INTRODUCTION TO MEASURE THEORY AND LEBESGUE INTEGRATION

Transcription

1 1 INTRODUCTION TO MEASURE THEORY AND LEBESGUE INTEGRATION Eduard EMELYANOV Ankara TURKEY 2007

2 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 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 Jordan decomposition of signed measure, theorems about convergence for Lebesgue integral, Cantor sets, the Radon Nikodym theorem, the theory of L p -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 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 3 Contents Introduction Chapter σ-algebras, Measurable Functions, Measures, the Egoroff Theorem, Exhaustion Argument 1.2. Vector-valued, Signed and Complex Measures, Variation of a Vector-valued Measure, Operations with Measures, the Jordan Decomposition Theorem, Banach Space of Signed Measures of Bounded Variation 1.3. Construction of the Lebesgue Integral, the Monotone Convergence Theorem, the Dominated Convergence Theorem, Chapter The Caratheodory Theorem, Lebesgue Measure on R, Lebesgue Stieltjes Measures, the Product of Measure Spaces, the Fubini Theorem 2.2. Lebesgue Measure on R n, Lebesgue Integral in R n, the Lusin Theorem, Cantor Sets Chapter The Radon Nikodym Theorem, Continuity of a Measure with Respect to another Measure, the Hahn Decomposition Theorem 3.2. Hölder s and Minkowski s Inequalities, Completeness, L p -Spaces, Duals 3.3. The Liapounoff Convexity Theorem Chapter Vector Spaces of Functions on R n, Convolutions 4.2. Radon Measures, the Riesz Representation Theorem Bibliography Index

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5 Chapter 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 σ-algebras. The following notion is the principal in the measure theory. Definition 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; (c) given a sequence (A k ) k A, we have A k A. k It follows from this definition that the empty set belongs to A since X A and = X \ X. Further, given a sequence (A k ) k A, we have A k = X \ (X \ A k ) A. k k Finally, the difference A \ B := A (X \ B) and the symmetric difference A B := (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 6 CHAPTER 1. Proposition Let {Ω α } α A be a nonempty family of σ-algebras in P(X), then Ω = α Ω α is also a σ-algebra. This proposition leads to the following definition. Definition 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 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) 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 We say that f is measurable, if f 1 ([a, b]) Ω for any reals a < b. The following three propositions are obvious. Proposition 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). Proposition 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.

7 1.1. σ-algebras, MEASURABLE FUNCTIONS, MEASURES 7 Proposition Let (f n ) n=1 be a sequence of measurable functions, then are measurable. lim sup f n and lim inf f n n n Measures. The main definition here is the following. Definition 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 (c) µ( k A k) = k µ(a k) for any sequence (A k ) k of pairwise disjoint sets from A, that is A i A j = for i 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 Let µ be a measure on a σ-algebra A, A n A, and A n A. Then A A and µ(a) = lim µ(a n ). In particular, if (B n ) n=1 is a decreasing n sequence of elements of A such that n=1 B n =, then µ(b n ) Measure spaces. We consider a fixed but arbitrary σ-algebra with a measure. Definition 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 8 CHAPTER 1. Now we present several examples of measure spaces. Example Let X = {x 1,..., x N } be a finite set, A be the σ-algebra of all subsets of X and a measure is defined by setting to each x i X a nonnegative number, say p i. This follows that the measure of a subset {x α1,..., x αk } X is just p α1 + + p αk. If p i = 1 for all i, then the measure is called a counting measure because it counts the number of elements in a set. Example 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 = R n, 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 general. In almost all applications, the following specific class of measure spaces is adequate. Definition A measure space (X, A, µ) is called σ-finite if there is a sequence (A k ) k=1, A k A, satisfying X = A k and µ(a k ) < 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 (X n ) n=1 of subsets of finite measure such that X = n=1 X n. If X = R, then A k may be chosen as intervals [ k, k]. In R d, they may be chosen as balls of radius k, etc. Definition 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(a B) < ε. A measure µ is called complete if there holds: [ µ(a) = 0 & B A ] B A.

9 1.1. σ-algebras, MEASURABLE FUNCTIONS, MEASURES 9 Every Borel measure on R (or on [0, 1], or on R n, etc.) possesses a unique completion, which is called a Lebesgue measure Convergence of functions and the Egoroff theorem. For measurable functions, there are several natural types of convergence. Some of them are given by the following definition. Definition Let (f n ) n=1 be a sequence of R-valued functions defined on X. We say that: (a) f n f pointwise, if f n (x) f(x) for all x X; (b) f n f almost everywhere (a.e.), if f n (x) f(x) for all x X except a set of measure 0; (c) f n f uniformly, if for any ε > 0 there is n(ε) such that for all n n(ε). sup{ f n (x) f(x) : x X} ε Theorem (Egoroff s theorem) Suppose that µ(x) <, {f n } n=1 and f are measurable functions on X such that f n f a.e. Then, for every ε > 0, there exists E X such that µ(e) < ε and f n f uniformly on E c = X \ E. Proof: Without loss of generality, we may assume that f n f everywhere on X and (by replacing f n with f n f) that f 0. For k, n N, let E n (k) := {x : f m (x) k 1 }. m=n Then, for a fixed k, E n (k) decreases as n increases, and n=1 E n(k) =. Since µ(x) <, we conclude that µ(e n (k)) 0 as n. Given ε > 0 and k N, choose n k such that µ(e nk (k)) < ε 2 k, and set E := E nk (k). k=1

