that satisfies j.A/j  .A/

for all A 2 A:

(5.20)

If such a measure exists, it must satisfy .E; F 2 A; E \ F D ;/ H) .E/

.F /  .E [ F /

Thus a lower bound for .A/ is the supremum of the numbers .E/ .F / over all decompositions of A into pairwise disjoint measurable sets E and F . The next theorem shows that this supremum defines the smallest measure that satisfies (5.20). Theorem 5.12. Let W A ! R be a signed measure and define 8 ˇ 9 ˇ E; F 2 A; = < ˇ jj.A/ WD sup .E/ .F / ˇˇ E \ F D ;; for A 2 A: : ˇ E [F DA;

(5.21)

Then j.A/j  jj.A/ < 1 for all A 2 A and jjW A ! Œ0; 1/ is a measure, called the total variation of .

198

5. The Radon–Nikodým theorem

Proof. We prove that jj is a measure. If follows directly from the definition that jj.;/ D 0 and jj.A/  j.A/j  0 for all A 2 A. We must prove that the function jjW A ! Œ0; 1 is -additive. Let Ai 2 A be a sequence of pairwise disjoint measurable sets and put A WD

1 [

Ai :

i D1

Let E; F 2 A are measurable sets such that E \ F D ;; Then ED

1 [

.E \ Ai /;

E [ F D A: 1 [

F D

iD1

(5.22)

.F \ Ai /:

i D1

Hence .E/

.F / D

1 X

.E \ Ai /

iD1

D

1 X

1 X

.F \ Ai /

i D1

..E \ Ai /

.F \ Ai //

iD1



1 X

jj.Ai /:

iD1

Take the supremum over all pairs of measurable sets E; F satisfying (5.22) to obtain jj.A/ 

1 X

jj.Ai /

(5.23)

i D1

To prove the converse inequality, fix a constant " > 0. Then there are sequences of measurable sets Ei ; Fi 2 A such that Ei \ Fi D ;;

Ei [ Fi D Ai ;

.Ei /

.Fi / > jj.Ai /

for all i 2 N. The sets E WD

1 [ iD1

Ei

and

F WD

1 [ i D1

Fi

" 2i

5.3. Signed measures

199

satisfy (5.22) and so jj.A/  .E/

.F / D

1 X

..Ei /

.Fi // >

jj.A/ >

1 X

jj.Ai /

":

i D1

iD1

Hence

1 X

jj.Ai /

"

for all " > 0.

iD1

Thus jj.A/ 

1 X

jj.Ai /

iD1

and so jj.A/ D

1 X

jj.Ai /

iD1

by (5.23). This shows that jj is a measure. It remains to prove that jj.X/ < 1. Suppose, by contradiction, that jj.X/ D 1: Claim. Let A 2 A such that jj.X n A/ D 1. Then there exists a measurable set B 2 A such that A  B, j.B n A/j  1, and jj.X n B/ D 1. There exist measurable sets E; F such that E \ F D ;, E [ F D X n A, and .E/

.F /  2 C j.X n A/j;

.E/ C .F / D .X n A/: Take the sum, respectively the difference, of these (in)equalities to obtain 2.E/  2 C j.X n A/j C .X n A/  2; 2.F /  .X n A/

2

j.X n A/j 

2;

and hence j.E/j  1 and j.F /j  1. Since jj.E/ C jj.F / D jj.X n A/ D 1, it follows that jj.E/ D 1 or jj.F / D 1. If jj.E/ D 1 choose B WD A [ F , and if jj.F / D 1 choose B WD A [ E. This proves the claim. It follows from the claim by induction that there exists a sequence of measurable sets ; WD A0  A1  A2 


such that j.An n An 1 /j  1 for all n 2 N. Hence En WD An n An 1 is a sequence of pairwise disjoint measurable sets such that P1 nD1 j.En /j D 1, in contradiction to Definition 5.10. This contradiction shows that the assumption that jj.X/ D 1 was wrong. Hence jj.X/ < 1 and thus jj.A/ < 1 for all A 2 A. This proves Theorem 5.12. 

200

5. The Radon–Nikodým theorem

Definition 5.13. Let W A ! R be a signed measure and let jjW A ! Œ0; 1/ the measure in Theorem 5.12. The Jordan decomposition of  is the representation of  as the difference of two measures ˙ whose sum is equal to jj. The measures ˙ W A ! Œ0; 1/ are defined by ˙ .A/ WD

jj.A/ ˙ .A/ D sup¹˙.E/ j E 2 A; E  Aº 2

(5.24)

for A 2 A and they satisfy C

 D ;

C C  D jj:

(5.25)

Exercise 5.14. Let .X; A; / be a measure space, let f 2 L1 ./, and define Z .A/ WD f d for A 2 A. A

Prove that  is a signed measure and Z Z jj.A/ D jf j d; ˙ .A/ D f ˙ d for all A 2 A: A

(5.26)

A

Definition 5.15. Let W A ! Œ0; 1 be a measure and let ; 1 ; 2 W A ! R be signed measures. (i)  is called absolutely continuous with respect to  (notation “  ”) if .E/ D 0 implies .E/ D 0 for all E 2 A. (ii)  is called concentrated on A 2 A if .E/ D .E \ A/ for all E 2 A. (iii)  is called singular with respect to  (notation “ ? ”) if there exists a measurable set A such that .A/ D 0 and  is concentrated on A. (iv) 1 and 2 are called mutually singular (notation “1 ? 2 ”) if there are measurable sets A1 ; A2 such that A1 \ A2 D ;, A1 [ A2 D X, and i is concentrated on Ai for i D 1; 2. Lemma 5.16. Let  be a measure on A and let ; 1 ; 2 be signed measures on A. Then the following holds: (i)    if and only if jj  ; (ii)  ?  if and only if jj ? ; (iii) 1 ? 2 if and only if j1 j ? j2 j.

5.3. Signed measures

201

Proof. The proof has four steps. Step 1. Let A 2 A. Then jj.A/ D 0 if and only if .E/ D 0 for all measurable sets E  A. If jj.A/ D 0, then j.E/j  jj.E/  jj.A/ D 0 for all measurable sets E  A. The converse implication follows directly from the definition. Step 2.  is concentrated on A 2 A if and only if jj.X n A/ D 0. The signed measure  is concentrated on A if and only if .E/ D .E \ A/ for all E 2 A, or equivalently .E n A/ D 0 for all E 2 A. By Step 1, this holds if and only if jj.X n A/ D 0. Step 3. We prove (i). Assume   . If E 2 A satisfies .E/ D 0, then every measurable set F 2 A with F  E satisfies .F / D 0 and so .F / D 0; hence jj.E/ D 0 by Step 1. Thus jj  . The converse follows from the fact that   jj. Step 4. We prove (ii) and (iii).  ?  if and only if there is a measurable sets A 2 A such that .A/ D 0 and  is concentrated on A. By Step 2, the latter holds if and only if jj.X n A/ D 0 or, equivalently, jj ? . This proves (ii). Assertion (iii) follows from Step 2 by the same argument and this proves Lemma 5.16.  Theorem 5.17 (Lebesgue decomposition). Let .X; A; / be a -finite measure space and let W A ! R be a signed measure. Then there exists a unique pair of signed measures a ; s W A ! R such that  D a C s ;

a  ;

s ? :

(5.27)

Proof. We prove existence. Let ˙ W A ! Œ0; 1/ be the measures defined by (5.24). ˙ By Theorem 5.3, there exist measures ˙ a W A ! Œ0; 1/ and s W A ! Œ0; 1/ such ˙ ˙ ˙ ˙ ˙ that a  , s ? , and  D a C s . Hence the signed measures a WD C a satisfy (5.27).

a

and

s WD C s

s

202

5. The Radon–Nikodým theorem

We prove uniqueness. Assume  D a C s D 0a C 0s where a ; s ; 0a ; 0s are signed measures on A such that a ; 0a   and s ; 0s ? . Then ja j; j0a j   and js j; j0s j ?  by Lemma 5.16. This implies ja jCj0a j   and js jCj0s j ?  by parts (i) and (ii) of Lemma 5.2. Moreover, ja

0a j  ja j C j0a j;

j0a

a j D js

0s j  js j C j0s j:

Hence ja 0a j   and ja 0a j ?  by part (iii) of Lemma 5.2. Thus, by part (iv) of Lemma 5.2, ja 0a j D 0, and therefore a D 0a and s D 0s . This proves Theorem 5.17.  Theorem 5.18 (Radon–Nikodým). Let .X; A; / be a  -finite measure space and let W A ! R be a signed measure. Then    if and only if there exists a -integrable function f W X ! R such that Z .A/ D f d for all A 2 A: (5.28) A

f is determined uniquely by (5.28) up to equality -almost everywhere. Proof. If  is given by (5.28) for some f 2 L1 ./, then    by part (vi) of Theorem 1.44. Conversely, assume    and let jj; C ;  W A ! Œ0; 1/ be the measures defined by (5.21) and (5.24). Then jj   by part (i) of Lemma 5.16, and so ˙  . Hence it follows from TheoremR 5.4 that there exist -integrable functions f ˙ W A ! Œ0; 1/ such that ˙ .A/ D A f ˙ d for all A 2 A. Hence the function f WD f C f 2 L1 ./ satisfies (5.28). The uniqueness of f , up to equality -almost everywhere, follows from Lemma 1.49. This proves Theorem 5.18.  Theorem 5.19 (Hahn decomposition). Let W A ! R be a signed measure. Then there exists a measurable set P 2 A such that .A \ P /  0;

.A n P /  0

for all A 2 A:

Moreover, there exists a measurable function hW X ! ¹1; 1º such that Z .A/ D h d jj for all A 2 A:

(5.29)

(5.30)

A

Proof. By Theorem 5.12, the function  WD jjW A ! Œ0; 1/ in (5.21) is a finite measure and satisfies j.A/j  .A/ for all A 2 A. Hence    and it follows from Theorem 5.18 that there exists a function h 2 L1 ./ such that (5.30) holds.

5.3. Signed measures

203

We prove that h.x/ 2 ¹1; 1º for -almost every x 2 X. To see this, fix a real number 0 < r < 1 and set Ar WD ¹x 2 X j jh.x/j  rº: If E; F 2 A such that E \ F D ; and E [ F D Ar , then Z Z .E/ .F / D h d h d E

F

Z

Z



jhj d C E

jhj d F

 r.Ar /: Take the supremum over all pairs E; F 2 A such that E \ F D ; and E [ F D Ar to obtain .Ar /  r.Ar /, and hence .Ar / D 0. Since this holds for all r < 1, it follows that jhj  1 -almost everywhere. Modifying h on a set of measure zero, if necessary, we may assume without loss of generality that jh.x/j  1 for all x 2 X. Define P WD ¹x 2 X j h.x/  1º; N WD ¹x 2 X j h.x/ 

1º:

Then P \ N D ;, P [ N D X, and Z .P /  h d D .P /  .P /; P

Z .N /  .N / D

h d 

.N /:

N

Hence Z .h

1/ d D .P /

.P / D 0;

P

Z .h C 1/ d D .N / C .N / D 0: N

By Lemma 1.49 this implies h D 1 -almost everywhere on P and h D 1 -almost everywhere on N . Modify h again on a set of measure zero, if necessary, to obtain h.x/ D 1 for all x 2 P and h.x/ D 1 for all x 2 N . This proves Theorem 5.19. 

204

5. The Radon–Nikodým theorem

Theorem 5.20 (Jordan decomposition). Let .X; A/ be a measurable space, let W A ! R be a signed measure, and let ˙ W A ! Œ0; 1/ be finite measures such that  D C  . Then the following are equivalent: (i) C C  D jj; (ii) C ?  ; (iii) there exists a measurable set P 2 A such that C .A/ D .A \ P / and  .A/ D .A n P / for all A 2 A. Moreover, for every signed measure , there is a unique pair of measures ˙ satisfying  D C  and these equivalent conditions. Proof. (i) H) (ii). By Theorem 5.19, there exists a measurable function hW X ! ¹˙1º such that

Z .A/ D

h d jj A

for all A 2 A. Set P WD ¹x 2 X j h.x/ D 1º: Then it follows from (i) that Z jj.P c / C .P c / 1Ch D d jj D 0; 2 2 Pc Z 1 h jj.P / .P /  .P / D D d jj D 0: 2 2 P

C .P c / D

Hence C ?  . (ii) H) (iii). By (ii), there exists a measurable set P 2 A such that C .P c / D 0 and

 .P / D 0:

Therefore, C .A/ D C .A \ P / D C .A \ P /  .A/ D  .A n P / D  .A n P / for all A 2 A.

 .A \ P / D .A \ P /; C .A n P / D

.A n P /

5.4. Radon–Nikodým generalized

205

(iii) H) (i). Assume (iii) and fix a set A 2 A. Then C .A/ C  .A/ D .A \ P /

.A n P /  jj.A/:

Now choose E; F 2 A such that E \ F D ; and E [ F D A. Then .E/

.F / D .E \ P / C .E n P /

.F \ P /

.F n P /

 .E \ P /

.E n P / C .F \ P /

.F n P /

D .A \ P /

.A n P /

D C .A/ C  .A/: Take the supremum over all such pairs E; F 2 A to obtain the inequality jj.A/  C .A/ C  .A/ for all A 2 A and hence jj D C C  : Thus we have proved that assertions (i), (ii), and (iii) are equivalent. Existence and uniqueness of ˙ now follows from (i) with ˙ D 21 .jj ˙ /. This proves Theorem 5.20. 

5.4 Radon–Nikodým generalized This section discusses an extension of the Radon–Nikodým theorem (Theorem 5.18) for signed measures to all measure spaces. Thus we drop the hypothesis that  is -finite. In this case Examples 5.5 and 5.6 show that absolute continuity of  with respect to  is not sufficient for obtaining the conclusion of the Radon–Nikodým theorem and a stronger condition is needed. In [4, Theorem 232B] Fremlin introduces the notion “truly continuous”, which is equivalent to “absolutely continuous” whenever  is  -finite. In [8] König reformulates Fremlin’s criterion in terms of “inner regularity of  with respect to ”. We shall discuss both conditions below, show that they are equivalent, and prove the generalized Radon–Nikodým theorem. As a warmup we rephrase absolute continuity in the familiar "-ı language of analysis. Standing assumption. Throughout this section .X; A; / is a measure space and W A ! R is a signed measure.

206

5. The Radon–Nikodým theorem

Lemma 5.21 (absolute continuity). The following are equivalent. (i)  is absolutely continuous with respect to . (ii) For every " > 0 there exists a constant ı > 0 such that A 2 A;

.A/ < ı H) j.A/j < ":

Proof. That (ii) implies (i) is obvious. Conversely, assume (i). Then jj   by Lemma 5.16. Assume by contradiction that (ii) does not hold. Then there exist a constant " > 0 and a sequence of measurable sets Ai 2 A such that .Ai /  2 i ; For n 2 N define Bn WD

j.Ai /j  " for all i 2 N: 1 [

Ai ;

B WD

1 \

Bn :

nD1

iDn

Then Bn  BnC1 ;

.Bn / 

1 2n 1

;

jj.Bn /  jj.An /  j.An /j  "

for all n 2 N. Hence .B/ D 0 and jj.B/ D limn!1 jj.Bn /  " by part (v) of Theorem 1.28. This contradicts the fact that jj   and shows that our assumption that (ii) does not hold was wrong. Thus (i) implies (ii) and this proves Lemma 5.21.  Definition 5.22. The signed measure  is called truly continuous with respect to  if, for every " > 0, there exist a constant ı > 0 and a measurable set E 2 A such that .E/ < 1 and A 2 A;

.A \ E/ < ı H) j.A/j < ":

(5.31)

Lemma 5.23. The following are equivalent: (i)  is truly continuous with respect to ; (ii) jj is truly continuous with respect to ; (iii) C and  are truly continuous with respect to . Proof. Assume (i), fix a constant " > 0, and choose ı > 0 and E 2 A such that .E/ < 1 and (5.31) holds. Let A 2 A such that .A \ E/ < ı. Then .B/ .A n B/ < 2" for every measurable set B  A, and hence jj.A/  2" by Theorem 5.12. This shows that (i) H) (ii). That (ii) H) (iii) and (iii) H) (i) follows directly from the definitions. 

5.4. Radon–Nikodým generalized

207

Definition 5.22 is due to Fremlin [4, Chapter 23]. If the measure space .X; A; / is -finite, then  is truly continuous with respect to  if and only if it is absolutely continuous with respect to  by Theorem 5.26 below. However, for general measure spaces the condition of true continuity is stronger than absolute continuity. The reader may verify that, when .X; A; / and  are as in part (i) of Example 5.5 or Example 5.6, the finite measure  is absolutely continuous with respect to , but is not truly continuous with respect to . Fremlin’s condition was reformulated by König [8] in terms of inner regularity of  with respect to . This notion can be defined in several equivalent ways. To formulate the conditions it is convenient to introduce the notation E WD ¹E 2 A j .E/ < 1º: Lemma 5.24. The following are equivalent: (i) for all A 2 A, .A \ E/ D 0 for all E 2 E H) .A/ D 0; (ii) for all A 2 A, jj.A \ E/ D 0 for all E 2 E H) jj.A/ D 0; (iii) for all A 2 A, jj.A/ D sup jj.A \ E/ D sup jj.E/: E 2E

E 2E E A

Proof. By Theorem 5.19, there exists a set P 2 A such that .A \ P /  0;

.A n P /  0;

jj.P / D .A \ P /

.A n P /

(5.32)

for all A 2 A. Such a measurable set P will be fixed throughout the proof. We prove that (i) implies (ii). Fix a set A 2 A such that jj.A \ E/ D 0 for all E 2 E. Then it follows from (5.32) that .A \ E \ P / D .A \ E n P / D 0 for all E 2 E. By (i), this implies .A \ P / D .A n P / D 0 and hence jj.A/ D 0 by (5.32). This shows that (i) implies (ii). We prove that (ii) implies (i). Fix a set A 2 A such that .A \ E/ D 0 for all E 2 E. Since E \ P 2 E and E n P 2 E for all E 2 E, this implies .A \ E \ P / D .A \ E n P / D 0 for all E 2 E. Hence it follows from (5.32) that jj.A \ E/ D 0 for all E 2 E. By (ii), this implies jj.A/ D 0 and hence .A/ D 0 because j.A/j  jj.A/. This shows that (ii) implies (i).

208

5. The Radon–Nikodým theorem

We prove that (ii) implies (iii). Fix a set A 2 A and define c WD sup jj.E/  jj.A/:

(5.33)

E 2E E A

Choose a sequence Ei 2 E such that Ei  A for all i and lim jj.Ei / D c:

i!1

For i 2 N define Fi WD E1 [ E2 [


[ Ei : Then

Fi 2 E;

Fi  Fi C1  A;

jj.Ei /  jj.Fi /  c

(5.34)

for all i , and hence lim jj.Fi / D c:

(5.35)

i !1

Put B WD A n F;

F WD

1 [

Fi :

(5.36)

iD1

Then jj.F / D lim jj.Fi / D c i!1

by part (iv) of Theorem 1.28 and hence jj.B/ D jj.A/

jj.F / D jj.A/

c:

(5.37)

Let E 2 E such that E  B. Then E \ Fi D ;, E [ Fi 2 E, and E [ Fi  A for all i by (5.36). Hence jj.E/ C jj.Fi / D jj.E [ Fi /  c for all i by (5.33). This implies jj.E/  lim .c i!1

jj.Fi // D 0

by (5.35). Hence jj.E/ D 0 for all E 2 E with E  B and it follows from (ii) that jj.B/ D 0. Therefore, (5.37) yields jj.A/ D c. This shows that (ii) H) (iii). That (iii) H) (ii) is obvious and this proves Lemma 5.24.  Definition 5.25. The signed measure  is called inner regular with respect to  if it satisfies the equivalent conditions of Lemma 5.24.

5.4. Radon–Nikodým generalized

209

Theorem 5.26 (generalized Radon–Nikodým theorem). Let .X; A; / be a measure space and let W A ! R be a signed measure. Then the following are equivalent: (i)  is truly continuous with respect to ; (ii)  is absolutely continuous and inner regular with respect to ; (iii) there exists a function f 2 L1 ./ such that (5.28) holds. If these equivalent conditions are satisfied, then the function f in (iii) is uniquely determined by (5.28) up to equality -almost everywhere. First proof of Theorem 5.26. This proof is due to König [8]. It has the advantage that it reduces the proof of the generalized Radon–Nikodým theorem to the standard Radon–Nikodým theorem (Theorem 5.18) for  -finite measure spaces. (i) H) (ii). To see that  is absolutely continuous with respect to , fix a measurable set A 2 A such that .A/ D 0 and fix a constant " > 0. Choose ı > 0 and E 2 A such that .E/ < 1 and (5.31) holds. Then we obtain .A \ E/  .A/ D 0 < ı, and hence j.A/j < " by (5.31). Thus j.A/j < " for all " > 0, and hence .A/ D 0. This shows that   . We prove that  is inner regular with respect to  by verifying that  satisfies condition (i) in Lemma 5.24. Fix a set A 2 A such that E 2 A;

.E/ < 1 H) .A \ E/ D 0:

We must prove that .A/ D 0. Let " > 0 and choose ı > 0 and E 2 A such that .E/ < 1 and (5.31) holds. Then ..A n E/ \ E/ D 0 < ı, hence j.A n E/j < " by (5.31), so j.A/j D j.A n E/ C .A \ E/j D j.A n E/j < ": This shows that j.A/j < " for all " > 0 and so .A/ D 0, as claimed. Thus we have proved that (i) H) (ii). (ii) H) (iii). Since  is inner regular with respect to , there exists a sequence of measurable sets Ei 2 A such that Ei  Ei C1 and .Ei / < 1 for all i 2 N and jj.X/ D lim jj.Ei /. Define i !1

X0 WD

1 [

Ei ;

A0 WD ¹A 2 A j A  X0 º;

0 WD jA0 ;

0 WD jA0 :

i D1

Then .X0 ; A0 ; 0 / is a -finite measure space and 0 W A0 ! R is a signed measure that is absolutely continuous with respect to 0 . Hence the Radon–Nikodým theorem (Theorem 5.18) for  -finite measure spaces asserts that there exists a function

210

5. The Radon–Nikodým theorem

f0 2 L1 .0 / such that Z 0 .A/ D

for all A 2 A0 :

f0 d0 A

Define f WX ! R by f jX0 WD f0

f jX nX0 WD 0:

and

Then f 2 L ./. Choose a measurable set A 2 A. Then it follows from part (v) of Theorem 1.28 that 1

j.A n X0 /j  jj.A n X0 /  jj.X n X0 / D lim jj.X n Ei / D 0: i !1

Hence

Z

Z

.A/ D 0 .A \ X0 / D

f0 d0 D A\X0

f d A

for all A 2 A. This shows that (i) H) (iii). The uniqueness of f up to equality -almost everywhere follows immediately from Lemma 1.49. (iii) H) (i). Choose f 2 L1 ./ such that (5.28) holds. Set Z c WD jj.X/ D jf j d X

and  jf .x/j  2n º; [ WD ¹x 2 X j f .x/ ¤ 0º D En :

En WD ¹x 2 X j 2 E1

n

n2N n

Then 2 .En /  jj.En /  c and hence .En /  2n c < 1 for all n 2 N. Moreover, c D jj.X/ D jj.E1 / D lim jj.En /: n!1

Now fix a constant " > 0. Choose n 2 N such that jj.En / > c ı WD 2

n 1

"=2 and put

":

If A 2 A such that .A \ En / < ı, then jj.A/ D jj.A n En / C jj.A \ En /  jj.X n En / C 2n .A \ En / " < C 2n ı D ": 2 This shows that  is truly continuous with respect to . This completes the first proof of Theorem 5.26. 

5.4. Radon–Nikodým generalized

211

Second proof of Theorem 5.26. This proof is due to Fremlin [4, Chapter 23]. It shows directly that (i) implies (iii) and has the advantage that it only uses the Hahn decomposition theorem (Theorem 5.19). It thus also provides an alternative proof of Theorem 5.18 (assuming the Hahn decomposition theorem) which is of interest on its own. By Lemma 5.23, it suffices to consider the case where W A ! Œ0; 1/ is a finite measure that is truly continuous with respect to . Consider the set ˇ ³ ² ˇ f is measurable and : F WD f W X ! Œ0; 1/ ˇˇ R A f d  .A/ for all A 2 A This set is nonempty because 0 2 F. Moreover, Z f d  .X/ < 1 for all f 2 F, X

and1 f; g 2 F H) max¹f; gº 2 F: Now put

Z c WD sup

f d  .X/

f 2F X

and choose a sequence gi 2 F such that Z lim gi d D c: i !1 X

Then it follows from (5.38) that fi WD max¹g1 ; g2 ; : : : ; gi º 2 F and

Z

Z gi d 

X 1

fi d  c

for all i 2 N.

X

Let f; g 2 F and A 2 A and define the sets Af WD ¹x 2 A j f .x/ > g.x/º

and Ag WD ¹x 2 A j g.x/  f .x/ºI then Af ; Ag 2 A, Af \ Ag D ;, and Af [ Ag D A; hence Z Z Z max¹f; gº d D f d C g d  .Af / C .Ag / D .A/: A

Af

Ag

(5.38)

212

5. The Radon–Nikodým theorem

Hence fi  fiC1 for all i and Z lim

i!1 X

fi d D c:

Define the function f W X ! Œ0; 1 by f .x/ WD lim fi .x/ for x 2 X . i!1

Then it follows from the Lebesgue monotone convergence Theorem 1.37 that Z Z f d D lim fi d D c; i !1 X

X

and

Z

Z f d D lim

i !1 A

A

fi d  .A/

for all A 2 A:

Hence f < 1 -almost everywhere by Lemma 1.47 and we may assume without loss of generality that f .x/ < 1 for all x 2 X . Thus f 2 F. We claim that Z f d D .A/ for all A 2 A. A

Suppose on the contrary that there exists a set A0 2 A such that Z f d < .A0 /: A0

Then the formula 0 .A/ WD .A/

Z

f d for A 2 A

(5.39)

A

defines a finite measure by Theorem 1.40. We prove that there is a measurable function hW X ! Œ0; 1/ such that Z Z h d > 0; h d  0 .A/ for all A 2 A: X

(5.40)

A

Define

0 .A0 / > 0: (5.41) 3 Since  is truly continuous with respect to , so is 0 . Hence there are a ı > 0 and a set E 2 A such that .E/ < 1 and " WD

A 2 A;

.A \ E/ < ı H) 0 .A/ < ":

(5.42)

5.4. Radon–Nikodým generalized

213

Take A WD X n E to obtain 0 .X n E/ < " and hence 0 .E/  0 .A0 \ E/ D 0 .A0 /

0 .A0 n E/ D 3"

0 .A0 n E/ > 2":

Then take A WD A0 . Since 0 .A0 / D 3"  " by (5.41), it follows from (5.42) that .E/  .A0 \ E/  ı > 0: Define the signed measure 00 W A ! R by 00 .A/ WD 0 .A/

"

.A \ E/ .E/

for A 2 A.

