Signed Measures. Chapter Basic Properties of Signed Measures. 4.2 Jordan and Hahn Decompositions

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1 Chapter 4 Signed Measures Up until now our measures have always assumed values that were greater than or equal to 0. In this chapter we will extend our definition to allow for both positive negative values. 4.1 Basic Properties of Signed Measures Definition Let (X, A) be a measurable space. A signed measure on (X, A) is a function µ : A R such that 1) µ takes on at most one of the values or. 2) µ( ) = 0 3) If { n } A is a sequence of pairwise disjoint sets, then µ( n ) = µ( n ). xample ) Assume that (X, A) is a measurable space that µ 1, µ 2 are measures on (X, A). If at least one of the µ i s is finite, then µ = µ 1 µ 2 is a signed measure on (X, A). 2) Let (X, A, µ) be a measure space let f be an integrable function. Then if f = f + f, we have that λ() = f dµ = f + dµ f dµ is a signed measure on (X, A). 4.2 Jordan Hahn Decompositions Problem We have just seen that if (X, A) is a measurable space µ 1, µ 2 are measures on (X, A) with at least one of the µ i s being finite, then µ = µ 1 µ 2 is a signed measure on (X, A). We can now ask: do all signed measures arise in this way? 61

2 The clue to answering the problem above comes from the second item in our previous example where we decomposed our integral function f into its positive negative parts. This decomposition, often called the Jordan decomposition of f, can in fact be extended to signed measures, something which we shall see shortly. We begin with the following natural definitions: Definition Let µ be a signed measure on (X, A). Let P, N, M A. Then we say that: 1) P is positive if µ( P ) 0 for all A. 2) N is negative if µ( N) 0 for all A. 3) M is null if µ( M) = 0 for all A. The following useful results follow almost immediately from the definitions above. As such the proofs will be left as exercises. Proposition Let µ be a signed measure on (X, A). 1) If P A is positive (negative) [null] if A is such that P, then is positive (negative) [null]. 2) Let { n } A, with each n being positive (negative) [null], then = (negative) [null]. 3) Let P be positive N be negative, then P N is null. n is also positive Lemma Let µ be a signed measure on (X, A). Let A be such that 0 < µ() <. Then there exists a positive set A A with A µ(a) > 0. Proof. If is positive we are done. So we may assume without loss of generality that is not positive. Let n 1 N be the smallest natural number such that there exists an 1 with µ( 1 ) < 1 n 1. If \ 1 is positive we are done. If not then we choose the smallest natural number n 2 N such that there exists an 2 \ 1 with µ( 2 ) < 1 n 2. Suppose that we have chosen { 1, 2,..., k 1 } natural numbers {n 1, n 2,..., n k } as above that \ k 1 j is not positive, then we choose the smallest natural number n k N such that there exists an j=1 k \ k 1 j with j=1 µ( k ) < 1 n k. Now either this process terminates, in which case we are done, or we are able to inductively construct a seqeunce { k } as above. In the latter case, let A = \ k. k=1 62

3 Since { k } is pairwise disjoint, each k is disjoint from A, we have that µ() = µ(a) + µ( k ) k=1 ( ) From ( ) we may conclude the following: 1) Since µ() is finite, the series µ( k ) converges. k=1 2) Since the series µ( k ) converges, so does the series k=1 3) Since each µ( k ) < 0, we have k=1 0 < µ() < µ(a). 1 n k, hence n k. Assume that there exists an 0 A with µ( 0 ) < 0. Then we can find a k large enough so that But we have that µ( 0 ) < 1 n k 1 k 1 0 A \ so this would contradict how we went about choosing n k. It follows that A is positive. j=1 j Theorem [Hahn Decomposition Theorem] Let µ be a signed measure on (X, A). The there is a positive set A A a negative set B A so that X = A B A B =. Proof. Since µ cannot take on both ± we may assume without loss of generality that µ never takes on the value. Let α = sup{µ() A, is positive}. Since is positive, we have that α 0. We can also find a sequence of positve sets {A i } so that Next let α = lim i µ(a i ). A = Then A is also positive since A i A for each i we have A i µ(a) = α. Let B = X \ A. Assume that B is not negative. Then there is an B with µ() > 0. From the previous lemma we can find a positive set A 0 B with µ(a 0 ) > 0. But then if A = A A 0 we have that A is also positive since A A 0 =, which is impossible. µ(a ) = µ(a) + µ(a 0 ) > α 63