10 10 CHAPTER 1. Then µ(e) < ε, and we have f n (x) < k 1 ( n > n k, x E). Thus f n 0 uniformly on X \ E Exhaustion argument. Let (X, Σ, µ) be a σ-finite measure space. Given a sequence (U n ) n=1 Σ, a set A Σ is called (U n ) n -bounded if there exists n such that A U n µ-almost everywhere. Theorem (Exhaustion theorem) Let (Y n ) n=1 Σ be a sequence satisfying Y n X and µ(y n ) < for all n. Let P be some property of (Y n ) n -bounded measurable sets, such that A P iff B P for all B, µ(a B) = 0. Suppose that any (Y n ) n -bounded set A, µ(a) > 0, has a subset B Σ, µ(b) > 0 with the property P. Moreover, assume that either (a) A 1 A 2 P for every A 1, A 2 P, or (b) n B n P for every at most countable family (B n ) n of pairwise disjoint sets possessing the property P. Then there exists a sequence (X n ) n=1 Σ such that X n X, and P X n Y n for all n. Moreover, there exists a pairwise disjoint sequence (A n ) n=1 Σ such that A n = X and A n P for all n. n=1 Proof: Let A be a (Y n ) n -bounded set with µ(a) > 0. Denote P A := {B P : B A} & m(a) := sup{µ(b) : B P A }. I(a) Suppose P satisfies (a). Then there exists a sequence (F n ) n=1 P A such that m(a) = lim µ(f n ), We may assume, that F n. By Proposition 1.1.5, the n set F = n=1f n 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 F 0 P. Then F n F 0 P A and µ(f n F 0 ) > m(a) for a sufficiently large n, which contradicts to the definition of m(a). Therefore, µ(a) = m(a). Now we apply this for A = Y n. Thus, there exists a sequence (X n) n Σ such that X n Y n, X n P, and µ(y n \ X n) < n 1 for all n. By (a), we may assume that X n. The set X 0 = n=1x n satisfies Y n \X 0 Y n \X n, so µ(y n \X 0) < n 1

11 1.1. σ-algebras, MEASURABLE FUNCTIONS, MEASURES 11 for all n. Then µ(y n \X 0) = 0, and µ(( n=1y n )\X 0) = 0, or µ(x \X 0) = 0. Let X n := (X n (X \X 0)) Y n, then the sequence (X n ) n has the required properties. The desired pairwise disjoint sequence (A n ) n=1 is given recurrently by A 1 = X 1 and A k+1 = X k+1 \ k i=1a i. I(b) Suppose P satisfies (b). Let F A be the family of all pairwise disjoint families of elements of P A of nonzero measure. Then F A is ordered by inclusion and, obviously, satisfies the conditions of the Zorn lemma. Therefore, we have a maximal element in F A, say. Then is at most countable family, say = {D n } n. By (b), its union D := n D n is an element of P A 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 F A which is strictly greater then. The obtained contradiction, shows that A P for every (Y n ) n -bounded set A. So, we may take X n := Y n for each n. Now we apply this for A = Z m := Y m \ m 1 k=1 Y k. Let Z m = n Dn m be a pairwise disjoint union, where Dn m P for all n, m. The family {Dn m } n,m is an at most countable disjoint decomposition of X, say {Dn m } n,m = (A n ) n=1. The sequence (A n ) n=1 satisfies the required properties Exercises to Section 1.1. Exercise Prove that the cardinality of any σ-algebra is either finite or uncountable. Exercise Let f and g be measurable R-valued functions. Show that f g is measurable. Exercise For each X N and for each k N define: n k X := card(x {1,..., k}). Let A be a family of all X N such that the number µ(x) = lim σ-algebra? Is µ a measure on A? Exercise Prove Proposition k k 1 n k X exists. Is A a Exercise 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 for the definition)?