(5.43)

Then 00 .E/ D 0 .E/

"  ":

By the Hahn decomposition theorem (Theorem 5.19), there exists a measurable set P 2 A such that 00 .A \ P /  0;

00 .A n P /  0

for all A 2 A:

Since 00 .E n P /  0, it follows that "  00 .E/  00 .E \ P /  0 .E \ P /: Hence .E \ P /  ı by (5.42). Now put h WD Then

"  : .E/ E \P

Z h d > 0: X

Moreover, if A 2 A, then 00 .A \ P /  0 and so, by (5.43), Z .A \ P \ E/ 0  .A \ P /  " D h d: .E/ A Thus

Z

h d  0 .A/

A

and so h satisfies (5.40), as claimed.

for all A 2 A,

(5.44)

214

5. The Radon–Nikodým theorem

It follows from (5.40) that Z

Z .f C h/ d 

A

f d C 0 .A/ D .A/

A

for all A 2 A and hence f C h 2 F. Since Z

Z .f C h/ d D c C

X

h d > c; X

R this contradicts the definition of c. Thus we have proved that A f d D .A/ for all A 2 A, and hence f satisfies (5.28). This completes the second proof of Theorem 5.26. 

5.5 Exercises Exercise 5.27. Let .X; A; / be a measure space such that .X/ < 1. Define .A; B/ WD .A n B/ C .B n A/ for A; B 2 A:

(5.45)

Define an equivalence relation on A by A  B if .A; B/ D 0. Prove that  descends to a function W A=  A= ! Œ0; 1/ (denoted by the same letter) and that the pair .A=; / is a complete metric space. Prove that the function Z A ! RW A 7 ! f d A

descends to a continuous function on A= for every f 2 L1 ./. Exercise 5.28 (Rudin [17, page 133]). Let .X; A; / be a measure space. A subset F  L1 ./ is called uniformly integrable if, for every " > 0, there is a constant ı > 0 such that, for all E 2 A and all f 2 F, ˇ ˇZ ˇ ˇ ˇ .E/ < ı H) ˇ f dˇˇ < ": E

5.5. Exercises

215

Prove the following. (i) Every finite subset of L1 ./ is uniformly integrable. Hint. Lemma 5.21. (ii) Vitali’s theorem. Assume .X/ < 1, let f W X ! R be measurable, and let fn 2 L1 ./ be a uniformly integrable sequence that converges almost everywhere to f . Then f 2 L1 ./ and Z lim jf fn j d D 0: n!1 X

Hint. Use Egoroff’s theorem in Exercise 4.54. (iii) The hypothesis .X/ < 1 cannot be omitted in Vitali’s theorem. Hint. Consider the Lebesgue measure on R. Find a uniformly integrable sequence fn 2 L1 .R/ that converges pointwise to the constant function f  1. (iv) Vitali’s theorem implies the Lebesgue dominated convergence theorem (Theorem 1.45) under the assumption .X/ < 1. (v) Find an example where Vitali’s theorem applies although the hypotheses of the Lebesgue dominated convergence theorem are not satisfied. (vi) Find an example of a measure space .X; A; / with .X/ < 1 and a sequence fn 2 L1 ./ that is not uniformly integrable, converges pointwise to zero, and satisfies Z lim fn d D 0: n!1 X

Hint. Consider the Lebesgue measure on X D Œ0; 1. (vii) Converse of Vitali’s theorem. Assume R .X/ < 1 and let fn be a sequence in L1 ./ such that the limit lim A fn d exists for all A 2 A. n!1

Then the sequence fn is uniformly integrable. Hint. Let " > 0. Prove that there exist a constant ı > 0, an integer n0 2 N, and a measurable set E0 2 E such that, for all E 2 A and all n 2 N, ˇZ ˇ ˇ ˇ ˇ .E; E0 / < ı; n  n0 H) ˇ .fn fn0 / dˇˇ < ": (5.46) E

(Here .E; E0 / is defined by (5.45) as in Exercise 5.27.)

216

5. The Radon–Nikodým theorem

If A 2 A satisfies .A/ < ı, then the sets E WD E0 n A

and

E WD E0 [ A

both satisfy .E; E0 / < ı. Deduce that, for all A 2 A and all n 2 N, ˇZ ˇ ˇ ˇ ˇ .A/ < ı; n  n0 H) ˇ .fn fn0 / dˇˇ < 2": (5.47) A

Now use part (i) to find a constant ı 0 > 0 such that, for all A 2 A, ˇZ ˇ ˇ ˇ 0 ˇ .A/ < ı H) sup ˇ fn dˇˇ < 3": n2N

(5.48)

A

Exercise 5.29 (Rudin [17, page 134]). Let .X; A; / be a measure space such that .X/ < 1 and fix a real number p > 1. Let f W X ! R be a measurable function and let fn 2 L1 ./ be a sequence that converges pointwise to f and satisfies Z sup jfn jp d < 1: n2N X

Prove that f 2 L1 ./;

Z lim

n!1 X

jf

fn j d D 0:

Hint. Use Vitali’s theorem in Exercise 5.28. Exercise 5.30. Let X WD R, denote by B  2X the Borel -algebra, and let W B ! Œ0; 1 be the restriction of the Lebesgue measure to B. Let W B ! Œ0; 1 be a measure. Prove the following. (1) If B 2 B and 0 < c < .B/, then there exists a Borel set A  B such that .A/ D c. Hint. Show that the function f .t/ WD .B \Œ t; t/ is continuous. (2) If there exists a constant 0 < c < 1 such that .B/ D c H) .B/ D c: for all B 2 B, then   . Exercise 5.31. Let X WD R, denote by B  2X the Borel  -algebra, let W B ! Œ0; 1 be the restriction of the Lebesgue measure to B, and let W B ! Œ0; 1 be the counting measure.

5.5. Exercises

217

Prove the following. (i)   . (ii)  is not inner regular with respect to . (iii) There does not exist any measurable function f W X ! Œ0; 1 such that Z .B/ D f d for all B 2 B. B

Exercise 5.32. Let X WD Œ1; 1/, denote by B  2X the Borel -algebra, and let W B ! Œ0; 1 be the restriction of the Lebesgue measure to B. Let W B ! Œ0; 1 be a Borel measure such that .B/ D ˛.˛B/ for all ˛  1 and all B 2 B: Prove that there exists a real number c  0 such that Z .B/ WD f d for all B 2 B;

(5.49)

(5.50)

B

where f W Œ1; 1/ ! Œ0; 1/ is the function given by f .x/ WD

c x2

for x  1:

(5.51)

Hint. Show that .Œ1; 1// < 1, and then that   . Exercise 5.33. Let X WD Œ0; 1/, denote by B  2X the Borel -algebra, and let W B ! Œ0; 1 be the restriction of the Lebesgue measure to B. Define the measures 1 ; 2 W B ! Œ0; 1 by2 Z 1 X 1 1 .B/ WD x dx; n3 B\Œn;nC1 nD1 Z 2 .B/ WD B\Œ1;1/

1 dx x2

for B 2 B. 2

Here we denote by

Z B

Z f .x/ dx WD

f d B

the Lebesgue integral of a Borel measurable function f W Œ0; 1/ ! Œ0; 1/ over a Borel set B 2 B.

218

5. The Radon–Nikodým theorem

Prove that 1 and 2 are finite measures that satisfy 1  ;

2  ;

1  2 ;

2  1 ;

and  6 1 ;

 6 2 :

Exercise 5.34. Let .X; A; / be a measure space. Show that the signed measures W A ! R form a Banach space M D M.X; A/ with the norm kk WD jj.X/: Show that the map

L1 ./ ! MW Œf  7 ! f

defined by (5.53) is an isometric linear embedding and hence L1 ./ is a closed subspace of M. Exercise 5.35. Let .X; U/ be a compact Hausdorff space such that every open subset of X is -compact, and denote by B  2X its Borel -algebra. Denote by C.X/ WD Cc .X/ the space of continuous real-valued functions on X . This is a Banach space equipped with the supremum norm kf k WD sup jf .x/j: x2X

Let M.X/ denote the space of signed Borel measures as in Exercise 5.34. For  2 M.X/ define the linear functional ƒ W C.X/ ! R by

Z ƒ .f / WD

f d: X

Prove the following. (i) kƒ k D kk. Hint. Use the Hahn decomposition theorem (Theorem 5.19) and the fact that every Borel measure on X is regular by Theorem 3.18. (ii) Every bounded linear functional on C.X/ is the difference of two positive linear functionals. Hint. For f 2 C.X/ with f  0 prove that ƒC .f / WD sup¹ƒ.hf / j h 2 C.X/; 0  h  1º D sup¹ƒ.g/ j g 2 C.X/; 0  g  f º:

(5.52)

5.5. Exercises

219

Here the second supremum is obviously greater than or equal to the first. To prove the converse inequality, show that, for all g 2 C.X/ with 0  g  f and all " > 0 there is an h 2 C.X/ such that 0  h  1 and jƒ.g hf /j < ". Namely, find  2 C.X/ such that 0    1, .x/ D 0 when f .x/  "=2 kƒk and .x/ D 1 when f .x/  "= kƒk; then define h WD g=f . Once (5.52) is established, show that ƒC extends to a positive linear functional on C.X/. (iii) The map M.X/ ! C.X/ W  7! ƒ is bijective. Hint. Use the Riesz representation theorem (Theorem 3.15). (iv) The hypothesis that every open subset of X is -compact cannot be removed in part (i). Hint. Consider Example 3.6. Exercise 5.36. Let .X; A; / be a measure space and let f W X ! Œ0; 1/ be a measurable function. Define the measure f W A ! Œ0; 1 by Z f .A/ WD f d for A 2 A: (5.53) A

(See Theorem 1.40.) Prove the following. (i) If  is -finite, then so is f . (ii) If  is semi-finite, then so is f . (iii) If  is localizable, then so is f . Note. See Theorem 5.4 for (i) and [4, Proposition 234N] for (ii) and (iii). It is essential that f does not take on the value 1. Find an example of a measure space .X; A; / and a measurable function f W X ! Œ0; 1 that violates assertions (i), (ii), and (iii). Hint 1. To prove (ii), fix a set A 2 A, define Af WD ¹x 2 A j f .x/ > 0º, and choose a measurable set E 2 A such that E  Af and 0 < .E/ < 1. Consider the sets En WD ¹x 2 E j f .x/  nº. Hint 2. To prove (iii), let E  A be any collection of measurable sets and choose a measurable -envelope H 2 A of E. Prove that the set Hf WD ¹x 2 H j f .x/ > 0º is a measurable f -envelope of E. In particular, if N 2 A is a measurable set such that f .E \ N / D 0 for all E 2 E, define Nf WD ¹x 2 N j f .x/ > 0º, show that .H \ Nf / D 0, and deduce that f .Hf \ N / D f .H \ Nf / D 0.

Chapter 6

Differentiation

This chapter returns to the Lebesgue measure on Euclidean space Rn introduced in Chapter 2. The main result is the Lebesgue differentiation theorem (Section 6.3). It implies that if f W Rn ! R is a Lebesgue integrable function, then for almost every element x 2 Rn , the mean value of f over a ball centered at x converges to f .x/ as the radius tends to zero. Essential ingredients in the proof are the Vitali covering lemma and the Hardy–Littlewood maximal inequality (Section 6.2). One of the consequences of the Lebesgue differentiation theorem is the fundamental theorem of calculus for absolutely continuous functions of one real variable (Section 6.4). The Lebesgue differentiation theorem also plays a central role in the proof of the Calderón–Zygmund inequality (Section 7.7). The chapter begins with a discussion of weakly integrable functions on general measure spaces.

6.1 Weakly integrable functions Assume throughout that .X; A; / is a measure space. Let f W X ! R be a measurable function. Define f W Œ0; 1/ ! Œ0; 1 by f .t/ WD .t; f / WD .A.t; f //;

A.t; f / WD ¹x 2 X j jf .x/j > tº;

(6.1)

for t  0. The function f is nonincreasing and hence Borel measurable. Define the function f  W Œ0; 1/ ! Œ0; 1 by f  .˛/ WD inf¹t  0 j .t; f /  ˛º

for 0  ˛ < 1:

(6.2)

Thus f  .0/ D kf k1 and f  is nonincreasing and hence Borel measurable. By definition, the infimum of the empty set is infinity. Thus f  .˛/ D 1 if and only if .A.t; f // > ˛ for all t > 0. When f  .˛/ < 1, it is the smallest number t such that the domain A.t; f / (on which jf j > t ) has measure at most ˛. This is spelled out in the next lemma.

222

6. Differentiation

Lemma 6.1. Let 0  ˛ < 1 and 0  t < 1. Then the following holds: (i) f  .˛/ D 1 if and only if f .s/ > ˛ for all s  0; (ii) f  .˛/ D t if and only if f .t/  ˛ and f .s/ > ˛ for 0  s < t; (iii) f  .˛/  t if and only if f .t/  ˛. Proof. It follows directly from the definition of f  in (6.2) that f  .˛/ D 1 if and only if .s; f / > ˛ for all s 2 Œ0; 1/, and this proves (i). To prove (ii), fix a constant 0  t < 1. Assume first that .t; f /  ˛

and .s; f / > ˛

for 0  s < t.

Since f is nonincreasing, this implies .s; f /  .t; f /  ˛

for all s  t ,

and hence f  .˛/ D t by definition. Conversely, suppose that f  .˛/ D t. Then it follows from the definition of f  that .s; f /  ˛

for s > t

and .s; t/ > ˛

for 0  s < t.

We must prove that .t; f /  ˛. To see this, observe that A.t; f / D

1 [

A.t C 1=n; f /:

nD1

Hence it follows from part (iv) of Theorem 1.28 that f .t/ D .A.t; f // D lim .A.t C 1=n; f // n!1

D lim .t C 1=n; f / n!1

 ˛: This proves (ii). If f  .˛/  t, then f .t/  f .f  .˛//  ˛ by (ii). If f .t/  ˛, then f .˛/  t by the definition of f  . This proves (iii) and Lemma 6.1.  

6.1. Weakly integrable functions

223

Lemma 6.2. Let f; gW X ! R be measurable functions and let c 2 R. Then kf k1;1 WD sup ˛f  .˛/ D sup tf .t/  kf k1 ;

(6.3)

kcf k1;1 D jcj kf k1;1 ;

(6.4)

˛>0

t>0

kf k1;1 kgk1;1 C for 0 <  < 1;  1  q q q kf C gk1;1  kf k1;1 C kgk1;1 :

kf C gk1;1 

(6.5) (6.6)

Moreover, kf k1;1 D 0 if and only if f vanishes almost everywhere. The inequality (6.6) is called the weak triangle inequality. Proof. For 0 < t; c < 1 it follows from part (iii) of Lemma 6.1 that 1

t.t; f /  c () .t; f /  ct () f  .ct () ct

1

1

/t

f  .ct

1

/  c:

This shows that sup t.t; f / D sup ˛f  .˛/: t>0

˛>0

Moreover, Z t.t; f / D t.A.t; f // 

Z jf j d 

A.t;f /

jf j d

for all t > 0.

X

This proves (6.3). For c > 0 equation (6.4) follows from the fact that A.t; cf / D A.t=c; f / and hence .t; cf / D .t=c; f / for all t > 0. Since k f k1;1 D kf k1;1 by definition, this proves (6.4). To prove (6.5), observe that A.t; f C g/  A.t; f / [ A..1 /t; g/, hence .t; f C g/  .t; f / C ..1

/t; g/;

and so t.t; f C g/  t.t; f / C t..1

/t; g/ 

kf k1;1 kgk1;1 C  1 

for all t > 0. Take the supremum over all t > 0 to obtain (6.5).

(6.7)

224

6. Differentiation

The inequality (6.6) follows from (6.5) and the identity r p p b a C D a C b for a; b  0: inf  1  0 0, for x  0,

is weakly integrable, but not integrable. Theorem 6.5. The metric space .L1;1 ./; d1;1 / is complete. Proof. Choose a sequence of weakly integrable functions fi W X ! R whose equivalence classes Œfi  form a Cauchy sequence in L1;1 ./ with respect to the metric (6.9). Then there is a subsequence i1 < i2 < i3 <




226

6. Differentiation

such that kfik

fikC1 k1;1 < 2

2k

for all k 2 N.

For k; ` 2 N put k

Ak WD A.2

; fik

fikC1 /

and E` WD

1 [

Ak ;

1 \

E WD

kD`

E` :

`D1

Then 2

k

.Ak /  fik

hence .E` / 

fikC1 1;1 < 2

1 X

.Ak / 

kD`

1 X

2

2k

k

for all k 2 N,

D 21

`

kD`

for all ` 2 N, and so .E/ D 0. If x 2 X n E, then there exists an ` 2 N such that x … Ak for all k  ` and so jfik .x/

fikC1 .x/j  2

k

for all k  `.

This shows that the limit f .x/ WD lim fik .x/ k!1

exists for all x 2 X n E. Extend f to a measurable function on X by setting f .x/ WD 0

for x 2 E.

lim kfi

f k1;1 D 0

We prove that i!1

and hence also f 2 L1;1 ./. To see this, fix a constant " > 0 and choose an integer i0 2 N such that

.i; j 2 N; i; j  i0 / H) 4 fi fj 1;1 < ": Now fix a constant t > 0 and choose ` 2 N such that i`  i0 ;

22 ` t  ";

22

`

 t:

6.2. Maximal functions

227

If x … E` , then x … Ak for all k  `, hence jfik .x/

fikC1 .x/j  2

k

for k  `,

and so jfi` .x/

f .x/j 

1 X

jfik .x/

fikC1 .x/j 

tfi

`

f

k

2

D 21

`

 t=2:

kD`

kD`

This shows that A.t=2; fi`

1 X

f /  E` and hence f //  t.E` /  t21

.t=2/ D t.A.t=2; fi`

`

 "=2:

With this understood, it follows from (6.7) with  D 1=2 that tfi

f

.t/  tfi

fi` .t=2/

C tfi

`

f

.t=2/  2kfi

fi` k1;1 C "=2 < "

for all i 2 N with i  i0 . Hence kfi

f k1;1 D sup tfi t>0

f

.t/  "

for every integer i  i0 and this proves Theorem 6.5.



6.2 Maximal functions Let .R; A; m/ be the Lebesgue measure space on R. In particular, the length of an interval I  R is m.I /. As a warmup we characterize the differentiability of a function that is obtained by integrating a signed measure. Theorem 6.6. Let W A ! R be a signed measure and define f WR ! R by f .x/ WD .. 1; x// for x 2 R:

(6.10)

Fix two real numbers x; A 2 R. Then the following are equivalent: (i) f is differentiable at x and f 0 .x/ D A. (ii) For every " > 0 there is a ı > 0 such that, for every open interval U  R, ˇ ˇ ˇ .U / ˇ ˇ x 2 U; m.U / < ı H) ˇ Aˇˇ  ": (6.11) m.U /

228

6. Differentiation

Proof. (i) H) (ii). Fix a constant " > 0. Since f is differentiable at x and we have f 0 .x/ D A, there exists a constant ı > 0 such that, for all y 2 R, ˇ ˇ ˇ f .x/ f .y/ ˇ ˇ 0 < jx yj < ı H) ˇ Aˇˇ  ": (6.12) x y Let a; b 2 R such that a < x < b and b ˇ ˇ ˇ f .x/ f .a/ ˇ ˇ Aˇˇ  "; ˇ x a

a < ı. Then, by (6.12), ˇ ˇ ˇ f .b/ f .x/ ˇ ˇ Aˇˇ  "; ˇ b x

or, equivalently, ".x

a/  f .x/

f .a/

A.x

a/  ".x

a/;

".b

x/  f .b/

f .x/

A.b

x/  ".b

x/:

f .a/

A.b

a/  ".b

a/:

Add these inequalities to obtain a/  f .b/

".b Since .Œa; b// D f .b/

Replace a by a C 2

k

f .a/ and m.Œa; b// D b a it follows that ˇ ˇ ˇ ˇ .Œa; b// ˇ  ": ˇ A ˇ ˇ m.Œa; b//

and take the limit k ! 1 to obtain ˇ ˇ ˇ ..a; b// ˇ ˇ ˇ  ": A ˇ m..a; b// ˇ

Thus we have proved that (i) H) (ii). (ii) H) (i). Fix a constant " > 0. Choose ı > 0 such that (6.11) holds for every open interval U  R. Choose y 2 R such that x < y < x C ı. Choose k 2 N such that y x C 2 k < ı. Then Uk WD .x

2

k

; y/

is an open interval of length m.Uk / < ı containing x and hence ˇ ˇ ˇ .Uk / ˇ ˇ ˇ ˇ m.U / Aˇ  " k by (6.11). Take the limit k ! 1 to obtain ˇ ˇ ˇ ˇ ˇ ˇ ˇ .Œx; y// ˇ ˇ ˇ f .y/ f .x/ ˇDˇ ˇ D lim ˇ .Uk / ˇ A A ˇ ˇ ˇ ˇ ˇ y x m.Œx; y// k!1 m.Uk /

ˇ ˇ Aˇˇ  ":

Thus (6.12) holds for x < y < x C ı and an analogous argument proves the inequality for x ı < y < x. Thus (ii) implies (i) and this proves Theorem 6.6. 

6.2. Maximal functions

229

The main theorem of this chapter will imply that, when  is absolutely continuous with respect to m, the derivative of the function f in (6.10) exists almost everywhere, defines a Lebesgue integrable function f 0 W R ! R, and that Z .A/ D f 0 dm A

for all Lebesgue measurable sets A 2 A. It will then follow that an absolutely continuous function on R can be written as the integral of its derivative. This is the fundamental theorem of calculus in measure theory (Theorem 6.19). The starting point for this program is the assertion of Theorem 6.6. It suggests the definition of the derivative of a signed measure W A ! R at a point x 2 R as the limit of the quotients .U /=m.U / over all open intervals U containing x as m.U / tends to zero, provided that the limit exists. This idea carries over to all dimensions and leads to the concept of a maximal function, which we explain next. Notation. Fix a natural number n 2 N. Let .Rn ; A; m/ denote the Lebesgue measure space and let n B  2R denote the Borel  -algebra of Rn with the standard topology. Thus L1 .Rn / denotes the space of Lebesgue integrable functions f W Rn ! R. An element of L1 .Rn / need not be Borel measurable, but differs from a Borel measurable function on a Lebesgue null set by Theorem 2.14 and part (v) of Theorem 1.55. For x 2 Rn and r > 0, denote the open ball of radius r, centered at x, by Br .x/ WD ¹y 2 Rn j jx Here jj WD

q

yj < rº:

12 C


C n2

denotes the Euclidean norm of  D .1 ; : : : ; n / 2 Rn . Definition 6.7 (Hardy–Littlewood maximal function). Let W B ! Œ0; 1/ be a finite Borel measure. The maximal function of  is the function MW Rn ! R defined by

.Br .x// : r>0 m.Br .x//

.M/.x/ WD sup

(6.13)

230

6. Differentiation

The maximal function of a signed measure W B ! R is defined as the maximal function M  WD M jjW Rn ! R of its total variation jjW B ! Œ0; 1/. Theorem 6.8 (Hardy–Littlewood maximal inequality). Let W B ! R be a signed Borel measure. Then the function M W Rn ! R in Definition 6.7 is lower semicontinuous, i.e., the pre-image of the open interval .t; 1/ under M  is open for all t 2 R. Hence M  is Borel measurable. Moreover, kM k1;1  3n jj.Rn /;

(6.14)

and so M  2 L1;1 .Rn /. 

Proof. See page 232. The proof of Theorem 6.8 relies on the following two lemmas.

Lemma 6.9. Let W B ! Œ0; 1/ be a finite Borel measure. Then the maximal function MW Rn ! R is lower semi-continuous and hence is Borel measurable. Proof. Let t > 0 and define U t WD A.t; M/ D ¹x 2 Rn j .M/.x/ > tº:

(6.15)

We prove that U t is open. Fix an element x 2 U t . Since .M/.x/ > t, there exists a number r > 0 such that .Br .x// : t< m.Br .x// Choose ı > 0 such that t

.r C ı/n .Br .x// < : n r m.Br .x//

Further, choose y 2 Rn such that jy

xj < ı. Then Br .x/  BrCı .y/, and hence

.BrCı .y//  .Br .x// .r C ı/n m.Br .x// rn .r C ı/n Dt m.Br .y// rn >t

D t
m.BrCı .y//:

6.2. Maximal functions

This implies .M/.y/ 

231

.BrCı .y// > t; m.BrCı .y//

and hence y 2 U t . This shows that U t is open for all t > 0. It follows that S n U0 D t >0 U t is open and U t D R is open for t < 0. Thus M is lower semi-continuous, as claimed. This proves Lemma 6.9.  The Hardy–Littlewood estimate on the maximal function M is equivalent to an upper bound for the Lebesgue measure of the set U t in (6.15). The proof relies on the next lemma about coverings by open balls. Lemma 6.10 (Vitali’s covering lemma). Let ` 2 N and, for i D 1; : : : ; `, let xi 2 Rn and ri > 0. Define W WD

` [

Bri .xi /:

iD1

Then there exists a set S  ¹1; : : : ; `º such that Bri .xi / \ Brj .xj / D ; for all i; j 2 S with i ¤ j

(6.16)

and W 

[

B3ri .xi /:

(6.17)

i2S

Proof. Abbreviate Bi WD Bri .xi / and choose the ordering such that r1  r2 


 r` : Choose i1 WD 1 and let i2 > 1 be the smallest index such that Bi2 \ Bi1 D ;. Continue by induction to obtain a sequence 1 D i1 < i2 <


< ik  ` such that Bij \ Bij 0 D ; for j ¤ j 0 and Bi \ .Bi1 [


[ Bij / ¤ ; for ij < i < ij C1 (respectively, for i > ik when j D k).

232

6. Differentiation

Then, Bi  B3ri1 .xi1 / [


[ B3rij .xij / and hence W D

` [

Bi 

iD1

k [

for ij < i < ij C1

B3rij .xij /:

j D1

With S WD ¹i1 ; : : : ; ik º this proves (6.17) and Lemma 6.10.



Proof of Theorem 6.8. Fix a constant t > 0. Then the set U t WD A.t; M / is open by Lemma 6.9. Choose a compact set K  U t . If x 2 K  U t , then .M /.x/ > t and so there exists a number r.x/ > 0 such that jj.Br.x/ .x// > t: m.Br.x/ .x//

(6.18)

Since K is compact, there exist finitely many points x1 ; : : : ; x ` 2 K such that K

` [

Bri .xi /;

iD1

where ri WD r.xi /. By Lemma 6.10, there is a subset S  ¹1; : : : ; `º such that S the balls Bri .xi / for i 2 S are pairwise disjoint and K  i2S B3ri .xi /. Since m.B3r / D 3n m.Br / by Theorem 2.17, this gives m.K/  3n

X i2S

m.Bri .xi //
0 to obtain kM k1;1  3n jj.Rn /. This proves Theorem 6.8. 

6.3. Lebesgue points

233

Definition 6.11. Let f 2 L1 .Rn /. The maximal function of f is the function Mf W Rn ! Œ0; 1/ defined by 1 .Mf /.x/ WD sup m.B r .x// r>0

Z jf j d m

for x 2 Rn :

(6.20)

Br .x/

Corollary 6.12. Let f 2 L1 .Rn / and define the signed Borel measure f on Rn by Z f .B/ WD f dm B

for every Borel set B  Rn . Then Mf D Mf 2 L1;1 .Rn / and kMf k1;1  3n kf k1 : R Proof. The formula jf j.B/ D B jf j d m for B 2 B shows that Mf D Mf . Hence the assertion follows from Theorem 6.8.  Corollary 6.12 shows that the map f 7! Mf descends to an operator (denoted by the same letter) from the Banach space L1 .Rn / to the topological vector space L1;1 .Rn /. Corollary 6.12 also shows that the resulting operator M W L1 .Rn / ! L1;1 .Rn / is continuous (because jMf Mgj  M.f g/). Note that it is not linear. By Theorem 6.8, M extends naturally to an operator  7! M  from the Banach space of signed Borel measures on Rn to L1;1 .Rn /. (See Exercise 5.34.)

