1 INTRODUCTION TO MEASURE THEORY AND LEBESGUE INTEGRATION

Eduard EMELYANOV

Ankara — TURKEY

2007 2 FOREWORD

This book grew out of a one-semester course for graduate students that the author have taught at the Middle East Technical University of Ankara in 2004- 06. It is devoted mainly to the measure theory and integration. They form the base for many areas of mathematics, for instance, the probability theory, and at least the large part of the base of the functional analysis, and operator theory.

Under measure we understand a σ-additive function with values in R+ ∪ {∞} defined on a σ-algebra. We give the extension of this basic notion to σ-additive vector-valued measures, in particular, to signed and complex measures and refer for more delicate topics of the theory of vector-valued measures to the excellent book of Diestel and Uhl [3]. The σ-additivity plays a crucial role in our book. We do not discuss here the theory of finitely-additive functions defined on algebras of sets which are sometimes referred as finitely-additive measures. There are two key points in the measure theory. The Caratheodory extension theorem and construction of the Lebesgue integral. Other results are more or less technical. Nevertheless, we can also emphasize the importance of the Jor- dan decomposition of signed measure, theorems about convergence for Lebesgue integral, Cantor sets, the Radon – Nikodym theorem, the theory of Lp-spaces, the Liapounoff convexity theorem, and the Riesz representation theorem. We provide with proofs only basic results, and leave the proofs of the others to the reader, who can also find them in many standard graduate books on the measure theory like [1], [4], and [5]. Exercises play an important role in this book. We expect that a potential reader will try to solve at least half of them. Exercises marked with ∗ are more difficult, and sometimes are too difficult for the first reading. In the case if someone will use this book as a basic text-book for the analysis graduate course, the author recommends to study subsections marked with ◦, omitting the material of subsections marked with !. I am indebted to many. I thank Professor Safak Alpay for reading the manuscript and offering many valuable suggestions. I thank to our students of 2004-2006 METU graduate real analysis classes, especially Tolga Karayayla, who made many corrections. Finally, I thank to my wife Svetlana Gorokhova for helping in preparing of the manuscript. 3

Contents

Introduction

Chapter 1 1.1. σ-Algebras, Measurable Functions, Measures, the Egoroff Theorem, Ex- haustion Argument 1.2. Vector-valued, Signed and Complex Measures, Variation of a Vector-valued Measure, Operations with Measures, the Jordan Decomposition Theorem, Ba- nach Space of Signed Measures of Bounded Variation 1.3. Construction of the Lebesgue Integral, the Monotone Convergence Theo- rem, the Dominated Convergence Theorem,

Chapter 2

2.1. The Caratheodory Theorem, Lebesgue Measure on R, Lebesgue – Stieltjes Measures, the Product of Measure Spaces, the Fubini Theorem

2.2. Lebesgue Measure on Rn, Lebesgue Integral in Rn, the Lusin Theorem, Cantor Sets

Chapter 3 3.1. The Radon – Nikodym Theorem, Continuity of a Measure with Respect to another Measure, the Hahn Decomposition Theorem

3.2. H¨older’sand Minkowski’s Inequalities, Completeness, Lp-Spaces, Duals 3.3. The Liapounoff Convexity Theorem

Chapter 4

4.1. Vector Spaces of Functions on Rn, Convolutions 4.2. Radon Measures, the Riesz Representation Theorem

Bibliography

Index 4 Chapter 1

1.1 σ-Algebras, Measurable Functions, Measures

Here we define the notion of σ-algebra which plays the key role in the measure theory. We study basic properties of σ-algebras and measurable functions. In the end of this section we define and study the notion of measure.

1.1.1.◦ σ-Algebras. The following notion is the principal in the measure theory. Definition 1.1.1 A collection A of subsets of a set X is called a σ-algebra if (a) X ∈ A; (b) if A ∈ A then X \ A ∈ A; S (c) given a sequence (Ak)k ⊆ A, we have Ak ∈ A. k It follows from this definition that the empty set ∅ belongs to A since X ∈ A and ∅ = X \ X. Further, given a sequence (Ak)k ⊆ A, we have \ [ Ak = X \ (X \ Ak) ∈ A. k k Finally, the difference A \ B := A ∩ (X \ B) and the symmetric difference A4B := (A \ B) ∪ (B \ A) both belong to A. Any σ-algebra of subsets of a set X has at least two elements: ∅ and X itself. One of the main and mostly obvious examples of a σ-algebra is P(X) is the set of all subsets of X. The following simple proposition shows us how to construct new σ-algebras from a given family of σ-algebras.

5 6 CHAPTER 1.

Proposition 1.1.1 Let {Ωα}α∈A be a nonempty family of σ-algebras in P(X), then Ω = ∩αΩα is also a σ-algebra. 2

This proposition leads to the following definition.

Definition 1.1.2 Let G ⊆ P(X), then the set of all σ-algebras containing G is nonempty since it contains P(X). Hence we may talk about the minimal σ- algebra containing G. This σ-algebra is called the σ-algebra generated by G.

An important special case of this notion is the following.

Definition 1.1.3 Let X be a topological space and let G be the family of all open subsets of X. The σ-algebra generated by G is called the Borel algebra of X and denoted by B(X).

1.1.2.◦ Measurable functions. Let f be a real-valued function defined on a set X. We suppose that some σ-algebra Ω ⊆ P(X) is fixed.

Definition 1.1.4 We say that f is measurable, if f −1([a, b]) ∈ Ω for any reals a < b.

The following three propositions are obvious.

Proposition 1.1.2 Let f : X → R be a function. Then the following conditions are equivalent: (a) f is measurable; (b) f −1([0, b)) ∈ Ω for any real b; (c) f −1((b, ∞)) ∈ Ω for any real b; (d) f −1(B) ∈ Ω for any B ∈ B(R). 2

Proposition 1.1.3 Let f and g be measurable functions, then (a) α · f + β · g is measurable for any α, β ∈ R; (b) functions max{f, g} and f · g are measurable. In particular, functions f + := max{f, 0}, f − := (−f)+, and |f| := f + + f − are measurable. 2 1.1. σ-ALGEBRAS, MEASURABLE FUNCTIONS, MEASURES 7

∞ Proposition 1.1.4 Let (fn)n=1 be a sequence of measurable functions, then

lim sup fn and lim inf fn n→∞ n→∞ are measurable. 2

1.1.3.◦ Measures. The main definition here is the following.

Definition 1.1.5 Let A be a σ-algebra. A function µ : A → R ∪ {∞} is called a measure, if: (a) µ(∅) = 0; (b) µ(A) ≥ 0 for all A ∈ A; and S P (c) µ( k Ak) = k µ(Ak) for any sequence (Ak)k of pairwise disjoint sets from A, that is Ai ∩ Aj = ∅ for i 6= j.

The axiom (c) is called σ-additivity of the measure µ. As usual, we will also assume that any measure under consideration satisfies the following axiom: (d) for any subset A ∈ A with µ(A) = ∞, there exists B ∈ A such that B ⊆ A and 0 < µ(B) < ∞. This axiom allows to avoid a pathological case that for some A, with µ(A) = 0 it holds either µ(B) = 0 or µ(B) = ∞ for any B ⊆ A, B ∈ A. We will use often the following simple proposition, the proof of which is left to the reader.

Proposition 1.1.5 Let µ be a measure on a σ-algebra A, An ∈ A, and An → A. ∞ Then A ∈ A and µ(A) = lim µ(An). In particular, if (Bn)n=1 is a decreasing n→∞ T∞ sequence of elements of A such that n=1 Bn = ∅, then µ(Bn) → 0. 2

1.1.4.◦ Measure spaces. We consider a fixed but arbitrary σ-algebra with a measure.

Definition 1.1.6 If A is a σ-algebra of subsets of X and µ is a measure on A, then the triple (X, A, µ) is called a measure space. The sets belonging to A are called measurable sets because the measure is defined for them. 8 CHAPTER 1.

Now we present several examples of measure spaces.

Example 1.1.1 Let X = {x1, . . . , xN } be a finite set, A be the σ-algebra of all subsets of X and a measure is defined by setting to each xi ∈ X a nonnegative number, say pi. This follows that the measure of a subset {xα1 , . . . , xαk } ⊆ X is just pα1 + ··· + pαk . If pi = 1 for all i, then the measure is called a counting measure because it counts the number of elements in a set.

Example 1.1.2 If X is a topological space, then the most natural σ-algebra of subsets of X is the Borel algebra B(X). Any measure defined on a Borel algebra is called Borel measure. In Section 2.1, we will prove that on the Borel σ- algebra B(R) there exists a unique measure such that µ([a, b]) = b − a for any interval [a, b] ⊆ R.

Whenever we consider spaces X = [a, b], X = R, or X = Rn, we usually assume that the measure under consideration is the Borel measure. As presented in Definition 1.1.6, the notion of measure space is extremely gen- eral. In almost all applications, the following specific class of measure spaces is adequate.

Definition 1.1.7 A measure space (X, A, µ) is called σ-finite if there is a se- ∞ quence (Ak)k=1, Ak ∈ A, satisfying ∞ [ X = Ak and µ(Ak) < ∞ for all k. k=1

Obviously, any σ-finite measure satisfies the axiom (d). Moreover, a measure µ ∞ is σ-finite iff (= if and only if) there exists an increasing sequence (Xn)n=1 of S∞ subsets of finite measure such that X = n=1 Xn. If X = R, then Ak may be chosen as intervals [−k, k]. In Rd, they may be chosen as balls of radius k, etc.

Definition 1.1.8 A measure space (X, A, µ) is called finite if µ(X) < ∞. In particular, if µ(X) = 1, then the measure space is said to be probabilistic, and µ is said to be a probability. A measure µ is called separable if there exists a countable family R ⊆ A such that for any A ∈ A, µ(A) < ∞, and for any ε > 0 there exists B ∈ R satisfying µ(A4B) < ε. A measure µ is called complete if there holds: [ µ(A) = 0 & B ⊆ A ] ⇒ B ∈ A . 1.1. σ-ALGEBRAS, MEASURABLE FUNCTIONS, MEASURES 9

Every Borel measure on R (or on [0, 1], or on Rn, etc.) possesses a unique com- pletion, which is called a Lebesgue measure.

1.1.5.◦ Convergence of functions and the Egoroff theorem. For mea- surable functions, there are several natural types of convergence. Some of them are given by the following definition.

∞ Definition 1.1.9 Let (fn)n=1 be a sequence of R-valued functions defined on X. We say that:

(a) fn → f pointwise, if fn(x) → f(x) for all x ∈ X;

(b) fn → f almost everywhere (a.e.), if fn(x) → f(x) for all x ∈ X except a set of measure 0;

(c) fn → f uniformly, if for any ε > 0 there is n(ε) such that

sup{|fn(x) − f(x)| : x ∈ X} ≤ ε for all n ≥ n(ε).

∞ Theorem 1.1.1 (Egoroff’s theorem) Suppose that µ(X) < ∞, {fn}n=1 and f are measurable functions on X such that fn → f a.e. Then, for every ε > 0, c there exists E ⊆ X such that µ(E) < ε and fn → f uniformly on E = X \ E.

Proof: Without loss of generality, we may assume that fn → f everywhere on X and (by replacing fn with fn − f) that f ≡ 0. For k, n ∈ N, let

∞ [ −1 En(k) := {x : |fm(x)| ≥ k }. m=n T∞ Then, for a fixed k, En(k) decreases as n increases, and n=1 En(k) = ∅. Since µ(X) < ∞, we conclude that µ(En(k)) → 0 as n → ∞. Given ε > 0 and k ∈ N, choose nk such that −k µ(Enk (k)) < ε · 2 , and set ∞ [ E := Enk (k). k=1 10 CHAPTER 1.

Then µ(E) < ε, and we have

−1 |fn(x)| < k (∀n > nk, x 6∈ E).

Thus fn → 0 uniformly on X \ E. 2

1.1.6.◦ Exhaustion argument. ∞ Let (X, Σ, µ) be a σ-finite measure space. Given a sequence (Un)n=1 ⊆ Σ, a set A ∈ Σ is called (Un)n-bounded if there exists n such that A ⊆ Un µ-almost everywhere.

∞ Theorem 1.1.2 (Exhaustion theorem) Let (Yn)n=1 ⊆ Σ be a sequence satis- fying Yn ↑ X and µ(Yn) < ∞ for all n. Let P be some property of (Yn)n-bounded measurable sets, such that A ∈ P iff B ∈ P for all B, µ(A4B) = 0. Suppose that any (Yn)n-bounded set A, µ(A) > 0, has a subset B ∈ Σ, µ(B) > 0 with the property P. Moreover, assume that either (a) A1 ∪ A2 ∈ P for every A1,A2 ∈ P, or (b) ∪nBn ∈ P for every at most countable family (Bn)n of pairwise disjoint sets possessing the property P. ∞ Then there exists a sequence (Xn)n=1 ⊆ Σ such that Xn ↑ X, and P 3 Xn ⊆ Yn ∞ for all n. Moreover, there exists a pairwise disjoint sequence (An)n=1 ⊆ Σ such ∞ S that An = X and An ∈ P for all n. n=1

Proof: Let A be a (Yn)n-bounded set with µ(A) > 0. Denote

PA := {B ∈ P : B ⊆ A} & m(A) := sup{µ(B): B ∈ PA} .

∞ I(a) Suppose P satisfies (a). Then there exists a sequence (Fn)n=1 ⊆ PA such that m(A) = lim µ(Fn), We may assume, that Fn ↑ . By Proposition 1.1.5, the n→∞ ∞ set F = ∪n=1Fn satisfies µ(F ) = m(A). We show that µ(A) = m(A). If not then µ(A \ F ) > 0. The set A \ F has a subset of positive measure F0 ∈ P. Then Fn ∪ F0 ∈ PA and µ(Fn ∪ F0) > m(A) for a sufficiently large n, which contradicts to the definition of m(A). Therefore, µ(A) = m(A).

0 Now we apply this for A = Yn. Thus, there exists a sequence (Xn)n ⊆ Σ such 0 0 0 −1 that Xn ⊆ Yn, Xn ∈ P, and µ(Yn \ Xn) < n for all n. By (a), we may assume 0 0 ∞ 0 0 0 0 −1 that Xn ↑. The set X0 = ∪n=1Xn satisfies Yn \X0 ⊆ Yn \Xn, so µ(Yn \X0) < n 1.1. σ-ALGEBRAS, MEASURABLE FUNCTIONS, MEASURES 11

0 ∞ 0 0 for all n. Then µ(Yn \X0) = 0, and µ((∪n=1Yn)\X0) = 0, or µ(X \X0) = 0. Let 0 0 Xn := (Xn ∪(X \X0))∩Yn, then the sequence (Xn)n has the required properties. ∞ The desired pairwise disjoint sequence (An)n=1 is given recurrently by A1 = X1 k and Ak+1 = Xk+1 \ ∪i=1Ai.

I(b) Suppose P satisfies (b). Let FA be the family of all pairwise disjoint families of elements of PA of nonzero measure. Then FA is ordered by inclusion and, obviously, satisfies the conditions of the Zorn lemma. Therefore, we have a maximal element in FA, say ∆. Then ∆ is at most countable family, say ∆ = {Dn}n. By (b), its union D := ∪nDn is an element of PA as well. If D is a proper subset of A, then µ(A \ D) > 0. The set A \ D has a subset F ∈ P of the positive measure. Then ∆1 := ∆ ∪ {F } is an element of FA which is strictly greater then ∆. The obtained contradiction, shows that A ∈ P for every (Yn)n-bounded set A. So, we may take Xn := Yn for each n. m−1 m Now we apply this for A = Zm := Ym \ ∪k=1 Yk. Let Zm = ∪nDn be a pairwise m m disjoint union, where Dn ∈ P for all n, m. The family {Dn }n,m is an at most m ∞ countable disjoint decomposition of X, say {Dn }n,m = (An)n=1. The sequence ∞ (An)n=1 satisfies the required properties. 2

1.1.7.◦ Exercises to Section 1.1.

Exercise 1.1.1 Prove that the cardinality of any σ-algebra is either finite or uncountable.

Exercise 1.1.2 Let f and g be measurable R-valued functions. Show that f · g is measurable.

Exercise 1.1.3 For each X ⊆ N and for each k ∈ N define: k nX := card(X ∩ {1, ..., k}) .

−1 k Let A be a family of all X ⊆ N such that the number µ(X) = lim k nX exists. Is A a k→∞ σ-algebra? Is µ a measure on A?

Exercise 1.1.4 Prove Proposition 1.1.5.

Exercise 1.1.5 ∗ Let (Ω, Σ, µ) be a measure space. Show that the set

R(µ) := {µ(E): E ∈ Σ} is, in general, neither closed nor convex in R∪{∞}. Is R(µ) closed if, additionally, µ is finite or non-atomic (see Exercise 1.1.9 for the definition)? 12 CHAPTER 1.

Exercise 1.1.6 Show that each separable measure which satisfies Axiom 1.1.3 d) is σ-finite.

∗ Exercise 1.1.7 Give an example of a family (Ωα)α∈[0,2π] of σ-subalgebras of the Borel algebra B([0, 1]2) such that for any α, β ∈ [0, 2π], α 6= β: 2 Ωα ∩ Ωβ = {∅, [0, 1] } and (∀α)[α ∈ [0, π]](∀a)[0 ≤ a ≤ 1](∃B ∈ Ωα)[µ(B) = a].

∗∗ ∞ 2 Exercise 1.1.8 Construct a sequence (Σn)n=1 of σ-subalgebras of the Borel algebra B([0, 1] ) such that, for any n < m, Σm ⊆ Σn &Σn 6= Σm and (∀n)(∀a)[0 ≤ a ≤ 1](∃B ∈ Σn)[µ(B) = a]. Exercise 1.1.9 ∗∗ Let (Ω, Σ, µ) be a measure space equipped with a probabilistic measure µ such that µ has no atom (non-atomic measure), i.e., for any A ∈ Σ, µ(A) > 0 implies that there exists A0 ⊆ A such that 0 < µ(A0) < µ(A). (a) Show that, for any 0 ≤ α ≤ 1, there exists B ∈ Σ such that µ(B) = α. (b) Give an example of a measure ν on (N, P(N)) such that ν(N) = 1 and, for any 0 ≤ α ≤ 1, there exists B ∈ P(N) such that µ(B) = α. Exercise 1.1.10 (The Borel – Cantelli lemma) Let (X, Σ, µ) be a measure space. Show P∞ that if {An}n ⊆ Σ and n=1 µ(An) < ∞ then µ(lim sup An) = 0. n→∞ Exercise 1.1.11 Show that Egoroff’s theorem may fail if µ(X) = ∞.

Exercise 1.1.12 ∗∗∗ Let (X, Ω, µ) be a measure space and µ(X) < ∞. Say that A ∼ B if µ(A4B) = 0. (a) Show that ∼ is an equivalence relation on Ω; (b) Show that the function ρ : (Ω/ ∼)2 → R: ρ([A], [B]) := µ(A4B), is a metric, and (Ω/ ∼, ρ) is a complete metric space.

Exercise 1.1.13 ∗∗∗ Let (Ω/ ∼, ρ) be the metric space from Exercise 1.1.12. (a) Show that (Ω/ ∼, ρ) is compact, if (X, Ω, µ) is a purely atomic measure space. (b) Show that (Ω/ ∼, ρ) is not compact, if (X, Ω, µ) is not purely atomic. Here, the measure space (X, Ω, µ) is called purely atomic if for any A ∈ Ω, µ(A) < ∞, there ∞ P exists a sequence (an)n=1 of elements of A such that µ(A) = n µ({an}).

∗ Exercise 1.1.14 Show that in Proposition 1.1.4 pointwise limits lim sup fn and lim inf fn n→∞ n→∞ may be replaced by a.e.-limits lim sup fn and lim inf fn. n→∞ n→∞ 1.2. VECTOR-VALUED MEASURES 13

1.2 Vector-valued Measures

In this section, we extend the notion of a measure. Then we study the basic operations with signed measures and present the Jordan decomposition theo- rem. We refer for further delicate results about vector-valued measures to [3] and about spaces of signed measures to [7] and [12].

1.2.1.◦ Vector-valued, signed and complex measures. Let Σ be a σ-algebra of subsets of a set X, and let E = (E, k · k) be a Banach space.

Definition 1.2.1 A function µ :Σ → E ∪ {∞} is called a vector-valued measure (or E-valued measure) if (a) µ(∅) = 0; S P (b) µ( k Ak) = k µ(Ak) for any pairwise disjoint sequence (Ak)k ⊆ Σ; (c) for any A ∈ Σ, µ(A) = ∞, there exists B ∈ Σ such that B ⊆ A and 0 < kµ(B)k < ∞.

Example 1.2.1 Take Σ = P(N), and c0 is the Banach space of all convergent C-valued sequences with a fixed element (αn)n ∈ c0. Define for any A ⊆ N:

ψ(A) := (βn)n, where βn = αn if n ∈ A and βn = 0 if n 6∈ A. Then ψ is a c0-valued measure on P(N). Example 1.2.2 Let X be a set and let Ω be a σ-algebra in P(X). Then for any m m family {µk}k=1 of finite measures on Ω and for any family {wk}k=1 of vectors of Rn, the Rn-valued measure Ψ on Ω is defined by the formula m X Ψ(E) := µk(E) · wk (E ∈ Ω). k=1 Example 1.2.3 Let X be a set and let Ω be a σ-algebra in P(X). Then for m m any family {µk}k=1 of finite measures on Ω, for any family {Ak}k=1 of pairwise m n n disjoint sets in Ω, and for any family {wk}k=1 of vectors of R , the R -valued measure Φ on Ω is defined by the formula m X Φ(E) := µk(E ∩ Ak) · wk (E ∈ Ω). k=1 14 CHAPTER 1.

From Definition 1.2.1, it is quite easy to see (and we leave this as an exercise to the reader) that if µ is an E-valued measure, then for two linearly independent ∞ vectors x, y ∈ E, kxk = kyk = 1, it is impossible to find sequences (An)n=1 and ∞ (Bn)n=1 of elements of Σ such that

µ(An) µ(An) kµ(An)k → ∞ , lim = x , and kµ(Bn)k → ∞ , lim = y. n→∞ kµ(An)k n→∞ kµ(Bn)k This can be expressed as: any vector-valued measure cannot be infinite in two different directions. The next two definitions are special cases of Definition 1.2.1.