12 12 CHAPTER 1. Exercise Show that each separable measure which satisfies Axiom d) is σ-finite. Exercise Give an example of a family (Ω α ) α [0,2π] of σ-subalgebras of the Borel algebra B([0, 1] 2 ) such that for any α, β [0, 2π], α β: and Ω α Ω β = {, [0, 1] 2 } ( α)[α [0, π]]( a)[0 a 1]( B Ω α )[µ(b) = a]. Exercise Construct a sequence (Σ n ) n=1 of σ-subalgebras of the Borel algebra B([0, 1] 2 ) such that, for any n < m, Σ m Σ n & Σ n Σ m and ( n)( a)[0 a 1]( B Σ n )[µ(b) = a]. Exercise 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 A 0 A such that 0 < µ(a 0 ) < 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 (The Borel Cantelli lemma) Let (X, Σ, µ) be a measure space. Show that if {A n } n Σ and n=1 µ(a n) < then µ(lim sup A n ) = 0. n Exercise Show that Egoroff s theorem may fail if µ(x) =. Exercise Let (X, Ω, µ) be a measure space and µ(x) <. Say that A B if µ(a B) = 0. (a) Show that is an equivalence relation on Ω; (b) Show that the function ρ : (Ω/ ) 2 R: ρ([a], [B]) := µ(a B), is a metric, and (Ω/, ρ) is a complete metric space. Exercise Let (Ω/, ρ) be the metric space from Exercise (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 exists a sequence (a n ) n=1 of elements of A such that µ(a) = n µ({a n}). Exercise Show that in Proposition pointwise limits lim sup may be replaced by a.e.-limits lim sup f n and lim inf f n. n n n f n and lim inf n f n

13 1.2. VECTOR-VALUED MEASURES 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 theorem. We refer for further delicate results about vector-valued measures to [3] and about spaces of signed measures to [7] and [12] Vector-valued, signed and complex measures. Let Σ be a σ-algebra of subsets of a set X, and let E = (E, ) be a Banach space. Definition A function µ : Σ E { } is called a vector-valued measure (or E-valued measure) if (a) µ( ) = 0; (b) µ( k A k) = k µ(a k) for any pairwise disjoint sequence (A k ) k Σ; (c) for any A Σ, µ(a) =, there exists B Σ such that B A and 0 < µ(b) <. Example Take Σ = P(N), and c 0 is the Banach space of all convergent C-valued sequences with a fixed element (α n ) n c 0. Define for any A N: ψ(a) := (β n ) n, where β n = α n if n A and β n = 0 if n A. Then ψ is a c 0 -valued measure on P(N). Example Let X be a set and let Ω be a σ-algebra in P(X). Then for any family {µ k } m k=1 of finite measures on Ω and for any family {w k} m k=1 of vectors of R n, the R n -valued measure Ψ on Ω is defined by the formula m Ψ(E) := µ k (E) w k (E Ω). k=1 Example Let X be a set and let Ω be a σ-algebra in P(X). Then for any family {µ k } m k=1 of finite measures on Ω, for any family {A k} m k=1 of pairwise disjoint sets in Ω, and for any family {w k } m k=1 of vectors of Rn, the R n -valued measure Φ on Ω is defined by the formula m Φ(E) := µ k (E A k ) w k (E Ω). k=1

14 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, x = y = 1, it is impossible to find sequences (A n ) n=1 and (B n ) n=1 of elements of Σ such that µ(a n ), lim n µ(a n ) µ(a n ) = x, and µ(b n), µ(a n ) lim n µ(b n ) = y. 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 Definition A function µ : Σ R { } is called a signed measure if µ( ) = 0, µ( k A k) = k µ(a k) for any sequence (A k ) 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 A function µ : Σ C { } is called a complex measure if µ( ) = 0, µ( k A k) = k µ(a k) for any sequence (A k ) 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 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 (A k ) k, A k Σ, satisfying X = A k 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 µ(a B) < ε. The definitions of σ-finite, finite, and separable signed or complex measure are obvious.

15 1.2. VECTOR-VALUED MEASURES 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 by the following formula: n=1 µ : Σ R + { } n=j n=j µ (A) := sup{ µ(a n ) : A n Σ, A n = A, A k A p = if n=1 k p}. (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 Let µ be an E-valued measure on (X, Σ). Then µ is a measure. Proof: Suppose that µ is an E-valued measure. We have to show only the σ-additivity of the function µ : Σ R + { }. Let (A n ) n=1 be a pairwise disjoint sequence of elements of Σ and A := n A n. 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 µ(e A n ) µ(e) = µ(e A) = E π E π E π n µ(e A n ) = µ(e A n ) n n E π n E π Since it holds for any partition π, we obtain the inequality µ (A n ). µ ( n A n ) n µ (A n ). (1.2.2) Since µ is additive and monotone, then for each n: i=n n µ (A i ) = µ ( i=1 i=1 A i ) µ ( n A n ),

16 16 CHAPTER 1. and by taking the limit: µ (A i ) µ ( A n ). (1.2.3) n i=1 Combining (1.2.2) and (1.2.3) give us σ-additivity of µ. It is also easy to see that µ is a σ-finite or finite E-valued measure iff µ is. Definition 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 is of bounded variation if α n = 2 n ; and, of unbounded variation if α n = 1/n 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 M b = Mb E (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., for every A Σ. (µ 1 + µ 2 )(A) = µ 1 (A) + µ 2 (A), (αµ)(a) = αµ(a) (1.2.4) Theorem M b is a Banach space with respect to the norm b : µ b := µ (X) ( µ M b ) which is called the total variation of a measure µ. Proof: By (1.2.1) and (1.2.4), the map b : M b R + satisfies all axioms of norm. To show that M b is complete under the norm b, it is enough to show that if µ n b < (1.2.5) n=1 for some sequence (µ n ) n=1 of E-valued measures, then µ := µ n M b. (1.2.6) n=1