6.3 Lebesgue points Definition 6.13. Let f 2 L1 .Rn /. An element x 2 Rn is a called a Lebesgue point of f if Z 1 lim jf f .x/j d m D 0: (6.21) r!0 m.Br .x// Br .x/ In particular, x is a Lebesgue point of f whenever f is continuous at x.

234

6. Differentiation

The next theorem is the main result of this chapter. Theorem 6.14 (Lebesgue differentiation theorem). Let f 2 L1 .Rn /. Then there exists a Borel set E  Rn such that m.E/ D 0 and every element of Rn n E is a Lebesgue point of f . Proof. For f 2 L1 .Rn / and r > 0 define the function Tr f W Rn ! Œ0; 1/ by

1 .Tr f /.x/ WD m.Br .x//

Z jf

f .x/j d m for x 2 Rn :

(6.22)

Br .x/

One can prove via an approximation argument that Tr f is Lebesgue measurable for every r > 0 and every f 2 L1 .Rn /. However, we shall not use this fact in the proof. For f 2 L1 .Rn / define the function Tf W Rn ! Œ0; 1 by

.Tf /.x/ WD lim sup.Tr f /.x/ for x 2 Rn : r!0

(6.23)

We must prove that Tf D 0 almost everywhere for every f 2 L1 .Rn /. To see this, fix a function f 2 L1 .Rn / and assume without loss of generality that f is Borel measurable. (See Theorem 2.14 and part (v) of Theorem 1.55.) By Theorem 4.15, there exists a sequence of continuous functions gi W Rn ! R with compact support such that kf

gi k 1
0. Take the limit superior as r tends to zero to obtain T hi  M hi C jhi j for all i . Moreover, it follows from the definition of Tr that Tr f D Tr .gi C hi /  Tr gi C Tr hi

for all i and all r > 0.

Take the limit superior as r tends to zero to obtain Tf  T gi C T hi D T hi  M hi C jhi j for all i . This implies A."; Tf /  A."=2; M hi / [ A."=2; hi /

for all i and all " > 0.

(6.24)

(See (6.1) for the notation A."; Tf /, etc.) Since hi and M hi are Borel measurable (see Theorem 6.8) the set Ei ."/ WD A."=2; M hi / [ A."=2; hi / is a Borel set. Since khi k1 < 2 i , we have m.A."=2; hi // 

2 2 1 khi k1;1  khi k1  i 1 " " 2 "

and, by Theorem 6.8, m.A."=2; M hi // 

2 2
3n 3n kM hi k1;1  khi k1  i 1 : " " 2 "

Thus m.Ei ."// 

3n C 1 : 2i 1 "

Since this holds for all i 2 N, it follows that the Borel set E."/ WD

1 \ iD1

has Lebesgue measure zero for all " > 0.

Ei ."/

(6.25)

236

6. Differentiation

Hence the Borel set E WD

1 [

E.1=k/

kD1

has Lebesgue measure zero. By (6.24) and (6.25), we have A.1=k; Tf /  E.1=k/ for all k 2 N, and hence ¹x 2 Rn j .Tf /.x/ ¤ 0º D

1 [

¹x 2 Rn j .Tf /.x/ > 1=kº

kD1

D

1 [

A.1=k; Tf /

kD1



1 [

E.1=k/

kD1

D E: This shows that .Tf /.x/ D 0

for all x 2 Rn n E,

and hence every element of Rn n E is a Lebesgue point of f . This proves Theorem 6.14.  The first consequence of Theorem 6.14 discussed here concerns a signed Borel measure  on Rn that is absolutely continuous with respect to the Lebesgue measure. The following theorem provides a formula for the function f in Theorem 5.18 (also called the Radon–Nikodým derivative of ). Theorem 6.15. Let W B ! R be a signed Borel measure on Rn that is absolutely continuous with respect to the Lebesgue R measure. Choose a Borel measurable func1 n tion f 2 L .R / such that .B/ D B f d m for all B 2 B. Then there exists a Borel set E  Rn such that m.E/ D 0 and f .x/ D lim

r!0

.Br .x// m.Br .x//

for all x 2 Rn n E:

(6.26)

6.3. Lebesgue points

237

Proof. By Theorem 6.14, there exists a Borel set E  Rn of Lebesgue measure zero such that every element of X n E is a Lebesgue point. Since ˇ ˇZ ˇ ˇ ˇ ˇ ˇ .Br .x// ˇ
1 ˇD ˇ ˇ ˇ f .x/ f f .x/ d m ˇ ˇ ˇ m.B .x// ˇ m.Br .x// Br .x/ r Z 1 jf f .x/j d m  m.Br .x// Br .x/ for all r > 0 and all x 2 Rn , it follows that (6.26) holds for all x 2 Rn n E. This proves Theorem 6.15.  Theorem 6.16. Let f 2 L1 .Rn / and let x 2 Rn be a Lebesgue point of f . Fix two constants 0 < ˛ < 1 and " > 0. Then there exists a ı > 0 such that, for every Borel set E 2 B and every r > 0, 8 ˆ ˛m.Br .x//

ˇ Z ˇ 1 ˇ f dm H) ˇ m.E/ E

ˇ ˇ f .x/ˇˇ < ":

(6.27)

Proof. Since x is a Lebesgue point of f , there exists a constant ı > 0 such that Z 1 m.Br .x// jf f .x/j d m < ˛" for 0 < r < ı. Br .x/

Assume 0 < r < ı and let E  Br .x/ be a Borel set such that m.E/  ˛m.Br .x//. Then ˇ ˇ Z Z ˇ 1 ˇ 1 ˇ ˇ f d m f .x/ˇ  jf f .x/j d m ˇ m.E/ m.E/ E E Z 1  jf f .x/j d m ˛m.Br .x// Br .x/ < ": This proves Theorem 6.16.



The Lebesgue differentiation theorem (Theorem 6.14) can be viewed as a theorem about signed Borel measures that are absolutely continuous with respect to the Lebesgue measure. The next theorem is an analogous result for signed Borel measures that are singular with respect to the Lebesgue measure.

238

6. Differentiation

Theorem 6.17 (singular Lebesgue differentiation). Let W B ! R be a signed Borel measure on Rn such that  ? m. Then there exists a Borel set E  Rn such that m.E/ D 0 D jj.Rn n E/ and jj.Br .x// D 0 for all x 2 Rn n E: (6.28) r!0 m.Br .x// Proof. The proof follows an argument in Heil [7, Section 3.4]. By assumption and Lemma 5.16, there exists a Borel set A  Rn such that lim

jj.Rn n A/ D 0:

m.A/ D 0;

For " > 0, define the set ˇ ± ° jj.Br .x// ˇ >" : A" WD x 2 Rn n A ˇ lim sup m.Br .x// r!0 We claim that A" is a Lebesgue null set for every " > 0. To see this, fix two constants " > 0 and ı > 0. Since the Borel measure jj is regular by Theorem 3.18, there exists an open set Uı  Rn such that A  Uı ;

jj.Uı / < ı:

For x 2 A" choose a radius r D r.x/ > 0 such that jj.Br.x/ .x// > "; m.Br.x/ .x//

Br.x/ .x/  Uı ;

and consider the open set Wı WD

[

Br.x/ .x/  Uı :

x2A"

Fix a compact subset K  Wı and cover K by finitely many of the balls Br.x/ .x/ with x 2 A" . By Vitali’s covering Lemma 6.10, there are elements x1 ; : : : ; xN 2 A" S such that the balls Br.xi / .xi / are pairwise disjoint and K  N i D1 B3r.xi / .xi /. Thus m.K/ 

N X

3n m.Br.xi / .xi //

i D1



1 [

A1=k

kD1

is a Lebesgue null set. It satisfies jj.Rn n E/ D 0 and (6.28) by definition, and this proves Theorem 6.17. 

6.4 Absolutely continuous functions Definition 6.18. Let I  R be an interval. A function f W I ! R is called absolutely continuous if for every " > 0 there exists a ı > 0 such that, for every finite sequence s1  t1  s2  t2 


 s`  t` in I , ` X

jsi

ti j < ı H)

i D1

` X

jf .si /

f .ti /j < ":

(6.29)

i D1

Every absolutely continuous function is continuous. The equivalence of (i) and (iii) in the following result is the fundamental theorem of calculus for Lebesgue integrable functions. The equivalence of (i) and (ii) is known as the Banach–Zarecki theorem. For functions of bounded variation see Exercise 6.20 below. Theorem 6.19 (fundamental theorem of calculus). Let I D Œa; b  R be a compact interval, let B  2I be the Borel  algebra, and let mW B ! Œ0; 1 be the restriction of the Lebesgue measure to B. Let f W I ! R be a function. Then the following are equivalent. (i) f is absolutely continuous. (ii) f is continuous, it has bounded variation, and if E  I is a Lebesgue null set, then so is f .E/. R (iii) There is a Borel measurable function gW I ! R such that I jgj d m < 1 and, for all x; y 2 I with x < y, Z y f .y/ f .x/ D g.t/ dt: (6.30) x

The right-hand side denotes the Lebesgue integral of g over the interval Œx; y. If (iii) holds, then there exists a Borel set E  I such that m.E/ D 0 and, for all x 2 I n E, f is differentiable at x and f 0 .x/ D g.x/.

240

6. Differentiation

Proof. We prove that (iii) implies the last assertion of the theorem. Thus assume that there exists a function g 2 L1 .I / that satisfies (6.30) for all x; y 2 I with x < y. Then Theorem 6.14 asserts that there exists a Borel set E  I of Lebesgue measure zero such that every element of I n E is a Lebesgue point of g. By Theorem 6.16 with ˛ D 1=2, every element x 2 I n E satisfies condition (ii) in Theorem 6.6 with A WD g.x/. Hence Theorem 6.6 asserts that the function f is differentiable at every point x 2 I n E and satisfies f 0 .x/ D g.x/ for x 2 I n E. (iii) H) (i). Thus assume f satisfies (iii) and define the signed measure W B ! R by

Z .B/ WD

g dm

(6.31)

B

for every Borel set B  I . Then  is absolutely continuous with respect to the Lebesgue measure and Z jj.B/ D jgj d m B

for ever Borel set B  I . Now let " > 0. Since jj  m, it follows from Lemma 5.21 that there exists a constant ı > 0 such that jj.B/ < " for every Borel set B  I with m.B/ < ı. Choose a sequence s1  t1 


 s`  t` in I such that

` X

jti

si j < ı

iD1

and define Ui WD .si ; ti / for i D 1; : : : ; `. Then the Borel set

` [

B WD

Ui

iD1

has Lebesgue measure m.B/ D

` X

jti

si j < ı:

i D1

Hence jj.B/ < ":

6.4. Absolutely continuous functions

Since

ˇZ ˇ f .si /j D ˇˇ

jf .ti /

Ui

241

ˇ Z ˇ g d mˇˇ  jgj d m D jj.Ui / Ui

for all i , it follows that ` X

jf .ti /

f .si /j 

` X

jj.Ui / D jj.B/ < ":

i D1

iD1

Hence f is absolutely continuous. (i) H) (ii). Assume f is absolutely continuous. It follows directly from the definition that f is continuous; that it has bounded variation is part (v) of Exercise 6.20. Now let E  I be a Lebesgue null set and assume without loss of generality that a; b … E. Fix any constant " > 0 and choose ı > 0 such that (6.29) holds. Since the Lebesgue measure is outer regular by Theorem 2.13, there exists an open set U  int.I / such that E  U and m.U / < ı. Choose a (possibly finite) sequence S of pairwise disjoint open intervals Ui  I such that U D i Ui . Choose si ; ti 2 Ui such that f .si / D inf Ui f and f .ti / D supUi f . Then it follows from (6.29) that P P f .si // < " for every finite sum. Take the limit to i m.f .Ui // D i .f .ti / P obtain m.f .U //  i m.f .Ui //  ". Since " > 0 was chosen arbitrary and the Lebesgue measure is complete, it follows that f .E/ is a Lebesgue measurable set and m.f .E// D 0. (ii) H) (iii). Assume f satisfies (ii). Then f has bounded variation and under this assumption Exercise 6.20 below outlines a proof that there exists a signed Borel measure W B ! R such that f .y/ f .x/ D ..x; y/ for x; y 2 I with x < y. Since C and  are regular by Theorem 3.18 and f is continuous, we have f .y/

f .x/ D .Œx; y/ D ..x; y/ D .Œx; y// D ..x; y//

(6.32)

for all x; y 2 I with x < y. By the Lebesgue decomposition theorem (Theorem 5.17), there exist two signed Borel measures a ; s W B ! R such that  D a C s ;

a  m;

s ? m:

(6.33)

Since a  m, it follows from Theorem 5.18 that there is an integrable function g 2 L1 .I / such that Z a .B/ D g dm (6.34) B

for every Borel set B  I . Define the functions fa ; fs W I ! R

242

6. Differentiation

by x

Z fa .x/ WD f .a/ C a .Œa; x/ D f .a/ C

fs .x/ WD s .Œa; x/:

g.t/ dt; a

Then f D fa C fs by (6.32) and (6.33). Since (iii) implies (i) and (i) implies (ii) (already proved), both functions f and fa satisfy (ii) and fa is absolutely continuous. Moreover fs D f fa is continuous. It remains to prove that fs  0. The proof given below follows an argument in Heil [7, Section 3.5.4]. By Theorem 6.17, there exists a Lebesgue null set Es  I such that a; b 2 Es (without loss of generality) and js j.I n Es / D 0;

lim

js ..x

r!0

r; x C r//j D0 r

for all x 2 I n Es :

This implies that every element x 2 I n Es satisfies condition (ii) in Theorem 6.6 with A D 0 and f replaced by fs . Hence fs is differentiable at every point x 2 I nEs and fs0 .x/ D 0 for x 2 I n Es . Since Es is a Lebesgue null set, so are the sets f .Es / and fa .Es /, because the functions f and fa satisfy (ii). Since the difference of two Lebesgue null sets in R is a Lebesgue null set, by Exercise 2.27, it follows that fs .Es / D ¹f .x/

fa .x/ j x 2 Es º  ¹f .x/

fa .y/ j x; y 2 Es º

is a Lebesgue null set. We claim that fs .I n Es / is also a Lebesgue null set. To see this, fix a constant " > 0. For n 2 N define the set ˇ ± ° 1 ˇ An WD x 2 I n Es ˇ y 2 I; jx yj < H) jfs .x/ fs .y/j  "jx yj : n Then An  AnC1 for all n 2 N and I n Es D

[

An :

n2N

Here the last assertion follows from the fact that fs is differentiable on I n Es with derivative zero. We prove that the Lebesgue outer measure of the set fs .I n Es / satisfies the estimate .fs .I n Es //  ".b a C "/: (6.35) To see this, cover the set An by at most countably many open intervals Ui , each of length less than 1=n, such that X m.Ui /  .An / C " i

6.5. Exercises

243

and each interval Ui contains an element of An . Then f .Ui / is contained in an interval of length at most "m.Ui /, by the definition of An . Hence X .f .An //  " m.Ui /  "..An / C "/  ".b a C "/: i

Since the Lebesgue outer measure is continuous from below by part (iii) of Theorem 2.13 and [ fs .I n Es / D fs .An /; n2N

it follows that .fs .I n Es // D lim .f .An //  ".b n!1

a C "/:

This proves (6.35). Since " > 0 was chosen arbitrary, this implies that fs .I n Es / is a Lebesgue null set, as claimed. Since fs .Es / is also a Lebesgue null set, as noted above, it follows that fs .I / is a Lebesgue null set. Finally, since fs is continuous and fs .0/ D 0 by definition, this implies fs  0. Hence f D fa is absolutely continuous and this proves Theorem 6.19. 

6.5 Exercises Exercise 6.20. Let I D Œa; b  R be a compact interval and let B  2I be the Borel  -algebra. A function f W I ! R is said to be of bounded variation if V .f / WD

sup

` X

jf .ti /

f .ti

1 /j

< 1:

(6.36)

aDt0 F .x/ for x < t < x C ıx : Since F .a/ D 0, there exists a maximal element x 2 Œa; b such that F .x/ D 0. If x < b, it follows from the previous discussion that F .t/ > 0 for x < t  b. In either case F .b/  0 and hence b

Z f .b/

f .a/ 

b

Z g.t/ dt C .b

f 0 .t/ dt C " C .b

a/
0 and all " > 0, it follows that b

Z f .b/

f 0 .t/ dt:

f .a/  a

Replace f by f to obtain the relation f .b/ that Z f .x/

f .a/ D

f .a/ D

Rb a

f 0 .t/ dt. Now deduce

x

f 0 .t/ dt

a

for all x 2 Œa; b. Example 6.24. (i) The Cantor function is the unique monotone function f W Œ0; 1 ! Œ0; 1 that satisfies

1 1  X ai  X ai f 2 D 3i 2i iD1

i D1

6.5. Exercises

247

for all sequences ai 2 ¹0; 1º. It is continuous and nonconstant, and its derivative exists and vanishes on the complement of the standard Cantor set n n 1 \ [ h X ai X ai 1i 2 ; 2 C : C WD 3i 3i 3n nD1 ai 2¹0;1º

iD1

iD1

This Cantor set has Lebesgue measure zero. Hence f is almost everywhere differentiable and its derivative is integrable. However, f is not equal to the integral of its derivative, and therefore is not absolutely continuous. (ii) The following construction was explained to me by Theo Buehler. Define the homeomorphisms gW Œ0; 1 ! Œ0; 2 and

hW Œ0; 2 ! Œ0; 1

by g.x/ WD f .x/ C x;

h WD g

1

:

The image g.Œ0; 1nC / is a countable union of disjoint open intervals of total length one and hence has Lebesgue measure one. Thus its complement K WD g.C /  Œ0; 2 is a modified Cantor set of Lebesgue measure one. Hence, by Theorem 6.19, g is not absolutely continuous. Moreover, by Lemma 2.15, there exists a set E  K which is not Lebesgue measurable. However, its image F WD h.E/  Œ0; 1 under h is a subset of the Lebesgue null set C and hence is a Lebesgue null set. Thus F is a Lebesgue measurable set and E D h 1 .F / is not Lebesgue measurable. This shows that the function hW Œ0; 2 ! Œ0; 1 is not measurable with respect to the Lebesgue  -algebras on both domain and target (i.e., it is not Lebesgue-Lebesgue measurable). (iii) Let I; J  R be intervals. Then it follows from Lemma 2.15 and Theorem 6.19 that every Lebesgue-Lebesgue measurable homeomorphism hW I ! J has an absolutely continuous inverse. (iv) Let hW Œ0; 2 ! Œ0; 1 and F  C  Œ0; 1 be as in part (ii). Then the characteristic function F W R ! R is Lebesgue measurable and hW Œ0; 2 ! R is continuous. However, the composition F ı hW Œ0; 2 ! R is not Lebesgue measurable because the set .F ı h/ 1 .1/ D E is not Lebesgue measurable. (v) By contrast, if I; J  R are intervals, f W J ! R is Lebesgue measurable, and hW I ! J is a C 1 diffeomorphism, then f ı hW I ! R is again Lebesgue measurable by Theorem 2.17.

Chapter 7

Product measures

The purpose of this chapter is to study products of two measurable spaces (Section 7.1), introduce the product measure (Section 7.2), and prove Fubini’s Theorem (Section 7.3). The archetypal example is the Lebesgue measure on RkC` D Rk  R` ; it is the completion of the product measure associated to the Lebesgue measures on Rk and R` (Section 7.4). Applications include the convolution (Section 7.5), Marcinkiewicz interpolation (Section 7.6), and the Calderón– Zygmund inequality (Section 7.7).

7.1 The product  -algebra Assume throughout that .X; A/ and .Y; B/ are measurable spaces. Definition 7.1. The product -algebra of A and B is defined as the smallest -algebra on the product space X  Y WD ¹.x; y/ j x 2 X; y 2 Y º that contains all subsets of the form A  B, where A 2 A and B 2 B. It will be denoted by A ˝ B  2X Y . x be an .A ˝ B/-measurable Lemma 7.2. Let E 2 A ˝ B and let f W X  Y ! R function. Then the following holds. x , defined by fx .y/ WD f .x; y/ for (i) For every x 2 X, the function fx W Y ! R y 2 Y , is B-measurable and Ex WD ¹y 2 Y j .x; y/ 2 Eº 2 B:

(7.1)

x , defined by f y .x/ WD f .x; y/ for (ii) For every y 2 Y , the function f y W X ! R x 2 X , is A-measurable and E y WD ¹x 2 X j .x; y/ 2 Eº 2 A:

(7.2)

250

7. Product measures

Proof. Define   2XY by  WD ¹E  X  Y j Ex 2 B for all x 2 X º: We prove that  is a -algebra. To see this, note first that X  Y 2 . Second, if E 2 , then Ex 2 B for all x 2 X , hence .E c /x D ¹y 2 Y j .x; y/ … Eº D .Ex /c 2 B for all x 2 X , and hence E c 2 . Third, if Ei 2  is a sequence and S S1 E WD 1 i D1 Ei , then Ex D iD1 .Ei /x 2 B for all x 2 X , and hence E 2 . This shows that  is a  -algebra. Since A  B 2  for all A 2 A and all B 2 B, it follows that A ˝ B  . This proves (7.1) for all x 2 X. x is open, then E WD f 1 .V / 2 A ˝ B and Now fix an element x 2 X . If V  R hence .fx / 1 .V / D Ex 2 B by (7.1). Thus fx is B-measurable. This proves (i). 

The proof of (ii) is analogous and this proves Lemma 7.2.

Definition 7.3. Let Z be a set. A collection of subsets M  2Z is called a monotone class if it satisfies the following two axioms: (a) if Ai 2 M for i 2 N such that Ai  Ai C1 for all i, then

S1

Ai 2 M;

(b) if Bi 2 M for i 2 N such that Bi  BiC1 for all i, then

T1

Bi 2 M.

i D1

i D1

Definition 7.4. A subset Q  X Y is called elementary if it is the union of finitely many pairwise disjoint subsets of the form A  B with A 2 A and B 2 B. The next lemma is a useful characterization of the product -algebra. Lemma 7.5. The product -algebra A ˝ B is the smallest monotone class in X  Y that contains all elementary subsets. Proof. Let E  2XY denote the collection of all elementary subsets and define M  2XY as the smallest monotone class that contains E. This is well defined because the intersection of any collection of monotone classes is again a monotone class. Since every -algebra is a monotone class and every elementary set is an element of A ˝ B, it follows that M  A ˝ B: Since E  M by definition, the converse inclusion follows once we know that M is a -algebra. We prove this in seven steps.

7.1. The product -algebra

251

Step 1. For every set P  X  Y the collection .P / WD ¹Q  X  Y j P n Q; Q n P; P [ Q 2 Mº is a monotone class. This follows immediately from the definition of monotone class. Step 2. If P; Q  X  Y , then Q 2 .P / if and only if P 2 .Q/. This follows immediately from the definition of .P / in Step 1. Step 3. If P; Q 2 E, then P \ Q; P n Q; P [ Q 2 E. For the intersection this follows from the fact that .A1  B1 / \ .A2  B2 / D .A1 \ A2 /  .B1 \ B2 / : For the complement it follows from the fact that .A1  B1 / n .A2  B2 / D ..A1 n A2 /  B1 / [ ..A1 \ A2 /  .B1 n B2 //: For the union this follows from the fact that P [ Q D .P n Q/ [ Q. Step 4. If P 2 E, then M  .P /. Let P 2 E. Then P n Q; Q n P; P [ Q 2 E  M for all Q 2 E by Step 3. Hence Q 2 .P / for all Q 2 E by the definition of .P / in Step 1. Thus we have proved that E  .P /. Since .P / is a monotone class by Step 1, it follows that M  .P /. This proves Step 4. Step 5. If P 2 M, then M  .P /. Fix a set P 2 M. Then P 2 .Q/ for all Q 2 E by Step 4. Hence Q 2 .P / for all Q 2 E by Step 2. Thus E  .P / and hence it follows from Step 1 that M  .P /. This proves Step 5. Step 6. If P; Q 2 M, then P n Q; P [ Q 2 M. If P; Q 2 M, then Q 2 M  .P / by Step 5, and hence P n Q; P [ Q 2 M by the definition of .P / in Step 1. Step 7. M is a -algebra. By definition, X  Y 2 E  M. If P 2 M, then P c D .X  Y / n P 2 M by S Step 6. If Pi 2 M for i 2 N, then Qn WD niD1 Pi 2 M for all n 2 N by Step 6 S S1 and hence 1 iD1 Pi D nD1 Qn 2 M because M is a monotone class. This proves Step 7 and Lemma 7.5. 

252

7. Product measures

Lemma 7.6. Let .X; UX / and .Y; UY / be topological spaces, let UX Y be the product topology on X  Y (see Appendix B), and let BX , BY , BX Y be the associated Borel -algebras. Then BX ˝ BY  BXY : (7.3) If .X; UX / is a second countable locally compact Hausdorff space, then BX ˝ BY D BX Y :

(7.4)

Proof. The projections X W X  Y ! X and Y W X  Y ! Y are continuous, and hence Borel measurable by Theorem 1.20. Thus X 1 .A/ D A  Y 2 BXY for all A 2 BX and Y 1 .B/ D X  B 2 BX Y for all B 2 BY . Hence A  B 2 BX Y for all A 2 BX and all B 2 BY , and this implies (7.3). Now assume .X; UX / is a second countable locally compact Hausdorff space and choose a countable basis ¹Ui j i 2 Nº of UX such that Uxi is compact for all i 2 N. Fix an open set W 2 UX Y and, for i 2 N, define Vi WD ¹y 2 Y j .x; y/ 2 W for all x 2 Ux i º: We prove that Vi is open. Let y0 2 Vi . Then .x; y0 / 2 W for all x 2 Ux i . Hence, for every x 2 Ux i , there exist open sets U.x/ 2 UX and V .x/ 2 UY such that .x; y0 / 2 U.x/V .x/  W . Since Ux i is compact, there are finitely many elements x1 ; : : : ; x` 2 Ux i such that Ux i  U.x1 / [


[ U.x` /. Define V WD V .x1 / \


\ V .x` /: Then V is open and Ux i  V  W , so y0 2 V  Vi . This shows that Vi is open for all i 2 N. Next we prove that W D

1 [

.Ui  Vi /:

(7.5)

iD1

Let .x0 ; y0 / 2 W . Then there exist open sets U 2 UX and V 2 UY such that .x0 ; y0 / 2 U  V  W : Since .X; UX / is a locally compact Hausdorff space, Lemma A.3 asserts that there exists an open set U 0  X such that x0 2 U 0  U 0  U . Since the sets Ui form a basis of the topology, there exists an integer i 2 N such that x0 2 Ui  U 0 and hence x0 2 Ux i  U 0  U . Thus Ux i  ¹y0 º  U  V  W ;

7.1. The product -algebra

253

hence y0 2 Vi , and so .x0 ; y0 / 2 Ui  Vi  W . Since the element .x0 ; y0 / 2 W was chosen arbitrarily, this proves (7.5). Thus we have shown that UX Y  BX ˝ BY and this implies BX Y  BX ˝ BY : Hence (7.4) follows from (7.3). This proves Lemma 7.6.