Definition 1.2.2 A function µ :Σ → R ∪ {∞} is called a signed measure if S P µ(∅) = 0, µ( k Ak) = k µ(Ak) for any sequence (Ak)k of pairwise disjoint sets from Σ, and, for any A ∈ Σ, µ(A) = ∞, there exists B ∈ Σ such that B ⊆ A and 0 < |µ(B)| < ∞.

Definition 1.2.3 A function µ :Σ → C ∪ {∞} is called a complex measure S P if µ(∅) = 0, µ( k Ak) = k µ(Ak) for any sequence (Ak)k of pairwise disjoint sets from Σ, and, for any A ∈ Σ, µ(A) = ∞, there exists B ∈ Σ such that B ⊆ A and 0 < |µ(B)| < ∞.

There are many interesting classes of vector-valued measures. In this book we consider only a few of them, in particular, the classes given by the following definition.

Definition 1.2.4 An E-valued measure µ is called finite if µ(A) ∈ E for every measurable A. An E-valued measure µ is called σ-finite if there is a sequence ∞ S (Ak)k, Ak ∈ Σ, satisfying X = Ak and µ|Ak is finite measure for all k ∈ N. k=1 An E-valued measure µ is called separable if there exists a countable family R ⊆ Σ such that, for any A ∈ Σ of a finite measure and for any ε > 0, there exists B ∈ R satisfying kµ(A4B)k < ε.

The definitions of σ-finite, finite, and separable signed or complex measure are obvious. 1.2. VECTOR-VALUED MEASURES 15

1.2.2.◦ The variation of a vector-valued measure. Now we show how in the natural way construct a new measure from a given vector-valued measure. Let µ be a E-valued measure on (X, Σ). We define a function

|µ| :Σ → R+ ∪ {∞} by the following formula:

n=j n=j X [ |µ|(A) := sup{ kµ(An)k : An ∈ Σ, An = A, Ak ∩ Ap = ∅ if k 6= p}. n=1 n=1 (1.2.1) This function is called the variation of µ. It is an exercise for the reader to show that |µ| is additive and therefore monotone.

Theorem 1.2.1 Let µ be an E-valued measure on (X, Σ). Then |µ| is a mea- sure.

Proof: Suppose that µ is an E-valued measure. We have to show only the σ-additivity of the function |µ| :Σ → R+ ∪ {∞}. ∞ S Let (An)n=1 be a pairwise disjoint sequence of elements of Σ and A := n An. If |µ|(A) = ∞, then there is nothing to proof. So we may assume that |µ|(A) < ∞. Let π be a finite partition of A into pairwise disjoint members of Σ. Then X X X X kµ(E)k = kµ(E ∩ A)k = k µ(E ∩ An)k ≤ E∈π E∈π E∈π n X X X X X kµ(E ∩ An)k = kµ(E ∩ An)k ≤ |µ|(An). E∈π n n E∈π n Since it holds for any partition π, we obtain the inequality [ X |µ|( An) ≤ |µ|(An). (1.2.2) n n Since |µ| is additive and monotone, then for each n:

i=n n X [ [ |µ|(Ai) = |µ|( Ai) ≤ |µ|( An), i=1 i=1 n 16 CHAPTER 1. and by taking the limit:

∞ X [ |µ|(Ai) ≤ |µ|( An). (1.2.3) i=1 n Combining (1.2.2) and (1.2.3) give us σ-additivity of |µ|. 2 It is also easy to see that µ is a σ-finite or finite E-valued measure iff |µ| is.

Definition 1.2.5 An E-valued measure µ is called a vector-valued measure of bounded variation if |µ| is finite.

The vector-valued measure constructed in Example 1.2.1 is of bounded variation −n if αn = 2 ; and, of unbounded variation if αn = 1/n.

1.2.3.◦ Operations with vector-valued measures. Let Σ be a σ-algebra of subsets of a non-empty set X and let E be a Banach space. The set Mb = E Mb (X, Σ) of all E-valued measures of bounded variation on Σ is a vector space with respect to the natural operations of addition and scalar multiplication, i.e.,

(µ1 + µ2)(A) = µ1(A) + µ2(A), (αµ)(A) = αµ(A) (1.2.4) for every A ∈ Σ.

Theorem 1.2.2 Mb is a Banach space with respect to the norm k · kb:

kµkb := |µ|(X)(∀µ ∈ Mb) which is called the total variation of a measure µ.

Proof: By (1.2.1) and (1.2.4), the map k · kb : Mb → R+ satisfies all axioms of norm. To show that Mb is complete under the norm k · kb, it is enough to show that if ∞ X kµnkb < ∞ (1.2.5) n=1 ∞ for some sequence (µn)n=1 of E-valued measures, then ∞ X µ := µn ∈ Mb. (1.2.6) n=1 1.2. VECTOR-VALUED MEASURES 17

For any A ∈ Σ, the series ∞ X µn(A) n=1 is convergent in E, because E is a Banach space, and there hold (1.2.5) and

kµn(A)k ≤ kµnkb (∀n).

So µ in (1.2.6) is well defined. We leave to the reader to show that µ ∈ Mb as an exercise. 2

In general, if µ1 and µ2 are not finite E-valued measures, the formula (1.2.4) does not define the sum of them. Thus, in general, the set of all E-valued measures is not a vector space.

Now we are going to study the important case of finite signed measures. Let Mb be the Banach space of all finite signed measures on (X, Σ). First, we define a partial ordering in Mb saying that µ1 ≤ µ2 whenever

µ1(A) ≤ µ2(A)(∀A ∈ Σ), (1.2.7) then Mb becomes a partially ordered vector space. We show that Mb is a vector lattice, which means, by the definition, that for every µ, ν ∈ Mb there exists sup{µ, ν} ∈ Mb.

For this purpose, we denote for µ ∈ Mb the number

kµk := sup{|µ(A)| : A ∈ Σ}.

From |(µ1 + µ2)(A)| ≤ |µ1(A)| + |µ2(A)| ≤ kµ1k + kµ2k holding for µ1 and µ2 in Mb and arbitrary A ∈ Σ, it follows that

kµ1 + µ2k ≤ kµ1k + kµ2k.

Furthermore, it is evident that kαµk = |α| · kµk for µ ∈ Mb and a real α. Thus, k · k is, clearly, a norm in Mb (it is equivalent to k · kb in Theorem 1.2.2).

We prove now that sup{µ1, µ2} exists in Mb for all µ1 and µ2 in Mb. For any A ∈ Σ, let the number ν(A) be defined by

ν(A) = sup{µ1(B) + µ2(A \ B): A ⊇ B ∈ Σ}. 18 CHAPTER 1.

It is not difficult to see that ν is a signed measure on Σ. Since the supremum on the right is less than or equal to kµ1k + kµ2k, we get

sup{|ν(A)| : A ∈ Σ} ≤ kµ1k + kµ2k.

It follows that ν is bounded and kνk ≤ kµ1k + kµ2k. Having shown that ν is an element of Mb, we prove now that ν = sup{µ1, µ2}. From the definition of ν, it is clear that ν(A) ≥ µ1(A) and ν(A) ≥ µ2(A) for every A ∈ Σ, so ν is an upper bound of µ1 and µ2. Let τ be another upper bound. Then, for any A ∈ Σ and A ⊇ B ∈ Σ, we have

τ(A) = τ(B) + τ(A \ B) ≥ µ1(B) + µ2(A \ B), so τ(A) ≥ sup{µ1(B) + µ2(A \ B): A ⊇ B ∈ Σ} = ν(A).

This shows that ν = sup{µ1, µ2} in Mb. Note now that inf{µ1, µ2} exists as well, because inf{µ1, µ2} = − sup{−µ1, −µ2}.

Hence Mb is a vector lattice. As a direct consequence of this we have the following theorem in the case of finite signed measures.

Theorem 1.2.3 (The Jordan decomposition theorem) If µ is a signed mea- sure then there exist unique positive measures ν and ξ such that µ = ν − ξ and, for any positive measures ν0 and ξ0 such that µ = ν0 − ξ0, the inequalities ν ≤ ν0 and ξ ≤ ξ0 hold.

Proof: In the case when µ is finite, it is enough to take ν := µ+ = sup{0, µ} and ξ := µ− = − inf{0, µ}. If µ is not finite (or even not σ-finite) the proof is more complicated. We will not treat it in this book, and refer the reader for the proof to [5] and [4]. 2

We list main properties of the Banach space Mb = (Mb, k · kb) of all signed measures of bounded variation on (X, Σ). Proofs are easy and left to the reader.

(i) The norm k · kb on Mb satisfies

|µ| ≤ |ν| ⇒ kµkb ≤ kνkb (∀µ, ν ∈ Mb), 1.2. VECTOR-VALUED MEASURES 19 where |µ| := sup{µ, −µ}. In other words, Mb is a Banach lattice. + (ii) The norm k · kb is additive on the positive cone Mb of Mb:

µ ≥ 0, ν ≥ 0 ⇒ kµ + νkb = kµkb + kνkb.

(iii) Mb is Dedekind complete in the following sense. For any order bounded nonempty family {µα}α∈A, the supremum supα∈A µα exists and can be evaluated by the formula

k j=k X [ sup µα(B) = sup{ µαj (Bj): Bj ∈ Σ, αj ∈ A, Bj = B,Bi∩Bj = ∅ if i 6= j} α∈A j=1 j=1 (1.2.8) for any B ∈ Σ.

1.2.4.◦ Exercises to Section 1.2.

Exercise 1.2.1 ∗∗ Prove the property of vector-valued measures given after Definition 1.2.1

Exercise 1.2.2 ∗ Let µ be a E-valued measure on (X, Σ). Show that

∞ ∞ X [ |µ|(A) := sup{ kµ(An)k : An ∈ Σ, An = A, Ak ∩ Ap = ∅ if k 6= p}. n=1 n=1

Exercise 1.2.3 Let µ be a E-valued measure on (X, Σ). Show that |µ| is additive and monotone in the following sense:

|µ|(A ∪ B) = |µ|(A) + |µ|(B) for any A, B ∈ Σ, A ∩ B = ∅, and |µ|(A) ≤ |µ|(B) for any A, B ∈ Σ, A ⊆ B.

Exercise 1.2.4 ∗∗ Show that a signed measure µ is finite iff its variation |µ| is finite. Show that a signed measure µ is σ-finite iff its variation |µ| is σ-finite.

Exercise 1.2.5 Show that the vector-valued measure constructed in Example 1.2.1 is of un- −n bounded variation if αn = 1/n; and, of bounded variation if αn = 2 . 20 CHAPTER 1.

Exercise 1.2.6 Let µ be a E-valued measure. Show that |µ| is a separable iff µ is separable. Exercise 1.2.7 Show that the function µ from Σ to E, which is defined in the proof of The- orem 1.2.2 by ∞ X µ(A) = µn(A)(∀A ∈ Σ), n=1 is an E-valued measure of bounded variation. Exercise 1.2.8 ∗∗∗ Let µ be a non-atomic signed measure of bounded variation on Σ(i.e. for any A ∈ Σ, µ(A) 6= 0, there exists B ∈ Σ, B ⊆ A, such that 0 < |µ(B)| < |µ(A)|). Using Exercise 1.1.9, show that, for any α, inf{µ(A): A ∈ Σ} ≤ α ≤ sup{µ(A): A ∈ Σ} , there is Aα ∈ Σ such that µ(Aα) = α. Exercise 1.2.9 Let µ be a signed non-atomic measure on (Ω, Σ). Show that the set R(µ) := {µ(E): E ∈ Σ}. is closed in R ∪ {∞}. ∗ Exercise 1.2.10 Let Mb be the Banach space of all finite signed measures on (X, Σ). Show that Mb is a partially ordered Banach space with respect to the order given in (1.2.7).

Exercise 1.2.11 Show that Mb satisfies 1.2.4(i).

Exercise 1.2.12 Show that Mb satisfies 1.2.4(ii).

∗∗ Exercise 1.2.13 Show that (1.2.8) defines the supremum supα∈A µα. Exercise 1.2.14 Show that the Rn-valued measure Ψ:Ω → Rn defined in Example 1.2.2 has a bounded variation and m X |Ψ|(X) ≤ µk(X) · kwkk. k=1 Exercise 1.2.15 ∗∗ Show that the Rn-valued measure Φ:Ω → Rn defined in Example 1.2.3 has a bounded variation and m X |Φ|(X) ≤ µk(Ak) · kwkk. k=1 Show that the range R(Φ) := {Φ(E): E ∈ Ω} n of Φ is a compact subset of R if all µk are non-atomic. Exercise 1.2.16 ∗∗ Take the Rn-valued measure Φ as in Exercise 1.2.15 and assume that any µk, which was used in the construction of Φ, is non-atomic on Ak. Show that R(Φ) is convex in Rn. 1.3. THE LEBESGUE INTEGRAL 21

1.3 The Lebesgue Integral

Historically, many approaches to integration have been used (see the book [2] of S.B. Chae for nice historical remarks). No doubt that the Lebesgue integra- tion is the most successful approach. In the following, we fix a σ-finite measure space (X, A, µ). To avoid unnecessary details, we consider only the case of R- valued functions and leave obvious generalizations to C- or Rn-valued cases to the reader.

1.3.1.◦ The construction of the Lebesgue integral. Recall that if a certain property involving the points of measure space is true, except a subset having measure zero, then we say that this property is true almost everywhere (abbreviated as a.e.). We denote f +(x) = max(0, f(x)), f −(x) = max(0, −f(x)), and |f|(x) = f +(x) + f −(x).

Let Ai ∈ A, i = 1, . . . , n, be such that µ(Ai) < ∞ for all i, and Ai ∩ Aj = ∅ for all i 6= j. The function

n X g(x) = λiχAi (x), λi ∈ R, i=1 is called a simple function. The Lebesgue integral of a simple function g(x) is defined as n Z X g dµ := λi · µ(Ai). X i=1 It is a simple exercise to show that the Lebesgue integral of a simple function is well defined. We leave this to the reader.

Definition 1.3.1 Suppose that µ is finite. Let f : X → R be an arbitrary ∞ nonnegative bounded measurable function and let (gn)n=1 be a sequence of simple functions which converges uniformly to f. Then the Lebesgue integral of f is Z Z f dµ := lim gn dµ. X n→∞ X 22 CHAPTER 1.

It can be easily shown that the limit in Definition 1.3.1 exists and does not ∞ ∞ depend on the choice of a sequence (gn)n=1, and moreover, the sequence (gn)n=1 can be chosen such that 0 ≤ gn ≤ f for all n.

Definition 1.3.2 Let f : X → R be an arbitrary nonnegative bounded measur- able function. Then the Lebesgue integral of f is Z Z f dµ := sup { f dµ : A ∈ A , µ(A) < ∞ } X A

Remark that the existence of the supremum in Definition 1.3.2 is guarantied by 1.1.3(d).

Definition 1.3.3 Let f : X → R be a nonnegative unbounded measurable. Put  f(x) if 0 ≤ f(x) ≤ M, f (x) = M M if M < f(x).

Then the Lebesgue integral of f is Z Z f dµ := lim fM dµ. X M→∞ X

Definition 1.3.4 Let f : X → R be a measurable function. Then the Lebesgue integral of f is defined by Z Z Z f dµ := f +dµ − f −dµ. X X X If both of these terms are finite then the function f is called integrable. In this case we write f ∈ L1.

1.3.2.◦ Elementary properties of the Lebesgue integral. We will use the following notation. For any A ∈ A: Z Z f dµ := f · χA dµ. A X Let us give several obvious properties of Lebesgue integral. 1.3. THE LEBESGUE INTEGRAL 23

Lemma 1.3.1 Let f : X → R be an arbitrary nonnegative measurable function then Z Z f dµ = sup{ φ dµ : φ is a simple function such that 0 ≤ φ ≤ f}. X X

Proof: Firstly consider the case of bounded f and finite µ. Then by the R definition of X f dµ there exists a sequence (see the remark after Definition ∞ 1.3.1) (gn)n=1 of simple functions such that 0 ≤ gn ≤ f for all n, and (gn) converges uniformly to f, and satisfies Z Z Z f dµ = lim gn dµ = sup gn dµ . X n→∞ X n X Now the using of Definitions 1.3.2 and 1.3.3 completes the proof. 2 L1. If f, g : X → R are measurable, g is integrable, and |f(x)| ≤ g(x), then f is integrable and Z Z

f dµ ≤ g dµ. X X R L2. X |f| dµ = 0 if and only if f(x) = 0 a.e.

L3. If f1, f2 : X → R are integrable then, for λ1, λ2 ∈ R, the linear combination λ1f1 + λ2f2 is integrable and Z Z Z [λ1f1 + λ2f2] dµ = λ1 f1 dµ + λ2 f2 dµ. X X X

If f is integrable function, we write f ∈ L1 = L1(X, A, µ).

L4. Let f ∈ L1, then the formula Z ν(A) := f dµ A defines a signed measure on the σ-algebra A.

◦ ∞ 1.3.3. Convergence. If (fn)n=1 is a sequence of integrable functions on a set X, the statement “fn → f as n → ∞” can be taken in many different senses, for example, for pointwise or uniform convergence (cf. the end of Section 1.1). One of them is L1-convergence. 24 CHAPTER 1.

∞ Definition 1.3.5 We say that a sequence (fn)n=1 of integrable functions L1- converges to f (or converges in L1) if Z |fn − f| dµ → 0 as n → ∞.

Of course, uniform convergence implies pointwise convergence, which in turn implies a.e.-convergence, but these modes of convergence do not imply L1-con- vergence. The reader should keep in mind the following examples:

−1 (i) fn = n χ(0,n); (ii) fn = χ(n,n+1); (iii) fn = nχ[0,1/n]; (iv) f1 = χ[0,1], f2 = χ[0,1/2], f3 = χ[1/2,1], f4 = χ[0,1/4], f5 = χ[1/4,1/2], and, in k k general, fn = χ[j/2k,(j+1)/2k], where n = 2 + j with 0 ≤ j < 2 .

In (i), (ii), and (iii), fn → 0 uniformly, pointwise, and a.e., respectively, but R R fn 6→ 0 in L1 (in fact, |fn| dµ = fn dµ = 1 for all n). In (iv), fn → 0 R −k k k+1 in L1 since |fn| dµ = 2 for 2 ≤ n < 2 , but fn(x) does not converge for any x ∈ [0, 1], since there are infinitely many n for which fn(x) = 0 and infinitely many, for which fn(x) = 1. On the other hand, if fn → f a.e. and |fn| ≤ g ∈ L1 for all n, then fn → f in L1. This will be clear from the domi- nated convergence theorem (see Theorem 1.3.2 below) since |fn − f| ≤ 2g. Also, we will see below that if fn → f in L1 then some subsequence converges to f a.e.

1.3.4.◦ The monotone convergence theorem. Now we state one of the most important results about convergence.

∞ Theorem 1.3.1 (The monotone convergence theorem) If (fn)n=1 is a se- + quence in L1 such that fj ≤ fj+1 for all j and f = supn fn then Z Z f dµ = lim fn dµ. X n→∞ X

R ∞ Proof: The limit of the increasing sequence ( fn dµ) (finite or infinite) R R X n=1 exists. Moreover by 1.3.2.(L1), X fn dµ ≤ X f dµ for all n, so Z Z lim fn dµ ≤ f dµ. n→∞ X X 1.3. THE LEBESGUE INTEGRAL 25

To establish the reverse inequality, fix α ∈ (0, 1), let φ be a simple function with ∞ 0 ≤ φ ≤ f, and let En = {x : fn(x) ≥ αφ(x)}. Then (En)n=1 is an increasing sequence of measurable sets whose union is X, and we have Z Z Z fn dµ ≥ fn dµ ≥ α φ dµ (∀n ≥ 1). X En En By 1.3.2.(L4) and by Proposition 1.1.5, lim R φ dµ = R φ dµ, and hence n→∞ En X Z Z lim fn dµ ≥ α φ dµ. n→∞ X X Since this is true for all α, 0 < α < 1, it remains true for α = 1: Z Z lim fn dµ ≥ φ dµ . n→∞ X X Using Lemma 1.3.1, we may take the supremum over all simple functions φ, 0 ≤ φ ≤ f. Thus Z Z lim fn dµ ≥ f dµ. 2 n→∞ X X Proofs of the following two corollaries of Theorem 1.3.1 are left to the reader.

∞ + P∞ Corollary 1.3.1 If (fn)n=1 is a sequence in L1 and f = n=1 fn pointwise then Z X Z f dµ = fn dµ. 2 n

∞ + + Corollary 1.3.2 If (fn)n=1 is a sequence in L1 , f ∈ L1 , and fn ↑ f a.e., then Z Z fn dµ ↑ f dµ. 2

1.3.5.◦ The Fatou lemma. The condition in the previous subsection that ∞ the sequence (fn)n=1 is increasing can be omitted in the following way.

∞ + Lemma 1.3.2 (Fatou’s lemma) If (fn)n=1 is any sequence in L1 then Z Z lim inf fn dµ ≤ lim inf fn dµ . n→∞ n→∞ 26 CHAPTER 1.

Proof: For each k ≥ 1, we have

inf fn ≤ fj (∀j ≥ k), n≥k hence Z Z inf fn dµ ≤ fj dµ (∀j ≥ k), n≥k hence Z Z inf fn dµ ≤ inf fj dµ. n≥k j≥k Now let k → ∞ and apply Theorem 1.3.1: Z Z Z lim inf fn dµ = lim inf fn dµ ≤ lim inf fn dµ . 2 n→∞ k→∞ n≥k n→∞

∞ + + Corollary 1.3.3 If (fn)n=1 is a sequence in L1 , f ∈ L1 , and fn → f a.e., then Z Z f dµ ≤ lim inf fn dµ. 2 n→∞

1.3.6.◦ The dominated convergence theorem. Now we present the following convergence theorem, which has many applications in PDEs, functional analysis, operator theory, etc.

Theorem 1.3.2 (The dominated convergence theorem) Let f and g be measurable, let fn be measurable for any n such that

|fn(x)| ≤ g(x) a.e., and fn → f a.e. If g is integrable then f and fn are also integrable and Z Z lim fn dµ = f dµ. n→∞

Proof: f is measurable by Exercise 1.1.14 and, since |f| ≤ g a.e., we have f ∈ L1. We have that g + fn ≥ 0 a.e. and g − fn ≥ 0 so, by Fatou’s lemma, Z Z Z Z Z g dµ + f dµ ≤ lim inf (g + fn) dµ = g dµ + lim inf fn dµ, n→∞ n→∞ 1.3. THE LEBESGUE INTEGRAL 27 Z Z Z Z Z g dµ − f dµ ≤ lim inf (g − fn) dµ = g dµ − lim sup fn dµ. n→∞ n→∞ Therefore Z Z Z lim inf fn dµ ≥ f dµ ≥ lim sup fn dµ, n→∞ n→∞ and the result follows. 2

1.3.7.! Convergence in measure. Another mode of convergence, that is ∞ frequently useful, is convergence in measure. We say that a sequence (fn)n=1 of measurable functions on (X, M, µ) is Cauchy in measure if, for every ε > 0,

µ({x : |fn(x) − fm(x)| ≥ ε}) → 0 as m, n → ∞,

∞ and that (fn)n=1 converges in measure to f if, for every ε > 0,

µ({x : |fn(x) − f(x)| ≥ ε}) → 0 as n → ∞.