17 1.2. VECTOR-VALUED MEASURES 17 For any A Σ, the series µ n (A) n=1 is convergent in E, because E is a Banach space, and there hold (1.2.5) and µ n (A) µ n b ( n). So µ in (1.2.6) is well defined. We leave to the reader to show that µ M b as an exercise. In general, if µ 1 and µ 2 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 M b be the Banach space of all finite signed measures on (X, Σ). First, we define a partial ordering in M b saying that µ 1 µ 2 whenever µ 1 (A) µ 2 (A) ( A Σ), (1.2.7) then M b becomes a partially ordered vector space. We show that M b is a vector lattice, which means, by the definition, that for every µ, ν M b there exists sup{µ, ν} M b. For this purpose, we denote for µ M b the number From µ := sup{ µ(a) : A Σ}. (µ 1 + µ 2 )(A) µ 1 (A) + µ 2 (A) µ 1 + µ 2 holding for µ 1 and µ 2 in M b and arbitrary A Σ, it follows that µ 1 + µ 2 µ 1 + µ 2. Furthermore, it is evident that αµ = α µ for µ M b and a real α. Thus, is, clearly, a norm in M b (it is equivalent to b in Theorem 1.2.2). We prove now that sup{µ 1, µ 2 } exists in M b for all µ 1 and µ 2 in M b. For any A Σ, let the number ν(a) be defined by ν(a) = sup{µ 1 (B) + µ 2 (A \ B) : A B Σ}.

18 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 µ 1 + µ 2, we get sup{ ν(a) : A Σ} µ 1 + µ 2. It follows that ν is bounded and ν µ 1 + µ 2. Having shown that ν is an element of M b, we prove now that ν = sup{µ 1, µ 2 }. From the definition of ν, it is clear that ν(a) µ 1 (A) and ν(a) µ 2 (A) for every A Σ, so ν is an upper bound of µ 1 and µ 2. Let τ be another upper bound. Then, for any A Σ and A B Σ, we have so τ(a) = τ(b) + τ(a \ B) µ 1 (B) + µ 2 (A \ B), τ(a) sup{µ 1 (B) + µ 2 (A \ B) : A B Σ} = ν(a). This shows that ν = sup{µ 1, µ 2 } in M b. Note now that inf{µ 1, µ 2 } exists as well, because inf{µ 1, µ 2 } = sup{ µ 1, µ 2 }. Hence M b is a vector lattice. As a direct consequence of this we have the following theorem in the case of finite signed measures. Theorem (The Jordan decomposition theorem) If µ is a signed measure then there exist unique positive measures ν and ξ such that µ = ν ξ and, for any positive measures ν and ξ such that µ = ν ξ, the inequalities ν ν and ξ ξ 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]. We list main properties of the Banach space M b = (M b, b ) of all signed measures of bounded variation on (X, Σ). Proofs are easy and left to the reader. (i) The norm b on M b satisfies µ ν µ b ν b ( µ, ν M b ),

19 1.2. VECTOR-VALUED MEASURES 19 where µ := sup{µ, µ}. In other words, M b is a Banach lattice. (ii) The norm b is additive on the positive cone M + b of M b : µ 0, ν 0 µ + ν b = µ b + ν b. (iii) M b is Dedekind complete in the following sense. For any order bounded nonempty family {µ α } α A, the supremum sup α A µ α exists and can be evaluated by the formula sup α A µ α (B) = sup{ for any B Σ. j=k k µ αj (B j ) : B j Σ, α j A, B j = B, B i B j = if i j} j=1 j=1 (1.2.8) Exercises to Section 1.2. Exercise Prove the property of vector-valued measures given after Definition Exercise Let µ be a E-valued measure on (X, Σ). Show that µ (A) := sup{ µ(a n ) : A n Σ, A n = A, A k A p = if n=1 n=1 k p}. Exercise Let µ be a E-valued measure on (X, Σ). monotone in the following sense: Show that µ is additive and µ (A B) = µ (A) + µ (B) for any A, B Σ, A B =, and for any A, B Σ, A B. µ (A) µ (B) Exercise 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 Show that the vector-valued measure constructed in Example is of unbounded variation if α n = 1/n; and, of bounded variation if α n = 2 n.