Lemma 7.7. Let .X; A/ be a measurable space such that the cardinality of X is greater than that of 2N . Then the diagonal  WD ¹.x; x/ j x 2 X º is not an element of A ˝ A. Proof. The proof has three steps. Step 1. Let Y be a set. For E  2Y denote by .E/  2Y the smallest  -algebra containing E. If D 2 .E/, then there exists a sequence Ei 2 E, i 2 N, such that D 2 .¹Ei j i 2 Nº/: The union of the sets .E0 / over all countable subsets E0  E is a  -algebra that contains E and is contained in .E/. Hence it is equal to .E/. Step 2. Let Y be a set, let E  2Y , and let D 2 .E/. Then there are a sequence Ei 2 E and a set I  2N such that  [\ \ DD Ei \ Eic : I 2I

i2I

i2NnI

By Step 1 there exists a sequence Ei 2 E such that D 2 .¹Ei j i 2 Nº/. For I  N define \   \  EI WD Ei \ Eic : i 2I

i2NnI

These sets form a partition of Y . Hence the collection ˇ °[ ± ˇ F WD EI ˇ I  2N I 2I

is a -algebra on Y . Since Ei 2 F for each i 2 N, it follows that D 2 F. This proves Step 2.

254

7. Product measures

Step 3.  … A ˝ A. Let E  2X X be the collection of all sets of the form A  B with A; B 2 A. Let D 2 A ˝ A. By Step 2, there are sequences Ai ; Bi 2 A and a set I  2N such that [ DD EI ; I 2I

where EI WD

\

\

.Ai  Bi / \

i 2I

 .Ai  Bi /c :

i2NnI

Thus EI D

[

AIJ  BIJ ;

J NnI

where AIJ WD

\

 \  Ai \ .X n Aj /

i2I

j 2J

and BIJ WD

\ i2I

  Bi \

\

 .X n Bj / :

j 2Nn.I [J /

If D  , then for all I and J , we have AIJ  BIJ   and so AIJ  BIJ is either empty or a singleton. Thus the cardinality of D is at most the cardinality of the set of pairs of disjoint subsets of N, which is equal to the cardinality of 2N . Since the cardinality of the diagonal is bigger than that of 2N , it follows that  … A ˝ A, as claimed. This proves Lemma 7.7.  Example 7.8. Let X be an uncountable set, of cardinality greater than that of 2N , and equipped with the discrete topology so that BX D UX D 2X . Then  is an open subset of X  X with respect to the product topology (which is also discrete because points are open). Hence  2 BXX D 2X X . However,  … BX ˝ BX by Lemma 7.7. Thus the product BX ˝ BX of the Borel -algebras is not the Borel -algebra of the product. In other words, the inclusion (7.3) in Lemma 7.6 is strict in this example. Note also that the distance function d W X  X ! R defined by d.x; y/ WD 1 for x ¤ y and d.x; x/ WD 0 is continuous with respect to the product topology, but is not measurable with respect to the product of the Borel -algebras.

7.2. The product measure

255

7.2 The product measure The definition of the product measure on the product -algebra is based on the following theorem. For a measure space .X; A; / and a measurable function W X ! Œ0; 1 we use the notation Z Z .x/ d.x/ WD  d: X

X

Theorem 7.9. Let .X; A; / and .Y; B; / be -finite measure spaces and let Q 2 A ˝ B. Then the functions X ! Œ0; 1W x 7 ! .Qx / and

Y

! Œ0; 1W y 7 ! .Qy /

(7.6)

are measurable and Z

Z

.Qy / d.y/:

.Qx / d.x/ D X

(7.7)

Y

Definition 7.10. Let .X; A; / and .Y; B; / be  -finite measure spaces. The product measure of  and  is the map  ˝ W A ˝ B ! Œ0; 1 defined by Z . ˝ /.Q/ WD

Z

.Qy / d.y/

.Qx / d.x/ D X

(7.8)

Y

for Q 2 A ˝ B. That  ˝  is  -additive, and hence is a measure, follows from P1 Theorem 1.38 and the fact that .Qx / D iD1 ..Qi /x / for every sequence of pairwise disjoint sets Qi 2 A ˝ B. The product measure satisfies . ˝ /.A  B/ D .A/
.B/ for A 2 A and B 2 B and hence is  -finite. Proof of Theorem 7.9. Define ˇ ² ³ ˇ the functions (7.6) are measurable  WD Q 2 A ˝ B ˇˇ : and equality (7.7) holds We prove in five steps that  D A ˝ B.

(7.9)

256

7. Product measures

Step 1. If A 2 A and B 2 B, then Q WD A  B 2 . By assumption Qx D

´ B

if x 2 A;

;

if x … A;

y

Q D

´ A

if y 2 B;

;

if y … B:

Define the function W X ! Œ0; 1 by .x/ WD .Qx / D .B/A .x/ for x 2 X and the function WY

! Œ0; 1

by .y/ WD .Qy / D .A/B .y/ for y 2 Y . Then ;

are measurable and Z Z  d D .A/.B/ D X

d: Y

Thus Q 2 . Step 2. If Q1 ; Q1 2  and Q1 \ Q2 D ;, then Q WD Q1 [ Q2 2 . Define i .x/ WD ..Qi /x /;

.x/ WD .Qx /;

(7.10a)

WD ..Qi /y /;

.y/ WD .Qy /

(7.10b)

i .y/

for x 2 X , y 2 Y and i D 1; 2. Then  D 1 C 2 and Z Z i d D i d X

Y

for i D 1; 2 because Qi 2 . Hence Z Z  d D X

and so Q 2 .

d Y

D

1

C

2.

Moreover,

7.2. The product measure

257

Step 3. If Qi 2  for i 2 N and Qi  QiC1 for all i, then Q WD Define i ; W X ! Œ0; 1 and Qx D

i;

1 [

S1

i D1

Qi 2 .

W Y ! Œ0; 1 by (7.10) for i 2 N. Since 1 [

y

Q D

.Qi /x ;

.Qi /y

i D1

iD1

and .Qi /x 2 B and .Qi /y 2 A for all i , it follows from Theorem 1.28 (iv) that .x/ D .Qx / D lim ..Qi /x / D lim i .x/ i !1

i !1

.y/ D .Qy / D lim ..Qi /y / D lim i !1

i!1

i .y/

for all x 2 X; for all y 2 Y:

By the Lebesgue monotone convergence theorem (Theorem 1.37) this implies Z Z Z Z  d D lim i d D lim d: i d D X

i !1 X

i !1 Y

Y

Thus Q 2  and this proves Step 3. Step 4. Let A 2 A and B 2 B such that .A/ < 1 and .B/ < 1. If Qi 2  T for i 2 N such that A  B  Q1  Q2 


, then Q WD 1 i D1 Qi 2 . Let i ; ; i ; be as in the proof of Step 3. Since .Qi /x  B and .B/ < 1, it follows from part (v) of Theorem 1.28 that i converges pointwise to . Moreover, i  .B/A for all i and the function .B/A W X ! Œ0; 1/ is integrable because .A/ < 1 and .B/ < 1. Hence it follows from the Lebesgue dominated convergence theorem (Theorem 1.45) that Z Z  d D lim i d: i!1 X

X

The same argument shows that Z

Z d D lim

Y

i!1 Y

Since Qi 2  for all i, this implies Z Z  d D X

and hence Q 2 . This proves Step 4.

d Y

i

d:

258

7. Product measures

Step 5.  D A ˝ B. Since .X; A; / and .Y; B; / are -finite, there exist sequences of measurable sets Xn 2 A and Yn 2 B such that Xn  XnC1 ; for all n 2 N and X D

Yn  YnC1 ;

S1

nD1

.Xn / < 1;

Xn and Y D

S1

nD1

.Yn / < 1

Yn . Define

M WD ¹Q 2 A ˝ B j Q \ .Xn  Yn / 2  for all n 2 Nº: Then M is a monotone class by Steps 3 and 4, E  M by Steps 1 and 2, and M  A ˝ B by definition. Hence it follows from Lemma 7.5 that M D A ˝ B. In other words, Q \ .Xn \ Yn / 2  for all Q 2 A ˝ B. By Step 3, this implies QD

1 [

for all Q 2 A ˝ B:


Q \ .Xn  Yn / 2 

nD1

Thus A ˝ B    A ˝ B and so  D A ˝ B, as claimed. This proves Step 5 and theorem (Theorem 7.9). 

Examples and exercises Example 7.11. Let X D Y D Œ0; 1, let A  2X be the  -algebra of Lebesgue measurable sets, let B WD 2Y , let W A ! Œ0; 1 be the Lebesgue measure, and let W B ! Œ0; 1 be the counting measure. Consider the diagonal  WD ¹.x; x/ j 0  x  1º D

1 [ n h \ i nD1 i D1

1 i i2 ; 2 A ˝ B: n n

Its characteristic function f WD  W X  Y

!R

is given by f .x; y/ WD

´ 1

if x D y;

0

if x ¤ y:

7.2. The product measure

259

Hence Z

y

. / D

f .x; y/ d.x/ D 0

for 0  y  1;

f .x; y/ d.y/ D 1

for 0  x  1;

X

Z .x / D Y

and so

Z

Z

y

. / d D 0 ¤ 1 D X

.x / d.x/: Y

Thus the hypothesis that .X; A; / and .Y; B; / are  -finite cannot be removed in Theorem 7.9. Example 7.12. Let X WD Y WD Œ0; 1, let A D B  2Œ0;1 be the -algebra of Lebesgue measurable sets, and let  D  be the Lebesgue measure. Claim 1. Assume the continuum hypothesis. Then there is a set Q  Œ0; 12 such that Œ0; 1 n Qx is countable for all x and Qy is countable for all y. Let Q be as in Claim 1 and define f WD Q W Œ0; 12 ! R: Then .Qy / D

Z f .x; y/ d.x/ D 0

for 0  y  1;

f .x; y/ d.y/ D 1

for 0  x  1;

X

Z .Qx / D Y

and hence

Z X

.Qy / d D 0 ¤ 1 D

Z .Q/ d.x/: Y

The sets Qx and Qy are all measurable and the integrals are finite, but the set Q is not A ˝ B-measurable. This shows that the hypothesis Q 2 A ˝ B in Theorem 7.9 cannot be replaced by the weaker hypothesis that the sets Qx and Qy are all measurable, even when the integrals are finite. It also shows that Lemma 7.2 does not have a converse. Namely, fx and f y are measurable for all x and y, but f is not A ˝ B-measurable. Claim 2. Assume the continuum hypothesis. Then there exists a bijection j W Œ0; 1 ! W with values in a well-ordered set .W; / such that the set ¹w 2 W j w  zº is countable for all z 2 W .

260

7. Product measures

Claim 2 H) Claim 1. Let j be as in Claim 2 and put Q WD ¹.x; y/ 2 Œ0; 12 j j.x/  j.y/º: Then the set Qy D ¹x 2 Œ0; 1 j j.x/  j.y/º is countable for all y 2 Œ0; 1 and the set Œ0; 1 n Qx D ¹y 2 Œ0; 1 j j.y/ 4 j.x/º is countable for all x 2 Œ0; 1.  Proof of Claim 2. By Zorn’s lemma, every set admits a well ordering. Choose any well ordering  on A WD Œ0; 1 and define B WD ¹b 2 A j the set ¹a 2 A j a  bº is uncountableº: If B D ; choose W WD A D Œ0; 1 and j D id. If B ¤ ;, then by the well ordering axiom, B contains a smallest element b0 . Since b0 2 B, the set W WD B n A D ¹w 2 A j w  b0 º is uncountable. Since W \ B D ; the set ¹w 2 W j w  zº is countable for all z 2 W . Since W is an uncountable subset of Œ0; 1, the continuum hypothesis asserts that there exists a bijection j W Œ0; 1 ! W . This proves Claim 2.  Example 7.13. Let X and Y be countable sets, let A D 2X and B D 2Y , and let W 2X ! Œ0; 1 and W 2Y ! Œ0; 1 be the counting measures. Then A ˝ B D 2X Y and  ˝ W 2XY ! Œ0; 1 is the counting measure. Example 7.14. Let .X; A; / and .Y; B; / be probability measure spaces, so that .X/ D .Y / D 1. Then  ˝ W A ˝ B ! Œ0; 1 is also a probability measure. A trivial example is A D ¹;; Xº and B D ¹;; Y º. In this case the product  -algebra is A ˝ B D ¹;; X  Y º and the product measure is given by . ˝ /.;/ D 0 and . ˝ /.X  Y / D 1. Exercise 7.15. Let .X; A; /, .Y; B; / be  -finite measure spaces and let W X ! X; be bijections. Define the bijection   . 

WY

!Y

W X  Y ! X  Y by

/.x; y/ WD ..x/; .y//

for x 2 X and y 2 Y . Prove that . 

/ .A ˝ B/ D  A ˝

 B;

. 

/ . ˝ / D   ˝

Hint. Use Theorem 1.19 to show that  A ˝ See also Exercise 2.34.

B

 . 

 :

/ .A ˝ B/.

7.3. Fubini’s Theorem

261

Exercise 7.16. For n 2 N let Bn  Rn be the Borel -algebra and let n W Bn ! Œ0; 1 be the restriction of the Lebesgue measure to Bn . Let k; ` 2 N and n WD k C `. Identify Rk  R` with Rn in the obvious manner. Then Bk ˝ B` D Bn by Lemma 7.6. Prove that the product measure k ˝ ` is translation-invariant and satisfies .k ˝ ` /.Œ0; 1n / D 1. Deduce that k ˝ ` D n : Hint. Use Exercise 7.15. We return to this example in Section 7.4.

7.3 Fubini’s Theorem There are three versions of Fubini’s Theorem. The first concerns nonegative functions that are measurable with respect to the product -algebra (Theorem 7.17), the second concerns real-valued functions that are integrable with respect to the product measure (Theorem 7.20), and the third concerns real valued functions that are integrable with respect to the completion of the product measure (Theorem 7.23). Theorem 7.17 (Fubini for positive functions). Let .X; A; /, .Y; B; / be -finite measure spaces and let  ˝ W A ˝ B ! Œ0; 1 be the product measure in Definition 7.10. Let f W X  Y ! Œ0; R 1 be an A ˝ B-measurable function. Then the function X ! RŒ0; 1W x 7! Y f .x; y/ d.y/ is A-measurable, the function Y ! Œ0; 1W y 7! X f .x; y/ d.x/ is B-measurable, and Z Z

Z



f d. ˝ / D X Y

f .x; y/ d.y/ X

Z Z D

(7.11)

 f .x; y/ d.x/

Y

d.x/

Y

d.y/:

X

Example 7.18. equation (1.20) is equivalent to equation (7.11) for the counting measure on X D Y D N.

262

7. Product measures

Proof of Theorem 7.17. Let fx .y/ WD f y .x/ WD f .x; y/ for .x; y/ 2 X  Y and define the functions W X ! Œ0; 1 by

and

W Y ! Œ0; 1

Z

Z

.x/ WD

fx d;

f y d

.y/ WD

Y

(7.12)

X

for x 2 X and y 2 Y . We prove in three steps that  is A-measurable, is B-measurable, and  and satisfy (7.11). Step 1. The assertion holds when f W X  Y ! Œ0; 1/ is the characteristic function of an A ˝ B-measurable set. Let Q 2 A ˝ B and f D Q . Then fx D Qx and f y D Qy , and so .x/ D .Qx /;

.y/ D .Qy /

for all x 2 X and all y 2 Y . Hence it follows from Theorem 7.9 that Z Z Z  d D d D . ˝ /.Q/ D f d. ˝ /: X

Y

X

Here the third equality follows from the definition of the measure ˝. This proves Step 1. Step 2. The assertion holds when f W X  Y ! Œ0; 1/ is an A ˝ B-measurable step-function. This follows immediately from Step 1 and the linearity of the integral. Step 3. The assertion holds when f W X  Y ! Œ0; 1 is A ˝ B-measurable. By Theorem 1.26, there exists a sequence of A ˝ B-measurable step-functions sn W X  Y ! Œ0; 1/ such that sn  snC1 for all n 2 N and sn converges pointwise to f . Define Z n .x/ WD sn .x; y/ d.y/ for x 2 X; Y

Z n .x/

WD

sn .x; y/ d.x/ X

for y 2 Y:

7.3. Fubini’s Theorem

263

Then n  nC1 ;

n



for all n 2 N

nC1

by part (i) of Theorem 1.35. Moreover, it follows from the Lebesgue monotone convergence theorem (Theorem 1.37) that .x/ D lim n .x/;

.y/ D lim

n!1

n!1

n .y/

for all x 2 X and all y 2 Y . Use the Lebesgue monotone convergence theorem (Theorem 1.37) again as well as Step 2 to obtain Z Z  d D lim n d n!1 X

X

Z D lim

n!1 X Y

sn d. ˝ /

Z D

f d. ˝ / XY

Z D lim

n!1 Y

n

d

Z D

d: Y



This proves Step 3 and Theorem 7.17.

A first application of Fubini’s Theorem 7.17 is Minkowski’s inequality for a measurable function on a product space that is p-integrable with respect to one variable such that the resulting Lp norms define an integrable function of the other variable. Theorem 7.19 (Minkowski). Fix a constant 1  p < 1. Let .X; A; / and .Y; B; / be -finite measure spaces and let f WX  Y

! Œ0; 1

be A ˝ B-measurable. Then Z Z p 1=p Z Z 1=p  f .x; y/p d.x/ d.y/: f .x; y/ d.y/ d.x/ X

Y

Y

X

In the notation fx .y/ WD f y .x/ WD f .x; y/; Minkowski’s inequality has the form Z 1=p Z p  kfx kL1 ./ d.x/ kf y kLp ./ d.y/: X

Y

(7.13)

264

7. Product measures

Proof. By Lemma 7.2, fx W Y ! Œ0; 1 is B-measurable for all x 2 X and fy W X ! Œ0; 1 is A-measurable for all y 2 Y . Moreover, by Theorem 7.17, p the function X ! Œ0; 1W x 7! kfx kL 1 ./ is A-measurable and the function y Y ! Œ0; 1W y 7! kf kLp ./ is B-measurable. Hence both sides of the inequality (7.13) are well defined. Theorem 7.17 also shows that for p D 1 equality holds in (7.13). Hence assume 1 < p < 1 and a choose 1 < q < 1 such that 1=p C 1=q D 1. It suffices to assume Z c WD Y

kf y kLp ./ d.y/ < 1:

Define W X ! Œ0; 1 by Z .x/ WD

fx d

for x 2 X

Y

and let g 2 Lq ./. Then the function X  Y ! Œ0; 1W .x; y/ 7! f .x; y/jg.x/j is A ˝ B-measurable. Hence it follows from Theorem 7.17 that Z Z

Z



jgj d D

f .x; y/jg.x/j d.y/

X

X

d.x/

Y

Z Z D

 f .x; y/jg.x/j d.x/

Y

Z  Y

d.y/

X

kf y kLp ./ kgkLq ./ d.y/

D c kgkLq ./ : Here the third step follows from Hölder’s inequality in Theorem 4.1. Since .X; A; / is semi-finite by part (ii) of Lemma 4.30, it follows from Lemma 4.34 that kkLp ./  c: This proves Theorem 7.19.



7.3. Fubini’s Theorem

265

Theorem 7.20 (Fubini for integrable functions). Let .X; A; / and .Y; B; / be -finite measure spaces, let  ˝ W A ˝ B ! Œ0; 1 be the product measure, and let f 2 L1 . ˝ /. Define fx .y/ WD f y .x/ WD f .x; y/ for x 2 X and y 2 Y . Then the following holds. (i) fx 2 L1 ./ for -almost every x 2 X and the map W X ! R defined by 8Z < fx d .x/ WD Y : 0

if fx 2 L1 ./;

(7.14)

if fx … L1 ./;

is -integrable. (ii) f y 2 L1 ./ for -almost every y 2 Y and the map WY

!R

defined by 8Z < f y d .y/ WD X : 0

if f y 2 L1 ./; if f

y

(7.15)

… L ./; 1

is -integrable. (iii) Let  2 L1 ./ and Z

2 L1 ./ be as in (i) and (ii). Then Z Z  d D f d. ˝ / D d:

X

X Y

Y

Proof. We prove part (i) and the first equality in (7.16). The functions f ˙ WD max¹˙f; 0ºW X  Y

! Œ0; 1/

are A  B-measurable by Theorem 1.24. Hence the functions fx˙ WD max¹˙fx ; 0º D f ˙ .x;
/W Y are B-measurable by Lemma 7.2.

! Œ0; 1/

(7.16)

266

7. Product measures

Define

ˆ˙ W Y

by

Z

˙

ˆ .x/ WD Y

! Œ0; 1 fx˙ d

for x 2 X:

By Theorem 7.17, the functions ˆ˙ W X ! Œ0; 1 are A-measurable and Z Z Z ˆ˙ d D f ˙ d. ˝ /  jf j d. ˝ / < 1: X

XY

(7.17)

X Y

Now Lemma 1.47 asserts that the A-measurable set ˇZ ² ³ ˇ ˇ E WD x 2 X ˇ jfx j d D 1 D ¹x 2 X j ˆC .x/ D 1 or ˆ .x/ D 1º Y

has measure .E/ D 0. Moreover, for all x 2 X, x 2 E () fx … L1 ./: Define

 ˙ W X ! Œ0; 1/

by ´ ˙

 .x/ WD

ˆ˙ .x/ if x … E; if x 2 E;

0

for x 2 X:

Then it follows from (7.17) that  ˙ 2 L1 ./ and Z Z ˙  d D f ˙ d. ˝ /: X

Hence  D  C Z

X Y

 2 L1 ./ and Z  d D  C d

X

X

 d X

Z D

Z

f

C

Z d. ˝ /

X Y

f

d. ˝ /

X Y

Z D

f d. ˝ /: X Y

This proves (i) and the first equality in (7.16). An analogous argument proves (ii) and the second equality in (7.16). This proves Theorem 7.20. 

7.3. Fubini’s Theorem

267

Example 7.21. Let .X; A; / D .Y; B; / be the Lebesgue measure space in the unit interval Œ0; 1 as in Example 7.12. Let gn W Œ0; 1 ! Œ0; 1/ be a sequence of smooth functions such that Z 1 gn .x/ dx D 1; 0

and gn .x/ D 0 for x 2 Œ0; 1 n Œ2

n 1

;2

n



for all n 2 N. Define f W Œ0; 12 ! R by f .x; y/ WD

1 X

.gn .x/

gnC1 .x//gn .y/:

nD1

The sum on the right is finite for every pair .x; y/ 2 Œ0; 12 . Then Z f .x; y/ dx D 0; X

Z f .x; y/ dy D Y

and hence Z 1 Z

1 0

.gn .x/

gnC1 .x// D g1 .x/;

nD1

 f .x; y/ dx

0

1 X

1

Z

1

Z

dy D 0 ¤ 1 D

 f .x; y/ dy

0

dx:

0

Thus the hypothesis f 2 L1 . ˝ / cannot be removed in Theorem 7.20. Example 7.22. This example shows that the product measure is typically not complete. Let .X; A; / and .Y; B; / be two complete -finite measure spaces. Suppose .X; A; / admits a nonempty null set A 2 A and B ¤ 2Y . Choose B 2 2Y nB. Then A  B … A ˝ B. However, A  B is contained in the  ˝ -null set A  Y and so belongs to the completion .A ˝ B/ . In the first version of Fubini’s Theorem integrability was not an issue. In the second version integrability of fx was only guaranteed for almost all x. In the third version the function fx may not even be measurable for all x.

268

7. Product measures

Theorem 7.23 (Fubini for the completion). Let .X; A; / and .Y; B; / be complete  -finite measure spaces, let .X  Y; .A ˝ B/ ; . ˝ / / denote the completion of the product space, and let f 2 L1 .. ˝ / /. Define fx .y/ WD f y .x/ WD f .x; y/ for x 2 X and y 2 Y . Then the following holds. (i) fx 2 L1 ./ for -almost every x 2 X and the map W X ! R defined by 8Z < fx d .x/ WD Y : 0

if fx 2 L1 ./;

(7.18)

if fx … L1 ./;

is -integrable. (ii) f y 2 L1 ./ for -almost every y 2 Y and the map WY

!R

defined by 8Z < f y d if f y 2 L1 ./; .y/ WD X : 0 if f y … L1 ./;

(7.19)

is -integrable. (iii) Let  2 L1 ./ and Z

2 L1 ./ be as in (i) and (ii). Then Z Z   d D f d. ˝ / D d:

X

X Y

(7.20)

Y

Proof. By part (v) of Theorem 1.55, there exists a function g 2 L1 . ˝ / such that the set N WD ¹.x; y/ 2 X  Y j f .x; y/ ¤ g.x; y/º 2 .A ˝ B/ has measure zero, i.e., . ˝ / .N / D 0. By, the definition of the completion, there exists a set Q 2 A ˝ B such that N  Q and . ˝ /.Q/ D 0. Thus Z Z .Qx / d.x/ D .Qy / d.y/ D 0: X

Y

Hence, by Lemma 1.49, .E/ D 0;

E WD ¹x 2 X j .Qx / ¤ 0º;

.F / D 0;

F WD ¹y 2 Y j .Qy / ¤ 0º:

7.3. Fubini’s Theorem

269

Since f D g on .X  Y / n Q, we have fx D gx on Y n Qx for all x 2 X and f y D g y on X n Qy for all y 2 Y . By Theorem 7.20, for g 2 L1 . ˝ / there are measurable sets E 0 2 A and F 0 2 B such that .E 0 / D .F 0 / D 0 and gx 2 L1 ./

for all x 2 X n E 0 ;

g y 2 L1 ./

for all y 2 Y n F 0 :

If x 2 X n .E [ E 0 /, then .Qx / D 0 and fx D gx on Y n Qx . Since .Y; B; / is complete and gx 2 L1 ./, every function that differs from gx on a set of measure zero is also B-measurable and -integrable. Hence fx 2 L1 ./ for all x 2 X n .E [ E 0 /. The same argument shows that f y 2 L1 ./ for all y 2 Y n .F [ F 0 /. Define the W X ! R by 8Z < fx d .x/ WD Y : 0

for x 2 X n .E [ E 0 /; for x 2 E [ E 0

and WY

!R

by 8Z < f y d .y/ WD X : 0 Since

Z .x/ D

gx d

for y 2 Y n .F [ F 0 /; for y 2 F [ F 0 :

for all x 2 X n .E [ E 0 /,

Y

it follows from part (i) of Theorem 7.20 for g that  2 L1 ./. The same argument, using part (ii) of Theorem 7.20 for g, shows that 2 L1 ./. Moreover, the three integrals in (7.20) for f agree with the corresponding integrals for g because .E [ E 0 / D .F [ F 0 / D . ˝ /.Q/ D 0: Hence equation (7.20) for f follows from part (iii) of Theorem 7.20 for g. This proves Theorem 7.23. 