For example, the sequences (i), (iii), and (iv) above converge to zero in measure, but (ii) is not Cauchy in measure.

Proposition 1.3.1 If fn → f in L1 then fn → f in measure.

Proof: Let En,ε = {x : |fn(x) − f(x)| ≥ ε}. Then Z Z |fn − f| dµ ≥ |fn − f| dµ ≥ εµ(En,ε), En,ε

−1 R so µ(En,ε) ≤ ε |fn − f| dµ → 0. 2

The converse of Proposition 1.3.1 is false, as Examples 1.3.3.(i) and 1.3.3.(iii) show.

∞ Theorem 1.3.3 Suppose that (fn)n=1 is Cauchy in measure. Then there is a measurable function f such that fn → f in measure, and there is a subsequence

(fnj )j that converges to f a.e. Moreover, if fn → g in measure then g = f a.e. 28 CHAPTER 1.

∞ Proof: We can choose a subsequence (gj)j = (fnj )j of (fn)n=1 such that if

−j Ej = {x : |gj(x) − gj+1(x)| ≥ 2 }

−j S∞ P∞ −j 1−k then µ(Ej) ≤ 2 . If Fk = j=k Ej then µ(Fk) ≤ k=1 2 = 2 , and if x 6∈ Fk we have for i ≥ j ≥ k

i−1 i−1 X X −l 1−j |gj(x) − gi(x)| ≤ |gl+1(x) − gl(x)| ≤ 2 ≤ 2 . (1.3.1) l=j l=j

c Thus (gj)j is pointwise Cauchy on Fk . Let

∞ \ F = Fk = lim sup Ej. j k=1

Then µ(F ) = 0, and if we set f(x) = lim gj(x) for x 6∈ F , and f(x) = 0 for x ∈ F , then f is measurable and gj → f a.e. By (1.3.1), we have that

1−j |gj(x) − f(x)| ≤ 2 for x 6∈ Fk and j ≥ k. Since µ(Fk) → 0 as k → ∞, it follows that gj → f in measure, because 1 1 {x : |f (x) − f(x)| ≥ ε} ⊆ {x : |f (x) − g (x)| ≥ ε} ∪ {x : |g (x) − f(x)| ≥ ε}, n n j 2 j 2 and the sets on the right both have small measure when n and j are large. Likewise, if fn → g in measure 1 1 {x : |f(x) − g(x)| ≥ ε} ⊆ {x : |f(x) − f (x)| ≥ ε} ∪ {x : |f (x) − g(x)| ≥ ε} n 2 n 2 for all n, hence µ({x : |f(x) − g(x)| ≥ ε}) = 0 for all ε > 0, and f = g a.e. 2

Theorem 1.3.4 Let fn → f in L1, then there is a subsequence (fnk )k such that fnk → f a.e.

Proof: Let En,ε = {x : |fn(x) − f(x)| ≥ ε}. 1.3. THE LEBESGUE INTEGRAL 29

Then Z Z |fn − f| dµ ≥ |fn − f| dµ ≥ εµ(En,ε), En,ε so µ(En,ε) → 0. Then, by Theorem 1.3.3, there is a subsequence (fnk )k such that fnk → f a.e. 2

1.3.8.◦ Exercises to Section 1.3.

Exercise 1.3.1 Check that the limit in Definition 1.3.1 exists and does not depend on the ∞ choice of a sequence (gn)n=1.

Exercise 1.3.2 Prove the properties L1 − L4 of the Lebesgue integral from 1.3.2.

∞ Exercise 1.3.3 Investigate simple properties of sequences (fn)n=1 in 1.3.3.(i)-(iv).

Exercise 1.3.4 Let [a, b] be a compact interval in R and µ be a standard Borel measure on [a, b]. Show that the Lebesgue integral from any continuous function f :[a, b] → R coincides with the Riemann integral from f.

Exercise 1.3.5 Show that the condition “µ(X) < ∞” in Egoroff’s theorem can be replaced by “|fn| ≤ g for all n, where g ∈ L1”.

Exercise 1.3.6 Prove the dominated convergence theorem by using of the Egoroff theorem.

∗ Exercise 1.3.7 Let fn → f in measure and gn → g in measure. Show that

fn · gn → f · g in measure if µ(X) < ∞, but not necessarily if µ(X) = ∞.

Exercise 1.3.8 Prove Corollary 1.3.1.

Exercise 1.3.9 Prove Corollary 1.3.2.

Exercise 1.3.10 Let fn → f in measure and fn ≥ 0 for all n. Show that Z Z f dµ ≤ lim inf fn dµ. n→∞ 30 CHAPTER 1.

Exercise 1.3.11 Prove Corollary 1.3.3.

∗∗ Exercise 1.3.12 Let χAn → f in measure. Show that f is a.e. equal to the characteristic function of a measurable set.

Exercise 1.3.13 Fix the Lebesgue measure on R. + (a) Given a nonempty set Ξ ⊆ L1 (R), show that n Z o Z inf ξ(t) dt : ξ ∈ Ξ ≤ f(t) dt for every f ∈ co(Ξ), where co(Ξ) is the convex hull of Ξ in the vector space L1(R), which is, by the definition, the intersection of all convex sets containing Ξ.

+ (b) Construct an example of a nonempty set Θ ⊆ L1 (R) such that n Z o Z inf ξ(t) dt : ξ ∈ Θ < f(t) dt for every f ∈ co(Θ).

+ (c) Show that a nonempty set ∆ ⊆ L1 (R) satisfying n Z o Z inf ξ(t) dt : ξ ∈ ∆ < f(t) dt for every f ∈ co(∆) can not be finite.

R t Exercise 1.3.14 Let f ∈ L1(R) and F (t) := −∞ f(s)ds. Show that F is continuous. Com- pare with Exercise 1.3.2. Chapter 2

2.1 The Extension of Measure

In this section we present the Caratheodory extension theorem which is the most important result in the measure theory. By using of this theorem, we can con- struct almost all important measures, in particular, the Lebesgue measure and Lebesgue – Stieltjes Measures on R. Then we consider the product of measure spaces and prove the Fubini theorem.

2.1.1.◦ Outer measures. Let X be a nonempty set. An outer measure (or submeasure) on X is a function

ξ : P(X) → [0, ∞] that satisfies: (a) ξ(∅) = 0; (b) ξ(A) ≤ ξ(B) if A ⊆ B;

∞ ! ∞ S P (c) ξ Aj ≤ ξ(Aj) for all sequences (Aj)j of subsets of X. j=1 j=1 The common way to obtain an outer measure is to start with a family G of “elementary sets” on which a notion of measure (= length, mass, charge, or volume) is defined (such as rectangles or cubes in Rn) and then approximate arbitrary sets from the outside by countable unions of members of G.

31 32 CHAPTER 2.

Proposition 2.1.1 Let G ⊆ P(X) and let ρ : G → [0, ∞] be such that ∅ ∈ G, X ∈ G, and ρ(∅) = 0. For any A ⊆ X, define

( ∞ ∞ ) ∗ X [ ξ(A) = ρ (A) = inf ρ(Gj): Gj ∈ G and A ⊆ Gj . (2.1.1) j=1 j=1 Then ξ is an outer measure.

S∞ Proof: For any A ⊆ X, we have A ⊆ j=1 X, so ξ is well defined. Obviously ∞ ξ(∅) = 0. To prove countable subadditivity, suppose {Aj}j=1 ⊆ P(X) and ε > 0. k ∞ S∞ k For each j, there exists {Gj }k=1 ⊆ G such that Aj ⊆ k=1 Gj and

∞ X k −j ρ(Gj ) ≤ ξ(Aj) + ε2 . k=1 S∞ Then if A = j=1 Aj, we have

∞ [ k X k X A ⊆ Gj & ρ(Gj ) ≤ ξ(Aj) + ε, j,k=1 j,k j P whence ξ(A) ≤ j=1 ξ(Aj) + ε. Since ε > 0 is arbitrary, we have done. 2

2.1.2.◦ The Caratheodory theorem. The principle step that leads from outer measures to measures is following. Let ξ be an outer measure on X.

Definition 2.1.1 A set A ⊆ X is called ξ-measurable if

ξ(B) = ξ(B ∩ A) + ξ(B ∩ (X \ A)) (∀B ⊆ X). (2.1.2)

Of course, the inequality

ξ(B) ≤ ξ(B ∩ A) + ξ(B ∩ (X \ A)) holds for any A and B. So, to prove that A is ξ-measurable, it suffices to prove the reverse inequality, which is trivial if ξ(B) = ∞. Thus, we see that A is ξ-measurable iff

ξ(B) ≥ ξ(B ∩ A) + ξ(B ∩ (X \ A)) (∀B ⊆ X, ξ(B) < ∞). (2.1.3) 2.1. THE EXTENSION OF MEASURE 33

Theorem 2.1.1 (Caratheodory’s theorem) Let ξ be an outer measure on X. Then the family Σ of all ξ-measurable sets is a σ-algebra, and the restriction of ξ to Σ is a complete measure.

Proof: First, we observe that Σ is closed under complements, since the def- inition of ξ-measurability of A is symmetric in A and Ac := X \ A. Next, if A, B ∈ Σ and E ⊆ X,

ξ(E) = ξ(E ∩ A) + ξ(E ∩ Ac)

= ξ(E ∩ A ∩ B) + ξ(E ∩ A ∩ Bc) + ξ(E ∩ Ac ∩ B) + ξ(E ∩ Ac ∩ Bc). But (A ∪ B) = (A ∩ B) ∪ (A ∩ Bc) ∪ (Ac ∩ B) so, by subadditivity,

ξ(E ∩ A ∩ B) + ξ(E ∩ A ∩ Bc) + ξ(E ∩ Ac ∩ B) ≥ ξ(E ∩ (A ∪ B)), and hence ξ(E) ≥ ξ(E ∩ (A ∪ B)) + ξ(E ∩ (A ∪ B)c). It follows that A ∪ B ∈ Σ, so Σ is an algebra. Moreover, if A, B ∈ Σ and A ∩ B = ∅,

ξ(A ∪ B) = ξ((A ∪ B) ∩ A) + ξ((A ∪ B) ∩ Ac) = ξ(A) + ξ(B), so ξ is finitely additive on Σ. To show that Σ is a σ-algebra, it suffices to show that Σ is closed under countable ∞ disjoint unions. If (Aj)j=1 is a sequence of disjoint sets in Σ, set

n ∞ [ [ Bn = Aj & B = Aj. j=1 j=1 Then, for any E ⊆ X,

c ξ(E ∩ Bn) = ξ(E ∩ Bn ∩ An) + ξ(E ∩ Bn ∩ An)

= ξ(E ∩ An) + ξ(E ∩ Bn−1), Pn so a simple induction shows that ξ(E ∩ Bn) = j=1 ξ(E ∩ Aj). Therefore

n c X c ξ(E) = ξ(E ∩ Bn) + ξ(E ∩ Bn) ≥ ξ(E ∩ Aj) + ξ(E ∩ B ) j=1 34 CHAPTER 2. and, letting n → ∞, we obtain

∞ ∞ ! X c [ c ξ(E) ≥ ξ(E ∩ Aj) + ξ(E ∩ B ) ≥ ξ (E ∩ Aj) + ξ(E ∩ B ) j=1 j=1

= ξ(E ∩ B) + ξ(E ∩ Bc) ≥ ξ(E). Thus the inequalities in this last calculation become equalities. It follows B ∈ Σ. P∞ Taking E = B we have ξ(B) = 1 ξ(Aj), so ξ is σ-additive on Σ. Finally, if ξ(A) = 0 then we have for any E ⊆ X

ξ(E) ≤ ξ(E ∩ A) + ξ(E ∩ Ac) = ξ(E ∩ Ac) ≤ ξ(E), so A ∈ Σ. Therefore ξ(E ∩ A) = 0 and ξ|Σ is a complete measure. 2 Combination of Proposition 2.1.1 and Theorem 2.1.1 gives the following corollary which is also called Caratheodory’s theorem.

Corollary 2.1.1 Let G ⊆ P(X) be such that ∅ ∈ G, X ∈ G, and let ρ : G → [0, ∞] satisfy ρ(∅) = 0. Then the family Σ of all ρ∗-measurable sets (where ρ∗ is given by (2.1.1)) is a σ-algebra, and the restriction of ρ∗ to Σ is a complete measure. 2

2.1.3.◦ Premeasures. Our main definition in this subsection is following.

Definition 2.1.2 Let A be an algebra of subsets of X, i.e. A contains ∅ and X, and A is closed under finite intersections and complements. A function ζ : A → [0, ∞] is called a premeasure if ζ(∅) = 0 and

∞ ∞ [ X ζ( Aj) = ζ(Aj) j=1 j=1 S∞ for any disjoint sequence (Aj)j of elements of A such that j=1 Aj ∈ A.

Theorem 2.1.2 If ζ is a premeasure on an algebra A ⊆ P(X) and

ζ∗ : P(X) → [0, ∞]

∗ ∗ is given by (2.1.1) then ζ |A = ζ and every A ∈ A is ζ -measurable. 2.1. THE EXTENSION OF MEASURE 35

Proof: First, ζ is obviously additive and hence monotone on A, i.e.

A ⊆ B ⇒ ζ(A) ≤ ζ(B)(∀A, B ∈ A).

S∞ Suppose that A ∈ A, An ∈ A, and A ⊆ n=1 An. Then

n−1 [ Bn := A ∩ (An \ Aj) ∈ A, j=1

∞ since A is an algebra. The sequence (Bn)n=1 is disjoint. This implies

∞ ∞ X X ζ(A) = ζ(Bn) ≤ ζ(An), n=1 n=1 and therefore ζ(A) ≤ ζ∗(A) (here we used the monotonity of ζ). The reverse inequality ζ∗(A) ≤ ζ(A) follows from the definition of ζ∗. To show that every A ∈ A is ζ∗-measurable, take A ∈ A, Y ⊆ X, and ε > 0. ∞ Then, by (2.1.1), there exists a sequence (An)n=1 in A such that

∞ ∞ [ X ∗ Y ⊆ An & ζ(An) ≤ ζ (Y ) + ε. n=1 n=1 S∞ S∞ Since ζ is additive on A, Y ∩ A ⊆ n=1(An ∩ A), and Y ∩ (X \ A) ⊆ n=1(An ∩ (X \ A)) then

∞ ∞ ∗ X X ∗ ∗ ζ (Y ) + ε ≥ ζ(An ∩ A) + ζ(An ∩ (X \ A)) ≥ ζ (Y ∩ A) + ζ (Y ∩ (X \ A)). n=1 n=1 Since ε > 0 is arbitrary and ζ∗ is an outer measure,

ζ∗(Y ) ≥ ζ∗(Y ∩ A) + ζ∗(Y ∩ (X \ A)) ≥ ζ∗(Y ).

Y ⊆ X is arbitrary, so A is ζ∗-measurable. 2

Theorem 2.1.3 If ζ is a premeasure on an algebra A ⊆ P(X) and Aˆ is the ˆ ˆ ˆ σ-algebra generated by A, then there exists a measure ζ on A such that ζ|A = ζ. Moreover, if ζ is σ-finite, then ζˆ is unique extension of ζ. 36 CHAPTER 2.

Proof: By Caratheodory’s theorem and Theorem 2.1.2, such an extension ζ∗ exists. We leave to the reader to show that measure ζˆ on Aˆ, which extends ζ from A, is unique if the premeasure ζ is σ-finite. 2

2.1.4.◦ The Lebesgue and Lebesgue – Stieltjes measure on R. The most important application of Caratheodory’s theorem is the construction of the Lebesgue measure on R. Take G as the set of all intervals [a, b], where a, b ∈ R∪{−∞, +∞} and [a, b] = ∅ if a > b. Define the function ρ : G → R∪{∞} by ρ([a, b]) := b − a (∀a ≤ b)& ρ([a, b]) = 0 (∀a > b). The function ρ has the obvious extension (which we denote also by ρ) to the algebra A generated by all finite or infinite intervals, and this extension is a premeasure on A. The σ-algebra Σ given by Corollary 2.1.1 is called the the Lebesgue σ-algebra in R, and the restriction of ρ∗ to Σ = Σ(R) is called the Lebesgue measure on R and is denoted by µ. By Theorem 2.1.3, µ is the unique extension of ρ. By the construction, B(R) ⊆ Σ(R). Hence the Lebesgue measure is a Borel measure. It can be shown that B(R) 6= Σ(R) (see Exercise 2.1.8) and that the Lebesgue measure can be obtained also as the completion of any Borel measure ω such that ω([a, b]) = b − a (∀a ≤ b). The notion of the Lebesgue measure on R has the following very natural gener- alization. Suppose that µ is a σ-finite Borel measure on R, and let F (x) := µ((−∞, x]) (∀x ∈ R). (2.1.4) Then F is increasing and right continuous (see Exercise 2.1.9). Moreover, if b > a,(−∞, b] = (−∞, a] ∪ (a, b], so µ((a, b]) = F (b) − F (a). Our procedure used above can be to turn this process around and construct a measure µ starting from an increasing, right-continuous function F . The special case F (x) = x will yield the usual Lebesgue measure. As building blocks we can use the left-open, right-closed intervals in R i.e. sets of the form (a, b] or (a, ∞) or ∅, where −∞ ≤ a < b < ∞. We call such sets h-intervals. The family A of all finite disjoint unions of h-intervals is an algebra, moreover, the σ-algebra generated by A is the Borel algebra B(R). We omit the direct and routine proof of the following elementary lemma. 2.1. THE EXTENSION OF MEASURE 37

Lemma 2.1.1 Given an increasing and right continuous function F : R → R, if (aj, bj](j = 1, . . . , n) are disjoint h-intervals, let

n ! n [ X µ0 (aj, bj] = [F (bj) − F (aj)] , 1 1 and let µ0(∅) = 0. Then µ0 is a premeasure. 2

Theorem 2.1.4 If F : R → R is any increasing, right continuous function, there is a unique Borel measure µF on R such that

µF ((a, b]) = F (b) − F (a)(∀a, b).

If G is another such function then µF = µG iff F − G is constant. Conversely, if µ is a Borel measure on R that is finite on all bounded Borel sets, and we define   µ((0, x]) if x > 0, F (x) = 0 if x = 0,  −µ((x, 0]) if x < 0, then F is increasing and right continuous function, and µ = µF .

Proof: Each F induces a premeasure on B(R) by Lemma 2.1.1. It is clear that F and G induce the same premeasure iff F − G is constant, and that these S∞ premeasures are σ-finite (since R = −∞(j, j + 1]). The first two assertions follow now from Exercise 2.1.11. As for the last one, the monotonicity of µ implies the monotonicity of F , and the continuity of µ from above and from below implies the right continuity of F for x ≥ 0 and x < 0. It is evident that µ = µF on A, and hence µ = µF on B(R) (accordingly to Exercise 2.1.11). 2 Lebesgue – Stieltjes measures possess some important and useful regularity prop- erties. Let us fix a complete Lebesgue – Stieltjes measure µ on R associated to an increasing, right continuous function F . We denote by Σµ the Lebesgue algebra correspondent to µ. Thus, for any E ∈ Σµ,

( ∞ ∞ ) X [ µ(E) = inf [F (bj) − F (aj)] : E ⊆ (aj, bj] j=1 j=1 38 CHAPTER 2.

( ∞ ∞ ) X [ = inf µ((aj, bj]) : E ⊆ (aj, bj] . j=1 j=1

Since B(R) ⊆ Σµ, we may replace in the second formula for µ(E) h-intervals by open intervals, namely.

Lemma 2.1.2 For any E ∈ Σµ, ( ∞ ∞ ) X [ µ(E) = inf µ((aj, bj)) : E ⊆ (aj, bj) . 2 j=1 j=1

Theorem 2.1.5 If E ∈ Σµ then µ(E) = inf{µ(U): U ⊇ E and U is open} = sup{µ(K): K ⊆ E and K is compact}.

Proof: By Lemma 2.1.2, for any ε > 0, there exist intervals (aj, bj) such that ∞ ∞ [ X E ⊆ (aj, bj)& µ(E) ≤ µ((aj, bj)) + ε. j=1 j=1 S∞ If U = j=1(aj, bj) then U is open, E ⊆ U, and µ(U) ≤ µ(E) + ε. On the other hand, µ(U) ≥ µ(E) whenever E ⊆ U so the first equality is valid. For the second one, suppose first that E is bounded. If E is closed then E is compact and the equality is obvious. Otherwise, given ε > 0, we can choose an open U, E¯ \ E ⊆ U, such that µ(U) ≤ µ(E¯ \ E) + ε. Let K = E¯ \ U. Then K is compact, K ⊆ E, and µ(K) = µ(E) − µ(E ∩ U) = µ(E) − [µ(U) − µ(U \ E)] ≥ µ(E) − µ(U) + µ(E¯ \ E) ≥ µ(E) − ε.

If E is unbounded, let Ej = E ∩ (j, j + 1]. By the preceding argument, for −j any ε > 0, there exist a compact Kj ⊆ Ej with µ(Kj) ≥ µ(Ej) − ε2 . Let Sj=n Hn = j=−n Kj. Then Hn is compact, Hn ⊆ E, and

j=n [ µ(Hn) ≥ µ( Ej) − ε. j=−n Sj=n Since µ(E) = lim µ( Ej), the result follows. 2 n→∞ j=−n The proofs of the following two theorems are left to the reader as exercises. 2.1. THE EXTENSION OF MEASURE 39

Theorem 2.1.6 If E ⊆ R, the following are equivalent: (a) E ∈ Σµ; (b) E = V \ N1, where V is a Gδ–set and µ(N1) = 0; (c) E = H ∪ N2, where H is an Fσ–set and µ(N2) = 0. 2

Theorem 2.1.7 If E ∈ Σµ and µ(E) < ∞ then, for every ε > 0, there is a set A that is a finite union of open intervals such that µ(E4A) < ε.2

◦ 2.1.5. Product measures. Let {(Xα, Σα, µα)}α∈∆ be a nonempty family of measure spaces. In this subsection we show how to construct a product-measure Q on Xα. First we define the family Ω of blocks: α∈∆ Y A(Aα1 ,Aα2 ,...,Aαn ) := Aα1 ×Aα2 ×· · ·×Aαn × Xα (Aαk ∈ Σαk );

α∈∆; α6=αk: k=1,...,n and define a function: µ :Ω → R+ ∪ {∞}: Y µ(A(Aα1 ,Aα2 ,...,Aαn )) := µα1 (Aα1 )·µα2 (Aα2 )·...·µαn (Aαn )· µα(Xα).