20 20 CHAPTER 1. Exercise Let µ be a E-valued measure. Show that µ is a separable iff µ is separable. Exercise Show that the function µ from Σ to E, which is defined in the proof of Theorem by µ(a) = µ n (A) ( A Σ), n=1 is an E-valued measure of bounded variation. Exercise Let µ be a non-atomic signed measure of bounded variation on Σ (i.e. for any A Σ, µ(a) 0, there exists B Σ, B A, such that 0 < µ(b) < µ(a) ). Using Exercise 1.1.9, show that, for any α, there is A α Σ such that µ(a α ) = α. inf{µ(a) : A Σ} α sup{µ(a) : A Σ}, Exercise Let µ be a signed non-atomic measure on (Ω, Σ). Show that the set is closed in R { }. R(µ) := {µ(e) : E Σ}. Exercise Let M b be the Banach space of all finite signed measures on (X, Σ). Show that M b is a partially ordered Banach space with respect to the order given in (1.2.7). Exercise Show that M b satisfies 1.2.4(i). Exercise Show that M b satisfies 1.2.4(ii). Exercise Show that (1.2.8) defines the supremum sup α A µ α. Exercise Show that the R n -valued measure Ψ : Ω R n defined in Example has a bounded variation and m Ψ (X) µ k (X) w k. k=1 Exercise Show that the R n -valued measure Φ : Ω R n defined in Example has a bounded variation and m Φ (X) µ k (A k ) w k. Show that the range k=1 R(Φ) := {Φ(E) : E Ω} of Φ is a compact subset of R n if all µ k are non-atomic. Exercise Take the R n -valued measure Φ as in Exercise and assume that any µ k, which was used in the construction of Φ, is non-atomic on A k. Show that R(Φ) is convex in R n.

21 1.3. THE LEBESGUE INTEGRAL 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 integration 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 R n -valued cases to the reader 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 and f + (x) = max(0, f(x)), f (x) = max(0, f(x)), f (x) = f + (x) + f (x). Let A i A, i = 1,..., n, be such that µ(a i ) < for all i, and A i A j = for all i j. The function n g(x) = λ i χ Ai (x), λ i R, i=1 is called a simple function. The Lebesgue integral of a simple function g(x) is defined as n g dµ := λ i µ(a i ). 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 Suppose that µ is finite. Let f : X R be an arbitrary nonnegative bounded measurable function and let (g n ) n=1 be a sequence of simple functions which converges uniformly to f. Then the Lebesgue integral of f is f dµ := lim g n dµ. X n X

22 22 CHAPTER 1. It can be easily shown that the limit in Definition exists and does not depend on the choice of a sequence (g n ) n=1, and moreover, the sequence (g n ) n=1 can be chosen such that 0 g n f for all n. Definition Let f : X R be an arbitrary nonnegative bounded measurable function. Then the Lebesgue integral of f is f dµ := sup { f dµ : A A, µ(a) < } X A Remark that the existence of the supremum in Definition is guarantied by 1.1.3(d). Definition Let f : X R be a nonnegative unbounded measurable. Put { f(x) if 0 f(x) M, f M (x) = M if M < f(x). Then the Lebesgue integral of f is f dµ := lim M X X f M dµ. Definition Let f : X R be a measurable function. Then the Lebesgue integral of f is defined by f dµ := f + dµ f dµ. X X If both of these terms are finite then the function f is called integrable. In this case we write f L 1. X Elementary properties of the Lebesgue integral. We will use the following notation. For any A A: f dµ := f χ A dµ. A X Let us give several obvious properties of Lebesgue integral.

23 1.3. THE LEBESGUE INTEGRAL 23 Lemma Let f : X R be an arbitrary nonnegative measurable function then 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 definition of f dµ there exists a sequence (see the remark after Definition X 1.3.1) (g n ) n=1 of simple functions such that 0 g n f for all n, and (g n ) converges uniformly to f, and satisfies f dµ = lim g n dµ = sup g n dµ. X n X n X Now the using of Definitions and completes the proof. L1. If f, g : X R are measurable, g is integrable, and f(x) g(x), then f is integrable and f dµ g dµ. X X L2. f dµ = 0 if and only if f(x) = 0 a.e. X L3. If f 1, f 2 : X R are integrable then, for λ 1, λ 2 R, the linear combination λ 1 f 1 + λ 2 f 2 is integrable and [λ 1 f 1 + λ 2 f 2 ] dµ = λ 1 f 1 dµ + λ 2 f 2 dµ. X If f is integrable function, we write f L 1 = L 1 (X, A, µ). L4. Let f L 1, then the formula ν(a) := defines a signed measure on the σ-algebra A. X A f dµ X Convergence. If (f n ) n=1 is a sequence of integrable functions on a set X, the statement f n 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 L 1 -convergence.

24 24 CHAPTER 1. Definition We say that a sequence (f n ) n=1 of integrable functions L 1 - converges to f (or converges in L 1 ) if f n 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 L 1 -convergence. The reader should keep in mind the following examples: (i) f n = n 1 χ (0,n) ; (ii) f n = χ (n,n+1) ; (iii) f n = nχ [0,1/n] ; (iv) f 1 = χ [0,1], f 2 = χ [0,1/2], f 3 = χ [1/2,1], f 4 = χ [0,1/4], f 5 = χ [1/4,1/2], and, in general, f n = χ [j/2 k,(j+1)/2 ], where n = 2 k + j with 0 j < 2 k. k In (i), (ii), and (iii), f n 0 uniformly, pointwise, and a.e., respectively, but f n 0 in L 1 (in fact, f n dµ = f n dµ = 1 for all n). In (iv), f n 0 in L 1 since f n dµ = 2 k for 2 k n < 2 k+1, but f n (x) does not converge for any x [0, 1], since there are infinitely many n for which f n (x) = 0 and infinitely many, for which f n (x) = 1. On the other hand, if f n f a.e. and f n g L 1 for all n, then f n f in L 1. This will be clear from the dominated convergence theorem (see Theorem below) since f n f 2g. Also, we will see below that if f n f in L 1 then some subsequence converges to f a.e The monotone convergence theorem. Now we state one of the most important results about convergence. Theorem (The monotone convergence theorem) If (f n ) n=1 is a sequence in L + 1 such that f j f j+1 for all j and f = sup n f n then f dµ = lim f n dµ. n X X Proof: The limit of the increasing sequence ( f X n dµ) n=1 (finite or infinite) exists. Moreover by (L1), f X n dµ f dµ for all n, so X f n dµ f dµ. lim n X X