4 Definition Let µ be a signed measure on (X, A). A pair {P, N} of elements in A for which P is positive N is negative, P N = X P N = is called a Hahn decomposition of X with respect to µ. Remark It is generally the case that the Hahn decomposition is not unique. In fact, let X = [0, 1] let A = P(X). If µ 1 is the point mass at 1 2 2, then if P = { 1 2 } N = [0, 1] \ { 1 2 }, then {P, N} is a Hahn decomposition of [0, 1] with respect to µ. However, P 1 = [0, 1 2 ] N 1 = ( 1 2, 1] is also a Hahn decomposition. In fact, if {P, N} is a Hahn decomposition of X with respect to µ if M A is null, then {P M, N \M} is a Hahn decomposition of X with respect to µ. Furthermore, if {P 1, N 2 } {P 2, N 2 } are Hahn decompositions of X with respect to µ, then M = P 1 P 2 = (P 1 N 2 ) (N 1 P 2 ) = N 1 N 2 is a null set. Furthermore, since P 1 \ P 2 P 1 P 2 P 2 \ P 1 P 1 P 2, it follows that for each A. Similarly, µ( P 1 ) = µ( P 1 P 2 ) = µ( P 2 ) µ( N 1 ) = µ( N 1 N 2 ) = µ( N 2 ) for each A. What is true however, as we shall see, is that every Hahn decomposition induced a decomposition of µ into the difference of two positive measures that any two Hahn decompositions induce the same decomposition of µ. Definition Let {P, N} be a Hahn decomposition of X with respect to µ. Define µ + () = µ( P ) µ () = µ( N). Remark It is easy to see that µ + µ are both positive measures that µ = µ + µ. The pair {µ +, µ } is called a Jordan decomposition of µ. Definition Two measures µ ν on (X, A) are said to be mutually singular if there are disjoint sets A, B A with X = A B µ(a) = 0 while ν(b) = 0. In this case, we write µ ν. Theorem [Jordan Decomposition Theorem] Let µ be a signed measure on (X, A). Then there exist two mutually singular positive measures µ + µ such that µ = µ + µ. Furthermore, if λ ν are any two positive measures with then for each A we have Finally, if λ ν, then λ = µ + ν = µ. µ = λ ν, λ() µ + () ν() µ (). 64

5 Proof. Let {µ +, µ } be a Jordan decomposition of µ arising from the Hahn decomposition {P, N}. Then clearly µ = µ + µ. Moreover, since we have that Let λ ν be any two positive measures with let A. Then µ + (N) = µ(p N) = µ (P ) = 0 µ + µ. µ = λ ν µ + () = µ(p ) = λ(p ) ν(p ) λ(p ) λ() A similar argument shows that ν() µ (). Finally, assume that λ ν. Let {A, B} be a partition of X so that λ(b) = 0 ν(a) = 0. For each A we have µ( A) = λ( A) ν( A) = λ( A) 0. That is A is positive. Similarly B is negative, so {A, B} is a Hahn decomposition. It follows that for an A, µ + () = µ( P ) = µ( A) = λ( A) = λ() µ () = µ( N) = µ( B) = ν( B) = ν(). Definition Let µ be a signed measure on (X, A). Then if {µ +, µ } is the Jordan decomposition of µ, then the measure µ def = µ + + µ is called the total variation of µ. xample Let (X, A, µ) be a measure space let f L(X, A, µ) be integrable. If λ() = f dµ then λ is a signed measure with Jordan decomposition λ + () = In particular, λ () = λ () = f + dµ f dµ. f + dµ + f dµ = f dµ. 65

6 Definition If µ ν are signed measures on (X, A), then we say that ν is absolutely continuous with respect to µ write ν µ if ν µ. We say that µ ν are mutually singular write if µ ν. µ ν Remark ) If ν is a signed measure µ is a positive measure on (X, A), then we claim that ν µ if only if whenever A with µ() = 0, we have ν() = 0. Clearly if ν µ µ() = 0, we have ν() = 0. To prove the converse assume that {P, N} is a Hahn-decomposition for ν. Let A be such that m() = 0. Then clearly It follows from our assumption that m( P ) = 0 = m( N). ν + () = ν( P ) = 0 ν () = ν( N) = 0. Therefore, we have that ν () = 0 ν µ as desired. 2) If ν is a signed measure µ is a positive measure on (X, A), then we claim that ν µ if only if there are disjoint sets A, B A with X = A B µ(a) = 0 while ν() = 0 for any B, A. If ν µ, then there are disjoint sets A, B A with X = A B µ(a) = 0 while ν (B) = 0. But then if B, A, we have ν () = 0 hence ν() = 0. To see that the converse holds let A, B A be disjoint sets with X = A B µ(a) = 0 while ν() = 0 for any B, A. Let {P, N} be a Hahn decomposition for ν. Let 1 = B P 2 = B n. Then µ + (B) = µ( 1 ) = 0 It follows that ν (B) = 0 that ν µ. µ (B) = µ( 2 ) = 0. The following proposition gives us an alternative characterization of absolute continuity for finite positive measures which will prove useful later. Proposition Let λ µ be finite positive measures on (X, A). Then the following are equivalent: 1) λ µ 2) For every ɛ > 0 there exist a δ > 0 such that if A µ() < δ, then λ() < ɛ. Proof. 1) 2). Suppose that 2 fails. Then we can find an ɛ 0 > 0 a sequence of sets { k } A such that µ( k ) < 1 but λ( 2 k k ) ɛ 0. Now let F n = k. k=n 66