α∈∆; α6=αk:k=1,...,n This function possesses an extension (by additivity) on the algebra A generated by Ω. It is an exercise to show that µ is a premeasure on A.

Definition 2.1.3 The measure µˆ on the σ-algebra Σ generated by A accordingly to Theorem 2.1.3 is called the product measure of {µα}α∈∆, and the triple Y ( Xα, Σ, µˆ). α∈∆ is called the product of measure spaces (Xα, Σα, µα). We denote the σ- N N algebra Σ by α∈∆ Σα, and the measure µˆ by α∈∆ µα.

This construction is essentially non-trivial only in the following two cases. The first one is when card(∆) < ∞. The second one is when every µα is a probability measure (i.e. µα(Xα) = 1 for all α). There are many interesting and important results about the product of measure spaces. The mostly used one is the Fubini theorem about changing the order 40 CHAPTER 2. of integration. Let us use the the following notations. If E ⊆ X1 × X2 and x1 ∈ X1, x2 ∈ X2, we define

x2 Ex1 := {x ∈ X2 :(x1, x) ∈ E} & E := {x ∈ X1 :(x, x2) ∈ E} .

x2 Also, if f : X1 ×X2 → R is a function, we define fx1 : X2 → R and f : X1 → R by x2 fx1 (x) := f(x1, x)& f (x) := f(x, x2) .

x2 In this notations, for example, (χ ) = χ and (χ ) = χ x2 . E x1 Ex1 E E

Theorem 2.1.8 (Fubini’s theorem) Let µ1, µ2 be σ-finite measures on (X1, Σ1) and (X2, Σ2),

(X1 × X2, Σ1 ⊗ Σ2, µ1 ⊗ µ2) = (X1, Σ1, µ1) × (X2, Σ2, µ2), and let f ∈ L1(X1 × X2, Σ1 ⊗ Σ2, µ1 ⊗ µ2). Then fx1 ∈ L1(X2, Σ2, µ2) µ1-a.e., x2 and f ∈ L1(X1, Σ1, µ1) µ2-a.e., and Z Z Z  Z Z  x2 f d(µ1 ⊗ µ2) = f dµ1 dµ2 = fx1 dµ2 dµ1 . X1×X2 X2 X1 X1 X2 To prove this theorem, we need several lemmas.

Lemma 2.1.3 Let (X1, Σ1, µ1) and (X2, Σ2, µ2) be measure spaces, E ∈ Σ1⊗Σ2, and let f be a Σ1 ⊗ Σ2-measurable function on X1 × X2, then: x2 (a) Ex1 ∈ Σ2 for all x1 ∈ X1 and E ∈ Σ1 for all x2 ∈ X2; x2 (b) fx1 is Σ2-measurable and f is Σ1-measurable for all x1 ∈ X1 and x2 ∈ X2.

Proof: Denote by A the collection of all A ⊆ X1 × X2 such that

x2 Ax1 ∈ Σ2 & A ∈ Σ1 (∀ x1 ∈ X1 , x2 ∈ X2) . The family A contains all rectangles. Thus, since ∞ ∞ ∞ ∞ [ [ [ x2 [ x2 [ An]x1 = [An]x1 , [ Bn] = [Bn] (2.1.5) n=1 n=1 n=1 n=1 and

x2 x2 [X1 × X2 \ A]x1 = X2 \ Ax1 , [X1 × X2 \ A] = X1 \ A , (2.1.6)

A is a σ-algebra. So Σ1 ⊗ Σ2 ⊆ A, and (a) is proved. Now the part (b) follows from (a) due to

−1 −1 x2 −1 −1 x2 fx1 (A) = [f (A)]x1 &[f ] (A) = [f (A)] (∀A ⊆ R) . 2 (2.1.7) 2.1. THE EXTENSION OF MEASURE 41

Definition 2.1.4 A family M ⊆ P(X) is called a monotone class if M is closed under countable increasing unions and countable decreasing intersections.

The proof of the following lemma is an elementary exercise and we leave it to the reader.

Lemma 2.1.4 If A ⊆ P(X) is an algebra then the monotone class generated by A coincides with the σ-algebra generated by A. 2

Lemma 2.1.5 Let (X1, Σ1, µ1) and (X2, Σ2, µ2) be measure spaces, E ∈ Σ1⊗Σ2. x2 Then the functions x1 → µ2(Ex1 ) and x2 → µ1(E ) are measurable on (X1, Σ1) and (X2, Σ2), and Z Z x2 µ1 ⊗ µ2(E) = µ1(E ) dµ2 = µ2(Ex1 ) dµ1 . (2.1.8) X2 X1

Proof: First we consider the case when µ1 and µ2 are finite. Let A be the family of all E ∈ Σ1 ⊗ Σ2 for which (2.1.8) is true. If E = A × B, then x2 µ1(E ) = µ1(A)χB(x2) and µ2(Ex1 ) = µ2(B)χA(x1), so E ∈ A. By additivity, it follows that finite disjoint unions of rectangles are in A so, by Lemma 2.1.4, ∞ it will suffice to show that A is a monotone class. If (En)n=1 is an increasing S∞ x2 sequence in A and E = n=1 En, then the function fn(x2) = µ1((En) ) are x2 measurable and increase pointwise to f(y) = µ1(E ). Hence f is measurable and, by the monotone convergence theorem, Z Z x2 x2 µ1(E ) dµ2 = lim µ1((En) ) dµ2 = lim µ1 × µ2(En) = µ1 × µ2(E). n→∞ n→∞ X2 X2 R ∞ Likewise µ1 × µ2(E) = µ2(Ex) dµ1, so E ∈ A. Similarly, if (En) is a X1 n=1 T∞ x2 decreasing sequence in A and E = n=1 En, the function x2 → µ1((E1) ) is in x2 L1(µ2) because µ1((E1) ) ≤ µ1(X1) < ∞ and µ2(X2) < ∞, so the dominated convergence theorem can be applied to show that E ∈ A. Thus, A is a monotone class, and the proof is complete for the case of finite measure spaces.

Finally, if µ1 and µ2 are σ-finite, we can write X1 × X2 as the union of an j j ∞ increasing sequence (X1 ×X2 )n=1 of rectangles of finite measure. If E ∈ Σ1 ⊗Σ2, j j the preceding argument applies to E ∩ (X1 × X2 ) for each j gives us Z Z j j x2 j j µ1 × µ2(E ∩ (X1 × X2 )) = µ1(E ∩ X1 ) dµ2 = µ2(Ex1 ∩ X2 ) dµ1 . j j X2 X1 The application of the monotone convergence theorem then yields the desired result. 2 42 CHAPTER 2.

Lemma 2.1.6 (Tonelli’s theorem) Let (X1, Σ1, µ1) and (X2, Σ2, µ2) be mea- sure spaces, and f : X1 × X2 → R+ be a Σ1 ⊗ Σ2-measurable function. Then the functions Z Z x2 fµ2 (x1) := fx1 dµ2 & fµ1 (x2) := f dµ1 X2 X1 are Σ1-measurable and Σ2-measurable, respectively, and Z Z Z  Z Z  x2 f dµ1 ⊗ µ2 = f dµ1 dµ2 = fx1 dµ2 dµ1 . (2.1.9) X1×X2 X2 X1 X1 X2

Proof: In the case when f is a characteristic function, the statement of this lemma follows from Lemma 2.1.5. Therefore, by linearity, it holds also for nonnegative simple functions. If a nonnegative measurable function f is arbitrary, there exists a sequence of nonnegative simple functions which increase ∞ pointwise to f, say (fn)n=1. By the monotone convergence theorem, Z Z Z n fµ2 dµ1 = lim fµ dµ1 = lim fn dµ1 ⊗ µ2, n→∞ 2 n→∞ X1 X1 X1×X2 and Z Z Z n fµ1 dµ2 = lim fµ dµ2 = lim fn dµ1 ⊗ µ2, n→∞ 1 n→∞ X2 X2 X1×X2 where Z Z n n x2 fµ2 (x1) := [fn]x1 dµ2 , fµ1 (x2) := [fn] dµ1 . X2 X1 This proves (2.1.9) and the lemma. 2

Proof of Theorem 2.1.8: Since an R-valued function f is Lebesgue inte- grable iff its positive f + and negative f − parts are integrable, it is sufficient to prove the theorem only for nonnegative function f ∈ L1(X1 × X2, Σ, µ1 ⊗ µ2). But this was exactly done in Lemma 2.1.6. 2

2.1.6.◦ Exercises to Section 2.1.

Exercise 2.1.1 Show that if {(aα, bα): α ∈ G} is a finite or countable family of intervals in S P R such that [0, 1] ⊆ α∈G(aα, bα) then α∈G |aα − bα| > 1. 2.1. THE EXTENSION OF MEASURE 43

Exercise 2.1.2 ∗ Show that the length of intervals in R is a premeasure on the subalgebra in P(R) generated by all intervals.

Exercise 2.1.3 ∗ Give an example of an outer measure on P(N) which is not a premeasure on any subalgebra A ⊆ P(N) of infinite cardinality.

Exercise 2.1.4 ∗∗ Give an example of an algebra A in P(N) which is not a σ-algebra, and an example of a function ρ : A → [0, 1], such that ρ(∅) = 0, and ρ(N) = 1, for which the outer measure ρ∗ constructed in Proposition 2.1.1 is identically equal to zero.

Exercise 2.1.5 Prove the uniqueness in Theorem 2.1.3

∗∗ S Exercise 2.1.6 Let G be a set of half-open intervals in R. Prove that g∈G g is Lebesgue measurable.

Exercise 2.1.7 Show that the Lebesgue measure can be obtained as the completion of any Borel measure ω such that ω([a, b]) = b − a (∀a ≤ b).

Exercise 2.1.8 ∗ Prove that card(B(R)) < card(Σ(R)). Remark that in Exercise 2.2.6 we will see that Σ(R) 6= P(R).

Exercise 2.1.9 Show that the distribution function F of the Borel measure µ given in (2.1.4) is increasing and right continuous.

Exercise 2.1.10 ∗ Prove Lemma 2.1.1.

Exercise 2.1.11 Let A ⊆ P(X) be an algebra, let µ0 be a σ-finite premeasure on A, and let Ω be the σ-algebra generated by A. Show that there exists a unique extension of µ0 to a measure µ on Ω.

Exercise 2.1.12 ∗∗ Prove Theorem 2.1.7.

Exercise 2.1.13 ∗∗ Show that the function µ defined in the subsection 2.1.5 is a premeasure on the algebra A.

Exercise 2.1.14 Show that if ∆ is infinite and R is taking with the Lebesgue measure then Q the measure µˆ constructed in 2.1.5 on R does not satisfy 1.1.3.(d). α∈∆ 44 CHAPTER 2.

Exercise 2.1.15 Prove the formulas in (2.1.5), (2.1.6) and (2.1.7).

Exercise 2.1.16 Prove Lemma 2.1.4.

Exercise 2.1.17 ∗∗ Prove the Steinhaus theorem: for any Lebesgue measurable set A ⊆ R such that µ(A) > 0 there exist nonempty open intervals I ⊆ A − A, J ⊆ A + A. Show that the conclusion of the theorem is true for the standard Cantor set in spite of it has measure zero.

Exercise 2.1.18 ∗ Let f : R → R be an additive function. Show that f is linear if f is bounded on a Lebesgue measurable set A ⊆ R, µ(A) > 0. Show that f is linear if f is bounded on a standard Cantor set. Show that f is linear if f is measurable on a nontrivial interval. Show that f is not Lebesgue measurable on any nontrivial interval if f is discontinuos at least at one point.

Exercise 2.1.19 ∗∗ Let A, B be Lebesgue measurable subsets of Rn such that µ(A) > 0 and µ(B) > 0. Prove that n µ({x ∈ R : µ(A ∩ (B − x)) > 0}) > 0.

Exercise 2.1.20 ∗∗ Let A be a subalgebra of P([0, 1]2) generated by all rectangles A × B where A and B are Lebesgue measurable subsets of [0, 1]. Let ζ : A → R be the unique additive extension of the function defined on rectangles by the formula:

ζ(A × B) = µ(A ∩ B) , where µ is the Lebesgue measure. Show that ζ is a premeasure on A.

Exercise 2.1.21 Show that the statement of Lemma 2.1.6 may fail without the positivity of the function f.

2.2 Lebesgue Measure and Integral in Rn

In this section, we study Rn and functions from Rn to R from the point of view of the Lebesgue measure and Lebesgue integration. All results presented below possess obvious C-valued analogs. Then we define and study Cantor sets which are very interesting from the point of view of the set topology and the measure theory. Cantor sets are closed Borel nowhere dense subsets of the interval [0, 1] or, more generally, of a Hausdorff space. The main advantage of Cantor sets is 2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 45 that they are used in constructions of many examples in analysis.

2.2.1.◦ The Lebesgue measure on Rn. The Lebesgue measure µn on Rn is the completion of the product of the Lebesgue measure on R according to Definition 2.1.3.

The domain Σn of µn (of course, B(Rn) ⊆ Σn) is the class of Lebesgue mea- surable sets in Rn. We write µ for µn and Z Z f(x) dx := f dµ.

We extend some of the results of previous section to the n-dimensional case. If Qn n E = 1 Ej is a block in R , we call sets Ej ⊆ R the sides of the block E.

Theorem 2.2.1 Let E ∈ Σn. Then (a) µ(E) = inf{µ(U): E ⊆ U, U open} = sup{µ(K): K ⊆ E,K compact};

(b) E = A1 ∪ N1 = A2 \ N2, where A1 is an Fσ set, A2 is a Gδ set, and µ(N1) = µ(N2) = 0; N (c) If µ(E) < ∞ then, for any ε > 0, there is a finite family {Rj}j=1 of disjoint SN blocks, whose sides are intervals such that µ(E4 j=1 Rj) < ε.

Proof: By the definition of product measures, if E ∈ Σn and ε > 0, there is a ∞ S∞ countable family {Tj}j=1 of blocks such that E ⊆ j=1 Tj and

∞ X µ(Tj) ≤ µ(E) + ε. j=1

For each j, by applying Theorem 2.1.6 to the sides of Rj, we can find blocks −j S∞ Uj ⊇ Fj whose sides are open sets such that µ(Uj) < µ(Tj)+ε2 . If U = j=1 Uj then U is open and ∞ X µ(U) ≤ µ(Uj) ≤ µ(E) + 2ε. j=1 This proves the first equation in part (a). The second equation and part (b) follow as in the proofs of Theorems 2.1.6 and 2.1.7. 46 CHAPTER 2.

Next, if µ(E) < ∞ then µ(Uj) < ∞ for all j. Since the sides of Uj are countable unions of open intervals, by taking suitable finite subunions, we obtain blocks −j Vj ⊆ Uj whose sides are finite unions of intervals such that µ(Vj) ≥ µ(Uj)−ε2 . If N is sufficiently large, we have

N ! N ! ∞ ! [ [ [ µ E \ Vj ≤ µ Uj \ Vj + µ Uj < 2ε j=1 j=1 j=N+1 and N ! ∞ ! [ [ µ Vj \ E ≤ µ Uj \ E < ε, j=1 j=1 SN SN so µ(E4 j=1 Vj) < 3ε. Since j=1 Vj can be expressed as a finite disjoint union of rectangles whose sides are intervals, we have proved (c). 2

2.2.2.◦ Lebesgue integrable functions on Rn. Let µ be the Lebesgue measure in Rn. The set M(Rn, µ) of all real µ-measurable functions on Rn is a real vector space (addition and scalar multiplication are pointwise). By L1(Rn, µ) we denote its subspace of all Lebesgue integrable functions (with finite integral). Now write f ≈ g for f and g in M(Rn, µ), whenever f and g differ only on a µ-null set (a set of measure zero). It is easily seen that ≈ is an equivalence relation. Let L0 = L0(Rn, µ) be the set of equivalence classes of functions in M(Rn, µ). We denote the equivalence classes of f, g, . . . by [f], [g],.... The set L0 becomes a real vector space by defining [f] + [g] = [f + g] and α[f] = [αf] for a real α. Observe that these definitions do not depend on the choice of f and g in their equivalence classes. The same is true for the partial order in L0, if we define [f] ≤ [g] to mean f(x) ≤ g(x) for all x ∈ Rn except a null set. In practice, the elements of L0 = L0(Rn, µ) are usually denoted by f, g, . . . and treated as if they were functions instead of equivalence classes of functions.

Theorem 2.2.2 If f ∈ L1(µ) and ε > 0, there is a simple function φ = PN R j=1 αjχRj , where each Rj is a product of intervals such that |f − φ| dµ < ε, and there is a continuous function g vanishing outside of a bounded set such that R |f − g| dµ < ε.

Proof: By the definition of Lebesgue integrable functions, we can approximate f by simple functions in L1-norm. Then use Theorem 2.2.1 to approximate a simple function by a function φ of the desired form. Finally, use the Urysohn Lemma to approximate such φ by a continuous function. 2 2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 47

Theorem 2.2.3 The Lebesgue measure on Rn is translation-invariant. Namely, n n n let a ∈ R . Define the shift τa : R → R by τa(x) = x + a. n n (a) If E ∈ L then τa(E) ∈ L and µ(τa(E)) = µ(E);

n (b) If f : R → R is Lebesgue measurable then so is f ◦ τa. Moreover, if either f ≥ 0 or f ∈ L1(µ) then Z Z (f ◦ τa) dµ = f dµ.

Proof: Since τa and its inverse τ−a are continuous, they preserve the class of Borel sets. The formula µ(τa(E)) = µ(E) follows easily from the trivial one- dimensional variant of this result if E is a block. For a general Borel set E, the formula µ(τa(E)) = µ(E) follows from the previous step, since µ is determined by its action on blocks. Assertion (a) now follows immediately.

If f is Lebesgue measurable and B is a Borel set in R, we have f −1(B) = E ∪N, −1 −1 where E is Borel and µ(N) = 0. But τa (E) is Borel and µ(τa (N)) = 0, so

−1 n (f ◦ τa) (B) ∈ Σ and f is Lebesgue measurable. The equality Z Z (f ◦ τa) dµ = f dµ

reduces to the equality µ(τ−a(E)) = µ(E) when f = χE. It is true for simple functions by linearity, and hence for nonnegative measurable functions by the definition of integral. Taking positive and negative parts of real and imaginary parts, we obtain the result for f ∈ L1(µ). 2

2.2.3.◦ The Lusin theorem. This theorem says that any Lebesgue mea- surable function on Rn can be approximated by a (partially defined) continuous function. More precisely.

Theorem 2.2.4 (Lusin’s theorem) If f is a Lebesgue measurable function on Rn and ε > 0 then there exist a measurable set A ⊆ Rn such that µ(Rn \ A) ≤ ε and the restriction of f onto A is continuous. 48 CHAPTER 2.

n Proof: First, assume that f vanishes outside of some Ik = [−k, k] . There exists a sequence (ψm)m of simple functions vanishing outside Ik such that ψm → n f pointwise. By Egoroff’s theorem, there exist G ⊆ Ik such that µ(G) > (2k) − ε/2 and ψk → f uniformly on G. By Theorem 2.2.1, we can find for each m a −m−1 compact Cm ⊆ G such that µ(Cm \ G) < 2 ε and ψk is continuous on Ck. ∞ Take C = ∩m=1Cm, then

µ(Ik \ C) ≤ µ(Ik \ G) + µ(G \ C) < ε.

Each ψm is continuous on C, and the sequence (ψm)m converges uniformly on C. Then ψm|C → ψ, which is continuous on C, and since ψm → f pointwise then f|C is continuous. In the general case when f is a Lebesgue measurable function on Rn we denote n n the set (−k, k) by Jk. As it was shown, for any k, there exists Rk ∈ Σ such −k that f|Rk is continuous, Rk ⊆ Ik, and µ(Ik \ Rk) < 2 ε. Denote

Ak := (Jk \ Ik−1) ∩ Rk,

∞ where I0 = ∅. Then the restriction of f onto A := ∪k=1Ak is continuous and µ(Rn \ A) < ε. 2

Corollary 2.2.1 If f ∈ L1(Rn) and ε > 0 then there exists a continuous func- tion g : Rn → R such that Z |f − g| dµ ≤ ε. 2 n R

2.2.4.◦ Cantor sets. Each x ∈ [0, 1] has a base-3 expansion

∞ X −j x = xj3 , where xj ∈ {0, 1, 2}. j=1

This expansion is unique unless x is of the form m · 3−k for some integers m, k; in which case x has two expansions: one with xj = 0 for j > k and one with xj = 2 for j > k. Assuming m is not divisible by 3, one of these expansions will have xk = 1 and the other will have xk = 0 or 2. If we agree always to use the latter expansion, we see that

x1 = 1 ⇔ x ∈ (1/3, 2/3), 2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 49

x1 6= 1 and x2 = 1 ⇔ x ∈ (1/9, 2/9) ∪ (7/9, 8/9), and so forth. It will also be useful to observe that if

∞ ∞ X −j X −j x = xj3 & y = yj3 , j=1 j=1 then x < y iff there exists n such that xn < yn and xj = yj for 1 ≤ j < n.

Definition 2.2.1 The standard Cantor set C is the set of all x ∈ [0, 1] having P∞ −j a base-3 expansion x = j=1 xj3 with xj 6= 1 for all j.

T∞ Thus, C = n=1 ∆n, where

∆1 := [0, 1] \ (1/3, 2/3) , ∆2 := ∆1 \ ((1/9, 2/9) ∪ (7/9, 8/9)), etc.

Obviously, C is compact, nowhere dense, and totally disconnected (i.e. the only connected subsets of C are single points). Moreover, C has no isolated points. These simple properties of C we leave to the reader as an exercise.

2.2.5.◦ Basic properties of the standard Cantor set. Other basic properties of C are summarized as follows:

Proposition 2.2.1 Let C be the standard Cantor set, then C is a Borel set such that µ(C) = 0 & card(C) = c.