270

7. Product measures

Example 7.24. Assume .X; A; / is not complete. Then there exist a E 2 2X n A and N 2 A such that E  N and .N / D 0. In this case E  Y is a null set in the completion .X  Y; .A ˝ B/ ; . ˝ / /. Therefore, f WD E Y 2 L1 .. ˝ / /: However, the function f y D E is not measurable for every y 2 Y . This shows that the hypothesis that .X; A; / and .Y; B; / are complete cannot be removed in Theorem 7.23. Exercise 7.25. Continue the notation of Theorem 7.23 and suppose that the map f W X  Y ! Œ0; 1 is .A ˝ B/ -measurable. Prove that fx is B-measurable for -almost all x 2 X, that f y is A-measurable for -almost all y 2 Y , and that (7.11) continues to hold. Hint. The proof of Theorem 7.23 carries over verbatim to nonnegative measurable functions. We close this section with two remarks about the construction of product measures in the non  -finite case, where the story is considerably more subtle. These remarks are not used elsewhere in this book and can safely be ignored. Remark 7.26. Let .X; A; / and .Y; B; / be two arbitrary measure spaces. In [4, Chapter 251] Fremlin defines the function W 2X Y

! Œ0; 1

by .W / WD inf

²X 1 nD1

ˇ ³ ˇ A 2 A; Bn 2 B for n 2 N S1 .An /
.Bn / ˇˇ n and W  nD1 .An  Bn /

(7.21)

for W  X  Y and proves that it is an outer measure. He shows that the -algebra C  2XY of -measurable sets contains the product -algebra A ˝ B and calls the measure 1 WD jC W C ! Œ0; 1 the primitive product measure. By Carathéodory’s theorem (Theorem 2.4), the measure space .X  Y; C; 1 / is complete. By definition, 1 .A  B/ D .A/
.B/ for all A 2 A and all B 2 B. Fremlin then defines the complete locally determined (CLD) product measure 0 W C ! Œ0; 1

7.3. Fubini’s Theorem

271

by ˇ ² ³ ˇ E 2 A; F 2 B; ˇ 0 .W / WD sup 1 .W \ .E  F // ˇ : .E/ < 1; .F / < 1

(7.22)

He shows that .X  Y; C; 0 / is a complete measure space, that 0  1 , and 1 .W / < 1 H) 0 .W / D 1 .W / for all W 2 C. (See [4, Theorem 251I].) One can also prove that a measure W C ! Œ0; 1 satisfies .E  F / D .E/
.F / for all E 2 A and F 2 B with .E/
.F / < 1 if and only if 0    1 . With these definitions Fubini’s Theorem holds for 0 whenever the factor .Y; B; / (over which the integral is performed first) is -finite and the factor .X; A; / (over which the integral is performed second) S is either strictly localizable (i.e., there is a partition X D i 2I Xi into measurable sets with .Xi / < 1 such that a set A  X is A-measurable if and only P if A \ Xi 2 A for all i 2 I and, moreover, .A/ D i 2I .A \ Xi / for all A 2 A), or is complete and locally determined (i.e., it is semi-finite and a set A  X is A-measurable if and only if A \ E 2 A for all E 2 A with .E/ < 1). See Fremlin [4, Theorem 252B] for details. If the measure spaces .X; A; / and .Y; B; / are both  -finite, then the measures 0 and 1 agree and are equal to the completion of the product measure  ˝  on A ˝ B (see [4, Proposition 251K]). Remark 7.27. For topological spaces yet another approach to the product measure is based on the Riesz representation theorem (Theorem 3.15). Let .X; UX / and .Y; UY / be two locally compact Hausdorff spaces, denote by BX and BY their Borel  -algebras, and let X W BX ! Œ0; 1 and Y W BY ! Œ0; 1 be Borel measures. Define ƒW Cc .X  Y / ! R by Z Z



ƒ.f / WD

f .x; y/ d.y/ X

Z Z D

(7.23)

 f .x; y/ d.x/

Y

d.x/

Y

d.y/

X

for f 2 Cc .X  Y /. That the two integrals agree for every continuous function with compact support follows from Fubini’s Theorem (Theorem 7.20) for finite measure spaces. (To see this, observe that every compact set K  X  Y is contained in the product of the compact sets KX WD ¹x 2 X j .¹xº  Y / \ K ¤ ;º and

272

7. Product measures

KY WD ¹y 2 Y j .X  ¹yº/ \ K ¤ ;º.) Since ƒ is a positive linear functional, the Riesz representation theorem (Theorem 3.15) asserts that there exists a unique outer regular Borel measure 1 W BXY ! Œ0; 1 that is inner regular on open sets and a unique Radon measure 0 W BXY ! Œ0; 1 such that Z Z ƒ.f / D f d0 D f d1 XY

X Y

for all f 2 Cc .X  Y /. It turns out that in this situation the Borel -algebra BX Y is contained in the -algebra C  2X Y of Remark 7.26 and 0 D 0 jBXY ;

1 D 1 jBXY :

Recall from Lemma 7.6 that the product -algebra BX ˝ BY agrees with the Borel -algebra BXY whenever one of the spaces X or Y is second countable. If they are both second countable, then so is the product space .X  Y; UXY / (Appendix B). In this case 0 D 1 D X ˝ Y is the product measure of Definition 7.10 and 0 D 1 W C ! Œ0; 1 is its completion. (See Theorem 3.15 and Remark 7.26.)

7.4 Fubini and Lebesgue For n 2 N denote by .Rn ; An ; mn / the Lebesgue measure space on Rn and by Bn  An the Borel  -algebra on Rn with respect to the standard topology. For k; ` 2 N we identify RkC` with Rk  R` in the standard manner. Since Rn is second countable for all n, it follows from Lemma 7.6 and Theorem 2.1 that Bk ˝ B` D BkC` ;

.mk jBk / ˝ .m` jB` / D mkC` jBkC` :

(7.24)

(See Exercise 7.16.) Thus Theorem 7.17 has the following consequence. Theorem 7.28 (Fubini and Borel). Let k; ` 2 N and n WD kC`. Let f W Rn ! Œ0; 1 be Borel measurable. Then fx WD f .x;
/W R` ! Œ0; 1 and f y WD f .
; y/W Rk ! Œ0; 1 are Borel measurable for all x 2 Rk and all y 2 R` . Moreover, the functions Z k R ! Œ0; 1W x 7 ! f .x; y/ d m` .y/ R`

7.4. Fubini and Lebesgue

273

and R` ! Œ0; 1W y 7 !

Z Rk

f .x; y/ d mk .x/

are Borel measurable, and Z

Z

Z Rn

f d mn D

Rk

R`

Z

Z D



R`

Rk

f .x; y/ d m` .y/

d mk .x/

 f .x; y/ d mk .x/ d m` .y/:

(7.25)

Proof. The assertion follows directly from (7.24) and Theorem 7.17.



For Lebesgue measurable functions f W Rn ! Œ0; 1 the analogous statement is considerably more subtle. In that case the function fx , respectively f y , need not be Lebesgue measurable for all x, respectively all y. However, they are Lebesgue measurable for almost all x 2 Rk , respectively almost all y 2 R` , and the three integrals in (7.25) can still be defined and agree. The key result that one needs to prove this is that the Lebesgue measure on Rn D Rk  R` is the completion of the product of the Lebesgue measures on Rk and R` . Then the assertion follows from Exercise 7.25. Theorem 7.29. Let k; ` 2 N, define n WD k C `, and identify Rn with the product space Rk  R` in the canonical way. Denote the completion of the product space .Rk  R` ; Ak ˝ A` ; mk ˝ m` / by .Rk  R` ; .Ak ˝ A` / ; .mk ˝ m` / /. Then An D .Ak ˝ A` /

and mn D .mk ˝ m` / :

Proof. Define Cn WD ¹Œa1 ; b1 / 


 Œan ; bn / j ai ; bi 2 R and ai < bi for i D 1; : : : ; nº; n

so that Cn  Bn  An  2R for all n. We prove the assertion in three steps. Step 1. Bn  Ak ˝ A` and mn .B/ D .mk ˝ m` /.B/ for all B 2 Bn . By Lemma 7.6, we have Bn D Bk ˝B`  Ak ˝A` . It then follows from the uniqueness of a normalized translation-invariant Borel measure on Rn in Theorem 2.1 that mn jBn D .mk ˝ m` /jBn . Here is a more direct proof.

274

7. Product measures

First, assume B D E D Œa1 ; b1 / 


 Œan ; bn / 2 Cn . Define E 0 WD Œa1 ; b1 / 


 Œak ; bk /;

E 00 WD ŒakC1 ; bkC1 / 


 Œan ; bn /:

Thus E 0 2 Ck  Ak , E 00 2 C`  A` , and so E D E 0  E 00 2 Ak ˝ A` . Moreover mn .E/ D

n Y

ai / D mk .E 0 /
m` .E 00 / D .mk ˝ m` /.E/:

.bi

iD1

Second, assume B D U  Rn is open. Then there is a sequence of pairwise disjoint S sets Ei 2 Cn such that U D 1 i D1 Ei . Hence U 2 Ak ˝ A` and .mk ˝ m` /.U / D

1 X

.mk ˝ m` /.Ei / D

i D1

1 X

mn .Ei / D mn .U /:

iD1

Thus every open set is an element of Ak ˝ A` and so Bn  Ak ˝ A` . Third, assume B D K  Rn is compact. Then there is an open set U  Rn such that K  U and mn .U / < 1. Hence the set V WD U n K is open. This implies that K D U n V 2 Ak ˝ A` and .mk ˝ m` /.K/ D .mk ˝ m` /.U / D mn .U /

.mk ˝ m` /.V /

mn .V /

D mn .K/: Now let B  Rn be any Borel set. Then B 2 Ak ˝ A` , as we have seen above. Moreover, it follows from Theorem 2.13 that mn .B/ D

inf

U B U is open

mn .U / D

inf .mk ˝ m` /.U /  .mk ˝ m` /.B/

U B U is open

and mn .B/ D

inf

KB K is compact

mn .K/ D

inf

.mk ˝ m` /.K/  .mk ˝ m` /.B/:

KB K is compact

Hence, mn .B/ D .mk ˝ m` /.B/, and this proves Step 1. Step 2. Ak ˝ A`  An . We claim that

E 2 Ak H) E  R` 2 An :

(7.26)

To see this, fix a set E 2 Ak . Then there exist Borel sets A; B 2 Bk such that A  E  B and mk .B n A/ D 0. Let W Rn  Rk denote the projection onto

7.4. Fubini and Lebesgue

275

the first k coordinates. This map is continuous and hence Borel measurable by Theorem 1.20. Thus A  R` D  1 .A/ and B  R` D  1 .B/ are Borel sets in Rn . Moreover, by Step 1, mn ..B  R` / n .A  R` // D mn ..B n A/  R` / D .mk ˝ m` /..B n A/  R` / D mk .B n A/
m` .R` / D 0: Since A  R`  E  R`  B  R` , it follows that E  R` 2 An , as claimed in (7.26). A similar argument shows that F 2 A` H) Rk  F 2 An : Hence E  F D .E  R` / \ .Rk  F / 2 An for all E 2 Ak and all F 2 A` . Thus Ak ˝ A`  An and this proves Step 2. Step 3. .Ak ˝ A` / D An and .mk ˝ m` / D mn . Let A 2 An . Then there are Borel sets B0 ; B1 2 Bn such that B0  A  B1 and mn .B1 n B0 / D 0. By Step 1, B0 ; B1 2 Ak ˝ A` and .mk ˝ m` /.B1 n B0 / D 0. Hence A 2 .Ak ˝ A` / and .mk ˝ m` / .A/ D .mk ˝ m` /.B0 / D mn .B0 / D mn .A/: Thus we have proved that An  .Ak ˝ A` / ;

.mk ˝ m` / jAn D mn :

Since Ak ˝ A`  An by Step 2, it follows that mn jAk ˝A` D .mk ˝ m` / jAk ˝A` D mk ˝ m` : Now let A 2 .Ak ˝ A` / . Then there are sets A0 ; A1 2 Ak ˝ A` such that A0  A  A1 and .mk ˝ m` /.A1 n A0 / D 0. Hence A0 ; A1 2 An by Step 2 and mn .A1 n A0 / D 0. Since .Rn ; An ; mn / is complete, it follows that A n A0 2 An and so A D A0 [ .A n A0 / 2 An . Hence An D .Ak ˝ A` / . This proves Step 3 and Theorem 7.29. 

276

7. Product measures

The next result specializes Theorem 7.23 to the Lebesgue measure. Theorem 7.30 (Fubini and Lebesgue). Let k; ` 2 N and n WD kC`. Let f W Rn ! R be Lebesgue integrable and define fx .y/ WD f y .x/ WD f .x1 ; : : : ; xk ; y1 ; : : : ; y` / for x D .x1 ; : : : ; xk / 2 Rk and y D .y1 ; : : : ; y` / 2 R` . Then there are Lebesgue null sets E  Rk and F  R` such that the following holds. (i) fx 2 L1 .R` / for every x 2 Rk n E and the map W Rk ! R defined by .x/ WD

8Z
0, then Hölder’s inequality in Theorem 4.1 shows that Z h.x/ D jfx j jfx j1  jgj d m  kjfx j kq 0 kjfx j1  jgjkq ; Rn

y/. Since q 0 D p, this implies Z q=qZ0 q q 0 jfx j.1 /q jgjq d m h.x/  jfx j d m Rn Rn Z q D kf kp jf .x y/j.1 /q jg.y/jq d m.y/

where fx .y/ WD f .x

(7.34)

Rn

n

for all x 2 R . This continues to hold for  D 0. Now it follows from Minkowski’s inequality in Theorem 7.19 with the exponent s WD r=q  1 that Z q=r q r h dm khkr D Rn

Z D

qs

h

dm

Rn

 kf kpq  kf

1=s

kpq

Z

Z Rn

Rn

jf .x

Z

Z Rn

D kf kpq kf kp.1

Rn

jf .x

/q

y/j.1

/q

.1 /qs

y/j

s 1=s jg.y/jq d m.y/ d m.x/ qs

jg.y/j

1=s d m.x/ d m.y/

kgkqq :

Here the last equality follows from the fact that .1 This proves Theorem 7.33.

/qs D .1

/r D p. 

It follows from Theorem 7.33 and part (ii) of Theorem 7.32 that the convolution descends to a map L1 .Rn /  L1 .Rn / ! L1 .Rn /W .f; g/ 7 ! f  g:

(7.35)

This map is bilinear by Theorem 1.44, it is associative by part (v) of Theorem 7.32, and satisfies kf  gk1  kf k1 kgk1 by Young’s inequality in Theorem 7.33. Hence L1 .Rn / is a Banach algebra. By part (iv) of Theorem 7.32, L1 .Rn / is commutative, and, by Theorem 7.33 with q D 1 and r D p, it acts on Lp .Rn /. (A Banach algebra is a Banach space .X; k
k/ equipped with an associative bilinear map X  X ! XW .x; y/ 7! xy that satisfies the inequality kxyk  kxk kyk for all x; y 2 X.)

7.5. Convolution

283

Definition 7.34. Fix a constant 1  p < 1. A Lebesgue measurable function f W Rn ! R is called locally p-integrable if Z jf jp d m < 1 K

for every compact set K  Rn . It is called locally integrable if it is locally 1-integrable. Theorem 7.33 carries over to locally integrable functions as follows. If 1=p C 1=q D 1 C 1=r, f is locally p-integrable, and g 2 Lq .Rn / has compact support, then E.f; g/ is a Lebesgue null set and f  g is locally r-integrable. To see this, let K  Rn be any compact set and choose a compactly supported smooth function ˇ such that ˇjK  1. Then ˇf 2 Lp .Rn / and .ˇf /  g agrees with f  g on the set ¹x 2 Rn j x supp.g/  Kº. In the following theorem C01 .Rn / denotes the space of compactly supported smooth functions on Rn . Theorem 7.35. Let 1  p < 1 and 1 < q  1 such that 1=p C 1=q D 1. (i) If f W Rn ! R is locally p-integrable, then Z lim jf .x C / f .x/jp d m.x/ D 0 !0 B

for every bounded Lebesgue measurable subset B  Rn . If f 2 Lp .Rn / this continues to hold for B D Rn . (ii) If f 2 Lp .Rn / and g 2 Lq .Rn /, then f  g is uniformly continuous. If f is locally p-integrable and g 2 Lq .Rn / has compact support (or if f 2 Lp .Rn / has compact support and g is locally q-integrable), then f  g is continuous. (iii) If f W Rn ! R is locally integrable and g 2 C01 .Rn /, then f  g is smooth and @˛ .f  g/ D f  @˛ g for every multi-index ˛. (iv) C01 .Rn / is dense in Lp .Rn / for 1  p < 1. Proof. (i) Assume first that f 2 Lp .Rn / and fix a constant " > 0. By Theorem 4.15, there is a function g 2 Cc .Rn / such that kf gkp < "1=p =3. Since g is uniformly continuous, there is a ı > 0 such that, for all  2 Rn , jj < ı H) sup jg.x C / x2Rn

g.x/j
0. By part (i), there exists a ı > 0 such that, for all  2 Rn , Z  " p jj < ı H) jf .y C / f .y/jp d m.y/ < kgkq Rn Fix two elements x;  2 Rn such that jj < ı and denote fx .y/ WD f .x Then, by Hölder’s inequality in Theorem 4.1, ˇZ ˇ j.f  g/.x C / .f  g/.x/j D ˇˇ .fxC Rn

y/:

ˇ ˇ fx /g d mˇˇ

 kfxC fx kp kgkq Z 1=p D jf .y C / f .y/jp d m.y/ kgkq Rn

< ": This shows that f  g is uniformly continuous. If f is locally p-integrable and g 2 Lq .Rn / has compact support, continuity follows by taking the integral over a suitable compact set. In the opposite case continuity follows by taking the Lq -norm of g over a suitable compact set.

7.5. Convolution

285

(iii) Fix an index i 2 ¹1; : : : ; nº and denote by ei 2 Rn the i th unit vector. Fix an element x 2 Rn and choose a compact set K  Rn such that B1 .y/  K whenever x y 2 supp.g/. Let " > 0. Since @i g is continuous, there is a constant 0 < ı < 1 such that " j@i g.y C hei / @i g.y/j < Z jf j d m K n

for all y 2 R and all h 2 R with jhj < ı. Hence the fundamental theorem of calculus asserts that ˇ ˇ ˇ g.y C hei y/ g.y/ ˇ " ˇ sup ˇ @i g.y/ˇˇ < Z n h y2R jf j d m K

for all h 2 R with 0 < jhj < ı. Take h 2 Rn with 0 < jhj < ı. Then ˇ ˇ ˇ .f  g/.x C hei / .f  g/.x/ ˇ ˇ ˇ .f  @ g/.x/ i ˇ ˇ h ˇ ˇZ   g.x C he ˇ ˇ y/ g.x y/ i @i g.x y/ d m.y/ˇˇ D ˇˇ f .y/ h Rn ˇ ˇ Z ˇ g.x C hei y/ g.x y/ ˇ jf .y/j ˇˇ  @i g.x y/ d m.y/ˇˇ h Rn < ": By part (ii), the function @i .f  g/ D f  @i g is continuous for i D 1; : : : ; n. For higher derivatives the assertion follows by induction. (iv) Let f 2 Lp .Rn / and choose a compactly supported smooth function W Rn ! Œ0; 1/ such that Z supp./  B1 ;  d m D 1: Rn

Define

ı W R n ! R

by ı .x/ WD

1 x   ın ı

for ı > 0 and x 2 Rn . Then Z supp.ı /  Bı ;

Rn

ı d m D 1

286

7. Product measures

by Theorem 2.17. By part (iii), the function fı WD ı  f W Rn ! R is smooth for all ı > 0. Now fix a constant " > 0. By part (i), there exists a constant ı > 0 such that, for all y 2 Rn , Z jf .x y/ f .x/jp d m.x/ < "p : jyj < ı H) Rn

Hence, by Minkowski’s inequality in Theorem 7.19, ˇZ ˇ ˇ ˇ n

Z kfı

f kp D

R

y/

Rn

y/

1=p f .x/jp ı .y/p d m.x/ d m.y/

y/

1=p f .x/j d m.x/

Z

Z 

.f .x

Rn

ˇp 1=p ˇ ˇ f .x//ı .y/ d m.y/ˇ d m.x/

Rn

jf .x

Z  sup jyj 0 such that Bx" ./  Br . Choose  WD Br n Bx" ./;

u.x/ WD K .x/ WD K.

x/;

v WD f:

Then @ D @Br [ @B" ./ and the functions v and @v=@ vanish on @Br . Moreover, K  0. Hence Green’s formula (7.48) asserts that Z Z  @K @f   K d: (7.49) K f d m D f @ @ Rn nB" ./ @B" ./ Here the reversal of sign arises from the fact that the outward unit normal vector on @B" ./ is inward pointing with respect to . Moreover, .x/ D jx

j

1

.x

/ for x 2 @B" ./,

so @K =@.x/ D !n 1 "1

n

by (7.43). Also, by (7.42), K .x/ D

´ 2 .2

1

if n D 2;

log."/

n/

1

!n 1 "2

Hence it follows from (7.49) that Z 1 K f d m D n n ! " n R nB" ./

n

if n > 2;

DW

Z 1

for x 2 @B" ./:

."/

Z u d

."/

@B" ./

f d m:

(7.50)

B" ./

The last summand is obtained from (7.48) with u D 1, v D f ,  D B" ./. Now take the limit " ! 0. Then the first term on the right in (7.50) converges to f ./ and the second term converges to zero. This proves (7.45). It follows from Theorem 7.35 and equation (7.45) that u D .K  f / D K  f D f:

7.7. The Calderón–Zygmund inequality

295

To prove the second equation in (7.47), fix an index i 2 ¹1; : : : ; nº and a point  2 Rn . Then the divergence theorem on  WD Br n Bx" ./ asserts that Z Rn nB" ./

.Ki .

x/f .x/

K.

x/@i f .x// d m.x/

Z D

Rn nB" ./

..@i K /f C K @i f / d m

Z D

Rn nB" ./

@i .K f / d m

Z D

i K f d @B" ./

xi

Z D

."/ @B" ./

i "

f .x/ d.x/:

The last term converges to zero as " tends to zero. Hence .Ki  f /./ D .K  @i f /./ D @i .K  f /./ by Theorem 7.35. This proves Theorem 7.41.



Remark 7.42. Theorem 7.41 extends to compactly supported C 1 -functions f W Rn ! R and asserts that K  f is C 2 . However, this does not hold for continuous functions with compact support. A counterexample is u.x/ D jxj3 , which is not C 2 and satisfies f WD u D 3.n C 1/jxj. It then follows that K  ˇf (for any ˇ 2 C01 .Rn / equal to 1 near the origin) cannot be C 2 . Theorem 7.43 (Calderón–Zygmund). Fix both an integer n  2 and a number 1 < p < 1. Then there exists a constant c D c.n; p/ > 0 such that n X

@i @j u  c kuk p p

(7.51)

i;j D1

for all u 2 C01 .Rn /. Proof. See page 304. The proof is based on the exposition in Gilbarg and Trudinger [5]. 

296

7. Product measures

The Calderón–Zygmund inequality is a beautiful and deep result in the theory of partial differential equations. It extends to all functions u D K  f with f 2 C01 .Rn / and thus can be viewed as a result about the convolution operator f 7! K  f . Theorem 7.35 shows that a derivative of a convolution is equal to the convolution with the derivative. This extends to the case where the derivative only exists in the weak sense and is locally integrable. For the function K this is spelled out in equation (7.47) in Theorem 7.41. Thus the convolution of an Lp function with a function whose derivatives are integrable has derivatives in Lp . The same holds for second derivatives. (The precise formulation of this observation requires the theory of Sobolev spaces.) The remarkable fact is that the second derivatives of the fundamental solution K of Laplace’s equation are not locally integrable and, nevertheless, the Calderón–Zygmund inequality still asserts that the second derivatives of its convolution u D K  f with a p-integrable function f are p-integrable. Despite this subtlety, the proof is elementary in the case p D 2. Denote by ru WD .@1 u; : : : ; @n u/W Rn ! Rn the gradient of a smooth function uW Rn ! R. Lemma 7.44. Fix an integer n  2 and let f 2 C01 .Rn /. Then kr.Kj  f /k2  kf k2

for j D 1; : : : ; n:

(7.52)

Proof. Define u WD Kj  f . This function is smooth by Theorem 7.35, but it need not have compact support. By the divergence theorem, Z X Z Z Z n @u 2 u d (7.53) uu d m D @i .u@i u/ d m D jruj d m C Br @Br @ Br Br iD1

for all r > 0. By Poisson’s identities (7.45) and (7.47), we have u D .Kj  f / D @j .K  f / D @j .K  f / D @j f: Since f has compact support, it follows from (7.43) that there is a constant c > 0 such that ju.x/j C j@u=@.x/j  cjxj1 n for jxj sufficiently large. Hence the integral on the right in (7.53) tends to zero as r tends to infinity. Thus Z 2 jruj2 d m kruk2 D Rn

Z D

Rn

u@j f d m

Z D This proves Lemma 7.44.

Rn

.@j u/f d m  kruk2 kf k2 : 

7.7. The Calderón–Zygmund inequality

297

By Theorem 7.35, the space C01 .Rn / is dense in L2 .Rn /. Thus Lemma 7.44 shows that the linear operator f 7! @k .Kj  f / extends uniquely to a bounded linear operator from L2 .Rn / to L2 .Rn /. The heart of the proof of the Calderón– Zygmund inequality is the following delicate argument which shows that this operator also extends to a continuous linear operator from the Banach space L1 .Rn / to the topological vector space L1;1 .Rn / of weakly integrable functions introduced in Section 6.1. This argument occupies the next six pages. Recall the definition kf k1;1 WD sup tf .t/; t>0

where f .t/ WD m.A.t; f //;

A.t; f / WD ¹x 2 Rn j jf .x/j > tº:

(See equation (6.1).) Lemma 7.45. Fix an integer n  2. Then there is a constant c D c.n/ > 0 such that k@k .Kj  f /k1;1  ckf k1 (7.54) for all f 2 C01 .Rn / and all indices j; k D 1; : : : ; n. Proof. Fix two integers j; k 2 ¹1; : : : ; nº and let T W L2 .Rn / ! L2 .Rn / be the unique bounded linear operator that satisfies Tf D @k .Kj  f /

(7.55)

for f 2 C01 .Rn /. This operator is well defined by Lemma 7.44. We prove in three steps that there is a constant c D c.n/ > 0 such that kTf k1;1  c kf k1 for all f 2 L1 .Rn / \ L2 .Rn /. Throughout we abuse notation and use the same letter f to denote a function in L2 .Rn / and its equivalence class in L2 .Rn /. Step 1. There is a constant c D c.n/  1 such that the following holds. If B  Rn is a countable union of closed cubes Qi  Rn with pairwise disjoint interiors and if h 2 L2 .Rn / \ L1 .Rn / satisfies Z n hjR nB  0; h d m D 0 for all i 2 N; (7.56) Qi

then

for all t > 0.