7 It follows that µ(f n ) < 1 2 n 1 while λ(f n ) ɛ 0. From here, since {F n } is decreasing, we get that while µ( λ( F n ) = lim n µ(f n) = 0, F n ) = lim n λ(f n) ɛ 0. This shows that λ is not absolutely continuous with respect to µ. 2) 1). Suppose 2) holds µ() = 0. Then for any ɛ > 0 we have that µ() < δ where δ > 0 is chosen as in 2). It follows that λ() < ɛ. But since ɛ > 0 was arbitrary, we get that λ() = Radon-Nikodym Theorem Remark We had previously asked about when, given a measure space (X, A, µ), any measure λ on A, does there exists an f M + (X, A) with the property that for every A, λ() = f dµ for all A. We have also seen that if such an f exists then it must be the case that λ µ. At this point we are in a position to use what we have learned about decompositions of signed measures to establish the converse of this result in the case of σ-finite measures. Theorem [Radon-Nikodym Theorem] Let λ µ be σ-finite measures on (X, A). Suppose that λ is absloutely continuous with respect to µ. Then there exists f M + (X, A) such that λ() = f dµ for every A. Moreover f is uniquely determined µ-almost everywhere. Proof. Case 1: We assume that λ, µ are finite. For each c > 0 let {P (c), N(c)} be an Hahn decomposition for the signed measure λ cµ. Let A 1 = N(c) for each k N let A k+1 = N((k + 1)c) \ It follows that {A i } is pairwise disjoint k N(ic) = Consequently, we have k A i. k A i. k 1 A k = N(kc) P (ic). If A A k, then N(kc) P ((k 1)c). As such we have (k 1)cµ() λ() kcµ(). ( ) 67

8 Next, let Since B P (kc) for all k N, we get that B = X \ A i = X \ N(ic) = P (ic). 0 kcµ(b) λ(b) λ(x) < for each k N. Therefore, µ(b) = 0 since λ µ we have that λ(b) = 0, as well. We now define for each c > 0, { (k 1)c if x A k, f c (x) = 0 if x B. For each A, we have = ( B) ( ( A k )). Applying ( ) to each of the component pieces above, we have that f c dµ λ() (f c + c) dµ f c dµ + cµ(x). Now for each n N let We get g n dµ λ() g n = f 1 2 n. If we let m n then ( ) tells us that g n dµ λ() g m dµ + µ(x) 2 m Combining these two give us that g n dµ + µ(x) 2 n ( ). g n dµ g m dµ µ(x) 2 n g m dµ λ() g n dµ + µ(x) 2 n. for each A. In particular this holds for 1 = {x X g n g m 0} 2 = {x X g n g m < 0}. This allows us to deduce that g n g m dµ 2µ(X) 2 n = µ(x) 2 n 1 X hence that {g n } is Cauchy in L 1 (X, A, µ). Assume that g n f in L 1 (X, A, µ). Since g n M + (X, A) we can also assume that f M + (X, A). Moreover, for any A we have g n dµ f dµ g n f dµ g n f 1 0. It then follows from ( ) that λ() = lim g n dµ = f dµ. n Suppose now that f, h M + (X, A) are such that f dµ = λ() = h dµ 68