1 Proof: C is obtained from [0, 1] by removing one interval of length 3 , two 1 intervals of length 9 , and so forth. Thus,

∞ X 2j 1 1 µ(C) = 1 − = 1 − · = 0. 3j+1 3 1 − (2/3) j=0

P∞ −j Suppose x ∈ C, so that x = j=1 xj3 , where xj = 0 or 2 for all j. Let

∞ X −j f(x) = yj2 , (2.2.1) j=1 50 CHAPTER 2. where yj = xj/2. The series defining f(x) is the base-2 expansion of a number in [0, 1], and any number in [0, 1] can be obtained in this way. Hence f maps C onto [0, 1] and card(C) is continuum. 2

2.2.6.◦ Generalized Cantor sets. The construction of the standard Cantor set which is starting with [0, 1] and successively removing open middle thirds of intervals has an obvious generalization. If I is a bounded interval and α ∈ (0, 1), let us call the open interval with the same midpoint as I, and length equal to α ∞ times the length of I the “open middle αth” of I. If (αj)j=1 is any sequence of ∞ numbers in (0, 1) then we define a decreasing sequence (∆j)j=1 of closed sets as follows: ∆0 = [0, 1], and ∆j is obtained by removing the open middle αjth from each of the intervals that make up Kj−1. The resulting limiting set

∞ \ K = ∆j j=1 is called a generalized Cantor set. Generalized Cantor sets possess all prop- erties of the standard Cantor set C, but they do not have always the Lebesgue measure zero. As for their Lebesgue measure, clearly

µ(∆j) = (1 − αj)µ(∆j−1), so ∞ Y µ(K) = (1 − αj). j=1

If the αj are all equal to a fixed α ∈ (0, 1), we have µ(K) = 0. However, if αj → 0 rapidly enough, µ(K) will be strictly positive and, for any β ∈ [0, 1), one can choose αj so that µ(K) = β.

2.2.7.◦ Cantor functions. The map f : C → [0, 1] given by (3.1.1) has the following property.

If x, y ∈ C, x < y, then f(x) < f(y) unless x and y are endpoints of one of the intervals removed from [0, 1] in obtaining C (in this case, f(x) = m·2−k for some nonnegative integers m, k; and f(x) and f(y) are the two base-2 expansions of this number). 2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 51

Therefore we can extend f to a map fˆ : [0, 1] → [0, 1] by setting it to be constant on each interval removed from C. The map fˆ, which is obviously increasing and continuous, is called the standard Cantor function. This notion can be naturally extended to a generalized Cantor function by using of the notion of the generalized Cantor set. 2.2.8.◦ Applications of Cantor sets and functions in constructing of counterexamples. Generalized Cantor sets are examples of nowhere dense subsets of [0, 1] with Lebesgue measure 1 − ε, where ε > 0 is arbitrarily. Let Ck 1 S∞ be a generalized Cantor set such that µ(Ck) > 1 − k , then A = k=1 Ak is a set of the first category (i.e. the countable union of nowhere dense sets) and µ(A) = 1. By the Baire theorem, [0, 1] is a set of the second category (i.e. the set which is not of the first category) as a complete metric space. Therefore, [0, 1] \ A is an example of a dense subset of [0, 1] of the second category of Lebesgue measure zero. By addition and multiplication, we define a Cantor like set K ⊆ [a, b] in an interval [a, b] as: K := a + (b − a) · G, where G is a some generalized Cantor set. The similar idea is used in the definition of a Cantor like function on an interval [a, b]. Now it is obvious how (with the help of Cantor like sets) to construct an example of a dense subset D ⊆ R of the second category of Lebesgue measure zero. Cantor like functions give us the following quite simple construction of a con- tinuous function f : R → [0, 1] which is nowhere differentiable. Let {an}n be a countable dense subset of R, and suppose that an 6= an+1 for all n. For each n, take a Cantor like function fn on the interval [an, an+1], then the function

∞ X −n f := 2 fn j=1 possesses desired properties.

2.2.8.◦ Exercises to Section 2.2.

Exercise 2.2.1 ∗ Show that the Lebesgue measure in Rn is separable.

Exercise 2.2.2 ∗ Prove Corollary 2.2.1. 52 CHAPTER 2.

Exercise 2.2.3 ∗ Show that the Lebesgue measure in Rn is rotation invariant.

Exercise 2.2.4 Show that the family of all maps from Rn to Rn preserving the Lebesgue measure in Rn is a group.

Exercise 2.2.5 ∗∗ Give an example of a transformation S : Rn → Rn, such that S is not equal a.e. to any affine transformation of R and S preserves the Lebesgue measure in Rn. Can such a transformation be continuous?

Exercise 2.2.6 ∗ Let Γ be a unit circle {λ ∈ C : |λ| = 1} endowed with the Lebesgue measure. Then any λ ∈ Γ is of the form ei 2π α, where α ∈ [0, 1). (a) Show that ≈ defined below is an equivalence relation on Γ:

i 2π α1 i 2π α1 e ≈ e if α1 − α2 ∈ Q . S (b) Let Γ = κ Γκ be a representation of Γ as the union of equivalence classes with respect to ≈ . By using of the axiom of choice, define a set Ω ⊆ Γ such that |Ω ∩ Γκ| = 1 for any κ. Let for any α ∈ R i 2π α Ωα := {e ω : ω ∈ Ω}.

Show that, for any two rationals r1 6= r2 from [0, 1), Ωr1 ∩ Ωr2 = ∅, and [ Γ = Ωr. r∈Q ∩[0,1) (c) By using (b), show that Ω is not Lebesgue measurable subset of Γ. (d) Construct non-measurable subsets of [0, 1], R, and Rn.

Exercise 2.2.7 Show that any generalized Cantor set is compact, nowhere dense, totally dis- connected, and has no isolated points.

Exercise 2.2.8 ∗ Show that for any β ∈ [0, 1) there is a generalized Cantor set K ⊆ [0, 1] such that µ(K) = β.

Exercise 2.2.9 ∗ Give an example of a subset in [0, 1] which is of the second category and has Lebesgue measure one.

Exercise 2.2.10 ∗ Give an algorithm for constructing of a generalized Cantor function and show that it is continuous.

Exercise 2.2.11 ∗∗ Construct a function f : [0, 1] → [0, 1] such that f 0 = 0 a.e. and f is constant on no nonempty subinterval of [0, 1]. 2.2. LEBESGUE MEASURE AND INTEGRAL IN RN 53

Exercise 2.2.12 ∗ Give an example of a set D ⊆ Rn of Lebesgue measure zero such that D ∩ U is of the second category for any nonempty open U ⊆ Rn.

Exercise 2.2.13 ∗ Construct an example of a dense subset D ⊆ Rn of the second category and of Lebesgue measure zero.

Exercise 2.2.14 ∗∗∗ Construct a Borel subset A ⊆ R such that for each interval [a, b], where a < b: µ(A ∩ [a, b]) > 0 & µ((R \ A) ∩ [a, b]) > 0.

∗∗∗ ∞ Exercise 2.2.15 Construct a sequence (Ak)k=1 of pairwise disjoint Borel subsets Ak ⊆ R such that for each interval [a, b], a < b, and k ∈ N µ(Ak ∩ [a, b]) > 0.

∗∗∗ Exercise 2.2.16 Construct a sequence (Ak)k∈N of pairwise disjoint Borel subsets Ak ⊆ Rn such that, for each rectangle

H = [a1, b1] × · · · × [an, bn](∀j ∈ {1, . . . , n} aj < bj) and for each k ∈ N, we have that µ(Ak ∩ H) > 0.

Exercise 2.2.17 ∗∗∗ Construct a nonnegative Lebesgue measurable function f on R, which R b is finite everywhere and a f dµ = ∞ for all a < b.

∗∗∗ ∞ Exercise 2.2.18 Construct a sequence (fn)n=1 of pairwise disjoint nonnegative Lebesgue measurable functions on R such that each fn is finite everywhere and Z b fn dµ = ∞ a for all a < b and n ∈ N.

∗∗ Exercise 2.2.19 Let f ∈ L1(R, µ) such that for every open U ⊆ R Z Z f dµ = f dµ. U U¯ Show that f = 0 a.e. 54 CHAPTER 2. Chapter 3

3.1 The Radon – Nikodym Theorem

In this section, we prove the Radon – Nikodym theorem and present several its applications, in particular, the Hahn decomposition theorem.

3.1.1.◦ The Radon – Nikodym theorem. Let (X, B, µ) be a measure space and let f : X → R be a nonnegative integrable function. Then µf defined by Z µf (A) = f(x) dµ (∀A ∈ A) A is a finite measure accordingly to 1.3.2 L4. There is a converse of the last property. Namely, the following theorem is of great importance. We give its direct proof accordingly to A. Schep [10].

Theorem 3.1.1 (The Radon – Nikodym theorem) Let ν and µ be finite measures on (X, B). Then there exist D ∈ B with µ(D) = 0 and a nonnegative µ-integrable function f0 such that Z ν(E) = ν(E ∩ D) + f0 dµ (∀E ∈ B) . E First, we need the following lemma.

Lemma 3.1.1 Let ν and µ be finite measures on (X, B) satisfying 0 ≤ ν ≤ µ on B. Then there exists a measurable function f0 , 0 ≤ f0 ≤ 1, such that Z ν(E) = f0 dµ (∀E ∈ B) . E

55 56 CHAPTER 3.

Proof: Let Z H = {f : f is measurable , ν(E) ≥ f dµ for all E ∈ B}. E

Note that H 6= ∅, since 0 belongs to H. Moreover, when f1, f2 ∈ H, then max{f1, f2} ∈ H. Indeed, if A = {x : f1(x) ≥ f2(x)} then Z Z Z c max{f1, f2} dµ = f1 dµ + f2 dµ ≤ ν(E ∩ A) + ν(E ∩ A ) = ν(E). E E∩A E∩Ac Let n Z o M := sup f dµ : f ∈ H . X ∞ Then 0 ≤ M < ∞, so there exist sequence (fn)n=1 ⊆ H with fn ↑ such that R −1 fn dµ > M − n . Let f0 = lim fn then f0 is a measurable function taking X R values in [0, 1]. By the monotone convergence theorem, f0 ∈ H and f0 dµ ≥ R X M. Hence X f0 dµ = M. R To complete the proof, we show that ν(E) = E f0 dµ for all E ∈ B. If not, there exists E ∈ B satisfying Z ν(E) > f0 dµ . E

Denote E1 = {x ∈ E : f0(x) = 1} and E0 = E \ E1. It follows from Z Z Z ν(E0) + ν(E1) > f0 dµ = f0 dµ + µ(E1) ≥ f0 dµ + ν(E1) E E0 E0 R that ν(E0) > f0 dµ. Thus we may assume that f0(x) < 1 for all x ∈ E E0 −1 (otherwise, we replace E with E0). Let Fn := {x ∈ E : f0(x) < 1 − n }. Then Fn ↑ E, and ν(Fn) → ν(E). Therefore, there exists n0 such that Z ν(F ) > (f + n−1χ ) dµ . n0 0 0 Fn0 Fn0 −1 We assert that f0 + n0 χF ∈ H for some measurable F ⊆ Fn0 , µ(F ) > 0.

If not, then every measurable subset F of Fn0 with positive µ-measure contains R −1 R −1 a measurable subset G with G(f0 + n0 χF ) dµ = G(f0 + n0 χG) dµ > ν(G). Denote this property by P, namely, Z −1 A ∈ P ⇔ A ∈ B ,A ⊆ Fn0 & (f0 + n0 χA) dµ > ν(A) . A 3.1. THE RADON – NIKODYM THEOREM 57

The property P satisfies conditions of Theorem 1.1.2. Namely, every measurable subset of Fn0 of non-zero measure contains a subset of non-zero measure with the property P, and P is closed under at most countable unions of pairwise disjoint families. Therefore by Theorem 1.1.2, Fn0 ∈ P, and Z ν(F ) > (f + n−1χ ) dµ > ν(F ), n0 0 0 Fn0 n0 Fn0

−1 which is a contradiction. Thus f0 + n0 χF ∈ H for some measurable subset F of Fn0 with a positive µ-measure. However, Z −1 −1 (f0 + n0 χF ) dµ = M + n0 µ(F ) > M .

R This contradiction yields ν(E) = E f0 dµ for all E ∈ B. 2 Proof of Theorem 3.1.1: Let λ = µ+ν. Then 0 ≤ ν ≤ λ so, by Lemma 3.1.1, R there exists a Lebesgue integrable g : X → [0, 1], such that ν(E) = E g dλ for all E in B. Then µ(E) = R (1 − g) dλ for all E ∈ B. Let D = {x : g(x) = 1}, R E then µ(D) = D 0 dλ = 0. Now, the equality Z Z ν(E) = g dν + g dµ E E implies Z Z (1 − g) dν = g dµ E E for all E ∈ B. Hence R φ(1 − g) dν = R φg dµ for all nonnegative simple φ R X R X and, thus, X f(1 − g) dν = X fg dµ for all nonnegative measurable f. Taking n f = (1 + g + ··· + g )χE, we have Z Z (1 − gn+1) dν = (1 + g + ··· + gn)g dµ E E for all E in B and all n ≥ 1. Since 0 ≤ g(x) < 1 on Dc, it follows from the monotone convergence theorem that Z Z ν(E ∩ Dc) = lim (1 − gn+1) dν = lim (1 + g + ··· + gn)g dµ n→∞ E∩Dc n→∞ E∩Dc Z Z −1 = g(1 − g) dµ = f0 dµ, E∩Dc E 58 CHAPTER 3.

−1 where f0 = g(1 − g) χDc . The proof is completed. 2 In some textbooks the Radon – Nikodym theorem is given in a slightly different form. We need the following definition.

Definition 3.1.1 Let µ and ν be signed measures on a σ-algebra A. Then ν is called µ-continuous if ν(A) = 0 for every A ∈ A such that µ(A) = 0. In this case, we write ν << µ.

Theorem 3.1.2 Let ν and µ be σ-finite signed measures on (X, B), let µ be positive and ν << µ. Then there exists a locally µ-integrable function f (i.e. R A f dµ < ∞ for every A ∈ B, µ(A) < ∞) such that Z ν(E) = f dµ (∀E ∈ B) . 2 E

The function f above is called the Radon – Nikodym derivative of ν with ∂ν respect to µ and denoted by ∂µ .

3.1.2.◦ Applications of the Radon – Nikodym theorem. We give two important applications of the Radon – Nikodym theorem. First one is the proof of the Hahn decomposition theorem in the case of a finite signed measure. We need the following definition.

Definition 3.1.2 Let µ be a signed measure on a σ-algebra A. Then P ∈ A is called a positive set for µ if, for any B ∈ A such that B ⊆ P , µ(B) ≥ 0.A set N ∈ A is called a negative set for µ if, for any B ∈ A such that B ⊆ N, µ(B) ≤ 0.

Theorem 3.1.3 (The Hahn decomposition theorem) If ν is a signed mea- sure on (X, A), there exist a positive set P and a negative set N for ν such that P ∪ N = X and P ∩ N = ∅.

Here we give a simple proof of this theorem based on Theorem 3.1.2 in the case of finite measure ν. 3.1. THE RADON – NIKODYM THEOREM 59

Proof: Let ν = ν+ − ν− be the Jordan decomposition of ν. Then obviously ν+ and ν− are |ν|-continuous measures, where |ν| = ν+ + ν−. Applying Theorem 3.1.2 to them gives us two functions f1, f2 ∈ L1(|ν|) such that Z Z ν+(E) = f1 d|ν| , ν−(E) = f2 d|ν| (∀E ∈ A). E E

Then P := {x ∈ X : f1(x) > 0} and N := X \ P are required sets. 2

The next one is an application in the operator theory. It is related to the definition of one of the most important classes of Markov operators. A transformation τ :Ω → Ω is called measurable if τ −1(A) ∈ Σ for all A ∈ Σ. A measurable transformation τ :Ω → Ω is called nonsingular if µ(τ −1(A)) = 0 for all A ∈ Σ such that µ(A) = 0.

Let τ be a nonsingular transformation and let f ∈ L1(µ). Then the function ν :Σ → , R Z ν(E) := f dµ, τ −1(E) is a finite signed µ-continuous measure on Σ. Then, by the Radon – Nikodym theorem, there exists Pf ∈ L1(µ) such that Z Z P f dµ = f dµ (∀A ∈ Σ). A τ −1(A)

The map f → Pf is a linear operator from L1(µ) := L1(µ)/{f : f = 0 a.e. } to itself. This operator is called the Frobenius – Perron operator corresponding to τ. The elementary properties of Frobenius – Perron operators are collected in Exercises 3.1.10 and 3.1.13.

3.1.3.◦ Exercises to Section 3.1.

Exercise 3.1.1 Let ν be a signed measure. Show that Z |ν|(A) = sup{| f d|ν| | : |f| ≤ 1} A for any measurable A. 60 CHAPTER 3.

Exercise 3.1.2 ∗∗∗ Let ν be a σ-finite signed measure on (X, Σ), which is µ-continuous for another σ-finite measure µ on (X, Σ). Show that Z ∂ν

|ν|(A) := dµ (∀A ∈ Σ). A ∂µ

Exercise 3.1.3 Show that a signed measure ν is µ-continuous iff, for any decreasing sequence ∞ of measurable sets (An)n=1, µ(An) → 0 implies ν(An) → 0.

Exercise 3.1.4 ∗ Let ν be a finite signed measure and let µ be a positive measure on (X, Σ). Show that ν is µ-continuous iff, for any ε > 0, there exists δ > 0 such that |ν(A)| ≤ ε whenever µ(A) ≤ δ. Give an example which shows that this assertion may fail if ν is not finite.

Exercise 3.1.5 Let ν and µ be σ-finite positive measures and let ν << µ. Show that if µ is separable then ν is also separable.

Exercise 3.1.6 Let (γλ)λ∈Λ be a countable family of signed measures of bounded variation on (X, Σ). Show that there exists a measure γ on (X, Σ) such that every γλ is γ-continuous, and γ ∧ |γλ| > 0 for any λ ∈ Λ.

Exercise 3.1.7 Obtain the Hahn decomposition theorem for σ-finite signed measure as a corollary of this theorem for finite signed measure.

Exercise 3.1.8 ∗ Prove the Hahn decomposition theorem for arbitrary signed measure.

Exercise 3.1.9 Obtain the Jordan decomposition theorem as a corollary of the Hahn decom- position theorem.

∗ Exercise 3.1.10 Let S be a nonsingular transformation of (X, Σ, µ) and let PS be the corresponding Frobenius – Perron operator PS : L1(µ) → L1(µ). Show that:

(a) PS is well defined; (b) PS is a linear operator; (c) PS preserves the L1-norm of any nonnegative f ∈ L1(µ).

Exercise 3.1.11 Let S be a nonsingular transformation of (X, Σ, µ). Show that Sn is non- n singular transformation for any nonnegative integer n and that PSn = PS for n ≥ 0.

Exercise 3.1.12 ∗∗ Let S be a transformation of [0, 1] with the Lebesgue measure. Find the Frobenius – Perron operator if: (a) S(x) = 4x(1 − x); (b) S(x) = 4x2(1 − x2); (c) S(x) = sin πx; (d) S(x) = tan(x + 1). 3.2. LP -SPACES 61

∗ Exercise 3.1.13 Let S be a nonsingular transformation of (X, Σ, µ) and let PS be the corre- sponding Frobenius – Perron operator PS : L1(µ) → L1(µ). Prove that, for every nonnegative f, g ∈ L1(µ), the condition supp(f) ⊆ supp(g) implies supp(PSf) ⊆ supp(PSg) a.e., where supp(ψ) := {x : ψ(x) 6= 0} for ψ ∈ L1(µ), and supp(f) is defined for f ∈ L1 as follows:

supp(f) := A such that ψ(x) 6= 0 a.e. on A , ψ(x) = 0 a.e. on X \ A for any ψ ∈ [f].

Exercise 3.1.14 ∗∗ In notations of Exercise 2.1.20, show that the unique extension ζ of the premeasure ζ (which exists due to Caratheodory theorem) to the product σ-algebra Σ([0, 1]) × Σ([0, 1])) is not µ×µ-continuous. Although the premeasure ζ on the subalgebra A ⊂ Σ([0, 1])× Σ([0, 1])) generated by all rectangles with measurable sides is µ×µ-continuous in the following sense: µ × µ(W ) = 0 ⇒ ζ(W ) = 0 (∀ W ∈ A) .

3.2 Lp-Spaces

Here we present basic results about Lp-spaces. We assume in this section that all measurable functions are C-valued and all linear spaces are complex.

◦ 3.2.1. Lp-spaces for 0 ≤ p ≤ ∞. Let (X, A, µ) be a measure space and let p be a fixed real number 0 ≤ p < ∞. We denote by Lp = Lp(X, A, µ) the set of all measurable functions f : X → C such that Z |f|p dµ < ∞.

For any f ∈ Lp, we define a function kfkp : Lp → R+ as Z 1  p  p kfkp := |f| dµ .

This function introduces the natural topology on Lp(X, A, µ) by the base

{U(f, ε): f ∈ Lp, ε > 0}, where U(f, ε) := {g ∈ Lp : kf − gkp < ε}. 62 CHAPTER 3.

Here we don’t give the proof of this result. Below, in the next subsection, we state the important case by showing that if 1 ≤ p < ∞ then Lp is a linear space and the function k · kp is a seminorm.

The most popular examples of Lp-spaces are: Lp[0, 1] and Lp(R), where we consider the Lebesgue measure on [0, 1] and on R. Another type of examples is following. Let X be a nonempty set, let A = P(X) be the power set of X (i.e. the σ-algebra of all subsets of X), and let µ be the counting measure on X. Every function f : X → C is measurable. A commonly used notation of Lp(X, A, µ) in this case is `p(X). It is easy to verify that a P p function f : X → C is in `p(X) iff x∈X |f(x)| < ∞, in the sense that the finite subsumes have a finite upper bound.

The choices X = N and X = Z are of a special interest; these yield the space of absolutely p-summable sequences (resp., bilateral sequences). If X = N then we denote Lp(N) by `p. In the case p = ∞, we cannot directly generalize the definition of our function k · kp, and we treat this case as follow. Set L∞ = L∞(X, A, µ), the set of all measurable functions f : X → C such that there exists a constant M ∈ R,

|f|(x) ≤ M a.e.

It is obvious that L∞(X, A, µ) is a topological linear space for the topology given by the seminorm k · k∞ defined as

kfk∞ = esssup(|f|) = inf{M ∈ R : |f|(x) ≤ M a.e.}.