  1 T h .t/  c m.B/ C khk1 t

(7.57)

298

7. Product measures

For i 2 N define hi W Rn ! R by hi .x/ WD

´ h.x/ if x 2 Qi ; if x … Qi :

0

Denote by qi 2 Qi the center of the cube Qi and by 2ri > 0 its side length. Then p jx qi j  nri for all x 2 Qi . Fix an element x 2 Rn n Qi . Then Kj is smooth on x Qi and so Theorem 7.35 asserts that .T hi /.x/ D .@k Kj  hi /.x/ Z D @k Kj .x y/

@k Kj .x


qi / hi .y/ d m.y/:

(7.58)

Qi

This identity is more delicate than it looks at first glance. To see this, note that equation (7.55) only holds for compactly supported smooth functions, but is not meaningful for all L2 functions f because Kj  f may not be differentiable. The function hi is not smooth, so care must be taken. Since x … Qi D supp.hi / one can approximate hi in L2 .Rn / by a sequence of compactly supported smooth functions that vanish near x (by using the mollifier method in the proof of Theorem 7.35). For the approximating sequence part (iii) of Theorem 7.35 asserts that the partial derivative with respect to the kth variable of the convolution with Kj is equal to the convolution with @k Kj near x. Now the first equality in (7.58) follows by taking the limit. The second equality follows from (7.56). It follows from (7.58) that Z j.T hi /.x/j  j@k Kj .x y/ @k Kj .x qi /jjhi .y/j d m.y/ Qi

 sup j@k Kj .x

y/

@k Kj .x

y2Qi



p nri sup jr@k Kj .x y2Qi

 c1 ri sup y2Qi

D

jx

y/j khi k1

1 khi k1 yjnC1

c1 ri khi k1 : d.x; Qi /nC1

qi /j khi k1

7.7. The Calderón–Zygmund inequality

Here d.x; Qi / WD inf jx y2Qi

299

yj and sup jyjnC1 jr@k Kj .y/j 

c1 D c1 .n/ WD max

j;k y2Rn n¹0º

n.n C 3/ : !n

Here the last inequality follows by differentiating equation (7.43). Now define p qi j < 2 nri º  Qi :

Pi WD ¹x 2 Rn j jx Then d.x; Qi /  jx

qi j

p

nri

for all x 2 Rn n Pi .

Hence 1

Z

Z Rn nPi

jT hi j d m  c1 ri

Rn nPi

.jx

qi j

Z D c1 ri

p jyj>2 nri

.jyj

p

nri /nC1

d m.x/ khi k1

1 d m.y/ khi k1 p nri /nC1

1

!n s n 1 ds p khi k1 p nri /nC1 2 nri .s p Z 1 .s C nri /n 1 ds D c1 !n ri p khi k1 s nC1 nri Z 1 ds n 1  c1 2 !n ri p khi k1 2 nri s Z

D c1 ri

D c2 khi k1 : p Here c2 D c2 .n/ WD c1 .n/2n 1 !n n  2n 1 n3=2 .n C 3/. The third step in the above computation follows from Fubini’s Theorem in polar coordinates (Exercise 7.47). Thus we have proved that Z jT hi j d m  c2 khi k1 for all i 2 N: (7.59) Rn nPi

Recall that T h and T hi are only equivalence classes in L2 .Rn /. Choose square integrable functions on Rn representing these equivalence classes and denote them by the same letters T h; T hi 2 L2 .Rn /. We prove that there is a Lebesgue null set E  Rn such that jT h.x/j 

1 X iD1

jT hi .x/j

for all x 2 Rn n E:

(7.60)

300

7. Product measures

P To see this, note that the sequence `iD1 hi converges to h in L2 .Rn / as ` tends to P` infinity. So the sequence iD1 T hi converges to T h in L2 .Rn / by Lemma 7.44. By Corollary 4.10, a subsequence converges almost everywhere. Hence there exist a Lebesgue null set E  Rn and a sequence of integers 0 < `1 < `2 < `3 <


P T hi .x/ converges to T h.x/ as  tends to infinity for such that the sequence `i D1 P1 P` n n all x 2 R n E. Since j iD1 T hi .x/j  iD1 jT hi .x/j for all x 2 R , this proves (7.60). Now put 1 [ A WD Pi : i D1

Then it follows from (7.59), (7.60), and Theorem 1.38 that Z Z 1 X jT hj d m  jT hi j d m Rn nA

Rn nA iD1

D

1 Z X i D1



Rn nA

jT hi j d m

1 Z X i D1

 c2

Rn nPi

1 X

jT hi j d m

khi k1

i D1

D c2 khk1 : Moreover, m.A/ 

1 X

m.Pi / D c3

i D1

where c3 D c3 .n/ WD

1 X

m.Qi / D c3 m.B/;

iD1

m.B2pn / m.Œ 1; 1n /

D m.Bpn / D !n nn=2

1

:

Hence tT h .t/  tm.A/ C tm.¹x 2 Rn n A j jT h.x/j > tº/ Z  tm.A/ C jT hj d m Rn nA

 c3 tm.B/ C c2 khk1  c4 .tm.B/ C khk1 /

7.7. The Calderón–Zygmund inequality

301

for all t > 0, where c4 D c4 .n/ WD max¹c2 .n/; c3 .n/º  max¹2n

1 3=2

n

.n C 3/; !n nn=2

1

º:

This proves Step 1. Step 2 (Calderón–Zygmund decomposition). Let f 2 L2 .Rn /\L1 .Rn / and t > 0. Then there exists a countable collection of closed cubes Qi  Rn with pairwise disjoint interiors such that Z 1 jf j  2n m.Qi / for all i 2 N (7.61) m.Qi / < t Qi and jf .x/j  t where B WD

S1

iD1

for almost all x 2 Rn n B;

(7.62)

Qi .

For  2 Zn and ` 2 Z define Q.; `/ WD ¹x 2 Rn j 2 ` i  xi  2 ` .i C 1/º: Let

Q WD ¹Q.; `/ j  2 Zn ; ` 2 Zº

and define the subset Q0  Q by R ˇ ³ ² ˇ tm.Q/ < Q jf j d m and, for all Q0 2 Q; R : Q0 WD Q 2 Q ˇˇ Q ¨ Q0 H) Q0 jf j d m  tm.Q0 / Then every decreasing sequence of cubes in Q contains at most one element of Q0 . Hence every element of Q0 satisfies (7.61) and any two cubes in Q0 have disjoint S interiors. Define B WD Q2Q0 Q. We prove that Z 1 x 2 Rn n B; x 2 Q 2 Q H) jf j d m  t: (7.63) m.Q/ Q Suppose, by contradiction, that there existsRan element x 2 Rn n B and a cube Q 2 Q such that x 2 Q and tm.Q/ < Q jf j d m. Then, since kf k1 < 1, R there is a maximal cube Q 2 Q such that x 2 Q and tm.Q/ < Q jf j d m. Such a maximal cube would be an element of Q0 and hence x 2 B, a contradiction. This proves (7.63). Now Theorem 6.14 asserts that there exists a Lebesgue null set E  Rn n B such that every element of Rn n .B [ E/ is a Lebesgue point of f . By (7.63), every point x 2 Rn n .B [ E/ is the intersection point of a decreasing sequence of cubes over which jf j has mean value at most t. Hence it follows from Theorem 6.16 that jf .x/j  t for all x 2 Rn n .B [ E/. This proves Step 2.

302

7. Product measures

Step 3. Let c D c.n/  1 be the constant in Step 1. Then kTf k1;1  .2nC1 C 6c/ kf k1

(7.64)

for all f 2 L2 .Rn / \ L1 .Rn /. Fix a function f 2 L2 .Rn / \ L1 .Rn / and a constant t > 0. Let the Qi be as in Step 2 and define [ B WD Qi : i 1 t

R

j d m for all i by Step 2, and hence Z X 1 1X jf j d m  kf k1 : m.B/ D m.Qi /  t t Qi

Then m.Qi /
0 such that k@i .Kj  f /kp  ckf kp

(7.67)

for all f 2 C01 .Rn / and all i; j D 1; : : : ; n. Proof. For p D 2 this estimate was established in Lemma 7.44 with c D 1. Second, suppose 1 < p < 2 and let c1 .n/ be the constant of Lemma 7.45. For i; j D 1; : : : ; n denote by Tij W L2 .Rn / ! L2 .Rn / the unique bounded linear operator that satisfies Tij f D @i .Kj  f /

for f 2 C01 .Rn /.

Then kTij f k1;1  c1 .n/ kf k1 for all f 2 C01 .Rn / and all i; j by Lemma 7.45. Since C01 .Rn / is dense in L2 .Rn / \ L1 .Rn / by Theorem 7.35, it follows that kTij f k1;1  c1 .n/ kf k1 for all f 2 L2 .Rn / \ L1 .Rn /. Hence Theorem 7.37 (with q D 2) asserts that (7.67) holds with  c D c.n; p/ WD 2

.2

p p/.p

1=p 1/

c1 .n/2=p

1

:

Third, suppose 2 < p < 1 and choose 1 < q < 2 such that 1=p C 1=q D 1. Then it follows from Theorem 7.41, integration by parts, Hölder’s inequality, and from what we have just proved that, for all f; g 2 C01 .Rn /, Z Z .@i .Kj  f //g d m D .@i @j f /g d m Rn

Rn

Z D

Rn

f .@i @j g/ d m

Z D

Rn

f .@i .Kj  g// d m

 kf kp k@i .Kj  g/kq  c.n; q/ kf kp kgkq : Since C01 .Rn / is dense in Lq .Rn / by Theorem 7.35, and the Lebesgue measure is semi-finite, it follows from Lemma 4.34 that k@i .Kj  f /kp  c.n; q/ kf kp for all f 2 C01 .Rn /. This proves Theorem 7.46. 

304

7. Product measures

Proof of Theorem 7.43. Fix both an integer n  2 and a number 1 < p < 1. Let c D c.n; p/ be the constant of Theorem 7.46 and let u 2 C01 .Rn /. Then @j u D @j .K  u/ D Kj  u by Theorem 7.41. Hence it follow from Theorem 7.46 with f D u that k@i @j ujp D k@i .Kj  u/jp  c.n; p/ kukp for i; j D 1; : : : ; n. This proves Theorem 7.43.



7.8 Exercises Exercise 7.47 (Lebesgue measure on the sphere). For n 2 N, let .Rn ; An ; mn / the Lebesgue measure space, denote the open unit ball by B n WD ¹x 2 Rn j jnj < 1º, and the unit sphere by Sn For A  S n

1

1

WD @B n D ¹x 2 Rn j jxj D 1º:

define A˙ WD ¹x 2 B n

1

p j .x; ˙ 1

jxj2 / 2 Aº:

Prove that the collection AS WD ¹A  S n

1

j AC ; A 2 An

1º

is a -algebra and that the map W AS ! Œ0; 1 defined by 1

Z .A/ WD

AC

p 1

1

Z jxj2

d mn

1 .x/

C A

p 1

jxj2

d mn

1 .x/

for A 2 AS is a measure. Prove Fubini’s Theorem in polar coordinates as stated below. Use it to prove that !n WD .S n

1

/ < 1:

7.8. Exercises

305

Fubini’s Theorem for polar coo rdinates. Let f W Rn Lebesgue integrable. For r  0 and x 2 S n 1 define

! R be

f r .x/ WD fx .r/ WD f .rx/: Then there is a set E 2 AS such that .E/ D 0 and fx 2 L1 .Œ0; 1// for all x 2 S n

1

n E,

and there is a Lebesgue null set F  Œ0; 1/ such that f r 2 L1 ./

for all r 2 Œ0; 1/ n F .

Define gW S n

1

!R

and

hW Œ0; 1/ ! R

by g.x/ WD

8Z ˆ
0.

(7.69)

306

7. Product measures

Exercise 7.48 (divergence theorem). Let f W Rn ! R be a smooth function. Prove that Z Z @i f d mn D xi f .x/ d.x/: (7.70) Bn

Sn

1

Hint. Assume first that i D n. Use the fundamental theorem of calculus and Fubini’s Theorem (Theorem 7.30) for Rn D Rn 1  R. Exercise 7.49. Prove that Z 1=" Z 1 sin.x/  sin.x/ dx WD lim dx D : "!0 x x 2 " 0 Hint. Use the identity 1

Z

e

rt

dt D 1=r

for r > 0

0

and Fubini’s Theorem. Exercise 7.50. Define the function f W R2 ! R by

sign.xy/ f .x; y/ WD 2 ; x C y2

sign.z/ WD

8 ˆ 1 ˆ
0;

0

if z D 0;

ˆ ˆ :

1

if z < 0;

for .x; y/ ¤ 0,

and by f .0; 0/ WD 0: Prove that fx ; f y W R ! R R are Lebesgue integrable for allR x; y 2 R. Prove that the functions R ! RW x 7! R fx d m1 and R ! RW y 7! R f y d m1 are Lebesgue integrable and   Z Z Z Z f .x; y/ d m1 .x/ d m1 .y/ D f .x; y/ d m1 .y/ d m1 .x/: R

R

Prove that f is not Lebesgue integrable.

R

R

7.8. Exercises

307

Exercise 7.51. Let .X; A; / and .Y; B; / be two -finite measure spaces and let f 2 L1 ./ and g 2 L1 ./. Define hW X  Y

!R

by h.x; y/ WD f .x/g.y/;

for x 2 X and y 2 Y:

Prove that h 2 L1 . ˝ / and Z Z Z h d. ˝ / D f d g d: X Y

X

Y

Exercise 7.52. Let .X; A; / and .Y; B; / be two -finite measure spaces and let W A ˝ B ! R be any measure such that .A  B/ D .A/.B/ for all A 2 A and all B 2 B. Prove that  D  ˝ . Exercise 7.53. Define W R ! R by .x/ WD

´ 1

cos.x/ for 0  x  2;

0

otherwise:

Define the functions f; g; hW R ! R by f .x/ WD 1;

0

g.x/ WD  .x/;

Z

x

h.x/ WD

.t/ dt 1

for x 2 R. Prove that .f  g/  h D 0 and f  .g  h/ > 0. Thus the convolution need not be associative on nonintegrable functions. Compare this with part (v) of Theorem 7.32. Prove that E.jf j  jgj; jhj/ D E.jf j; jgj  jhj/ D R while E.f  g; h/ D E.f; g  h/ D ;: Exercise 7.54. Let .R; A; m/ be the Lebesgue measure space, let B  A be the Borel -algebra, and denote by M the Banach space of all signed Borel measures W B ! Œ0; 1/ with the norm kk WD jj.R/. (See Exercise 5.34.) The convolution of two signed measures ;  2 M is the map   W B ! R

308

7. Product measures

defined by .  /.B/ WD . ˝ /.¹.x; y/ 2 R2 j x C y 2 Bº/

(7.71)

for B 2 B, where . ˝ / WD C ˝  C C  ˝ 

C ˝ 

 ˝ C:

(See Definition 5.13 and Theorem 5.20.) Prove the following. (i) If ;  2 M, then    2 M and k  k  kk kk : (ii) There exists a unique element ı 2 M such that ıD for all  2 M. (iii) The convolution product on M is commutative, associative, and distributive. Thus M is a commutative Banach algebra with unit. (iv) If f 2 L1 .R/ and

f W B ! R

is defined by

Z f .B/ WD

f d m for B 2 B;

B

then f 2 M and kf k D kf k1 . (v) If f; g 2 L1 .R/, then f  g D f g : (vi) Let ; ;  2 M. Then  D    if and only if Z Z f d D f .x C y/d. ˝ /.x; y/ R2

R

for all bounded Borel measurable functions f W R ! R. (vii) If ;  2 M and B 2 B, then Z .  /.B/ D

.B R

t/ d.t/:

Chapter 8

The Haar measure

The purpose of this last chapter is to prove the existence and uniqueness of a normalized invariant Radon measure on any compact Hausdorff group. In the case of a locally compact Hausdorff group the theorem asserts the existence of a left-invariant Radon measure that is unique up to a scaling factor. An example is the Lebesgue measure on Rn . A useful exposition is the paper by Gert K. Pedersen [16], which also discusses the original references.

8.1 Topological groups Let G be a group, in multiplicative notation, with the group operation G  G ! GW .x; y/ 7 ! xy

(8.1)

the unit 1 2 G, and the inverse map G ! GW x 7 ! x

1

:

(8.2)

A topological group is a pair .G; U/ consisting of a group G and a topology U  2G such that the group multiplication (8.1) and the inverse map (8.2) are continuous. Here the continuity of the group multiplication (8.1) is understood with respect to the product topology on G  G (see Appendix B). A locally compact Hausdorff group is a topological group .G; U/ such that the topology is locally compact and Hausdorff (see page 95). Example 8.1. Let G be any group and define U WD ¹;; Gº. Then .G; U/ is a compact topological group, but is not Hausdorff unless G D ¹1º. Example 8.2 (discrete groups). Let G be any group. Then the pair .G; U/ with the discrete topology U WD 2G is a locally compact Hausdorff group, called a discrete group. Examples of discrete groups (where the discrete topology appears naturally) are the additive group Zn , the multiplicative group SL.n; Z/ of integer

310

8. The Haar measure

n  n-matrices with determinant one, the mapping class group of isotopy classes of diffeomorphisms of any manifold, and every finite group. Example 8.3 (Lie groups). Let G  GL.n; C/ be any subgroup of the general linear group of invertible complex n  n-matrices that is closed as a subset of GL.n; C/ with respect to the relative topology, i.e., GL.n; C/ n G is an open set in Cnn . Let U  2G be the relative topology on G, i.e., U  G is open if and only if there is an open subset V  Cnn such that U D G \ V . Then .G; U/ is a locally compact Hausdorff group. In fact, it is a basic result in the theory of Lie groups that every closed subgroup of GL.n; C/ is a smooth submanifold of Cnn and hence is a Lie group. Specific examples of Lie groups are the general linear groups GL.n; R/ and GL.n; C/, the special linear groups SL.n; R/ and SL.n; C/ of real and complex n  n-matrices with determinant one, the orthogonal group O.n/ of matrices x 2 Rnn such that x T x D 1, the special orthogonal group SO.n/ WD O.n/ \ SL.n; R/, the unitary group U.n/ of matrices x 2 Cnn such that x  x D 1, the special unitary group SU.n/ WD U.n/ \ SL.n; C/, the group Sp.1/ of the unit quaternions, the unit circle S 1 D U.1/ in the complex plane, the torus Tn WD S 1 


 S 1 (n times), or, for any multi-linear form  W .Cn /k ! C, the group of all matrices x 2 GL.n; C/ that preserve  . The additive groups Rn and Cn are also Lie groups. Lie groups form an important class of locally compact Hausdorff groups and play a central role in differential geometry. Example 8.4. If .V; k
k/ is a normed vector space (Example 1.11), then the additive group V is a Hausdorff topological group. It is locally compact if and only if V is finite-dimensional. Example 8.5. The rational numbers Q with the additive structure form a Hausdorff topological group with the relative topology as a subset of R. It is totally disconnected (every connected component is a single point), but does not have the discrete topology. It is not locally compact. Example 8.6 (p-adic integers). Fix a prime number p 2 N and denote by N0 WD N [ ¹0º

the set of nonnegative integers. For x; y 2 Z define dp .x; y/ WD jx

yjp WD inf¹p

k

j k 2 N0 ; x

y 2 p k Zº:

(8.3)

8.1. Topological groups

311

Then the function dp W Z  Z ! Œ0; 1 is a distance function and so .Z; dp / is a metric space. It is not complete. Its completion is denoted by Zp and called the ring of p-adic integers. Here is another description of the p-adic integers. Consider the sequence of projections kC1

k




! Z=p k Z ! Z=p k

1

k

3

1

2

1

Z !


! Z=p 2 Z ! Z=p Z ! ¹1º:

The inverse limit of this sequence of maps is the set of sequences Zp WD ¹x D .xk /k2N0 j xk 2 Z=p k Z; k .xk / D xk

1

for all k 2 Nº:

This set is a commutative ring with unit. Addition and multiplication are defined term by term, i.e., x C y WD .xk C yk /x2N0 ;

xy WD .xk yk /x2N0

for x D .xk /k2N0 2 Zp and y D .yk /k2N0 2 Zp . The ring of p-adic integers is a compact metric space with dp .x; y/ WD inf¹p

k

j k 2 N0 ; xk D yk º:

(8.4)

The inclusion of Z into the p-adic integers is given by p W Z ! Zp ;

p .x/ WD .x mod p k /k2N :

This is an isometric embedding with respect to the distance functions (8.3) and (8.4). The additive p-adic integers form an uncountable compact Hausdorff group (with the topology of a Cantor set) that is not a Lie group. Example 8.7 (p-adic Rationals). Fix a prime number p 2 N. Write a nonzero rational number x 2 Q in the form x D p k a=b where k 2 Z and the numbers a 2 Z and b 2 N are relatively prime to p, and define jxjp WD p k . For x D 0, define j0jp WD 0. Define the function dp W Q  Q ! Œ0; 1/ by dp .x; y/ WD jx

yjp ° WD inf p k

ˇ ˇ ˇ k 2 Z; x

± a y D p k ; a 2 Z; b 2 N n p N : b

(8.5)

312

8. The Haar measure

Then .Q; dp / is a metric space. The completion of Q with respect to dp is denoted by Qp and is called the field of p-adic rational numbers. It can also be described as the quotient field of the ring of p-adic integers in Example 8.6. The multiplicative group of nonzero p-adic rationals is a locally compact Hausdorff group that is not a Lie group. One can also consider groups of matrices whose entries are p-adic rationals. Such groups play an important role in number theory. Example 8.8 (infinite products). Let I be any index set and, for i 2 I , let Gi be a compact Hausdorff group. Then the product Y G WD Gi i2I

F is a compact Hausdorff group. Its elements are maps I ! i 2I Gi W i 7! xi such that xi 2 Gi for all i 2 I . Write such a map as x D .xi /i2I . The product topology on G is defined as the smallest topology such that the obvious projections Q i W G ! Gi are continuous. Thus the (infinite) products U D i2I Ui of open sets Ui  Gi , such that Ui D Gi for all, but finitely many i , form a basis for the topology of G. (See Appendix B for #I D 2.) The product topology is obviously Hausdorff and Tychonoff’s theorem asserts that it is compact (see Munkres [14]). An uncountable product of nontrivial groups Gi is not first countable. Example 8.9. Let .X; k
k/ be a Banach algebra with a unit 1 and the product X  X ! XW .x; y/ 7! xy. (See page 282.) Then the group of invertible elements G WD ¹x 2 X j there exists y 2 X such that xy D yx D 1º is a Hausdorff topological group. Examples include the group of nonzero quaternions, the general linear group of a finite-dimensional vector space, the group of bijective bounded linear operators on a Banach space, and the multiplicative group of nowhere vanishing real-valued continuous functions on a compact topological space. In general G is not locally compact.

8.2 Haar measures Throughout let G be a locally compact Hausdorff group, in multiplicative notation, and denote by B  2G its Borel -algebra. We begin our discussion with a technical lemma about continuous functions on G. Lemma 8.10. Let f 2 Cc .G/ and fix a constant " > 0. Then there exists an open neighborhood U of 1 such that, for all x; y 2 G, x

1

y 2 U H) jf .x/

f .y/j < ":

(8.6)

8.2. Haar measures

313

Proof. Choose an open neighborhood U0  G with compact closure and define K WD ¹ab

1

j a 2 supp.f /; b 2 Ux0 º:

This set is compact because the maps (8.1) and (8.2) are continuous. Also, x … K; x

1

y 2 U0 H) f .x/ D f .y/ D 0

(8.7)

for all x; y 2 G. (If y 2 supp.f / and x 1 y 2 U0 , then x D y.x 1 y/ 1 2 K.) Since f is continuous there exists, for each x 2 K, an open neighborhood V .x/  G of x such that y 2 V .x/ H) jf .x/

f .y/j
0 for every nonempty open set U  G. (ii) G admits a right Haar measure , unique up to a positive factor. Every such measure satisfies .U / > 0 for every nonempty open set U  G. (iii) Assume G is compact. Then G admits a unique invariant Haar measure  such that .G/ D 1. This measure satisfies .B 1 / D .B/ for all B 2 B and .U / > 0 for every nonempty open set U  G. Proof. See page 327.



8.2. Haar measures

315

Examples of Haar measures are the restriction to the Borel -algebra of the Lebesgue measure on Rn (where the group structure is additive), the restriction to the Borel -algebra of the measure  on S 1 D U.1/ or on S 3 D Sp.1/ in Exercise 7.47, the measure dt=t on the multiplicative group of the positive real numbers, and the counting measure on any discrete group. The proof of Theorem 8.12 rests on the Riesz representation theorem (Theorem 3.15) and the following result about positive linear functionals. Definition 8.13. Let G be a locally compact Hausdorff group. A linear functional ƒW Cc .G/ ! R is called left-invariant if ƒ.f ı Lx / D ƒ.f /

(8.12)

for all f 2 Cc .G/ and all x 2 G. It is called right-invariant ƒ.f ı Rx / D ƒ.f /

(8.13)

for all f 2 Cc .G/ and all x 2 G. It is called invariant if it is both left- and rightinvariant. It is called a left Haar integral if it is left-invariant, positive, and nonzero. It is called a right Haar integral if it is right-invariant, positive, and nonzero. Theorem 8.14 (Haar). Every locally compact Hausdorff group G admits a left Haar integral, unique up to a positive factor. Moreover, if ƒW Cc .G/ ! R is a left Haar integral and f 2 Cc .G/ is a nonnegative function that does not vanish identically, then ƒ.f / > 0. 

Proof. See page 318.

The proof of Theorem 8.14 given below follows the notes of Pedersen [16], which are based on a proof by Weil. Our exposition benefits from elegant simplifications due to Urs Lang [11]. In preparation for the proof it is convenient to introduce some notation. Let CcC .G/ WD ¹f 2 Cc .G/ j f  0; f 6 0º

(8.14)

be the space of nonnegative continuous functions with compact support that do not vanish identically. A function ƒW CcC .G/ ! Œ0; 1/ is called • additive if ƒ.f C g/ D ƒ.f / C ƒ.g/ for all f; g 2 CcC .G/, • subadditive if ƒ.f C g/  ƒ.f / C ƒ.g/ for all f; g 2 CcC .G/, • homogeneous if ƒ.cg/ D cƒ.f / for all f 2 CcC .G/ and all c > 0, • monotone if f  g implies ƒ.f /  ƒ.g/ for all f; g 2 CcC .G/, • left-invariant if ƒ.f ı Lx / D ƒ.f / for all f 2 CcC .G/ and all x 2 G.

