Signed Measures and Complex Measures

Transcription

1 Chapter 8 Signed Measures Complex Measures As opposed to the measures we have considered so far, a signed measure is allowed to take on both positive negative values. To be precise, if M is a σ-algebra of subsets of X, a signed measure ν on M is a function (8.1) ν : M [, ], allowed to assume at most one of the values ± satisfying (8.2) ν( ) = 0 ( ) (8.3) E j M disjoint sequence = ν E j = j j ν(e j ), the series j ν(e j) being absolutely convergent. A signed measure taking values in [0, ] is what we have dealt with in Chapters 2 7; sometimes we call this a positive measure. If µ 1 µ 2 are positive measures one of them is finite, then µ 1 µ 2 is a signed measure. The following result is easy to prove but useful. Proposition 8.1. If ν is a signed measure on (X, M), then for a sequence {E j } M, (8.4) E j E = ν(e) = lim j ν(e j ), (8.5) E j E, ν(e 1 ) finite = ν(e) = lim j ν(e j ). 107

2 Signed Measures Complex Measures Proof. Exercise. If ν is a signed measure on (X, M) E M, we say ν is E-positive if ν(f ) 0 for every F M, F E. Similarly we say ν is E-negative if ν(f ) 0 for every such F. We say ν is null on E if ν(f ) = 0 for all such F. Note that, if P j is a sequence of sets in M, then (8.6) ν P j -positive = ν P -positive, for P = j P j. To see this, let Q n = P n \ j n 1 P j. Then Q n P n, so ν is Q n -positive. Now, if F P j, then ν(f ) = ν(f Q n ) 0, as desired. The following key result is known as the Hahn Decomposition Theorem. Theorem 8.2. If ν is a signed measure on (X, M), then there exist P, N M such that ν is P -positive N-negative P N = X, P N =. Let us assume that ν does not take on the value +. The theorem can be proved via the following: Lemma 8.3. If ν(a) >, then there exists a measurable P A such that ν is P -positive ν(p ) ν(a). Given Lemma 8.3, we can prove Theorem 8.2 as follows. Let s = sup {ν(a) : A M}. By (8.2), s 0. Take A j M such that ν(a j ) s, ν(a j ) >. By the lemma, ν is P j -positive on a sequence of sets P j A j, such that ν(p j ) s. Set P = P j. By (8.6), ν is P -positive. Then we deduce that ν(p ) = s. In particular, under our hypothesis, s <. Set N = X \ P. We claim that ν is N-negative. If not, there is a measurable S N with ν(s) > 0, then ν(p S) = s + ν(s) > s, which is not possible. Thus we have the desired partition of X into P N. To prove Lemma 8.3, it is convenient to start with a weaker result: Lemma 8.4. If ν(a) >, then for all ε > 0, there exist measurable B A such that ν(b) ν(a) ν(e) ε, for all measurable E B. We show how Lemma 8.4 yields Lemma 8.3. We define a sequence of sets A j M inductively. Let A 1 = A. Given A j, j n 1, take A n A n 1 such that ν(a n ) ν(a n 1 ) ν(e) 1/n, for all measurable E A n. Lemma 8.4 says you can do this. Then let P = A j. Clearly ν is P -positive,

3 8. Signed Measures Complex Measures 109 (8.5) implies that ν(p ) = lim ν(a j ) ν(a). It remains to prove Lemma 8.4. We use a proof by contradiction. If Lemma 8.4 is false, then (considering B = A), we see that, for some ε > 0, there is a measurable E 1 A such that ν(e 1 ) ε. Thus ν(a \ E 1 ) ν(a) + ε. Next, considering B = A \ E 1, we have a measurable E 2 A \ E 1 such that ν(e 2 ) ε, so ν ( A\(E 1 E 2 ) ) ν(a)+2ε. Continue, producing a disjoint sequence of measurable E j A with ν(e j ) ε. Then A j = A \ (E 1 E j ) has ν(a j ) ν(a) + jε, hence, by (8.5), we must have A j F, ν(f ) = +. But we are working under the hypothesis that ν does not take on the value +, so this gives a contradiction, the proof is complete. We call the pair P, N produced by Theorem 8.2 a Hahn partition for ν. It is essentially unique, as the following result shows. Proposition 8.5. If ν, P N are as in Theorem 8.2 if also P, Ñ is a Hahn partition for ν, then ν is null on P P = N Ñ = (P Ñ) (N P ). Proof. Let E M, E P P. Write E = E 0 E 1, a disjoint union, where we take E 0 = E (P Ñ), E 1 = E (N P ). Then ν(e) = ν(e 0 ) + ν(e 1 ). But each ν(e j ) is simultaneously 0 0, so we have ν(e j ) = 0. If ν is a signed measure on (X, M) we have a Hahn partition of X into P N, we define two positive measures ν + ν by (8.7) ν + (E) = ν(p E), ν (E) = ν(n E), for E M. We have (8.8) ν = ν + ν. Now ν + is supported on P, i.e., ν + is null on X \P. Similarly ν is supported on N, so (8.9) ν + ν have disjoint supports. This is called the Hahn decomposition of the signed measure ν. By Proposition 8.5, it is unique. We can also form the positive measure (8.10) ν = ν + + ν,