25 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 E n = {x : f n (x) αφ(x)}. Then (E n ) n=1 is an increasing sequence of measurable sets whose union is X, and we have f n dµ X f n dµ α E n φ dµ E n ( n 1). By (L4) and by Proposition 1.1.5, lim E n φ dµ = φ dµ, and hence X lim f n dµ α φ dµ. n X X n Since this is true for all α, 0 < α < 1, it remains true for α = 1: f n dµ φ dµ. lim n X Using Lemma 1.3.1, we may take the supremum over all simple functions φ, 0 φ f. Thus lim f n dµ f dµ. n X X Proofs of the following two corollaries of Theorem are left to the reader. Corollary If (f n ) n=1 is a sequence in L + 1 and f = n=1 f n pointwise then f dµ = f n dµ. n X Corollary If (f n ) n=1 is a sequence in L + 1, f L + 1, and f n f a.e., then f n dµ f dµ The Fatou lemma. The condition in the previous subsection that the sequence (f n ) n=1 is increasing can be omitted in the following way. Lemma (Fatou s lemma) If (f n ) n=1 is any sequence in L + 1 lim inf f n dµ lim inf f n dµ. n n then

26 26 CHAPTER 1. Proof: For each k 1, we have inf n k f n f j ( j k), hence hence inf f n dµ n k f j dµ inf f n dµ inf n k j k ( j k), f j dµ. Now let k and apply Theorem 1.3.1: lim inf f n dµ = lim inf f n dµ lim inf n k n k n f n dµ. Corollary If (f n ) n=1 is a sequence in L + 1, f L + 1, and f n f a.e., then f dµ lim inf f n dµ. n The dominated convergence theorem. Now we present the following convergence theorem, which has many applications in PDEs, functional analysis, operator theory, etc. Theorem (The dominated convergence theorem) Let f and g be measurable, let f n be measurable for any n such that f n (x) g(x) a.e., and f n f a.e. If g is integrable then f and f n are also integrable and f n dµ = f dµ. lim n Proof: f is measurable by Exercise and, since f g a.e., we have f L 1. We have that g + f n 0 a.e. and g f n 0 so, by Fatou s lemma, g dµ + f dµ lim inf (g + f n ) dµ = g dµ + lim inf f n dµ, n n

27 1.3. THE LEBESGUE INTEGRAL 27 Therefore g dµ f dµ lim inf n lim inf n and the result follows. f n dµ (g f n ) dµ = f dµ lim sup n g dµ lim sup n f n dµ, f n dµ ! Convergence in measure. Another mode of convergence, that is frequently useful, is convergence in measure. We say that a sequence (f n ) n=1 of measurable functions on (X, M, µ) is Cauchy in measure if, for every ε > 0, µ({x : f n (x) f m (x) ε}) 0 as m, n, and that (f n ) n=1 converges in measure to f if, for every ε > 0, µ({x : f n (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 If f n f in L 1 then f n f in measure. Proof: Let E n,ε = {x : f n (x) f(x) ε}. Then f n f dµ f n f dµ εµ(e n,ε ), E n,ε so µ(e n,ε ) ε 1 f n f dµ 0. The converse of Proposition is false, as Examples (i) and (iii) show. Theorem Suppose that (f n ) n=1 is Cauchy in measure. Then there is a measurable function f such that f n f in measure, and there is a subsequence (f nj ) j that converges to f a.e. Moreover, if f n g in measure then g = f a.e.

28 28 CHAPTER 1. Proof: We can choose a subsequence (g j ) j = (f nj ) j of (f n ) n=1 such that if E j = {x : g j (x) g j+1 (x) 2 j } then µ(e j ) 2 j. If F k = j=k E j x F k we have for i j k then µ(f k ) k=1 2 j = 2 1 k, and if i 1 i 1 g j (x) g i (x) g l+1 (x) g l (x) 2 l 2 1 j. (1.3.1) l=j Thus (g j ) j is pointwise Cauchy on F c k. Let F = k=1 l=j F k = lim sup E j. j Then µ(f ) = 0, and if we set f(x) = lim g j (x) for x F, and f(x) = 0 for x F, then f is measurable and g j f a.e. By (1.3.1), we have that g j (x) f(x) 2 1 j for x F k and j k. Since µ(f k ) 0 as k, it follows that g j f in measure, because {x : f n (x) f(x) ε} {x : f n (x) g j (x) 1 2 ε} {x : g j(x) f(x) 1 2 ε}, and the sets on the right both have small measure when n and j are large. Likewise, if f n g in measure {x : f(x) g(x) ε} {x : f(x) f n (x) 1 2 ε} {x : f n(x) g(x) 1 2 ε} for all n, hence µ({x : f(x) g(x) ε}) = 0 for all ε > 0, and f = g a.e. Theorem Let f n f in L 1, then there is a subsequence (f nk ) k such that f nk f a.e. Proof: Let E n,ε = {x : f n (x) f(x) ε}.