316

8. The Haar measure

Every additive functional ƒW CcC .G/ ! Œ0; 1/ is necessarily homogeneous and monotone. Moreover, every positive linear functional on Cc .G/ restricts to an additive functional ƒW CcC .G/ ! Œ0; 1/ and, conversely, every additive functional ƒW CcC .G/ ! Œ0; 1/ extends uniquely to a positive linear functional on Cc .G/. This is the content of the next lemma. Lemma 8.15. Let ƒW CcC .G/ ! Œ0; 1/ be an additive functional. Then there is a unique positive linear functional on Cc .G/ whose restriction to CcC .G/ agrees with ƒ. If ƒ is left-invariant, then so is its linear extension to Cc .G/. Proof. We prove that ƒ is monotone. Let f; g 2 CcC .G/ such that f  g. If f ¤ g, then g f 2 CcC .G/ and hence ƒ.f /  ƒ.f / C ƒ.g

f / D ƒ.g/

by additivity. If f D g there is nothing to prove. We prove that ƒ is homogeneous. Let f 2 CcC .G/. Then ƒ.nf / D nƒ.f / for all n 2 N by additivity and induction. If c D m=n is a positive rational number, then ƒ.f / D nƒ.f =n/ and hence ƒ.cf / D mƒ.f =n/ D cƒ.f /. If c > 0 is irrational, then it follows from monotonicity that aƒ.f / D ƒ.af /  ƒ.cf /  ƒ.bf / D bƒ.f / for all a; b 2 Q with 0 < a < c < b, and this implies ƒ.cf / D cƒ.f /. Now define ƒ.0/ WD 0 and, for f 2 Cc .G/, ƒ.f / WD ƒ.f C /

ƒ.f /:

If f; g 2 Cc .G/, then f C C g C C .f C g/ D f

C g C .f C g/C , hence

ƒ.f C / C ƒ.g C / C ƒ..f C g/ / D ƒ.f / C ƒ.g / C ƒ..f C g/C / by additivity, and hence ƒ.f / C ƒ.g/ D ƒ.f C g/. Moreover, . f /C D f . f / D f C , and so ƒ. f / D ƒ.f /

ƒ.f C / D

and

ƒ.f /:

Hence it follows from homogeneity that ƒ.cf / D cƒ.f / for all f 2 Cc .G/ and all c 2 R. This shows that the extended functional is linear. If the original functional ƒW CcC .G/ ! Œ0; 1/ is left-invariant, then so is the extended linear functional on Cc .G/ because .f ı Lx /˙ D f ˙ ı Lx for all f 2 Cc .G/ and all x 2 G. This proves Lemma 8.15. 

8.2. Haar measures

317

Consider the space ˇ ² ³ ˇ ƒ is subadditive, monotone, C L WD ƒW Cc .G/ ! .0; 1/ ˇˇ : homogeneous, and left-invariant

(8.15)

The strategy of the proof of Theorem 8.14 is to construct certain functionals ƒg 2 L associated to functions g 2 CcC .G/ supported near the identity element and to construct the required positive linear functional ƒW Cc .G/ ! R as a suitable limit where the functions g converge to the Dirac ı-function at the identity. The precise definition of the ƒg involves the following construction. Denote by P the set of all Borel measures W B ! Œ0; 1/ of the form  WD

k X

˛i ıxi

(8.16)

iD1

where k 2 N, ˛1 ; : : : ; ˛k are positive real numbers, x1 ; : : : ; xk 2 G, and ıxi is the Dirac measure at xi (see Example 1.31). The norm of a measure  2 P of the form (8.16) is defined by kk WD .G/ D

k X

˛i > 0:

(8.17)

i D1

P If  WD j`D1 ˇj ıyj 2 P is any other such measure, define the convolution product of  and  by k X ` X    WD ˛i ˇj ıxi yj : i D1 j D1

This product is not commutative in general. It satisfies k  k D kk kk. Associated to a measure  2 P of the form (8.16) are two linear operators L ; R W Cc .G/ ! Cc .G/ defined by .L f /.a/ WD

k X

˛i f .xi a/;

.R f /.a/ WD

i D1

k X

˛i f .axi /

(8.18)

iD1

for f 2 Cc .G/ and a 2 G. The next two lemmas establish some basic properties of the operators L and R . Denote by kf k1 WD sup jf .x/j x2G

the supremum norm of a compactly supported function f W G ! R.

318

8. The Haar measure

Lemma 8.16. Let ;  2 P, f 2 Cc .G/, and x 2 G. Then L ı L D L

f ı Lx D Lıx f;

kL f k1  kk kf k1 ;

R ı R D R ;

f ı Rx D Rıx f;

kR f k1  kk kf k1 ;

L ı R D R ı L : 

Proof. The assertions follow directly from the definitions.

Lemma 8.17. Let f; g 2 CcC .G/. Then there exists a  2 P such that f  L g: Proof. Fix an element y 2 G such that g.y/ > 0. For x 2 G define ˇ ˇ f .x/ C 1 Ux WD a 2 G ˇˇ f .a/ < g.yx g.y/ ²

1

³ a/

This set is an open neighborhood of x. Since f has compact support, there exist finitely many points x1 ; : : : ; xk 2 G such that supp.f /  Ux1 [


[ Uxk . Define  WD

k X f .xi / C 1 i D1

g.y/

ıyx

1 i

:

Then f .a/ 

k X f .xi / C 1 iD1

g.y/

g.yxi 1 a/ D .L g/.a/

for all a 2 supp.f / and hence f  L g. This proves Lemma 8.17.



Proof of Theorem 8.14. The proof has five steps. Step 1 is the main construction of a subadditive functional Mg W CcC .G/ ! .0; 1/ associated to a function g 2 CcC .G/. Step 2 shows that Mg is asymptotically linear as g concentrates near the unit 1. The heart of the convergence proof is Step 3 and is due to Cartan. Step 4 proves uniqueness and Step 5 proves existence.

8.2. Haar measures

319

Step 1. For f 2 CcC .G/ define Mg .f / WD M.f I g/ WD inf¹kk j  2 P; f  L gº:

(8.19)

Then the following holds. (i) M.f I g/ > 0 for all f; g 2 CcC .G/. (ii) For every g 2 CcC .G/, the functional Mg W CcC .G/ ! .0; 1/ is subadditive, homogeneous, monotone, and left-invariant, and hence is an element of L. (iii) Let ƒ 2 L. Then ƒ.f /  M.f I g/ƒ.g/

for all f; g 2 CcC .G/:

(8.20)

In particular, M.f I h/  M.f I g/M.gI h/ for all f; g; h 2 CcC .G/. (iv) M.f I f / D 1 for all f 2 CcC .G/. We prove part (ii). Monotonicity follows directly from the definition. Homogeneity follows from the identities Lc g D cL g and kck D c kk. To prove leftinvariance, observe that .L g/ ı Lx D Lıx g;

k  ıx k D kk

for all  2 P by Lemma 8.16. Since f  L g if and only if f ı Lx  .L g/ ı Lx , this proves left-invariance. To prove subadditivity, fix a constant " > 0 and choose ; 0 2 P such that f  L g;

f 0  L0 g;

" kk < M.f I g/ C ; 2

0

 < M.f 0 I g/ C " : 2

Then f C f 0  L g C L0 g D LC0 g and hence

M.f C f 0 I g/   C 0 D kk C 0 < M.f I g/ C M.f 0 I g/ C ": Thus M.f C f 0 I g/ < M.f I g/ C M.f 0 I g/ C " This proves subadditivity and part (ii).

for all " > 0.

320

8. The Haar measure

We prove part (iii). Fix a functional ƒ 2 L. We prove first that ƒ.L f /  kk ƒ.f / for all f 2 CcC .G/ and all  2 P. To see this write  D P L f D kiD1 ˛i .f ı Lxi /, and hence ƒ.L f / 

k X

(8.21) Pk

iD1

˛i ıxi . Then

ƒ.˛i .f ı Lxi //

i D1

D

k X

˛i ƒ.f ı Lxi /

i D1

D

k X

˛i ƒ.f /

i D1

D kk ƒ.f /: Here the first step follows from subadditivity, the second step follows from homogeneity, the third step follows from left-invariance, and the last step follows from the definition of kk. This proves (8.21). Now let f; g 2 CcC .G/. By Lemma 8.17, there is a  2 P such that f  L g. Since ƒ is monotone this implies ƒ.f /  ƒ.L g/  kk ƒ.g/ by (8.21). Now take the infimum over all  2 P such that f  L g to obtain ƒ.f /  M.f I g/ƒ.g/. We prove parts (i) and (iv). Since the map CcC .G/ ! .0; 1/W f 7! kf k1 ; is an element of L, it follows from part (iii) that kf k1  M.f I g/ kgk1

(8.22)

and hence M.f I g/ > 0 for all f; g 2 CcC .G/. Next observe that M.f I f /  1 by (8.22) and M.f I f /  1 because f D Lı1 f . This proves Step 1. Step 2. Let f; f 0 2 CcC .G/ and let " > 0. Then there is an open neighborhood U  G of 1 such that every g 2 CcC .G/ with supp.g/  U satisfies Mg .f / C Mg .f 0 / < .1 C "/Mg .f C f 0 /:

(8.23)

By Urysohn’s lemma (Theorem A.1), there is a function  2 Cc .G/ such that .x/ D 1 for all x 2 supp.f / [ supp.f 0 /. Choose a constant 0 < ı  1=2 such that 2ı C 2ıkf C f 0 k1 M.I f C f 0 / < ": (8.24)

8.2. Haar measures

321

Define h WD f C f 0 C ıkf C f 0 k1 : Then f = h and f 0 = h extend to continuous functions on G with compact support by setting them equal to zero on G n supp./. By Lemma 8.10, there exists an open neighborhood U  G of 1 such that ˇ ˇ ˇ ˇ ˇ f .x/ f .y/ ˇ ˇ f 0 .x/ f 0 .y/ ˇ 1 ˇCˇ ˇ 0. Then there is an open neighborhood U  G of 1 with the following significance. For every g 2 CcC .G/ such that supp.g/  U;

g.x/ D g.x

1

for all x 2 G;

/

(8.25)

there exists an open neighborhood W  G of 1 such that every h 2 CcC .G/ with supp.h/  W satisfies the inequality M.f I g/Mh .g/  .1 C "/Mh .f /:

(8.26)

This inequality continues to hold with Mh replaced by any left-invariant positive linear functional ƒW Cc .G/ ! R. By Urysohn’s lemma (Theorem A.1) there is a function  2 CcC .G/ such that .x/ D 1 for all x 2 K WD supp.f /. Choose "0 and "1 such that 1 C "0  1 C "; 1 "0

0 < "0 < 1;

"1 WD

"0 : 2M.I f /

(8.27)

By Lemma 8.10, there exists an open neighborhood U  G of 1 such that x

1

y 2 U H) jf .x/

f .y/j < "1

(8.28)

for all x; y 2 G. We prove that the assertion of Step 3 holds with this neighborhood U . Fix a function g 2 CcC .G/ that satisfies (8.25). Put "2 WD

"1 : 2M.f I g/

(8.29)

Use Lemma 8.10 to find an open neighborhood V  G of 1 such that xy

1

2 V H) jg.x/

g.y/j < "2

(8.30)

for all x; y 2 G. Then the sets xV for x 2 K form an open cover of K. Hence S` there exist finitely many points x1 ; : : : ; x` 2 K such that K  i D1 xi V . By Theorem A.4, there exist functions 1 ; : : : ; ` 2 CcC .G/ such that 0  i  1;

supp.i /  xi V;

` X

i jK  1:

(8.31)

i D1

It follows from Step 2 by induction that there exists an open neighborhood W  G of 1 such that every h 2 CcC .G/ with supp.h/  W satisfies ` X

Mh .i f / < .1 C "0 / Mh .f /:

iD1

We prove that every h 2 CcC .G/ with supp.h/  W satisfies (8.26).

(8.32)

8.2. Haar measures

323

For x 2 G define the function Fx 2 Cc .G/ by Fx .y/ WD f .y/g.y

1

x/

for y 2 G:

(8.33)

It follows from (8.25) and (8.28) that f .x/g.y 1 x/ f .y/g.y 1 x/  "1 g.y 1 x/ for all x; y 2 G. Since g.y 1 x/ D g.x 1 y/ D .gıLx 1 /.y/ by (8.25), this implies f .x/g ı Lx 1  Fx C "1 g ı Lx 1 : Hence, for all x 2 G and all h 2 CcC .G/, f .x/Mh .g/  Mh .Fx / C "1 Mh .g/

(8.34)

Now fix a function h 2 CcC .G/ with supp.h/  W . By (8.30) and (8.31), x/  i .y/f .y/.g.xi 1 x/ C "2 / P for all x; y 2 G and all i D 1; : : : ; `. Since Fx D i i Fx , this implies X Mh .Fx /  Mh .i Fx / i .y/Fx .y/ D i .y/f .y/g.y

1

i



X

Mh .i f /.g.xi 1 x/ C "2 /

i



X

Mh .i f /g.xi 1 x/ C 2"2 Mh .f /:

i

Here the last step uses (8.32). By (8.34) and (8.35), X f .x/Mh .g/  Mh .i f /g.xi 1 x/ C 2"2 Mh .f / C "1 Mh .g/ i



X

Mh .i f /g.xi 1 x/ C 2"1 Mh .g/:

i

Here the second step uses (8.29) and the inequality Mh .f /  M.f I g/Mh .g/: Thus .f

2"1 /C Mh .g/  L g;

where  WD

X

Mh .i f /ıx

1 i

:

i

This implies Mg ..f

2"1 /C /Mh .g/ 

X i

Mh .i f /  .1 C "0 /Mh .f /:

(8.35)

324

8. The Haar measure

Here the second step uses (8.32). Since f  .f Mg .f /Mh .g/  Mg ..f

2"1 /C C 2"1 , we have

2"1 /C /Mh .g/ C 2"1 Mg ./Mh .g/

 .1 C "0 /Mh .f / C 2"1 M.I f /Mg .f /Mh .g/ D .1 C "0 /Mh .f / C "0 Mg .f /Mh .g/: Hence

1 C "0 Mh .f /  .1 C "/Mh .f / 1 "0 and this proves Step 3 for Mh . This inequality and its proof carry over to any leftinvariant positive linear functional ƒW Cc .G/ ! R. Mg .f /Mh .g/ 

Step 4. We prove uniqueness. Let ƒ; ƒ0 W Cc .R/ ! R be two left-invariant positive linear functionals that do not vanish identically. Then there exists a function f 2 CcC .G/ such that ƒ.f / > 0 by Lemma 8.15. Hence ƒ.g/  M.f I g/ 1 ƒ.f / > 0 for all g 2 CcC .G/ by (8.20). The same argument shows that ƒ0 .g/ > 0 for all g 2 CcC .G/. Now fix two functions f; f0 2 CcC .G/ and a constant " > 0. Choose an open neighborhood U  G of 1 that satisfies the requirements of Step 3 for both f and f0 and this constant ". By Urysohn’s lemma (Theorem A.1), there exists a function g 2 CcC .G/ such that g.1/ > 0;

supp.g/  ¹x 2 G j x 2 U and x

1

2 U º:

Replacing g by the function x 7! g.x/ C g.x 1 /, if necessary, we may assume that g satisfies (8.25). Hence it follows from Step 1 and Step 3 that ƒ.f /  M.f I g/ƒ.g/  .1 C "/ƒ.f / and .1 C "/ƒ.f0 /  M.f0 I g/ƒ.g/  ƒ.f0 /: Take the quotient of these inequalities to obtain .1 C "/

1

ƒ.f / M.f I g/ ƒ.f /   .1 C "/ : ƒ.f0 / M.f0 I g/ ƒ.f0 /

The same inequality holds with ƒ replaced by ƒ0 . Hence .1 C "/

2

ƒ0 .f / ƒ.f / ƒ.f /  0  .1 C "/2 : ƒ.f0 / ƒ .f0 / ƒ.f0 /

8.2. Haar measures

325

Since this holds for all " > 0, it follows that ƒ0 .f / D cƒ.f /;

c WD

ƒ0 .f0 / : ƒ.f0 /

Since this holds for all f 2 CcC .G/, ƒ0 and cƒ agree on CcC .G/. Hence ƒ0 D cƒ by Lemma 8.15. This proves Step 4. Step 5. We prove existence. The proof follows the elegant exposition [11] by Urs Lang. Fix a reference function f0 2 CcC .G/ and, for g 2 CcC .G/, define ƒg W CcC .G/ ! .0; 1/ by ƒg .f / WD

M.f I g/ M.f0 I g/

for f 2 CcC .G/:

(8.36)

Then ƒg 2 L for all g 2 CcC .G/ by Step 1. It follows also from Step 1 that M.f0 I g/  M.f0 I f /M.f I g/ and M.f I g/  M.f I f0 /M.f0 I g/ and hence M.f0 I f /

1

 ƒg .f /  M.f I f0 /

(8.37)

for all f; g 2 CcC .G/. Fix a function f 2 CcC .G/ and a number " > 0. Define 8 ˇ 9 ˇ ƒ.f0 / D 1 and there exists a neighborhood = < ˇ L" .f / WD ƒ 2 L ˇˇ W  G of 1 such that for all h 2 CcC .G/ : : ˇ supp.h/  W H) ƒ.f /  .1 C "/ƒh .f / ; We prove that L" .f / ¤ ;. To see this let U  G be the open neighborhood of 1 constructed in Step 3 for f and ". Choose a function g 2 CcC .G/ that satisfies (8.25) and choose an open neighborhood W  G of 1 associated to g that satisfies the requirements of Step 3. Let h 2 CcC .G/ with supp.h/  W . Then M.f I g/M.gI h/  .1 C "/M.f I h/ and M.f0 I g/M.gI h/  M.f0 I h/ by Step 3 and Step 1. Take the quotient of these inequalities to obtain ƒg .f /  .1C"/ƒh .f /. Since ƒg .f0 / D 1, it follows that ƒg 2 L" .f /. This shows that L" .f / ¤ ;, as claimed. Next we observe that ƒ.f /  M.f I f0 /ƒ.f0 / D M.f I f0 / for all ƒ 2 L" .f / by Step 1. Hence the supremum ƒ" .f / WD sup¹ƒ.f / j ƒ 2 L" .f /º

(8.38)

326

8. The Haar measure

is a real number, bounded above by M.f I f0 /. Since L" .f / contains an element of the form ƒg for some g 2 CcC .G/, it follows from (8.37) that M.f0 I f /

1

 ƒ" .f /  M.f I f0 /

(8.39)

for all f 2 CcC .G/ and all " > 0. Moreover, the function " 7! ƒ" .f / is nondecreasing by definition. Hence the limit ƒ0 .f / WD lim ƒ" .f / D inf ƒ" .f / "!0

">0

(8.40)

exists and is a positive real number for every f 2 CcC .G/. We prove that the functional ƒ0 W CcC .G/ ! .0; 1/ is left-invariant. To see this, fix a function f 2 CcC .G/ and an element x 2 G. Then L" .f / D L" .f ı Lx / for all " > 0. Namely, if W  G is an open neighborhood of 1 such that ƒ.f /  .1 C "/ƒh .f / for all h 2 CcC .G/ with supp.h/  W , then the same inequality holds with f replaced by f ı Lx because both ƒ and ƒh are left-invariant. Hence ƒ" .f / D ƒ" .f ı Lx / for all " > 0, and so ƒ0 .f / D ƒ0 .f ı Lx /. We prove that the functional ƒ0 W CcC .G/ ! .0; 1/ is additive. To see this, fix two functions f; f 0 2 CcC .G/. We show that .1 C "/

1

ƒ" .f C f 0 /  ƒ" .f / C ƒ" .f 0 /  .1 C "/ƒ" .f C f 0 /

(8.41)

for all " > 0. To establish the first inequality in (8.41), choose any functional ƒ 2 L" .f C f 0 /. Then there exists an open neighborhood W  G of 1 such that ƒ.f Cf 0 /  .1C"/ƒh .f Cf 0 / for all h 2 CcC .G/ with supp.h/  W . Moreover, we have seen above that h 2 CcC .G/ can be chosen such that supp.h/  W and also ƒh 2 L" .f / \ L" .f 0 /. Any such h satisfies .1 C "/

1

ƒ.f C f 0 /  ƒh .f C f 0 /  ƒh .f / C h .f 0 /  ƒ" .f / C ƒ" .f 0 /:

Take the supremum over all ƒ 2 L" .f C f 0 / to obtain the first inequality in (8.41). To prove the second inequality in (8.41), fix a constant ˛ > 1 and choose two functionals ƒ 2 L" .f / and ƒ0 2 L" .f 0 /. Then there exists an open neighborhood W  G of 1 such that ƒ.f /  .1 C "/ƒh .f / and ƒ0 .f 0 /  .1 C "/ƒh .f 0 / for all h 2 CcC .G/ with supp.h/  W . By Step 2, the function h 2 CcC .G/ can be chosen such that supp.h/  W and also ƒh .f / C ƒh .f 0 /  ˛ƒh .f C f 0 / and ƒh 2 L" .f C f 0 /. Any such h satisfies .1 C "/

1

.ƒ.f / C ƒ0 .f 0 //  ƒh .f / C ƒh .f 0 /  ˛ƒh .f C f 0 /  ˛ƒ" .f C f 0 /:

8.2. Haar measures

327

Take the supremum over all pairs of functionals ƒ 2 L" .f / and ƒ0 2 L" .f 0 / to obtain .1 "/ 1 .ƒ" .f / C ƒ" .f 0 //  ˛ƒ" .f C f 0 / for all ˛ > 1: This proves the second inequality in (8.41). Take the limit " ! 0 in (8.41) to obtain that ƒ0 is additive. Moreover, it follows directly from the definition that ƒ0 .f0 / D 1. Hence by Lemma 8.15, ƒ0 extends to a nonzero left-invariant positive linear functional on Cc .G/. This proves Step 5 and Theorem 8.14.  If one is prepared to use some abstract concepts from general topology, then the existence proof in Theorem 8.14 is essentially complete after Step 2. This approach is taken in Pedersen [16]. In this language the space G WD ¹g 2 CcC .G/ j 0  g  1; g.1/ D 1º is a directed set equipped with a map g 7! ƒg that takes values in the space ˇ ² ³ ˇ M.f0 I f / 1  ƒ.f /  M.f0 I f / C ˇ L WD ƒW Cc .G/ ! R ˇ : for all f 2 CcC .G/ The map G ! LW g 7! ƒg is a net. A net can be thought of as an uncountable analogue of a sequence and a subnet as an analogue of a subsequence. The existence of a universal subnet is guaranteed by the general theory and its convergence for each f by the fact that the target space is compact. Instead Step 3 in the proof of Theorem 8.14 implies that the original net g 7! ƒg converges and so there is no need to choose a universal subnet. That this can be proved with a refinement of the uniqueness argument (ƒ in Step 3) is pointed out in Pedersen [16]. That paper also contains two further uniqueness proofs. One is based on Fubini’s Theorem and the other on the Radon–Nikodým theorem. Another existence proof for compact second countable Hausdorff groups is due to Pontryagin. It uses the Arzelà–Ascoli theorem to establish the existence of a sequence i 2 P with ki k D 1 such that Li f converges to a constant function whose value is then taken to be ƒ.f /. Proof of Theorem 8.12. Existence and uniqueness in (i) follow directly from Theorem 8.14 and the Riesz representation theorem (Theorem 3.15). That nonempty open sets have positive measure follows from Urysohn’s lemma (Theorem A.1). To prove (ii), consider the map W G ! G defined by .x/ WD x

1

for x 2 G.

328

8. The Haar measure

Since  is a homeomorphism, it preserves the Borel  -algebra B. Since  ı Rx D Lx

1

ı ;

a measure W B ! Œ0; 1 is a left Haar measure if and only if the measure W B ! Œ0; 1 defined by .B/ WD ..B// D .B

1

/

is a right Haar measure. Hence assertion (ii) follows from (i). We prove (iii). Assume G is compact and let W B ! Œ0; 1 be the unique left Haar measure such that .G/ D 1. For x 2 G define x W B ! Œ0; 1 by x .B/ WD .Rx .B// for B 2 B. Since Rx commutes with Ly for all y by (8.11), x is a left Haar measure. Since x .G/ D .Rx .G// D .G/ D 1, it follows that x D  for all x 2 G. Hence  is right-invariant. Therefore, the map B ! Œ0; 1W B 7 ! .B/ WD ..B// D .B

1

/

is a left Haar measure and, since .G/ D 1, it agrees with . This proves Theorem 8.12.  In the noncompact case the left and right Haar measures need not agree. The above argument then shows that the measure x differs from  by a positive factor. Thus there exists a unique map W G ! .0; 1/ such that .Rx .B// D .x/.B/

(8.42)

for all x 2 G and all B 2 B. The map W G ! .0; 1/ in (8.42) is a continuous group homomorphism, called the modular character. It is independent of the choice of . A locally compact Hausdorff group is called unimodular if its modular character is trivial or, equivalently, if its left and right Haar measures agree. Thus every compact Hausdorff group is unimodular.

8.2. Haar measures

329

Exercise 8.18. Prove that  is a continuous homomorphism. Exercise 8.19. Prove that the group of all real 2  2-matrices of the form   a b ; a; b 2 R; a > 0; 0 1 is not unimodular. Prove that the additive group Rn is unimodular. Prove that every discrete group is unimodular. Exercise 8.20. Let W B ! Œ0; 1 be a right Haar measure. Show that the modular character is characterized by the condition .Lx 1 .B// D .x/.B/ for all x 2 G and all B 2 B. Haar measures are extremely useful tools in geometry, especially when the group in question is compact. For example, if a compact Hausdorff group G acts on a topological space X continuously via G  X ! XW .g; x/ 7 ! g x

(8.43)

one can use the Haar measure to produce G-invariant continuous functions by averaging. Namely, if f W X ! R is any continuous function, and  is the Haar measure on G with .G/ D 1, then the function F W X ! R defined by Z F .x/ WD f .a x/ d.a/ (8.44) G

for x 2 X is G-invariant in that F .g x/ D F .x/ for all x 2 X and all g 2 G. Exercise 8.21. Give a precise definition of what it means for a topological group to act continuously on a topological space. Exercise 8.22. Show that the map F in (8.44) is continuous and G-invariant. Exercise 8.23. Let W G ! GL.V / be a homomorphism from a compact Hausdorff group to the general linear group of automorphisms of a finite-dimensional vector space. (Such a homomorphism is called a representation of G.) Prove that V admits a G-invariant inner product. This observation does not extend to noncompact groups. Show that the standard representation of SL.2; R/ on R2 does not admit an invariant inner product.

330

8. The Haar measure

Exercise 8.24. Show that the Haar measure on a discrete group is a multiple of the counting measure. Deduce that for a finite group the formula (8.44) defines F .x/ as the average (with multiplicities) of the values of f over the group orbit of x. Exercise 8.25. Let G be a locally compact Hausdorff group and let  be a left Haar measure on G. Define the convolution product on L1 ./. Show that L1 ./ is a Banach algebra. (See page 282.) Find conditions under which f  g D g  f . Show that the convolution is not commutative in general. Hint. See Section 7.5 for G D Rn . See also Step 3 in the proof of Theorem 8.14.

Appendices

Appendix A

Urysohn’s lemma

Theorem A.1 (Urysohn’s lemma). Let X be a locally compact Hausdorff space and let K  X be a compact set and U  X be an open set such that K  U: Then there exists a compactly supported continuous function f W X ! Œ0; 1 such that f jK  1;

supp.f / D ¹x 2 X j f .x/ ¤ 0º  U:

(A.1) 

Proof. See page 334.