∞ The main examples of L -spaces are L∞[0, 1], L∞(R), and `∞, which are de- fined similarly to Lp[0, 1], Lp(R), and `p.

3.2.2.◦ Inequalities. We begin with elementary analytic lemma.

Lemma 3.2.1 If a ≥ 0, b ≥ 0, and 0 < λ < 1 then

aλb1−λ ≤ λa + (1 − λ)b, with equality iff a = b. 3.2. LP -SPACES 63

Proof: The result is obvious if b = 0; otherwise, dividing both sides by b and setting t = a/b, we are reduced to showing that tλ ≤ λt + (1 − λ) with equality iff t = 1. The derivative f 0(t) = λtλ−1 −λ of the function f(t) = tλ −λt satisfies: f 0(t) > 0 if t < 1; f 0(1) = 0; f 0(t) < 0 if t > 1. Then f is strictly increasing for t < 1 and strictly decreasing for t > 1, so its maximum value, namely 1 − λ, occurs at t = 1. 2

p −1 −1 If p > 1, we define q = p−1 ; thus q > 0 and satisfies p +q = 1, i.e. p+q = pq. The following inequality which called the Elementary Young Inequality αp βq αβ ≤ + p q

1 is true for all α, β ≥ 0 and p > 1. For the proof, it is enough to substitute λ = p , 1 1 α = a p , and β = b q in Lemma 3.2.1.

Theorem 3.2.1 (H¨older’sinequality) Let p > 1, f ∈ Lp, and g ∈ Lq. Then f · g ∈ L1 and kf · gk1 ≤ kfkpkgkq. (3.2.1)

Proof: The result is trivial if kfkp = 0 or kgkq = 0, since f = 0 or g = 0 a.e. Moreover, we observe that if (3.2.1) holds for a some f and g then it also holds for all scalar multiples of f and g; for it, f and g are replaced by a · f and b · g, both sides of (3.2.1) change by a factor of |ab|. It suffices therefore to prove p q (3.2.1) when kfkp = kgkq = 1 with equality iff |f| = |g| a.e. To this end, we apply the Young inequality with α = |f(x)| and β = |g(x)| to obtain

|f(x)g(x)| ≤ p−1|f(x)|p + q−1|g(x)|q. (3.2.2)

Integration of the both sides yields Z Z −1 p −1 q −1 −1 kfgk1 ≤ p |f| dµ + q |g| dµ = p + q = 1 = kfkpkgkq.

Equality holds here iff it holds a.e. in (3.2.1) and, by Lemma 3.2.1, this happens exactly iff |f|p = |g|q a.e. 2

One of the most important application of the H¨olderinequality is the following inequality. 64 CHAPTER 3.

Theorem 3.2.2 (Minkowski’s inequality) Let p ≥ 1, f, g ∈ Lp. Then f + g ∈ Lp and kf + gkp ≤ kfkp + kgkp.

Proof: The result is obvious if p = 1 or f + g = 0 a.e.. Otherwise, we have

|f + g|p ≤ (|f| + |g|)|f + g|p−1 = |f||f + g|p−1 + |g||f + g|p−1 .

Taking the integral,

p p−1 p−1 k(f + g) k1 ≤ kf(f + g) k1 + kg(f + g) k1. (3.2.3)

Integration of the inequality

|f + g)|p ≤ 2p max{|f|, |g|}p ≤ 2p max{|f|p, |g|p} gives us f + g ∈ Lp. Since (p − 1)q = p when q is the conjugate to p and applying H¨older’sinequality to the functions

p−1 p φ = f ∈ Lp, ψ = (f + g) = (f + g) q ∈ Lq, we have

p−1 p−1 kf(f + g) k1 ≤ kφ · ψk1 ≤ kφkpkψkq = kfkpk(f + g) kq. (3.2.4)

Similarly p−1 p−1 kg(f + g) k1 ≤ kgkpk(f + g) kq. (3.2.5) Combining (3.2.3), (3.2.4), and (3.2.5) gives us

Z 1 R p  p  p |f + g| dµ kf + gkp = |f + g| dµ = 1 = (R |f + g|p dµ) q

R |f + g|p dµ k(f + g)pk 1 2 1 = p−1 ≤ kgkp + kfkp . (R |f + g|(p−1)q dµ) q k(f + g) kq

Corollary 3.2.1 Lp is a vector space with respect to pointwise linear operations. The function f → kfkp is a seminorm on Lp. 2 3.2. LP -SPACES 65

◦ 3.2.3. Completeness of Lp, where 1 ≤ p < ∞ . A basic property of Lp is given in the following theorem.

∞ Theorem 3.2.3 Let (fn)n=1 be a sequence in Lp = Lp(X, Σ, µ) satisfying kfm −fnkp → 0 as n, m → ∞. Then there exists f ∈ Lp with lim kfn −fkp = 0. n→∞

Proof: We prove this theorem for 1 ≤ p < ∞. The case p = ∞ is more elementary and it is left for the reader as an exercise. We may suppose, by passing to a subsequence, that

−n kfn+1 − fnkp ≤ 2 (∀n ∈ N).

P∞ Thus we have n=1 kfn+1 − fnkp ≤ 1. With the convention f0 = 0, define

n X gn := |f1| + |f2 − f1| + ··· + |fn − fn−1| = |fk − fk−1|. k=1

By Theorem 3.2.2, gn ∈ Lp and

n n X X kgnkp ≤ kfk − fk−1kp = kf1kp + kfk − fk−1kp ≤ kf1kp + 1 . k=1 k=2

p Thus gn ∈ L1 for all n and Z p p p |gn| dµ = kgnkp ≤ (kf1kp + 1) . (3.2.6) X

∞ p ∞ Since (gn)n=1 is an increasing sequence and p ≥ 1, the sequence ((gn) )n=1 is also increasing. It follows from (3.2.6) and the monotone convergence theorem p 1/p that there exists h ∈ L1, h ≥ 0, such that (gn) ↑ h a.e. Denoting g := h , we have g ∈ Lp and gn ↑ g a.e.

Let E ∈ Σ be such that µ(E) = 0 and gn(x) ↑ g(x) for all x ∈ X \ E. Then

n X g(x) = lim gn(x) = lim |fk(x) − fk−1(x)| (∀x ∈ X \ E) . n→∞ n→∞ k=1 66 CHAPTER 3.

P∞ Thus the series k=1 |fk(x) − fk−1(x)| is convergent to g(x) for all x ∈ X \ E. P∞ Therefore the series k=1(fk(x)−fk−1(x)) is also convergent for x ∈ X \E, and satisfies: n X lim fn(x) = lim (fk(x) − fk−1(x)) ≤ g(x)(∀x ∈ X \ E) . n→∞ n→∞ k=1 Define

f(x) := lim fn(x) for x ∈ X \ E, and f(x) := 0 for x ∈ E. n→∞

p p Then f is measurable, fn → f a.e., and |f| ≤ g. Therefore |f| ≤ g = h ∈ L1, p and |f| ∈ L1, i.e., f ∈ Lp.

It remains to show that kfn − fkp → 0. Given an ε > 0, choose Nε such that kfm − fnkp ≤ ε for all m, n ≥ Nε. Fix m ≥ Nε, then Z p p |fm − fn| dµ ≤ ε for all n ≥ Nε . X

p Since |fm − fn| ∈ L1 for all n ≥ Nε, it follows from the Fatou lemma that Z Z p p p (lim sup |fm − fn| ) dµ ≤ lim sup |fm − fn| dµ ≤ ε . (3.2.7) X n→∞ n→∞ X

p p p In particular, lim sup |fm − fn| ∈ L1. But |fm − fn| → |fm − f| a.e. as n→∞ n → ∞, thus p p lim sup |fm − fn| = |fm − f| a.e. n→∞ R p p p So X |fm − f| dµ ≤ ε by (3.2.7), i.e., kfm − fkp ≤ ε for all m ≥ Nε. Since ε > 0 has been chosen arbitrary, the proof is completed. 2

◦ 3.2.4. Lp-spaces for 1 ≤ p < ∞. Let now 1 ≤ p < ∞ and (X, A, µ) be a measure space. Take the following relation ”≈” on Lp(X, A, µ): f ≈ g ⇔ f(x) = g(x) a.e. (3.2.8)

Obviously ”≈” is an equivalence relation and {f : f ≈ 0} = {f : kfkp = 0} is a linear subspace of Lp. Thus we may consider a quotient space

Lp(X, A, µ) := Lp(X, A, µ)/{f : kfkp = 0}. 3.2. LP -SPACES 67

We will denote elements of Lp(X, A, µ) in the same way as functions of Lp(X, A, µ), so we will identify f ∈ Lp(X, A, µ) with [f] ∈ Lp(X, A, µ). The seminorm k · kp on Lp becomes the norm on Lp. We use the same symbol k·kp for its denotation. Accordingly to Corollary 3.2.1, all Lp-spaces are normed linear spaces. As a direct consequence of Theorem 3.2.3, we have the following theorem.

Theorem 3.2.4 Every Lp-space is a Banach space for 1 ≤ p < ∞, i.e. every Cauchy sequence in Lp has a norm limit. 2

As an application of the last theorem, we have the following important property ∞ P of Lp-spaces. Let 1 ≤ p < ∞. If fn ∈ Lp and kfikp < ∞ then there exists i=1 f ∈ Lp such that n X lim g − fi = 0 . n→∞ i=1 p ∞ P In this case, we write g = fi. i=1

In Lp-spaces, the norm convergence and the a.e. convergence are related to each other as follows.

∞ Theorem 3.2.5 Let 1 ≤ p < ∞, {fn}n=1 ⊆ Lp, and let g ∈ Lp satisfy

fn(x) → 0 a.e. & |fn(x)| ≤ g(x) a.e. (∀n).

Then kfnkp → 0. 2

The proof of this theorem is a direct application of the dominated convergence theorem.

∞ Theorem 3.2.6 Let 1 ≤ p < ∞, {fn}n=1 ⊆ Lp, and let M ∈ R satisfy

fn(x) ↑ a.e. & kfnkp ≤ M (∀n ∈ N).

Then there exists f ∈ Lp such that kfn − fk → 0. 2 68 CHAPTER 3.

The proof of this theorem is a direct application of the monotone convergence theorem.

! 3.2.5. Separable Lp-spaces. A Banach space X is called separable if there exists a countable family {xn}n ⊆ X such that its norm closure is equal to X. It is easy to see that L∞(µ) is separable iff its dimension is finite.

Theorem 3.2.7 If the measure µ is σ-finite and separable and 1 ≤ p < ∞ then Lp(µ) = Lp(X, Σ, µ) is separable.

Proof: We will prove that the countable collection Y of all simple functions, which are finite sums with rational complex coefficients of characteristic func- tions of measurable sets from a fixed countable family R (where R is a collection of sets as in Definition 1.1.8), is dense in Lp(µ). Let f ∈ Lp(µ) and ε > 0 be given. Since there exists an increasing sequence Xn ↑ X (each Xn of finite mea- 0 sure), we have |f − f · χXn | ↓ 0 pointwise. Hence, there exists a set X of finite 0 measure (we may assume that X = X1) such that if we define f1 := f · χX1 then kf − f1kp ≤ ε/3.

The function f1 can be written as f = (g1 − g2) + i(h1 − h2), where g1, g2, h1, h2 are non-negative members of Lp vanishing outside of X1. For g1, there exists (n) ∞ a sequence (g )n=1 of rational simple functions, non-negative and vanishing (n) (n) p outside of X1, such that g ↑ g1 pointwise. Then |g1 − g | ↓ 0, so

(n) kg1 − g kp ≤ ε/12 for n sufficiently large. Similarly for g2, h1, and h2. By linear combination, we obtain a rational simple function f2 vanishing outside X1 and such that kf1 − f2kp ≤ ε/3. Pk The function f2 can be written as f2 = j=1 αjχAj , where all αj are rational k complex, all Aj are mutually disjoint, and ∪j=1Aj = X1. For each Aj, there exists a set Bj ∈ R such that

p p µ(Aj4Bj) ≤ ε /(3k|αj|) , so 1/p kχAj − χBj kp = µ(Aj4Bj) ≤ ε/(3k|αj|) . 3.2. LP -SPACES 69

Pk Hence, writing f3 = j=1 αjχBj , we have

k X kf2 − f3kp ≤ |αj| · kχAj − χBj kp ≤ ε/3 . j=1

It follows that kf −f3kp ≤ ε. Since f3 ∈ Y , this shows that Lp(µ) is separable. 2

◦ 3.2.6. Hilbert spaces and L∞-spaces. Despite of the historically different definitions of Hilbert spaces and L2-spaces, they are the same object and it is a quite reasonable idea to define Hilbert space as an L2-space. The main tool in the investigation of Hilbert spaces is a notion of the scalar product: Z (f, g) := f · g¯ dµ (f, g ∈ L2).

1 The scalar product generates the norm k · k of L2 as kfk = |(f, f)| 2 . The scalar product leads to the notion of orthogonality of functions. Two func- tions f, g ∈ L2 are called orthogonal if (f, g) = 0. The family {fα}α∈A is called an orthonormal basis of L2 if kfαk = 1 for any α and (fα, fβ) = 0 for any α 6= β. The following theorem is a generalization of the well known Pythagorian theorem.

Theorem 3.2.8 Any L2-space has at least one orthonormal basis. If {fα}α∈A is an orthonormal basis of L2 then for any f ∈ L2 X f = (f, fα)fα . 2 α∈A

In Hilbert spaces, the H¨olderinequality is known as the Cauchy – Schwarz inequality: 1 1 |(f, g)| ≤ (f, f) 2 · (g, g) 2 (f, g ∈ L2).

L∞-space is the quotient space of L∞ with respect to the equivalence relation ”≈” defined in (3.2.8). The correspondent norm we will denote by k · k∞. The poof of the next theorem is direct and routine and we refer for them to any textbook on functional analysis. 70 CHAPTER 3.

Theorem 3.2.9 Every L∞-space is a Banach space. 2

It is an important remark that Theorem 3.2.5 and Theorem 3.2.6 fail for infinite dimensional L∞-spaces.

All Lp-spaces for 1 ≤ p ≤ ∞ are vector lattices and, since

|f| ≤ |g| ⇒ kfk ≤ kgk (∀f, g ∈ Lp), they are Banach lattices.

3.2.7.! The duality and the weak convergence. Let X be a complex Banach space. A function ψ : X → C is called a linear functional if

ψ(αx + βy) = αψ(x) + βψ(y) ∀x, y ∈ X, α, β ∈ C. A linear functional ψ is called continuous if

kxn − xk → 0 ⇒ ψ(xn − x) → 0. The set of all continuous linear functionals on the Banach space X is called the dual space (or adjoint) to X and denoted by X∗.

Theorem 3.2.10 The dual space X∗ of a Banach space X is a Banach space with respect to the norm

kψk = sup{|ψ(x)| : x ∈ X, kxk ≤ 1}. 2

There is a standard way for constructing continuous linear functionals on Lp- spaces. Namely, for 1 ≤ p ≤ ∞ let q be such that

1/p + 1/q = 1 .

Let g ∈ Lq, denote Z gˆ(f) = hf, gi := f · g dµ ∀f ∈ Lp.

By the H¨olderinequality, if p < ∞ (and, by the direct arguments, if p = ∞) gˆ is a continuous linear functional on Lp. Moreover, it can be evaluated that kgˆk = kgkq. If p < ∞ then this is the most general form of a continuous linear functional on Lp-space. 3.2. LP -SPACES 71

Theorem 3.2.11 Let 1 ≤ p < ∞. Then, for every continuous linear functional ψ on Lp, there exists a unique g ∈ Lq such that ψ =g ˆ. Moreover, the dual space ∗ Lp is isometrically isomorphic to Lq. 2

∗ It is convenient to write Lp = Lq. The structure of the dual of L∞ is more ∗ complicated. We say only that L∞ is much more bigger then L1 if dim(L∞) = ∞.

For measurable functions and, in particular, for functions from Lp-spaces, we have studied several modes of convergence: the convergence in measure, con- vergence in norm, uniform convergence, etc. Now we give one more type of convergence in Lp-spaces for 1 ≤ p < ∞.

∞ Definition 3.2.1 A sequence (fn)n=1 ⊆ Lp is weakly convergent to f ∈ Lp if Z Z lim fn · g dµ = f · g dµ (∀g ∈ Lq), n→∞ where, as usual, p−1 + q−1 = 1.

From the H¨olderinequality, we have

|hfn − f, gˆi| ≤ kfn − fkp · kgkq and, thus, if kfn − fkp converges to zero, so does hfn − f, gˆi. Hence the conver- gence in norm implies the weak convergence. In general, the weak convergence is strictly weaker then the norm convergence. There is only one interesting exception, namely.

∞ Theorem 3.2.12 Let (ρn)n=1 be a sequence of elements of `1 which converges weakly to ρ ∈ `1. Then kρn − ρk → 0. 2

The weak convergence generates the weak topology on Lp. This topology is, by the definition, the weakest topology on Lp in which all functionalsg ˆ (where −1 −1 g ∈ Lq, p +q = 1) are continuous. The following theorem is very important. Although the proof uses only ideas from the measure theory, it is quite involved, and we omit it in this book.

+ Theorem 3.2.13 Let g ∈ L1 , then the set [0, g] ⊆ L1 is weakly compact. 2 72 CHAPTER 3.

3.2.8.◦ Exercises to Section 3.2.

Exercise 3.2.1 Prove the Chebyshev inequality:

kfkp µ({x : |f(x)| > ε}) ≤ p , εp where f ∈ Lp for 0 < p < ∞, and ε > 0.

Exercise 3.2.2 ∗∗∗ Let 1 ≤ a < b < c ≤ ∞. Show that

α 1−α La ∩ Lc ⊆ Lb & kfkb ≤ kfka · kfkc , where α is defined by b−1 = αa−1 + (1 − α)c−1.

Exercise 3.2.3 ∗∗∗ Let µ be a finite measure and 1 ≤ a ≤ b ≤ ∞. Show that

(a−1−b−1) Lb ⊆ La & kfka ≤ kfkb · µ(X) .

This implies, in particular, that the strong convergence in Lb implies the strong convergence in ∞ La with the same limit. Suppose that µ is infinite, construct an example of a sequence (fn)n=1 in La ∩ Lb and f ∈ La ∩ Lb such that kfn − fkb → 0, but kfn − fka 6→ 0.

∗ Exercise 3.2.4 Let (X, Σ, µ) be a finite measure space and let f ∈ L∞(X, Σ, µ). Show that the function N (p) := kfkp , (1 ≤ p < ∞) is continuous and lim N (n) = kfk∞. n→∞

Exercise 3.2.5 Show that the linear subspace of all simple functions is dense in Lp-space for any 1 ≤ p ≤ ∞.

Exercise 3.2.6 Show that `p and Lp(R) are separable Banach spaces for any 1 ≤ p < ∞.

Exercise 3.2.7 Let fn = sin nx ∈ Lp[0, 1], where 1 ≤ p < ∞. Show that fn → 0 weakly, but fn 6→ 0 in measure.

Exercise 3.2.8 Let gn = nχ 1 ∈ Lp[0, 1], where 1 ≤ p < ∞. Show that gn → 0 in measure, [0, n ] but gn 6→ 0 weakly. 3.3. THE LIAPOUNOFF CONVEXITY THEOREM 73

Exercise 3.2.9 (Minkowski’s Inequality for Integrals. I.) ∗∗∗ Let 1 ≤ p < ∞, and let (X, M, µ), (Y, N , ν) be σ-finite measure spaces, and let f ≥ 0 be a M ⊗ N -measurable function on X × Y . Show that

h Z  p i1/p  Z 1/p f(x, y) dν(y) dµ(x) ≤ f(x, y)p dµ(x) dν(y).

Exercise 3.2.10 (Minkowski’s Inequality for Integrals. II.) ∗∗∗ Let 1 ≤ p ≤ ∞, f(·, y) ∈ Lp(µ) for a.e. y, and let the function y → kf(·, y)kp be in L1(ν). R Show that f(x, ·) ∈ L1(ν) for a.e. x; the function x → f(x, y) dν(y) is in Lp(µ); and Z Z f(·, y) dν(y) ≤ kf(·, y)kp dν(y). p

Exercise 3.2.11 ∗∗∗ Prove Theorem 3.2.12.

Exercise 3.2.12 ∗∗∗ Prove Theorem 3.2.13.

∗∗ Exercise 3.2.13 Let An is a measurable subset of [0, 1] for each n, χAn ∈ L1, and χAn → f weakly in L1. Show that f is not necessarily a characteristic function of some measurable set.

Exercise 3.2.14 Let (X, Σ, µ) be a measure space, and P be a Frobenius – Perron operator ∗ on L1(X, Σ, µ). Show that the dual operator P on L∞(X, Σ, µ) satisfies the condition that ∗ P (χA) is a characteristic function of some measurable set for every A ∈ Σ.

3.3 The Liapounoff Convexity Theorem

In Exercise 1.2.8, we have seen that if µ is a non-atomic signed measure of bounded variation on Σ then the range of µ is the closed interval

[inf{µ(A): A ∈ Σ}, sup{µ(A): A ∈ Σ}].

One of the central and nicest result about vector-valued measures is the Lia- pounoff convexity theorem, which says that the range of a non-atomic Rn-valued measure of bounded variation is compact and convex. In this section, we give a direct proof of this theorem, which is due to D.A. Ross [8]. 74 CHAPTER 3.

3.3.1.◦ One-dimensional case. For the convenience of the reader, who will not study the proof of main theorem of this section, we give here the proof of the result of Exercise 1.2.8, which is very partial, but important, case of the Liapounoff theorem. First by the Jordan decomposition theorem, we may suppose that µ has nonnega- tive values only. Also we may suppose, that sup{µ(A): A ∈ Σ} = 1. So we have to show that the range of µ coincides with the interval [0, 1]. It takes obviously the values 0 and 1. Let 0 < α < 1. Consider the family F of all pairs (A, B) of measurable subsets of X satisfying µ(A) ≤ α, µ(B) ≤ 1 − α. The family F is ordered by inclusion, and it is trivial to show that it satisfies conditions of the Zorn lemma. Therefore, there exists a maximal element, say (A0,B0) in F. If µ(A0) < α, then µ(X\(A0 ∪B0)) > 0, and there exists a subset Y of X\(A0 ∪B0) (due to non-atomic condition on µ) with 0 < µ(Y ) < α − µ(A0). Thus we get that a pair (A1,B0) is strictly greater then (A0,B0), where A1 := A0 ∪ Y . The obtained contradiction with the maximality of (A0,B0) shows that µ(A0) = α, what is required.