9 for all A. Let 1 = {x X f(x) > h(x)} 2 = {x X f(x) < h(x)}. Since f h dµ = 1 f dµ 1 h dµ = λ( 1 ) λ( 1 ) = 0 1 f h dµ = 2 f dµ 2 h dµ = λ( 2 ) λ( 2 ) = 0 2 we have that µ( 1 ) = µ( 2 ) = 0 hence that f = h µ-a.e. Case 2: Asume that λ µ are σ-finite. Let {X n } A be an increasing sequence such that X = λ(x n ) < µ(x n ) <. Following the first case, we get for each n N a function f n M + (X, A) such that f n X 0, if A with X c n, then n λ() = f n dµ. If m n, then X n X m, by our previous uniqueness result, f n = f m µ-a.e. Let F n = sup{f 1, f 2,..., f n }. Then {F n } is an increasing sequence of positive measurable functions with F n = f n µ-a.e F n (x) = 0 for all x X c n. Let f = lim n F n. If A, then λ( X n ) = f n dµ = F n dµ. Given that X n, continuity from below the Monotone Convergence Theorem shows us that λ() = lim λ( X n) = lim F n dµ = f dµ. n n The uniqueness of f is determined as in the finite case. X n, Remark A close look at the proof of the Radon-Nikodym Theorem shows that the construction of the function f resembles differentiation. The next example shows that in fact this is not simply a coincidence. xample Let F : R R be a continuously differentiable function with F (x) > 0. Then F is strictly increasing. Let µ F be the Lebesgue-Stieltjes measure on (R, B(R)) generated by F. Then µ F is σ-finite µ F m. Moreover, the Fundamental Theorem of Calculus shows that µ F ((a, b]) = F (b) F (a) = b From here we can deduce that if B(R), then µ F () = a F (x) dx = F dm. [a,b] F dm = (a,b] F dm. In particular, the function we would have obtained in via the Radon-Nikodym Theorem is m-a.e equal to F. Motivated by the previous example, we have the following definition. 69

10 Definition The function f whose existence was established in the Radon-Nikodym Theorem is called the Radon-Nikodym derivative of λ with respect to µ it is denoted by dλ dµ. Remark ) dλ dµ is integrable if only if λ is a finite measure. 2) In the case that λ is σ- finite, with X = that X n each X n being such that λ(x n ) <, we have λ(x n ) = X n dλ dµ dµ so dλ dµ must be finite µ-a.e on X n. As such, we may assume that dλ dµ is actually finite everywhere on X. 3) Let (X, A, λ) = (R, M(R), m) be the ususal Lebesgue measure space. We can define µ on (R, M(R)) to be the restriction of the counting measure on (R, P(R)) to (R, M(R)). Then m µ, since µ() = 0 implies =. However, there is no f M + (R, M(R)) such that m() = f dµ. This shows that the Radon-Nikodym Theorem can fail if µ is not σ-finite. However, we leave it as an exercise to show that the Radon-Nikodym Theorem can be extended to the case where λ is arbitrary. 4) Let µ, λ, ν be σ-finite measures on (X, A) with λ µ ν µ. (i) If c > 0, then cλ µ (ii) We have (λ + ν) µ (iii) If ν λ λ µ, then ν µ (iv) If µ λ λ µ, then d(cλ) dµ = c dλ dµ. d(λ + ν) dµ = dλ dµ + dν dµ. dν dµ = dν dλ dλ dµ. dν dµ = 1. dµ dν Note: All of the equalities above apply almost everywhere. The proofs are left as an exercise. 5) If λ is a signed measure µ is a σ-finite measure with λ µ, then λ + µ λ µ. In this case, we let dλ dµ def = dλ+ dµ dλ dµ. As an application of the Radon-Nikodym Theorem we present the following fundamental Decomposition Theorem. Theorem [Lebesgue Decomposition Theorem] Let λ µ be σ-finite measures on (X, A). Then there exists two measures λ 1 λ 2 on (X, A) such that λ = λ 1 + λ 2, λ 1 µ λ 2 µ. Moreover, these measures are unique. 70

11 Proof. Let ν = λ + µ. Then clearly ν is σ-finite, λ ν µ ν. It follows that there are functions f, g M + (X, A) such that λ() = f dν µ() = g dν. for every A. Let Then {A, B} is a partition of X. Let for every A. Clearly λ = λ 1 + λ 2. Since A = {x X g(x) = 0} B = {x X g(x) > 0}. λ 1 () = λ( A) λ 1 () = λ( B) µ(a) = µ() = we have that λ 1 µ. If µ() = 0, then g dν = 0 so g(x) = 0 for ν-almost every in. It follows that ν( B) = 0 hence that λ 2 () = λ( B) = 0 since λ ν. That is λ 2 µ. To see that λ 1 λ 2 are unique we first assume that both λ µ are finite. Assume also that we can find λ 1 λ 2 ν 1 ν 2 with λ = λ 1 + λ 2, λ 1 µ λ 2 µ, λ = ν 1 + ν 2, ν 1 µ ν 2 µ. Then g dν γ = λ 1 ν 1 = ν 2 λ 2. In addition γ is such that γ µ γ µ hence γ = 0. (It is an easy exercise to verify this claim). The case where λ µ are σ-finite is left as an exercise. 71