Lemma A.2. Let X be a topological space and let K  X be compact. Then the following holds: (i) every closed subset of K is compact; (ii) if X is Hausdorff, then for every y 2 X n K, there exist open sets U; V  X such that K  U , y 2 V , and U \ V D ;; (iii) if X is Hausdorff, then K is closed. Proof. (i) Let F  K be closed and let ¹Ui ºi2I be an open cover of F . Then the sets ¹Ui ºi 2I together with V WD X n F form an open cover of K. Therefore, there exist finitely many indices i1 ; : : : ; in such that the sets Ui1 ; : : : ; Uin ; V cover K. Hence F  Ui1 [


[ Uin . This shows that every open cover of F has a finite subcover, and so F is compact. (ii) Assume X is Hausdorff and let y 2 X n K. Define U WD ¹U  X j U is open and y … Ux º:

334

A. Urysohn’s lemma

Since X is Hausdorff, the collection U is an open cover of K. Since K is compact, there exists finitely many set U1 ; : : : ; Un 2 U such that K  U WD U1 [


[ Un : Since y … Uxi for all i , it follows that y 2 V WD X n Ux and U \ V D ;. Hence the sets U and V satisfy the requirements of (ii). (iii) Assume X is Hausdorff. Then it follows from part (ii) that, for every y 2 X n K, there exists an open set V  X such that y 2 V and V \ K D ;. Hence X n K is the union of all open sets in X that are disjoint from K. Thus X n K is open and so K is closed. This proves Lemma A.2.  Lemma A.3. Let X be a locally compact Hausdorff space and let K; U be subsets of X such that K is compact, U is open, and K  U . Then there exists an open set V  X such that Vx is compact and K  V  Vx  U:

(A.2)

Proof. We first prove the assertion in the case where K D ¹xº consist of a single element. Choose a compact neighborhood B  X of x. Then F WD B n U is a closed subset of B and hence is compact by part (i) of Lemma A.2. Since x … F , it follows from part (ii) of Lemma A.2 that there exist open sets W; W 0  X such that x 2 W , F  W 0 and W \ W 0 D ;. Hence V WD int.B/ \ W is an open neighborhood of x, its closure is a closed subset of B and hence compact, and S  B n W 0  B n F  U: Vx  B \ W This proves the lemma in the case #K D 1. Now consider the general case. By the first part of the proof, the open sets whose closures are compact and contained in U form an open cover of K. Since K is compact, there exist finitely any open sets V1 ; : : : ; Vn such that Vxi is a compact S S subset of U for all i and K  niD1 Vi . Hence V WD niD1 Vi is an open set Sn containing K and its closure Vx D i D1 Vxi is a compact subset of U . This proves Lemma A.3.  Proof of Theorem A.1. The proof has three steps. The first step requires the Axiom of Dependent Choice.1 1

See http://en.wikipedia.org/wiki/Axiom_of_dependent_choice.

335

Step 1. There exists a family of open sets Vr  X with compact closure, parametrized by r 2 Q \ Œ0; 1, such that K  V1  Vx1  V0  Vx0  U

(A.3)

s > r H) Vxs  Vr

(A.4)

and for all r; s 2 Q \ Œ0; 1. The existence of open sets V0 and V1 with compact closure satisfying (A.3) follows from Lemma A.3. Now choose a bijective map N0 ! Q \ Œ0; 1W i 7! qi such that q0 D 0 and q1 D 1. Suppose by induction that the open sets Vi D Vqi have been constructed for i D 0; 1; : : : ; n such that (A.4) holds for r; s 2 ¹q0 ; q1 ; : : : ; qn º. Choose k; ` 2 ¹0; 1; : : : ; nº such that qk WD max¹qi j 0  i  n; qi < qnC1 º; q` WD min ¹qi j 0  i  n; qi > qnC1 º: Then Vx`  Vk . Hence it follows from Lemma A.3 that there exists an open set VnC1 D VqnC1  X with compact closure such that Vx`  VnC1  VxnC1  Vk . This completes the induction argument and Step 1 then follows from the Axiom of Dependent Choice. (Denote by V the collection of all open sets V  X such that K  V  Vx  U . Denote by V the set of all finite sequences v D .V0 ; : : : ; Vn / in V that satisfy (A.3) and qi < qj H) Vxj  Vi for all i; j . Define a relation on V by v D .V1 ; : : : ; Vn /  v0 D .V10 ; : : : ; Vn00 / if n < n0 and Vi D Vi0 for i D 0; : : : ; n. Then V is nonempty and for every v 2 V there is a v0 2 V such that v  v0 . Hence, by the Axiom of Dependent Choice, there exists a sequence vj D .Vj;0 ; : : : ; Vj;nj / 2 V such that vj  vj C1 for all j 2 N. Define the map Q \ Œ0; 1 ! VW q 7! Vq by Vqi WD Vj;i for i; j 2 N with nj  i . This map is well defined and satisfies (A.3) and (A.4) by the definition of V and .) Step 2. Let Vr  X be as in Step 1 for r 2 Q \ Œ0; 1. Then f .x/ WD sup¹r 2 Q \ Œ0; 1 j x 2 Vr º D inf ¹s 2 Q \ Œ0; 1 j x … Vxs º

(A.5)

for all x 2 X . (Here the supremum of the empty set is zero and the infimum over the empty set is one.) We prove equality in (A.5). Fix a point x 2 X and define a WD sup¹r 2 Q \ Œ0; 1 j x 2 Vr º; b WD inf¹s 2 Q \ Œ0; 1 j x … Vxs º:

336

A. Urysohn’s lemma

We prove that a  b. If b D 1 this follows directly from the definitions. Hence assume b < 1 and choose an element s 2 Q \ Œ0; 1 such that x … Vxs : If r 2 Q \ Œ0; 1 is such that x 2 Vr , then Vr n Vxs ¤ ;, hence Vxs  Vr , and hence r  s. Thus we have proved that x 2 Vr H) r  s for all r 2 Q \ Œ0; 1. Take the supremum over all r 2 Q \ Œ0; 1 with x 2 Vr to obtain a  s. Then take the infimum over all s 2 Q \ Œ0; 1 with x … Vxs to obtain a  b. Now suppose, by contradiction, that a < b. Choose rational numbers r; s 2 Q \ Œ0; 1 such that a < r < s < b. Since a < r, it follows that x … Vr , since s < b, it follows that x 2 Vxs , and since r < s, it follows from Step 1 that Vxs  Vr . This is a contradiction and shows that our assumption that a < b was wrong. Thus a D b and this proves Step 2. Step 3. The function f W X ! Œ0; 1 in Step 2 is continuous and ´ 0 for x 2 X n V0 ; f .x/ D 1 for x 2 Vx1 :

(A.6)

Thus f satisfies the requirements of Theorem A.1. That f satisfies (A.6) follows directly from the definition of f in (A.5). We prove that f is continuous. To see this, fix a constant c 2 R. Then f .x/ < c if and only if there exists a rational number s 2 Q \ Œ0; 1 such that s < c and x … Vxs . Likewise, f .x/ > c if and only if there exists a rational number r 2 Q \ Œ0; 1 such that r > c and x 2 Vr . Hence, [ [ f 1 ..c; 1// D Vr ; f 1 .. 1; c// D .X n Vxs /: r2Q\.c;1

s2Q\Œ0;c/

This implies that the pre-image under f of every open interval in R is an open subset of X. Hence also the pre-image under f of every union of open intervals is open in X and so f is continuous. This proves Step 3 and Theorem A.1.  Theorem A.4 (partition of unity). Let X be a locally compact Hausdorff space, let U1 ; : : : ; Un  X be open sets, and let K  U1 [


[ Un be a compact set. Then there exist continuous functions f1 ; : : : ; fn W X ! R with compact support such that fi  0;

n X

fi  1;

i D1

for all i and

Pn

iD1

fi .x/ D 1 for all x 2 K.

supp.fi /  Ui

337

Proof. Define the set ² V WD V  X

ˇ ³ ˇ V is open; Vx is compact; and there exists ˇ ˇ an index i 2 ¹1; : : : ; nº such that Vx  Ui :

If x 2 K, then x 2 Ui for some index i 2 ¹1; : : : ; nº and, by Lemma A.3, there is an open set V  X such that Vx is compact and x 2 V  Vx  Ui . Thus V is an open cover of K. Since K is compact there exist finitely many open sets V1 ; : : : ; V` 2 V such that K  V1 [


[ V` : For i D 1; : : : ; n define [ Ki WD Vxj : 1j `; Vxj Ui

Then K  K1 [


[ Kn and Ki is a compact subset of Ui for each i. Hence it follows from Urysohn’s lemma (Theorem A.1) that, for each i, there exists a compactly supported continuous function gi W X ! R such that 0  gi  1;

supp.gi /  Ui ;

gi jKi  1:

Define f1 WD g1 ; f2 WD .1

g1 /g2 ;

f3 WD .1

g1 /.1

g2 /g3 ;

:: : fn WD .1

g1 /


.1

gn

1 /gn :

Then supp.fi /  supp.gi /  Ui for each i and 1

n X i D1

fi D

n Y

.1

gi /:

iD1

Since K  K1 [


[ Kn and the factor 1 gi vanishes on Ki , this implies Pn  iD1 fi .x/ D 1 for all x 2 K. This proves Theorem A.4.

Appendix B

The product topology

Let .X; UX / and .Y; UY / be topological spaces, denote the product space by X  Y WD ¹.x; y/ j x 2 X; y 2 Y º; and let X W X  Y ! X and Y W X  Y ! Y be the projections onto the first and second factor. Consider the following universality property for a topology U  2X Y on the product space. (F) Let .Z; UZ / be any topological space and let hW Z ! X  Y be any map. Then hW .Z; UZ / ! .X  Y; U/ is continuous if and only if the maps f WD X ı hW .Z; UZ / ! .X; UX /;

(B.1a)

g WD Y ı hW .Z; UZ / ! .Y; UY /

(B.1b)

are continuous. Theorem B.1. (i) There is a unique topology U on X  Y that satisfies (P). (ii) Let U  2X Y be as in (i). Then W 2 U if and only if there are open sets S Ui 2 UX and Vi 2 UY , indexed by any set I , such that W D i2I .Ui  Vi /. (iii) Let U  2X Y be as in (i). Then U is the smallest topology on X  Y with respect to which the projections X and Y are continuous. (iv) Let U  2XY be as in (i). Then the inclusion x W .Y; UY / ! .X  Y; U/;

x .y/ WD .x; y/

for y 2 Y;

is continuous for every x 2 X and the inclusion y W .X; UX / ! .X  Y; U/; is continuous for every y 2 Y .

y .x/ WD .x; y/

for x 2 X;

340

B. The product topology

Definition B.2. The product topology on X  Y is defined as the unique topology that satisfies (P) or, equivalently, as the smallest topology on X  Y such that the projections X and Y are continuous. It is denoted by UXY  2X Y : Proof of Theorem B.1. The proof has five steps. Step 1. If U  2X Y is a topology satisfying (P), then the projections X and Y are continuous. Take h WD idW X  Y ! X  Y so that f D X ı h D X and g D Y ı h D Y . Step 2. We prove uniqueness in (i). Let U; U0  2X Y be two topologies satisfying (P) and consider the map h WD idW .X  Y; U/ ! .X  Y; U0 /: Since f D X W .X  Y; U/ ! .X; UX / and g D Y W .X  Y; U/ ! .Y; UY / are continuous by Step 1, and U0 satisfies (P), it follows that h is continuous and hence U0  U. Interchange the roles of U and U0 to obtain U0 D U. Step 3. We prove (ii) and existence in (i). S Define U  2X Y as the collection of all sets of the form W D i 2I .Ui  Vi /, where I is any index set and Ui 2 UX , Vi 2 UY for i 2 I . Then U is a topology and the projections X W .X  Y; U/ ! .X; UX / and Y W .X  Y; U/ ! .Y; UY / are continuous. We prove that U satisfies (P). To see this, let .Z; UZ / be any topological space, let hW Z ! X  Y be any map, and define f WD X ı h and g WD Y ı h as in (B.1). If h is continuous, then so are f and g. Conversely, if f and g are continuous, then h 1 .U  V / D f 1 .U / \ g 1 .V / is open in Z for all U 2 UX and all V 2 UY , and hence it follows from the definition of U that h 1 .W / is open for all W 2 U. Thus h is continuous. Step 4. We prove (iii). Let U be the topology in (i) and let U0 be any topology on X  Y with respect to which X and Y are continuous. If U 2 UX and V 2 UY , then U  V D X 1 .U / \ Y 1 .V / 2 U0 . Hence U  U0 by (ii). Since X and Y are continuous with respect to U, it follows that U is the smallest topology on X  Y with respect to which X and Y are continuous.

341

Step 5. We prove (iv). Fix an element x 2 X and consider the map h WD x W Y ! X  Y . Then the map f WD X ı hW Y ! X is constant and g WD Y ı hW Y ! Y is the identity. Hence f and g are continuous and so is h by condition (P). An analogous argument shows that y is continuous for all y 2 Y . 

Appendix C

The inverse function theorem

This appendix contains a proof of the inverse function theorem. The result is formulated in the setting of continuously differentiable maps between open sets in a Banach space. Readers who are unfamiliar with bounded linear operators on Banach spaces may simply think of continuously differentiable maps between open sets in finite-dimensional normed vector spaces. The inverse function theorem is used on page 81 in the proof of Lemma 2.19, which is a key step in the proof of the transformation formula for the Lebesgue measure (Theorem 2.17). Assume throughout that .X; k
k/ is a Banach space. When ˆW X ! X is a bounded linear operator denote its operator norm by kˆk WD kˆkL.X/ WD

kˆxk : x2X n¹0º kxk sup

For x 2 X and r > 0 denote by Br .x/ WD ¹y 2 X j kx radius r about x. For x D 0 abbreviate Br WD Br .0/.

yk < rº the open ball of

Theorem C.1 (inverse function theorem). Fix an element x0 2 X and two real numbers r > 0 and 0 < ˛ < 1. Let W Br .x0 / ! X be a continuously differentiable map such that kd .x/

1kL.X/  ˛

for all x 2 Br .x0 /:

(C.1)

Then B.1 Moreover, the map differentiable, and d

˛/r .

.x0 // 

.Br .x0 //  B.1C˛/r . .x0 //:

(C.2)

1

is continuously

is injective, its image is open, the map

1

.y/ D d .

1

.y//

1

for all y 2

.Br .x0 //.

344

C. The inverse function theorem

Proof. Assume without loss of generality that x0 D Step 1.

.x0 / D 0. .Br /  B.1C˛/r .

is a homeomorphism onto its image and

Define  WD id

W Br ! X:

Then kd.x/k  ˛ for all x 2 Br . Hence k.x/

.y/k  ˛kx

yk:

(C.3)

for all x; y 2 Br , and so .1

˛/ kx

yk  k .x/

.y/k  .1 C ˛/ kx

yk :

(C.4)

The second inequality in (C.4) shows that .Br /  B.1C˛/r and the first inequality 1 in (C.4) shows that is injective and is Lipschitz continuous. Step 2. B.1

˛/r



.Br /.

Let  2 B.1 ˛/r and define " > 0 by kk DW .1 ˛/.r "/. Then, by (C.3) with y D 0, we have k.x/k  ˛kxk for all x 2 Br . If kxk  r " this implies k.x/ C k  r ". Thus the map x 7! .x/ C  is a contraction of the closed ball Bxr " . By the contraction mapping principle, it has a unique fixed point x and the fixed point satisfies .x/ D x .x/ D . Hence  2 .Br /. Step 3.

.Br / is open.

Let x 2 Br and define y WD .x/. Choose " > 0 such that B" .x/  Br . Then, by Step 2, B.1 ˛/" . .x//  .B" .x//  .Br /. Step 4.

1

is continuously differentiable.

Let x0 2 Br and define y0 WD .x0 / and ‰ WD d .x0 /. Then k1 ‰k  ˛, so ‰ P 1 is invertible, ‰ 1 D 1 ‰/k , and k‰ 1 k  .1 ˛/ 1 . We prove that kD0 .1 1 1 is differentiable at y0 and d .y0 / D ‰ . Let " > 0. Since is differentiable at x0 and d .x0 / D ‰, there is a constant ı > 0 such that, for all x 2 Br with kx x0 k < ı.1 ˛/ 1 , we have k .x/

.x0 /

x0 /k  ".1

‰.x

˛/2 kx

x0 k:

Shrinking ı, if necessary, we may assume, by Step 3, that Bı .y0 /  .Br /. 1 Now suppose ky y0 k < ı and denote x WD .y/ 2 Br . Then, by (C.4), kx

x0 k  .1

˛/

1

ky

y0 k < ı.1

˛/

1

;

345

and hence k

1

.y/

1

.y0 /

‰

1

.y

y0 /k D k‰ 

Hence Thus d

1

1

1 1

˛

.y

y0

k .x/

 ".1

˛/ kx

 "ky

y0 k:

‰.x .x0 /

x0 //k ‰.x

x0 /k

x0 k

is differentiable at y0 and d 1 .y0 / D ‰ 1 D d . 1 is continuous by Step 1. This proves Theorem C.1.

1

.y0 //

1

. 

Bibliography

[1] R. B. Ash, Basic probability theory. John Wiley & Sons, New York, 1970. Reprint, Dover Publications, Mineola, N,Y., 2008. John Wiley & Sons 1970, Dover Publications, 2008. MR 0268923 Zbl 1234.60001 Zbl 0206.18001 (reprint) http://www.math.uiuc.edu/~r-ash/BPT/BPT.pdf [2] V. I. Bogachev, Measure Theory. Volumes I and II. Springer, Berlin etc., 2006. MR 2267655 Zbl 1120.28001 [3] W. F. Eberlein, Notes on integration I: The underlying convergence theorem. Comm. Pure Appl. Math. 10 (1957), 357–360. MR 0088537 Zbl 0077.31302 [4] D. H. Fremlin, Measure theory. Volumes 1–5. Torres Fremlin, Colchester, 2000–2008. MR 2462519 (I) MR 2462280 (II) MR 2459668 (III) MR 2462372 (IV) Zbl 1162.28001 (I) Zbl 1165.28001 (II) Zbl 1165.28002 (III) Zbl 1166.28001 (IV) Zbl 1166.28002 (V) https://www.essex.ac.uk/maths/people/fremlin/mt.htm [5] D. Gilbarg and N. S. Trudinger, Elliptic partial differential equations of second order. Second edition. Springer, Berlin etc., 1983. MR 0737190 Zbl 0562.35001 [6] P. R. Halmos, Measure theory. University Series in Higher Mathematics. D. Van Nostrand, New York, and Macmillan & Co., London, 1950. Reprint, Graduate Texts in Mathematics, 18. Springer, Berlin etc., 1974. MR 0033869 Zbl 0040.16802 Zbl 0283.28001 (reprint) [7] C. E. Heil, Lecture notes on real analysis. Georgia Institute of Technology, Atlanta, Fall 2007. http://people.math.gatech.edu/~heil/6337/fall07/section3.4.pdf http://people.math.gatech.edu/~heil/6337/fall07/section3.5b.pdf

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Index

—A— absolutely continuous function, 239, 245 measure, 179 signed measure, 200, 206 almost everywhere, 36 —B— Baire measurable function, 129 set, 129 Baire measure, 129 Baire -algebra, 129 Banach algebra, 282 Banach space, 8 Banach–Zarecki theorem, 239 basis of a topological space, 124 bilinear form positive definite, 147 symmetric, 147 Boolean algebra, 6 Borel measurable function, 12 set, 9 Borel measure, 95 extended, 107 left-invariant, 314 right-invariant, 314 Borel outer measure, 108 Borel set, 9 Borel -algebra, 9 bounded linear functional, 149 bounded variation, 243

—C— Calderón–Zygmund inequality, 295 Carathéodory criterion, 61 Cauchy–Schwarz inequality, 147 characteristic function, 12 closed set, 6 compact set, 10 compactly supported function, 113 complete measure space, 43 complete metric space, 8 completion of a measure space, 43 continuous map, 11 continuum hypothesis, 147 convergence in measure, 173 convolution, 277 of signed measures, 307 counting measure, 21 cuboid, 64

—D— dense subset, 141 Dieudonné’s measure, 96 Dirac measure, 21 divergence theorem, 306 double arrow space, 126 dual space of a Hilbert space, 149 of a normed vector space, 149 of C.X /, 124, 218 of L2 ./, 152 of Lp ./, 159 of `1 , 160, 176

352

Index —E—

Egoroff’s Thorem, 172 elementary set, 250 envelope, 152 essential supremum, 138 —F— first countable, 124 Fubini’s Theorem for integrable functions, 265 for positive functions, 261 for the completion, 268 for the Lebesgue measure, 276 in polar coordinates, 305 fundamental solution of Laplace’s equation, 292 fundamental theorem of calculus, 239 —G— Green’s formula, 293 group discrete, 309 Lie, 310 topological, 309 unimodular, 328

—I— inner product, 147 on L2 ./, 148 inner regular, 96, 208 on open sets, 101 integrable function Lebesgue, 32 locally, 283 partially defined, 46 Riemann, 87 weakly, 224 integral Haar, 315 Lebesgue, 22, 32 Riemann, 87 Riemann–Stieltjes, 244 invariant linear functional, 315 measure, 314 inverse limit, 311 —J— Jensen’s inequality, 168 Jordan decomposition, 200, 204 Jordan measurable set, 87 Jordan measure, 87 Jordan null set, 65

—L— —H— Haar integral, 315 Haar measure, 314 Hahn decomposition, 202 Hahn–Banach theorem, 160 Hardy’s inequality, 171 Hardy–Littlewood maximal inequality, 230 Hausdorff dimension, 93 Hausdorff measure, 92 Hausdorff space, 11, 95 Hilbert space, 148 Hölder inequality, 134

Laplace operator, 292 Lebesgue differentiation theorem, 234, 238 Lebesgue decomposition, 180 for signed measures, 201 Lebesgue dominated convergence theorem, 35 Lebesgue integrable, 32 Lebesgue integral, 22, 32 Lebesgue measurable function, 57, 69 set, 57, 69 Lebesgue measure, 57, 69 on the sphere, 304

Index Lebesgue monotone convergence theorem, 25 Lebesgue null set, 65 Lebesgue outer measure, 65 continuous from below, 70 regularity, 70 Lebesgue point, 233 left-invariant linear functional, 315 measure, 314 lexicographic ordering, 126 Lie group, 310 linear functional left/right-invariant, 315 localizable, 152, 174 strictly, 271 locally compact, 95 Hausdorff group, 309 locally determined, 270 locally integrable, 283 lower semi-continuous, 245 lower sum, 87 Lp ./, 136 Lp .Rn /, 137 L1 ./, 138 `p , 137 `1 , 159

—M— Marcinkiewicz interpolation, 289 maximal function, 229, 233 measurable function, 12 Baire, 129 Borel, 12 Lebesgue, 57, 69 partially defined, 46 measurable set, 5 Baire, 129 Borel, 9 Jordan, 87 Lebesgue, 57, 69 w.r.t. an outer measure, 58 measurable space, 5

353 measure, 19 absolutely continuous, 179 Baire, 129 Borel, 95 Borel outer, 108 counting, 21 Dirac, 21 Haar, 314 Hausdorff, 92 inner regular, 96, 101 Jordan, 87 Lebesgue, 57, 69 Lebesgue outer, 65 left-invariant, 314 localizable, 152, 174 locally determined, 271 nonatomic, 147 outer, 58 outer regular, 95 probability, 147 product, 255 Radon, 96 regular, 96 right-invariant, 314 semi-finite, 152 -finite, 152 signed, 197 singular, 179 strictly localizable, 271 translation-invariant, 57 truly continuous, 206 measure space, 19 complete, 43 Lebesgue, 57, 69 localizable, 152, 174 locally determined, 271 semi-finite, 152 -finite, 152 strictly localizable, 271 metric space, 7 Minkowski inequality, 134, 263 modular character, 328 mollifier, 286 monotone class, 250 mutually singular, 179 signed measures, 200

354

Index —N—

neighborhood, 95 nonatomic measure, 147 norm of a bounded linear functional, 149 normed vector space, 7 null set, 36 Jordan, 65 Lebesgue, 65

—O— one-point compactification, 127 open ball, 7 open set, 6 in a metric space, 7 outer measure, 58 Borel, 108 Lebesgue, 65 translation-invariant, 65 outer regular, 95

—P— p-adic integers, 311 p-adic rationals, 312 partition of a set, 5 partition of unity, 336 perfectly normal, 126 metric spaces are, 128 Poisson identity, 293 positive linear functional on Cc .X /, 113 on Lp ./, 161 pre-image, 11 probability theory, 51–55, 147 product measure, 255 complete locally determined, 270 primitive, 270 product -algebra, 249 product topology, 312, 340 pushforward of a measure, 50 of a -algebra, 13, 50

—R— Radon measure, 96 Radon–Nikodým derivative, 236 Radon–Nikodým theorem, 180 for signed measures, 202 generalized, 209 regular measure, 96 Riemann integrable, 87 Riemann integral, 87 Riemann–Stieltjes integral, 244 Riesz representation theorem, 115 right-invariant linear functional, 315 measure, 314 —S— second countable, 124 semi-finite, 152 separability of Lp ./, 142 separable, 141 set of measure zero, 36 -additive measure, 19 signed measure, 197 -algebra, 5 Baire, 129 Borel, 9 Lebesgue, 57, 69 product, 249 -compact, 95 -finite, 152 signed measure, 197 absolutely continuous, 200, 206 concentrated, 200 Hahn decomposition, 202 inner regular w.r.t. a measure, 208 Jordan decomposition, 200, 204 Lebesgue decomposition, 201 mutually singular, 200 Radon–Nikodým theorem, 202 total variation, 197 truly continuous, 206 simple function, 17 singular measure, 179

Index smooth function, 292 Sorgenfrey line, 126, 128 step function, 17 Stone–ech compactification of N, 176 —T— topological group, 309 invariant measure, 314 left-invariant measure, 314 locally compact Hausdorff, 309 right-invariant measure, 314 topology, 6 basis, 124 first countable, 124 group, 309 Hausdorff, 11, 95 locally compact, 95 perfectly normal, 126 product, 312, 340 second countable, 124 separable, 141 -compact, 95 x, 8 standard on R standard on R, 8 total variation of a signed measure, 197

355 transformation formula, 77 translation-invariant measure, 57 outer measure, 65 triangle inequality, 7, 147 weak, 223 truly continuous, 206 Tychonoff’s theorem, 312 —U— uniformly integrable, 214 upper semi-continuous, 245 upper sum, 87 Urysohn’s lemma, 333 —V— Vitali’s covering lemma, 231 Vitali’s theorem, 215 Vitali–Carathéodory theorem, 245 —W— weak triangle inequality, 223 weakly integrable, 224 —Y— Young’s inequality, 133, 281

Dietmar A. Salamon

Dietmar A. Salamon

The book is intended as a companion to a one semester introductory lecture course on measure and integration. After an introduction to abstract measure theory it proceeds to the construction of the Lebesgue measure and of Borel measures on locally compact Hausdorff spaces, Lp spaces and their dual spaces and elementary Hilbert space theory. Special features include the formulation of the Riesz Representation Theorem in terms of both inner and outer regularity, the proofs of the Marcinkiewicz Interpolation Theorem and the Calderon–Zygmund inequality as applications of Fubini’s theorem and Lebesgue differentiation, the treatment of the generalized Radon–Nikodym theorem due to Fremlin, and the existence proof for Haar measures. Three appendices deal with Urysohn’s Lemma, product topologies, and the inverse function theorem. The book assumes familiarity with first year analysis and linear algebra. It is suitable for second year undergraduate students of mathematics or anyone desiring an introduction to the concepts of measure and integration.

ISBN 978-3-03719-159-0

www.ems-ph.org

Salamon Cover | Font: Frutiger_Helvetica Neue | Farben: Pantone 116, Pantone 287 | RB 31 mm

Measure and Integration

Measure and Integration

Textbooks in Mathematics

Dietmar A. Salamon

Measure and Integration