29 1.3. THE LEBESGUE INTEGRAL 29 Then f n f dµ f n f dµ εµ(e n,ε ), E n,ε so µ(e n,ε ) 0. Then, by Theorem 1.3.3, there is a subsequence (f nk ) k such that f nk f a.e Exercises to Section 1.3. Exercise Check that the limit in Definition exists and does not depend on the choice of a sequence (g n ) n=1. Exercise Prove the properties L1 L4 of the Lebesgue integral from Exercise Investigate simple properties of sequences (f n ) n=1 in (i)-(iv). Exercise 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 Show that the condition µ(x) < in Egoroff s theorem can be replaced by f n g for all n, where g L 1. Exercise Prove the dominated convergence theorem by using of the Egoroff theorem. Exercise Let f n f in measure and g n g in measure. Show that f n g n f g in measure if µ(x) <, but not necessarily if µ(x) =. Exercise Prove Corollary Exercise Prove Corollary Exercise Let f n f in measure and f n 0 for all n. Show that f dµ lim inf f n dµ. n

30 30 CHAPTER 1. Exercise Prove Corollary Exercise Let χ An f in measure. Show that f is a.e. equal to the characteristic function of a measurable set. Exercise Fix the Lebesgue measure on R. (a) Given a nonempty set Ξ L + 1 (R), show that { } 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 L 1 (R), which is, by the definition, the intersection of all convex sets containing Ξ. (b) Construct an example of a nonempty set Θ L + 1 (R) such that { } inf ξ(t) dt : ξ Θ < f(t) dt for every f co(θ). (c) Show that a nonempty set L + 1 (R) satisfying { } inf ξ(t) dt : ξ < f(t) dt for every f co( ) can not be finite. f(s)ds. Show that F is continuous. Com- Exercise Let f L 1 (R) and F (t) := t pare with Exercise

31 Chapter 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 construct 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 Outer measures. Let X be a nonempty set. An outer measure (or submeasure) on X is a function that satisfies: (a) ξ( ) = 0; ξ : P(X) [0, ] (b) ξ(a) ξ(b) if A B; ( ) (c) ξ A j ξ(a j ) for all sequences (A j ) 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 R n ) and then approximate arbitrary sets from the outside by countable unions of members of G. 31

32 32 CHAPTER 2. Proposition Let G P(X) and let ρ : G [0, ] be such that G, X G, and ρ( ) = 0. For any A X, define { } ξ(a) = ρ (A) = inf ρ(g j ) : G j G and A G j. (2.1.1) Then ξ is an outer measure. j=1 Proof: For any A X, we have A j=1 X, so ξ is well defined. Obviously ξ( ) = 0. To prove countable subadditivity, suppose {A j } j=1 P(X) and ε > 0. For each j, there exists {G k j } k=1 G such that A j k=1 Gk j and ρ(g k j ) ξ(a j ) + ε2 j. k=1 Then if A = j=1 A j, we have A j,k=1 G k j & j,k ρ(g k j ) j j=1 ξ(a j ) + ε, whence ξ(a) j=1 ξ(a j) + ε. Since ε > 0 is arbitrary, we have done The Caratheodory theorem. The principle step that leads from outer measures to measures is following. Let ξ be an outer measure on X. Definition A set A X is called ξ-measurable if Of course, the inequality ξ(b) = ξ(b A) + ξ(b (X \ A)) ( B X). (2.1.2) 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)

33 2.1. THE EXTENSION OF MEASURE 33 Theorem (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 definition of ξ-measurability of A is symmetric in A and A c := X \ A. Next, if A, B Σ and E X, ξ(e) = ξ(e A) + ξ(e A c ) = ξ(e A B) + ξ(e A B c ) + ξ(e A c B) + ξ(e A c B c ). But (A B) = (A B) (A B c ) (A c B) so, by subadditivity, and hence ξ(e A B) + ξ(e A B c ) + ξ(e A c B) ξ(e (A B)), ξ(e) ξ(e (A B)) + ξ(e (A B) c ). It follows that A B Σ, so Σ is an algebra. A B =, Moreover, if A, B Σ and ξ(a B) = ξ((a B) A) + ξ((a B) A c ) = 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 (A j ) j=1 is a sequence of disjoint sets in Σ, set Then, for any E X, B n = n A j & B = A j. j=1 j=1 ξ(e B n ) = ξ(e B n A n ) + ξ(e B n A c n) = ξ(e A n ) + ξ(e B n 1 ), so a simple induction shows that ξ(e B n ) = n j=1 ξ(e A j). Therefore ξ(e) = ξ(e B n ) + ξ(e B c n) n ξ(e A j ) + ξ(e B c ) j=1