4 Signed Measures Complex Measures called the total variation measure of ν. Given a signed measure ν, we can define f dν for a suitable class of measurable functions f. We can use the Hahn decomposition (8.8) to do this, setting (8.11) f dν = f dν + f dν, when ν satisfies (8.8). In particular this is well defined for f L 1 (X, ν ). Let µ be a positive measure on (X, M). We say that a signed measure ν on (X, M) is absolutely continuous with respect to µ ( write ν << µ) provided that, for E M, (8.12) µ(e) = 0 = ν(e) = 0. This definition was made in (4.45) in the case when ν is also a positive measure. It is useful to have the following. Proposition 8.6. If µ is a positive measure ν a signed measure on (X, M), with Hahn decomposition (8.8), then (8.13) ν << µ = ν + << µ ν << µ. Proof. If E M µ(e) = 0, then µ(e P ) = µ(e N) = 0, so ν + (E) = ν(e P ) = 0 ν (E) = ν(e N) = 0. We can therefore apply Theorem 4.10 to ν + ν to obtain the following extension of the Radon-Nikodym Theorem. Theorem 8.7. Let µ be a positive finite measure ν a finite signed measure on (X, M), such that ν << µ. Then there exists h L 1 (X, µ) such that (8.14) F dν = F h dµ, for all F L 1 (X, ν ). X Proof. Note that ν + (X) = ν(p ) < ν (X) = ν(n) <. Apply Theorem 4.10 to ν ±, obtaining h ± L 1 (X, µ), let h = h + h. The only difference between (4.47) (8.14) is that this time h need not be 0. X

5 8. Signed Measures Complex Measures 111 It is also useful to consider the concept of a complex measure on (X, M), which is defined to be a function (8.15) ρ : M C, satisfying (8.16) ρ( ) = 0 ( ) (8.17) E j M disjoint sequence = ρ E j = j j ρ(e j ). In this case, we do not allow ρ to assume any infinite values. It is clear that we can set (8.18) ν 0 (S) = Re ρ(s), ν 1 (S) = Im ρ(s), ν j are signed measures, which do not take on either + or as a value. We have (8.19) ρ(s) = ν 0 (S) + iν 1 (S), we can define (8.20) F dρ = F dν 0 + i F dν 1, where in turn F dν j are defined as in (8.11). There is an obvious extension of the Radon-Nikodym Theorem: if µ is a finite positive measure ν a complex measure on (X, M), such that ν << µ, then (8.14) continues to hold for some complex-valued h. Exercises 1. Write down a proof of Proposition 8.1. In Exercises 2 3, let µ ν be finite positive measures on (X, F). For τ [0, ), set ϕ τ = ν τµ. For each such τ, ϕ τ is a signed measure.

6 Signed Measures Complex Measures 2. Show that there exists a family P τ a family N τ of elements of F such that P 0 = X, for each τ [0, ), P τ, N τ is a partition of X, τ 1 < τ 2 = P τ1 P τ2, ϕ τ is P τ -positive N τ -negative, τ [0, ). Hint. To get the nesting property, make a preliminary Hahn decomposition P α, Ñ α for α Q + = Q [0, ), set P τ = { P α : α Q +, α τ}. 3. For each k Z + = {0, 1, 2,... }, set h k (x) = sup {τ 2 k Z + : x P τ }. Show that h k h M + (X) ν = hµ + ρ, ρ µ, obtaining another proof of the Lebesgue-Radon-Nikodym Theorems Suppose µ 1 µ 2 are finite positive measures on (X, M), form the signed measure ν = µ 1 µ 2. Show that ν (E) µ 1 (E) + µ 2 (E), E M. Hint. First look at E P E N, with P N as in Theorem Take ν as in Exercise 4. With f dν defined by (8.11), show that for f L 1 (X, µ 1 + µ 2 ), f dν = f dµ 1 f dµ 2. Hint. ν + + µ 2 = ν + µ Let M(X, F) denote the set of finite signed measures on (X, F). Show that this is a linear space, in fact, a Banach space, with norm (8.21) ν = ν (X). 7. Suppose ν j are finite signed measures on (X j, F j ). Show that ν 1 ν 2 is a well-defined, finite signed measure on (X 1 X 2, F 1 F 2 ), ν 1 ν 2 = ν 1 ν Let E be a Banach space. Define the notion of an E-valued measure on (X, M), develop some basic properties.