3.3.2.◦ Examples. First, let us show that the Liapounoff convexity theorem may fail for E-valued measures, if E is an infinite dimensional Banach space (see, [3]).

Example 3.3.1 (J.J. Uhl, Jr.) A non-atomic vector-valued measure of bounded variation, whose range is closed and is neither convex nor compact. Let B([0, 1]) be the Borel algebra on [0, 1] and µ be the Lebesgue measure. Define

G : B([0, 1]) → L1(µ) by G(E) := χE. If π ⊆ B([0, 1]) is a partition of [0, 1], it is evident that X X kG(E)k1 = µ(E) = 1. E∈π E∈π

By an easy application of the dominated convergence theorem, G(Σ) is closed in 1 L1(µ). Show that G(Σ) is not a convex set. Indeed, 2 χ[0,1] ∈ coG(Σ), but 1 1 kG(E) − χ k = [µ([0, 1] \ E) + µ(E)]/2 = (∀E ∈ Σ). 2 [0,1] 1 2 3.3. THE LIAPOUNOFF CONVEXITY THEOREM 75

n To see that G(Σ) is not compact, let En = {t ∈ [0, 1] : sin(2 πt) > 0} for each positive integer n. An easy computation shows that 1 kG(E ) − G(E )k = (∀m 6= n). m n 4 Hence G(Σ) is not compact.

Example 3.3.2 (A. Liapounoff) An `2-valued non-atomic measure of bounded variation whose range is not convex. Let Σ = B[0, 2π] with the Lebesgue measure µ. Accordingly to Exercise 3.3.4, select a complete orthogonal ∞ system (wn)n=0 in L2[0, 2π] such that each wn assumes only the values ±1 and R 2π w0 = χ[0,2π] while 0 wn(t) dt = 0 for n ≥ 1. For each n, define λn on Σ by Z −n λn(E) = 2 [(1 + wn(t))/2] dt, E

E ∈ Σ. Define G :Σ → `2 by

G(E) = (λ0(E), λ1(E), . . . , λn(E),...).

By Exercise 3.3.6, G is a non-atomic `1-valued measure. G has bounded varia- tion, since kG(E)k`2 ≤ 2µ(E) for each E ∈ Σ. Now consider G([0, 2π]) = (2π, π/2, π/4,...) and suppose there is E ∈ Σ such that

G(E) = G([0, 2π])/2.

Then we have Z π = λ0(E) = dt = µ(E) E and for n ≥ 1 Z −n−1 −n −n 2 π = λn(E) = 2 [(1 + wn(t))/2] dt = 2 µ(E ∩ Un), E where Un = {s ∈ [0, 2π]: wn(s) = 1}. From this and the identities

µ(Un) = µ(E) = π, 76 CHAPTER 3. it follows

µ(E ∩ Un) = µ(E \ Un) = µ(Un \ E) = µ([0, 2π] \ (E ∪ Un)) = π/2 for n > 0. Define f = χE − χ[0,2π]\E. Then we have

Z 2π f(t) · w0(t) dt = π − π = 0 0 and, for n > 0, we have Z 2π f(t)·wn(t) dt = µ(E∩Un)+µ([0, 2π]\(E∪Un)) = µ(E\Un)−µ(Un \E) = 0. 0

∞ Since f ∈ L2([0, 2π]) and f 6= 0, this contradicts the completeness of (wn)n=0 and shows that G(Σ) is not convex.

3.3.3.! Liapounoff’s convexity theorem. Suppose that Ψ is an Rn-valued measure on (X, Σ). The following theorem was stated by A. Liapounoff in [6].

Theorem 3.3.1 (Liapounoff convexity theorem) If Ψ is non-atomic and has the bounded variation then

R(Ψ) := {Ψ(A): A ∈ Σ} is a closed convex subset of Rn.

n We deduce this theorem from its special case, when R(Ψ) ⊆ R+. Suppose that µ1, µ2, . . . , µn are finite non-atomic measures on (X, Σ). Denote by Φ = n (µ1, . . . , µn) the corresponding R -valued measure.

n Lemma 3.3.1 The set {Φ(E): E ∈ Σ} ⊆ R+ is closed and convex.

The proof of this lemma is postponed until the next subsection.

Proof of Theorem 3.3.1: Let Ψ be non-atomic and have bounded variation. Then Ψ = (ν1+ − ν1−, . . . , νn+ − νn−) 3.3. THE LIAPOUNOFF CONVEXITY THEOREM 77 and, by the Hahn decomposition theorem (Theorem 3.1.3), there exists a family n {Pk}k=1 ⊆ Σ such that Pk is a positive set for νk and X \ Pk is a negative set for 0 1 νk. Denote Pk := Pk and Pk := X \ Pk for all k ∈ 1, n. It follows from Lemma 3.3.1 that, for any function f : 1, n → {0, 1}, the set

n \ f(k) Rf (Ψ) := {Ψ(A): A ∈ Σ,A ⊆ Pk } k=1 is closed and convex in Rn. Then our range R(Ψ) is a closed and convex subset of Rn, as a finite sum of closed convex subsets of Rn: X R(Ψ) = Rf (Ψ). 2 f is a function from 1,n to {0,1}

3.3.4.! Proof of Lemma 3.3.1. This lemma is an application of the following, more elaborate, lemma the proof of which is postponed until the next subsection.

Lemma 3.3.2 Let µ1, µ2, . . . , µn be finite non-atomic measures on (X, Σ). De- n note by Φ = (µ1, . . . , µn) the corresponding R -valued measure. (LT1) For each E in Σ and r ∈ [0, 1], there is a subset A of E with A in Σ and Φ(A) = r · Φ(E). (LT2) For each E in Σ, there is an r ∈ (0, 1) and a subset A of E with A in Σ and Φ(A) = r · Φ(E). (LT3) For each E in Σ, there is a subfamily {Ar}r∈[0,1] of Σ such that Ar ⊆ As ⊆ E whenever 0 ≤ r ≤ s ≤ 1 and Φ(Ar) = r · Φ(E) for each r in [0, 1]. Moreover, the conditions (LT1), (LT2), and (LT3) are equivalent

To see that Lemma 3.3.1 follows from Lemma 3.3.2, let E and F belong to Σ, and let 0 ≤ r ≤ 1. By (LT1), there are subsets E1 of E \ F and F1 of F \ E with Φ(E1) = r · Φ(E \ F ) and Φ(F1) = (1 − r) · Φ(F \ E). Then

Φ(E1 ∪ [E ∩ F ] ∪ F1) = r · Φ(E) + (1 − r) · Φ(F ).

This shows that R(Φ) is convex. Now we show that R(Φ) is closed. It is known that any bounded closed convex subset of Rn is the convex hull of the set of all its extreme points. Thus R(Φ) = 78 CHAPTER 3. co(Ex(R(Φ))), and to show that R(Φ) is closed it is enough to show that z ∈ R(Φ) for any z ∈ Ex(R(Φ)). Let z ∈ Ex(R(Φ)). There is (and unique up to n scalar multiplication) yz ∈ R such that

(z, yz) = max{(x, yz): x ∈ R(Φ)} = sup{(x, yz): x ∈ R(Φ)}. (3.3.1)

Consider a signed measure µz on (X, Σ) defined by

µz(A) := (Φ(A), yz)(A ∈ Σ).

Then, by (3.3.1), (z, yz) = sup{µz(A): A ∈ Σ}. (3.3.2) It follows from (3.3.2) that there is a set E ∈ Σ such that

(z, yz) = µz(E) = (Φ(E), yz).

Since z is an extreme point of R(Φ) and yz satisfies (3.3.1), z = Φ(E) ∈ R(Φ), what is required. 2

3.3.5.! Proof of Lemma 3.3.2. The proof is carried out by an induction on the number n of non-atomic measures. It is convenient to assume (LT3), the strongest of the three statements, as the induction hypothesis for n measures, then prove the weakest of the statements, (LT2), for n+1 measures. We therefore need to show first that (LT1), (LT2), and (LT3) are equivalent, in the sense that a vector-valued measure Φ satisfying one of the statements must satisfy all three. The implications (LT3) ⇒ (LT1) ⇒ (LT2) are clear. Assume then that (LT2) holds, and fix a set E in Σ. If A is a measurable subset of E, r is in [0, 1], and Φ(A) = r · Φ(E), then write rA = r. Let E be the family of all such A. Order E by A < B if A ⊆ B and rA < rB. Let C be a maximal chain in E with ∅ and E belonging to C. Put I = {rA : A ∈ C}. It suffices to show that I = [0, 1] (since we can take {Ar}r∈[0,1] = C). Suppose (in search of a contradiction) that a ∈ (0, 1) \ I. Put

a∞ = sup(I ∩ [0, a)) ≤ a & b∞ = inf(I ∩ (a, 1]) ≥ a.

There exist An and Bn in E for n = 1, 2, 3,... such that rA increases to a∞ S T n and rBn decreases to b∞. Put A∞ = n An and B∞ = n Bn. It is easy to see 3.3. THE LIAPOUNOFF CONVEXITY THEOREM 79 that A∞ and B∞ are members of E with a∞ = rA∞ and b∞ = rB∞. Since C S T is totally ordered, A∞ = {A ∈ C : rA < a} and B∞ = {A ∈ C : rA > a}. Because C is maximal, A∞ and B∞ belong to C and (a∞, b∞)∩I = ∅. By (LT2), there is a subset C of B∞ \ A∞ with Φ(C) = r · Φ(B∞ \ A∞). Then A∞ ∪ C is in E and a∞ < r(A∞∪C) < b∞, so C ∪ {A∞ ∪ C} is a chain in E that properly extends C, a contradiction. Thus

LT3 ⇔ LT1 ⇔ LT2 .

Given the measures µ1, . . . , µn, put

0 Φ = (µ1, µ1 + µ2, µ1 + µ2 + µ3, . . . , µ1 + ... + µn). If A belongs to Σ and Φ0(A) = r · Φ0(E), then one can easy verify that Φ(A) = r · Φ(E). So we may always assume µ1  µ2  ...  µn.

When n = 1, (LT2) is trivial since µ1 is non-atomic (see Exercise 1.1.9), so Lemma 3.3.2 is immediate. We proceed by induction on n. Suppose that Lemma 3.3.2 holds for up to n measures, and that E ∈ Σ and measures µ1, µ2, . . . , µn+1 are given. Put ν = (µ2, . . . , µn+1). By two applications of (LT1), there are disjoint sets E1, E2, and E3 in Σ with E = E1 ∪E2 ∪E3 and ν(Ei) = (1/3)·ν(E) i for each i. By (LT3), for each i, there is a subfamily {Ar}r∈[0,1] of Σ such that i i i Ar ⊆ As ⊆ E whenever 0 ≤ r ≤ s ≤ 1 and

i i ν(Ar) = r · ν(E ) = (r/3) · ν(E).

i i We may assume that A1 = E . There are now three cases. i i Case 1: µ1(E ) = (r/3) · µ1(E) for some i. Since also ν(E ) = (1/3) · ν(E), (LT2) holds with r = 1/3.

Case 2: for some choice of i1, i2, and i3,

i1 i3 i2 µ1(E ) ≥ µ1(E ) > (1/3) · µ1(E) > µ1(E ).

For definiteness, take i1 = 1, i2 = 2, and i3 = 3. Define a subfamily {Ar}r∈[0,1] of Σ by  1 A3r if 0 ≤ r ≤ 1/3;  1 2 Ar = A1 ∪ A3r−1 if 1/3 < r ≤ 2/3; 1 2 3  A1 ∪ A1 ∪ A3r−2 if 2/3 < r ≤ 1.

One readily verifies that ν(Ar) = r·ν(E) for r in [0, 1]. If µ1(E) = 0 then Case 1 applies. Otherwise the function φ given on [0, 1] by φ(r) = µ1(Ar)/µ1(E) is well 80 CHAPTER 3. defined. Note that φ is increasing, and, the assumption µ1  µ2 ensures that φ is continuous. Moreover, µ (A1) µ (E1) φ(1/3) = 1 1 = 1 > 1/3, µ1(E) µ1(E) and µ (A1 ∪ A2) µ (E1 ∪ E2) µ (E) − µ (E3) φ(2/3) = 1 1 1 = 1 = 1 1 < 1 − 1/3 = 2/3. µ1(E) µ1(E) µ1(E) By the intermediate value theorem, φ(r) = r for some r ∈ (1/3, 2/3). In other words, µ1(Ar) = r · µ1(E). Since already ν(Ar) = r · ν(E), A = Ar ensures that (LT2) holds.

Case 3: for some choice of i1, i2, and i3,

i1 i3 i2 µ1(E ) ≤ µ1(E ) < (1/3) · µ1(E) < µ1(E ).

In this case, the argument from Case 2 applies without change, except now φ(1/3) < 1/3 and φ(2/3) > 2/3. 1 2 3 This exhausts the cases (since µ1(E) = µ1(E ) + µ1(E ) + µ1(E )) and proves the theorem. 2

3.3.6.! Exercises to Section 3.3.

∗ 2 2 Exercise 3.3.1 Let C := {x = (x1, x2) ∈ R : |x1| ≤ 1 & |x2| ≤ 1}. Construct a R -valued measure ν on B[0, 1] such that R(ν) = C.

∗∗∗ 2 2 2 2 Exercise 3.3.2 Let D := {x = (x1, x2) ∈ R : x1 + x2 ≤ 1} be the closed unit disk in R . Construct an R2-valued measure ν on the Borel algebra B[0, 1] whose range R(ν) is equal to D.

Exercise 3.3.3 ∗ Construct two R2-valued measures µ and ν of bounded variation on B[0, 1] such that 2 R(ν) ∪ R(µ) ⊆ {x = (x1, x2) ∈ R : x1 ≥ 0 & x2 ≥ 0}, and R(ν) + R(µ) 6= R(ν + µ).

∗∗ ∞ Exercise 3.3.4 Show that in L2([0, 2π], µ) there exists a complete orthogonal system (wn)n=0 R 2π of {−1, 1}-valued functions such that w0 = χ[0,2π] while 0 wn dµ = 0 for n ≥ 1. 3.3. THE LIAPOUNOFF CONVEXITY THEOREM 81

Exercise 3.3.5 ∗∗∗ Show that the result of Exercise 3.3.4 can be extended to the case of a non-atomic measure space with finite measure. Namely, show that if (X, Σ, µ) is a measure space with a non-atomic finite measure µ, then in L2(µ) there exists a complete orthogonal ∞ R system (wn)n=0 of {−1, 1}-valued functions such that w0 ≡ 1 and X wn dµ = 0 for n ≥ 1.

Exercise 3.3.6 Show that the function G : B[0, 2π] → `2 constructed in Example 3.3.2 is a non-atomic `2-valued measure.

Exercise 3.3.7 Let µ be a finite signed measure on (X, Σ). Show that there exists A ∈ Σ such that µ(A) = sup{µ(B): b ∈ Σ} .

Prove the following facts about convex subsets of Rn which were used in the proof of Lemma 3.3.1.

Exercise 3.3.8 Let A be a bounded closed convex subset of Rn. Show that A is the convex hull of all extreme points of A.

Exercise 3.3.9 Let A be a bounded convex subset of Rn and let a ∈ A be an extreme point of the closure A of A. Show that there is (and unique up to scalar multiplication) y ∈ Rn such that (a, y) = max{(x, y): x ∈ A} = sup{(x, y): x ∈ A}. Let A, a, and y be as before, and let b ∈ A. Show that

(b, y) = max{(x, y): x ∈ A} ⇒ b = a . 82 CHAPTER 3. Chapter 4

4.1 Function Spaces on Rn

In this section we describe some elementary notions which play an important role in the Fourier analysis and its applications to PDEs. Our aim is to give only a few basic constructions from measure theory concerning function spaces on Rn. We do not give any introduction to Fourier transform and to the theory of distributions; for this propose, we send the reader to many special text-books (see, for example, [4] and [9]).

4.1.1.◦ Multi-indexes. Let U be an open set in Rn and m ≥ 0, we de- note by Cm(U) the space of all real or complex valued functions on U whose partial derivatives of order ≤ m exist and are continuous. We set C∞(U) = T∞ m n ∞ m=1 C (U). Furthermore, for any E ⊆ R , we denote by C00 (E) the space of all C∞-functions on Rn whose support is compact and contained in E. If n n n E = R or U = R , we usually write without any confusion Lp = Lp(R ), n ∞ ∞ n C00 = C00(R ), and C = C (R ). The proof of the next simple lemma is left to the reader as an exercise (Exercise 4.1.2).

n Lemma 4.1.1 If f ∈ C00(R ) then f is uniformly continuous. 2

n n n If x = (xk)k=1, y = (yk)k=1 ∈ R , we set

n X p (x , y) := xk yk , kxk := (x , x). k=1

83 84 CHAPTER 4.

It will be convenient to have a compact notation for partial derivatives. We ∂ write ∂j = , and we use multi-index notation for higher-order derivatives. A ∂xj multi-index is an ordered n-tuple of nonnegative integers. If α = (α1, . . . , αn) is a multi-index, we set

n n α α X Y  ∂  1  ∂  n |α| = α , α! = α !, ∂α = ◦ · · · ◦ ; j j ∂x ∂x j=1 j=1 1 n

n n α Q αk and if x ∈ R , then x = xk ∈ R. k=1 Two subspaces of C∞(Rn) are of the great importance. The first one is the space ∞ ∞ C00 of C -functions with compact support. The existence of nonzero functions ∞ in C00 is not quite obvious; the standard construction is based on the fact that −1/t ∞ the function η(t) = e · χ(0,∞)(t) is C -function even at the origin. If we set  exp[(kxk2 − 1)−1)] if kxk < 1, ψ(x) = η(1 − kxk2) = (4.1.1) 0 if kxk ≥ 1,

We have ψ ∈ C∞ and supp(ψ) is the closed unit ball in Rn. The second impor- tant subspace of C∞ is the Schwartz space S consisting of those C∞-functions which, together with all their derivatives, vanish at infinity faster than any power of kxk. More precisely, for any nonnegative integer N and any multi-index α, we define N α kfk(N,α) = sup (1 + kxk) |∂ f(x)|; n x∈R then ∞ S = {f ∈ C : kfk(N,α) < ∞ for all N, α}. α −kxk2 Examples of functions in S are fα(x) = x e , where α is any multi-index. ∞ Also, clearly, C00 ⊆ S. Another important observation is that if f ∈ S then α α −N ∂ f ∈ Lp for all α and all p ∈ [1, ∞]. Indeed, |∂ f(x)| ≤ CN (1 + kxk) for all −N N, and (1 + kxk) ∈ Lp for all N > n/p.

Proposition 4.1.1 S is a Fr´echet space (= complete linear metric space) with the topology defined by the seminorms {k · k(N,α)}N,α.

∞ Proof: It is enough to prove the completeness of S. If (fk)k=1 is a Cauchy sequence in S then kfj − fkk(N,α) → 0 for all N, α. In particular, for each multi- α ∞ index α, the sequence (∂ fk)k=1 converges uniformly to a continuous function 4.1. FUNCTION SPACES ON RN 85 gα. Denoting by ej the vector (0...., 1,..., 0) with the 1 in the j-th position, we have Z t fk(x + tej) − fk(x) = ∂jfk(x + sej) ds. 0 Letting k → ∞, we obtain

Z t

g0(x + tej) − g0(x) = gej (x + sej) ds. 0

Here we consider the vector ej as a multi-index. Then gej = ∂jg0, and an α induction on |α| yields gα = ∂ g0 for all α. Now it is easy to check that kfk − g0k(N,α) → 0 for all α. 2

4.1.2.◦ Shift operation and convolution. Let f be a function on Rn and y ∈ Rn. We define the shift of f with respect to the vector y by

(τyf)(x) := f(x − y) .

Observe that kτyfkp = kfkp for 1 ≤ p ≤ ∞. It is almost obvious that a continuous function f is uniformly continuous iff kτyf − fk∞ → 0 as kyk → 0.

Proposition 4.1.2 If 1 ≤ p < ∞, the translation is continuous in the Lp norm; n that is, if f ∈ Lp and z ∈ R , then limy→0 kτy+zf − τzfkp = 0.

Proof: Since τy+z = τyτz, replacing f by τzf it suffices to assume that z = 0. First, if g ∈ C00, for kyk ≤ 1 the functions τyg are all supported in a common compact set K. So, by Lemma 4.1.1, Z p p |τyg − g| dµ ≤ kτyg − gk∞ µ(K) → 0 as kyk → 0 . (4.1.2) n R

Now suppose f ∈ Lp. If ε > 0 then, by the Urysohn lemma, there exists g ∈ C00 with kg − fkp < ε/3, so 2 kτ f − fk ≤ kτ (f − g)k + kτ g − gk + kg − fk < ε + kτ g − gk , y p y p y p p 3 y p and kτyg − gkp < ε/3 if y is small enough accordingly to (4.1.2). 2 86 CHAPTER 4.

Let f and g be measurable functions on Rn. The convolution of f and g is the function f ∗ g defined by Z (f ∗ g)(x) := f(x − y)g(y) dy n R for all x ∈ Rn such that the integral exists. Various conditions can be imposed on f and g to guarantee that f ∗ g is defined at least almost everywhere. For ex- R ample, if f ∈ C00, g can be any locally integrable function i.e. A |g| dµ < ∞ for any compact A.

4.1.3.◦ Basic properties of convolutions. The elementary properties of convolutions are summarized in the following proposition.

Proposition 4.1.3 Assuming that all integrals below exist, we have: (a) f ∗ g = g ∗ f; (b)(f ∗ g) ∗ h = f ∗ (g ∗ h); n (c) For z ∈ R , τz(f ∗ g) = (τzf) ∗ g = f ∗ (τzg); (d) If A is the closure of {x + y : x ∈ supp(f), y ∈ supp(g)} then supp(f ∗ g) ⊆ A.

Proof: (a): is proved by the substitution z = x − y: Z Z f ∗ g(x) = f(x − y)g(y) dy = f(z)g(x − z) dz = g ∗ f(x).