34 34 CHAPTER 2. and, letting n, we obtain ( ) ξ(e) ξ(e A j ) + ξ(e B c ) ξ (E A j ) + ξ(e B c ) j=1 j=1 = ξ(e B) + ξ(e B c ) ξ(e). Thus the inequalities in this last calculation become equalities. It follows B Σ. Taking E = B we have ξ(b) = 1 ξ(a j), 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 A c ) = ξ(e A c ) ξ(e), so A Σ. Therefore ξ(e A) = 0 and ξ Σ is a complete measure. Combination of Proposition and Theorem gives the following corollary which is also called Caratheodory s theorem. Corollary 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 Premeasures. Our main definition in this subsection is following. Definition 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 ζ( A j ) = j=1 ζ(a j ) j=1 for any disjoint sequence (A j ) j of elements of A such that j=1 A j A. Theorem 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.

35 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). Suppose that A A, A n A, and A n=1 A n. Then B n := A (A n \ n 1 j=1 A j ) A, since A is an algebra. The sequence (B n ) n=1 is disjoint. This implies ζ(a) = ζ(b n ) n=1 ζ(a n ), 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 (A n ) n=1 in A such that n=1 Y n=1 A n & ζ(a n ) ζ (Y ) + ε. n=1 Since ζ is additive on A, Y A n=1 (A n A), and Y (X \ A) n=1 (A n (X \ A)) then ζ (Y ) + ε ζ(a n A) + ζ(a n (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. Theorem If ζ is a premeasure on an algebra A P(X) and  is the σ-algebra generated by A, then there exists a measure ˆζ on  such that ˆζ A = ζ. Moreover, if ζ is σ-finite, then ˆζ is unique extension of ζ.

36 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 ˆζ on Â, which extends ζ from A, is unique if the premeasure ζ is σ-finite 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 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) 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 generalization. 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). b > a, (, b] = (, a] (a, b], so µ((a, b]) = F (b) F (a). Moreover, if 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.

37 2.1. THE EXTENSION OF MEASURE 37 Lemma Given an increasing and right continuous function F : R R, if (a j, b j ] (j = 1,..., n) are disjoint h-intervals, let ( n ) µ 0 (a j, b j ] = and let µ 0 ( ) = 0. Then µ 0 is a premeasure. 1 n [F (b j ) F (a j )], Theorem If F : R R is any increasing, right continuous function, there is a unique Borel measure µ F on R such that µ F ((a, b]) = F (b) F (a) ( a, b). If G is another such function then µ F = µ G 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 = µ F. Proof: Each F induces a premeasure on B(R) by Lemma It is clear that F and G induce the same premeasure iff F G is constant, and that these premeasures are σ-finite (since R = (j, j + 1]). The first two assertions follow now from Exercise 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 = µ F on A, and hence µ = µ F on B(R) (accordingly to Exercise ). Lebesgue Stieltjes measures possess some important and useful regularity properties. Let us fix a complete Lebesgue Stieltjes measure µ on R associated to an increasing, right continuous function F. We denote by Σ µ the Lebesgue algebra correspondent to µ. Thus, for any E Σ µ, { } µ(e) = inf [F (b j ) F (a j )] : E (a j, b j ] j=1 1 j=1

38 38 CHAPTER 2. { } = inf µ((a j, b j ]) : E (a j, b j ]. j=1 Since B(R) Σ µ, we may replace in the second formula for µ(e) h-intervals by open intervals, namely. Lemma For any E Σ µ, { } µ(e) = inf µ((a j, b j )) : E (a j, b j ). j=1 Theorem If E Σ µ then Proof: j=1 j=1 µ(e) = inf{µ(u) : U E and U is open} = sup{µ(k) : K E and K is compact}. By Lemma 2.1.2, for any ε > 0, there exist intervals (a j, b j ) such that E (a j, b j ) & µ(e) µ((a j, b j )) + ε. j=1 If U = j=1 (a j, b j ) 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, Ē \ E U, such that µ(u) µ(ē \ E) + ε. Let K = Ē \ U. Then K is compact, K E, and j=1 µ(k) = µ(e) µ(e U) = µ(e) [µ(u) µ(u \ E)] µ(e) µ(u) + µ(ē \ E) µ(e) ε. If E is unbounded, let E j = E (j, j + 1]. By the preceding argument, for any ε > 0, there exist a compact K j E j with µ(k j ) µ(e j ) ε2 j. Let H n = j=n j= n K j. Then H n is compact, H n E, and µ(h n ) µ( j=n j= n E j ) ε. Since µ(e) = lim n µ( j=n j= n E j), the result follows. The proofs of the following two theorems are left to the reader as exercises.