(b): follows from (a) and Fubini’s theorem. (c): Z Z τz(f ∗ g)(x) = f(x − z − y)g(y) dy = τzf(x − y)g(y) dy = ((τzf) ∗ g)(x) and, by (a), τz(f ∗ g) = τz(g ∗ f) = (τzg) ∗ f = f ∗ (τzg). (d): Observe that if x 6∈ A then, for any y ∈ supp(g), we have x − y 6∈ supp(f); hence f(x − y)g(y) = 0 for all y, so (f ∗ g)(x) = 0. 2

Theorem 4.1.1 (Young’s inequality) If f ∈ L1 and g ∈ Lp (1 ≤ p ≤ ∞) then (f ∗ g)(x) exists almost everywhere, f ∗ g ∈ Lp, and kf ∗ gkp ≤ kfk1kgkp. 4.1. FUNCTION SPACES ON RN 87

Proof: By Minkowski’s inequality for integrals (see Exercise 3.2.10), Z Z kf ∗ gkp = f(y) · g(·, −y) dy ≤ |f(y)| · kτygkp dy = kfk1 · kgkp . 2 n p n R R −1 −1 Proposition 4.1.4 If p + q = 1, f ∈ Lp, and g ∈ Lq then f ∗ g(x) exists n for every x ∈ R , f ∗ g is bounded and uniformly continuous, and kf ∗ gk∞ ≤ n kfkp kgkq. If 1 < p < ∞ then f ∗ g ∈ C0(R ).

Proof: The existence of f ∗g and the estimate for kf ∗gk∞ follow immediately from H¨older’s inequality. In view of Propositions 4.1.2 and 4.1.3, so does the uniform continuity of f ∗ g: If 1 ≤ p < ∞,

kτy(f ∗ g) − f ∗ gk∞ = k(τyf − f) ∗ g)k∞ ≤ kτyf − fkpkgkq → 0 as y → 0. (If p = ∞, interchange the roles of f and g.) Finally, if 1 < p, q < ∞, choose ∞ ∞ sequences (fn)n=1 and (gn)n=1 of continuous functions with compact supports such that kfn − fkp → 0 & kgn − fkq → 0.

By Proposition 4.1.3 (d), fn ∗ gn ∈ C00. But

kfn ∗ gn − f ∗ gk∞ ≤ kfn − fkpkgnkq + kfkpkgn − gkq → 0, so f ∗ g ∈ C0 by Exercise 4.1.1. 2

4.1.4.◦ Approximate identity. The following theorem underlies many important applications of convolutions on Rn. If φ is any function on Rn and t > 0, we set −n −1 φt(x) = t φ(t x). (4.1.3) R Remark that if φ ∈ L1 then n φt(x) dx is independent on t, thus, R Z Z Z −n −1 φt(x) dx = t φ(t x) dx = φ(y) dy . n n n R R R The proof of the following theorem is left to the reader as an exercise. R Theorem 4.1.2 Suppose φ ∈ L1 and n φ(x) dx = 1. R (a) If f ∈ Lp, where 1 ≤ p < ∞ then lim kf ∗ φt − fkp = 0 . t→0 (b) If f is bounded and uniformly continuous then lim kf ∗ φt − fk∞ = 0. t→0 (c) If f ∈ L∞ and f is continuous on an open set U then f ∗ φt → f uniformly on compact subsets of U as t → 0. 2 88 CHAPTER 4.

The family {φt}t≥0 as in Theorem 4.1.2 is called an approximate identity. In some sense, an approximate identity plays the role of an identity operator whenever we are working with convolution as a multiplication operation.

∞ Proposition 4.1.5 C00 (and hence also S) is dense in Lp (for 1 ≤ p < ∞) and in C0.

Proof: Given f ∈ Lp and ε > 0, there exists g ∈ C00 with kf − gkp < ε/2 by ∞ R Exercise 4.2.5. Let φ be a function in C00 such that n φ dµ = 1 (for example, R −1 ∞R take φ = ( n ψ dµ) ψ as in (4.1.1)). Then g ∗ φt ∈ C by Proposition 4.1.3(d) R 00 and Exercise 4.1.9, and kg ∗ φt − gkp < ε/2 for sufficiently small t by Theorem 4.1.2. The same argument works if Lp is replaced by C0, and k · kp, by k · k∞. 2

Theorem 4.1.3 (The C∞-Urysohn lemma) If K ⊆ Rn is compact and U is ∞ an open set containing K, there exists f ∈ C00 such that 0 ≤ f ≤ 1, f = 1 on K, and supp(f) ⊆ U.

Proof: Let δ be the distance from K to U c (δ is positive since K is compact) and let n V = {x ∈ R : ρ(x,K) < δ/3}. ∞ R Choose a nonnegative φ ∈ C00 such that n φ dx = 1 and φ(x) = 0 for kxk > R −1 R δ/3 (for example, ( n ψ dx) · ψδ/3 with ψ as in (4.1.1)), and set f = χV ∗ φ. ∞ R Then f ∈ C00 by Proposition 4.1.3 (d) and Exercise 4.1.9. It is easy to see that 0 ≤ f ≤ 1, f = 1 on K, and supp(f) ⊆ {x : ρ(x,K) ≤ 2δ/3} ⊆ U. 2

Theorem 4.1.2 does not say anything about the pointwise convergence f ∗ φt to f. Indeed, the situation with the pointwise convergence is much more delicate. To illustrate problems which arise here, we give without a proof the following theorem.

Theorem 4.1.4 Suppose |φ(x)| ≤ C(1 + kxk)−n−ε for some C, ε > 0. Then R φ ∈ L1 and n φ(x) dx = 1. If f ∈ Lp (1 ≤ p < ∞) then f ∗ φt(x) → f(x) as R t → 0 for every x in the Lebesgue set of f. In particular, for almost every x and for every x at which f is continuous. 2 4.1. FUNCTION SPACES ON RN 89

n By the Lebesgue set Lf of a locally integrable function f : R → R (for short n f ∈ Lloc(R )) we understood here Z n n 1 o Lf := x ∈ R : lim −1 |f(y) − f(x)|dy = 0 . t→∞ µ(B(x, t )) B(x,t−1)

4.1.5.◦ Exercises to Section 4.1.

n Exercise 4.1.1 Show that the space C0(R ) of functions vanishing at infinity is uniform ∞ closure of C00 . n Exercise 4.1.2 Show that any f ∈ C0(R ) is uniformly continuous. Exercise 4.1.3 ∗ Show that for x, y ∈ R: X k! (x + y)k = xα1 yα2 α = (α , α ). α! 1 2 |α|=k

Exercise 4.1.4 ∗ Show that for x ∈ Rn: X k! (x + x + ... + x )k = xα. 1 2 n α! |α|=k

Exercise 4.1.5 ∗ Show that for x, y ∈ Rn: X α! (x + y)α = xβ · yγ . β! · γ! β+γ=α Exercise 4.1.6 ∗ Prove the product rule for partial derivatives:

α β β α X δ γ β α α β X 0 δ ∂ (x f) = x ∂ f + cγδx ∂ f, x ∂ f = ∂ (x f) + cγδ∂(x f), 0 where cγδ = cγδ = 0 unless |γ| < |α| and |δ| < |β|. ∗∗ Exercise 4.1.7 Let f ∈ L1(R) and f be nonnegative. Use the result of Exercise 2.1.19 to show that: (a) if f ∗ f = 0 then f(t) = 0 a.e. on R; (b) if f ∗ f = f then f(t) = 0 for all t ∈ R; (c) the equality (f ∗ f)(t) = 1 a.e. is impossible. Exercise 4.1.8 ∗∗ Prove Theorem 4.1.2.

∗∗ k α Exercise 4.1.9 Let f ∈ L1, g ∈ C and let ∂ g be bounded for |α| ≤ k. Show that f ∗ g ∈ Ck and ∂α(f ∗ g) = f ∗ (∂αg) for |α| ≤ k. Exercise 4.1.10 ∗∗∗ Prove Theorem 4.1.4. 90 CHAPTER 4.

4.2 The Riesz Representation Theorem

In this section, we study the structure of the dual space to the spaces of continu- ous functions on a locally compact Hausdorff space. We treat only the R-valued case. The complex case is essentially the same.

◦ 4.2.1. The space C00(X) and positive functionals. Let X be a locally compact Hausdorff space (LCH-space, for short). This means that, for every x 6= y in X, there exist open neighborhoods U of x and V of y such that U¯ and V¯ are compact and U¯ ∩V¯ = ∅. From the point-set topology, it is known that any LCH-space is normal, so we may use the Urysohn lemma in its investigation.

We denote by C00(X) the space of continuous functions on X with compact support, i.e. f ∈ C00(X) if f is continuous and supp(f) = cl{x : f(x) 6= 0} is compact. A linear functional Ψ on C00(X) is called positive if Ψ(f) ≥ 0 whenever f ≥ 0. It is an important fact that every positive linear functional on C00(X) is continuous with respect to the uniform convergence on compact subsets of X. There are many ways for proving this, one of them is given below.

Lemma 4.2.1 If Ψ is a positive linear functional on C00(X). Then, for each compact K ⊆ X, there is CK < ∞ such that |Ψ(f)| ≤ CK kfk∞ for all f ∈ C00(X) with supp(f) ⊆ K.

Proof: Given a compact K, choose φ ∈ C00(X) with values in [0, 1] such that φ|K ≡ 1 by Urysohn’s lemma. If supp(f) ⊆ K, we have

kfk∞φ − f ≥ 0 & kfk∞φ + f ≥ 0.

Thus

kfk∞Ψ(φ) − Ψ(f) ≥ 0 & kfk∞Ψ(φ) + Ψ(f) ≥ 0, so |Ψ(f)| ≤ CK kfk∞ with CK = Ψ(φ). 2

If µ is a Borel measure on X such that µ(K) < ∞ for every compact K ⊆ X, 1 then C00(X) ⊆ L (µ), so Z f → f dµ X 4.2. THE RIESZ REPRESENTATION THEOREM 91 is a positive linear functional on C00(X). The main result of this section says that every positive linear functional on C00(X) arises in this way. Moreover, we give some regularity conditions on µ, under which µ is unique. Let µ be a Borel measure on X and let E be a Borel subset of X. The measure µ is called outer regular on E if

µ(E) = inf{µ(U): E ⊆ U, U open};

µ is called inner regular on E if

µ(E) = sup{µ(K): K ⊆ E,K compact}.

If µ is outer and inner regular on all Borel sets, µ is called regular (compare with 2.1.5). The regularity of a Borel measure sometimes is too strong condition.

Definition 4.2.1 A Borel measure that is finite on compact sets, outer regular on Borel sets, and inner regular on open sets is called a Radon measure.

◦ 4.2.2. The Riesz representation theorem for C00(X). This theorem historically is one of the first results in functional analysis. Originally it was stated by F. Riesz in 1909 for C[0, 1]. Then it was generalized by several authors for C(K), where K is a compact Hausdorff space, and for C00(X), where X is an LCH-space. Here we begin with the general case and derive from it the Riesz representation theorem for C(K). We use the following notation. If U is open in X and f ∈ C00(X), 0 ≤ f ≤ 1, and supp(f) ⊆ U, then we write f ≺ U.

Theorem 4.2.1 If Ψ is a positive linear functional on C00(X), where X is an LCH-space. Then there exists a unique Radon measure µ on X such that Z Ψ(f) = f dµ (∀f ∈ C00(X)) . (4.2.1) X Moreover,

µ(U) = sup{Ψ(f): f ∈ C00(X), f ≺ U} for all open U ⊆ X (4.2.2) and

µ(K) = inf{Ψ(f): f ∈ C00(X), f ≥ χK } for all compact K ⊆ X. (4.2.3) 92 CHAPTER 4.

Sketch of the proof: First, we prove uniqueness. Let µ be a Radon measure satisfying (4.2.1). Let U ⊆ X be an open subset of X. Then Ψ(f) ≤ µ(U) whenever f ≺ U. If K is a compact subset of U, by Urysohn’s lemma, there is f ∈ C00(X) such that f ≺ U and f|K ≡ 1. Thus Z µ(K) ≤ f dµ = Ψ(f) ≤ µ(U) . X This proves (4.2.2) since µ is inner regular on U. Thus µ is uniquely determined by Ψ on open sets, and hence on all Borel sets because of outer regularity. This proves the uniqueness of µ and suggests how to prove existence. We define ν(U) := sup{Ψ(f): f ∈ C00(X), f ≺ U} for any open U, and we define ν∗(E) for an arbitrary E ⊆ X by

ν∗(E) := inf{ν(U): U ⊇ E,U open}.

Clearly, ν(U) ≤ ν(V ) if U ⊆ V , and hence ν∗(U) = ν(U) if U is open. The outline of the proof is now as follows. We only fix several important steps and leave the proofs of them to the reader. (1) Show that ν∗ is an outer measure on P(X). (2) Show that every open set is ν∗-measurable. Now it follows from Caratheodory’s theorem that every Borel set is ν∗-measurable ∗ ∗ and µ := ν |BX is a Borel measure (note that µ(U) = ν (U) = ν(U)) for any open U). Moreover, µ is outer regular and satisfies (4.2.2) by the definition. (3) Show that µ satisfies (4.2.3). Since X is an LCH-space, µ is finite on compact sets. Prove the inner regularity on open sets. Let U be an open set and µ(U) > α. Fix ε > 0, µ(U) > α + ε. Choose f ∈ C00(X) such that f ≺ U and Ψ(f) > α + ε, and let

K := supp(f) ⊆ U.

If g ∈ C00(X) and g ≥ χK then g ≥ f and hence Ψ(g) ≥ Ψ(f) > α + ε. But then µ(K) ≥ α + ε > α by (4.2.3), so µ is inner regular on U. (4) Show that µ satisfies (4.2.1). 4.2. THE RIESZ REPRESENTATION THEOREM 93

With this last step, the proof of the theorem is completed. 2

4.2.3.◦ The Riesz representation theorem for C(K) . Let K be a compact Hausdorff space and let C(K) be a vector space of all continuous R-valued functions on K. It is well known that C(K) is a Banach space with respect to the sup-norm. The following theorem, which is historically also called the Riesz representation theorem describes the structure of its dual space space C(K)∗. We prove this theorem by using of Theorem 4.2.1.

Theorem 4.2.2 Let K be a compact Hausdorff space and let ψ ∈ C(K)∗. Then there are two unique Radon measures µ1 and µ2 on X such that Z Z ψ(f) = f dµ1 − f dµ2 (∀f ∈ C(K)) . K K

Moreover, µ1 = µ+ and µ2 = µ− for some signed measure µ of bounded variation on the Borel algebra B(K).

∗ Proof: Let ψ ∈ C(K) . Then the function ϑ : C+(K) → R+, which is defined by the formula

ϑ(f) := sup{ψ(g) : 0 ≤ g ≤ f} (f ∈ C+(K)) , (4.2.4) is additive on C+(K) (see Exercise 4.2.14) . Every such an additive function ϑ : C+(K) → R+ possesses a unique extension ψ+ on C(K), which is a positive linear functional (see Exercise 4.2.15) . This extension is given by the formula:

ψ+(f) := ϑ(f+) − ϑ(f−)(f ∈ C(K)).

Let ψ− be the positive functional correspondent to −ψ, i.e., ψ− := (−ψ)+. ∗ ∗ Now we apply Theorem 4.2.1 to ψ+ ∈ C(K) and ψ− ∈ C(K) . Then we have Z ψ+(f) = f dµ1 (∀f ∈ C(K)) K and Z ψ−(f) = f dµ2 (∀f ∈ C(K)), K where µ1 and µ2 are unique Radon measures on X. It is routine to show that

µ1 = µ+ & µ2 = µ− (4.2.5) for a signed Borel measure µ := µ1 − µ2. 2 94 CHAPTER 4.

4.2.4.◦ Exercises to Section 4.2.

Exercise 4.2.1 ∗∗ Let X be a locally compact Hausdorff space. Show that, for any compact ∞ K ⊆ X such that K = ∩n=1Un and Un is open for any n ∈ N, there exists f ∈ C00(X) such that K = f −1({1}).

2n Exercise 4.2.2 Let fn(t) = t be the sequence of R-valued functions on R. These functions define Borel measures µn on R as follows Z 2n µn(A) := t dt (∀A ∈ B(R)). A

(a) Show that µn is a Radon measure for each n.

(b) Show that µk is µj-continuous for every k, j ∈ N.

(c) Find the Radon – Nikodym derivative of µk with respect to µj.

(d) Let A ∈ B(R) with the Lebesgue measure µ(A) 6= 0. Show that there is no constant M ∈ R such that µn(A) ≤ M for all n.

Exercise 4.2.3 Show that any Radon measure µ is inner regular on each σ-finite subset of X, where A is called σ-finite if there exists a sequence An of measurable sets such that µ(An) < ∞ ∞ for any n and A = ∪n=1An.

Exercise 4.2.4 Let Y be the one-point compactification of a set X with the discrete topology and let µ be a Radon measure on Y . Show that supp(µ) is countable, where

supp(µ) = Y \ [∪{U ⊆ Y : µ(U) = 0}] .

∗ Exercise 4.2.5 Show that C00(X) is dense in any Lp(µ) for any 1 ≤ p < ∞ and any Radon measure µ on X.

Exercise 4.2.6 Prove the properties (1) − (4) in the proof of Theorem 4.2.1.

∗∗ ∞ Exercise 4.2.7 Show that a sequence (fn)n=1 in C0(X) converges weakly to 0 iff sup kfnk∞ < ∞ and fn → 0 pointwise.

Exercise 4.2.8 Let µ be a Radon measure on X such that µ({x}) = 0 for all x ∈ X and a Borel set A ∈ B(X) satisfies µ(A) < ∞. Show that, for any α, 0 ≤ α ≤ µ(A), there exists B ∈ B(X), B ⊆ A, such that µ(B) = α. 4.2. THE RIESZ REPRESENTATION THEOREM 95

Exercise 4.2.9 Let ν be a non-atomic Radon measure on Rn and let µ be another measure on Rn such that µ is ν-continuous. Show that µ is a Radon measure.

Exercise 4.2.10 ∗ Let ν be a Radon measure on Rn and let µ be another measure on Rn such that µ is ν-continuous. Show that µ is a Radon measure.

Exercise 4.2.11 ∗ Let ν be a σ-finite Radon measure on a LCH-space and let µ be another measure such that µ is ν-continuous. By using of the Radon – Nikodym theorem, show that µ is a Radon measure.

Exercise 4.2.12 ∗∗∗ Let ν be a Radon measure on a LCH-space and let µ be another measure such that µ is ν-continuous. Show that µ is a Radon measure.

∗ ∼ Exercise 4.2.13 Let X = N with the discrete topology. Show that C0(X) = `1.

Exercise 4.2.14 Show that the function ψ+ in (4.2.4) is well defined and additive on C+(K).

Exercise 4.2.15 Show that every additive on C+(K) R-valued function possesses a unique extension on C(K) which is a positive linear functional.

Exercise 4.2.16 Prove the formula (4.2.5). 96 CHAPTER 4. Bibliography

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[12] A.C. Zaanen, Introduction to operator theory in Riesz spaces, Springer- Verlag, Berlin, (1997), xii+312 pp. Index

(Xn)n-bounded sequence, 10 Caratheodory extension theorem, 31 C∞-Urysohn lemma, 88 Caratheodory theorem, 33 E-valued measure, 13 Cauchy – Schwarz inequality, 69 L1-convergence, 24 Cauchy in measure sequence, 27 µ-continuous measure, 58 Chebyshev inequality, 72 µ-integrable function, 58 compact support, 90 σ-additivity of measure, 7 complete measure, 8 σ-algebra, 5 complex measure, 14 σ-algebra generated by G, 6 continuous linear functional, 70 σ-finite measure, 14 convergence in measure, 27 σ-finite measure space, 8 convex hull, 30 σ-finite premeasure, 43 convolution, 86 σ-finite subset, 94 counting measure, 8 ξ-measurable set, 32 Dedekind completeness, 19 a.e., 21 difference of sets, 5 additive function, 19 distribution function, 43 adjoint space, 70 dominated convergence theorem, 26 algebra of subsets, 34 dual space, 70 almost everywhere, 21 Egoroff theorem, 9 almost everywhere convergence, 9 elementary Young inequality, 63 approximate identity, 88 exhaustion argument, 10 exhaustion theorem, 10 Baire theorem, 51 Banach lattice, 19, 70 Fatou lemma, 25 block, 39 finite measure, 14 Borel algebra of X, 6 finite measure space, 8 Borel measure, 8 Frobenius – Perron operator, 59 Fubini theorem, 31, 40 Cantor like function, 51 Cantor like set, 51 generalized Cantor function, 51

99 100 INDEX generalized Cantor set, 50 negative set, 58 non-atomic measure, 11, 12, 20 h-interval, 36 H¨olderinequality, 63 orthogonal functions, 69 Hahn decomposition theorem, 58 orthonormal basis, 69 outer measure, 31 inner regular measure, 91 outer regular measure, 91 integrable function, 22 partial ordering, 17 Jordan decomposition theorem, 18 partially ordered vector space, 17 pointwise convergence, 9 LCH-space, 90 positive linear functional, 90 Lebesgue σ-algebra, 36 positive set, 58 Lebesgue – Stieltjes Measure, 31 premeasure, 34 Lebesgue integral, 21 probabilistic measure space, 8 n Lebesgue measurable set in R , 45 probability, 8 Lebesgue measure, 9, 31 product measure, 39 Lebesgue measure on R, 36 product of measure spaces, 39 n Lebesgue measure on R , 45 purely atomic measure, 12 Lebesgue set, 89 Liapounoff, 73, 75 Radon – Nikodym derivative, 58 Liapounoff convexity theorem, 76 Radon – Nikodym theorem, 55 linear functional, 70 Radon measure, 91 locally compact Hausdorff space, 90 regular measure, 91 locally integrable function, 86 Riesz representation theorem, 91, 93 Lusin theorem, 47 Ross, 73 measurable function, 6 scalar product, 69 measurable set, 7 Schep, 55 measurable transformation, 59 Schwartz space, 84 measure, 7 separable Banach space, 68 measure of bounded variation, 16 separable measure, 8, 14 measure space, 7 set of the first category, 51 Minkowski inequality, 64, 73 set of the second category, 51 monotone class, 41 shift, 47, 85 monotone convergence theorem, 24 side of block, 45 monotone function, 19 signed measure, 14 monotone function on A, 35 simple function, 21 multi-index, 84 standard Cantor function, 51 INDEX 101 standard Cantor set, 49 submeasure, 31 symmetric difference of sets, 5

Tonelli theorem, 42 total variation of measure, 16 totally disconnected set, 49

Uhl, Jr., 74 uniform convergence, 9 Urysohn lemma, 46, 85, 90, 92 variation of measure, 15 vector lattice, 17 vector-valued measure, 13 weak topology on Lp, 71 weakly convergent sequence, 71

Young inequality, 86