Signed Measures and Complex Measures

Chapter 8

As opposed to the measures we have considered so far, a signed measure is allowed to take on both positive and 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 ∞ and satisfying (8.2) ν(∅) = 0 and

(8.3) Ej ∈ M disjoint sequence =⇒ ν Ej = ν(Ej), [j Xj the series j ν(Ej) being absolutely convergent. A signed measure taking values in [0P, ∞] is what we have dealt with in Chapters 2–7; sometimes we call this a positive measure. If µ1 and µ2 are positive measures and one of them is ﬁnite, then µ1 − µ2 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 {Ej} ⊂ M,

(8.4) Ej % E =⇒ ν(E) = lim ν(Ej), j→∞ and

(8.5) Ej & E, ν(E1) ﬁnite =⇒ ν(E) = lim ν(Ej). j→∞

107 108 8. Signed Measures and Complex Measures

Proof. Exercise.

If ν is a signed measure on (X, M) and 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 Pj is a sequence of sets in M, then

(8.6) ν Pj-positive =⇒ ν P -positive, for P = Pj. [j

To see this, let Qn = Pn \ j≤n−1 Pj. Then Qn ⊂ Pn, so ν is Qn-positive. Now, if F ⊂ Pj, then ν(FS) = ν(F ∩ Qn) ≥ 0, as desired. The followingS key result is knoPwn 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 and N-negative and 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 and 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 Aj ∈ M such that ν(Aj) % s, ν(Aj) > −∞. By the lemma, ν is Pj-positive on a sequence of sets Pj ⊂ Aj, such that ν(Pj) → s. Set P = Pj. By (8.6), ν is P -positive. Then we deduce that ν(P ) = s. In particular,S 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, and then ν(P ∪ S) = s + ν(S) > s, which is not possible. Thus we have the desired partition of X into P and 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) and ν(E) ≥ −ε, for all measurable E ⊂ B.

We show how Lemma 8.4 yields Lemma 8.3. We deﬁne a sequence of sets Aj ∈ M inductively. Let A1 = A. Given Aj, j ≤ n−1, take An ⊂ An−1 such that ν(An) ≥ ν(An−1) and ν(E) ≥ −1/n, for all measurable E ⊂ An. Lemma 8.4 says you can do this. Then let P = Aj. Clearly ν is P -positive, T 8. Signed Measures and Complex Measures 109

and (8.5) implies that ν(P ) = lim ν(Aj) ≥ 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 E1 ⊂ A such that ν(E1) ≤ −ε. Thus ν(A \ E1) ≥ ν(A) + ε. Next, considering B = A \ E1, we have a measurable E2 ⊂ A \ E1 such that ν(E2) ≤ −ε, so ν A\(E1 ∩E2) ≥ ν(A)+2ε. Continue, producing a disjoint sequence of measurable Ej ⊂A with ν(Ej) ≤ −ε. Then Aj = A \ (E1 ∪ · · · ∪ Ej) has ν(Aj) ≥ ν(A) + jε, and hence, by (8.5), we must have

Aj & F, ν(F ) = +∞.

But we are working under the hypothesis that ν does not take on the value +∞, so this gives a contradiction, and 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 and N are as in Theorem 8.2 and if also P , N is a Hahn partition for then is null on ν, ν e e P M P = N M N = (P ∩ N) ∪ (N ∩ P ). e e e e Proof. Let E ∈ M, E ⊂ P M P . Write E = E0∪E1, a disjoint union, where we take E0 = E ∩ (P ∩ N), E1 e= E ∩ (N ∩ P ). Then ν(E) = ν(E0) + ν(E1). But each ν(Ej) is simultaneouslye ≥ 0 and ≤e 0, so we have ν(Ej) = 0. If ν is a signed measure on (X, M) and we have a Hahn partition of X into P ∪ N, we deﬁne two positive measures ν+ and 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+ and ν− have disjoint supports.

This is called the Hahn decomposition of the signed measure ν. By Propo- sition 8.5, it is unique. We can also form the positive measure

(8.10) |ν| = ν+ + ν−, 110 8. Signed Measures and 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 HahnR decomposition (8.8) to do this, setting

(8.11) f dν = f dν+ − f dν , Z Z Z − when ν satisﬁes (8.8). In particular this is well deﬁned for f ∈ L1(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 (and 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 and ν a signed measure on (X, M), with Hahn decomposition (8.8), then

(8.13) ν << µ =⇒ ν+ << µ and ν− << µ.

Proof. If E ∈ M and µ(E) = 0, then µ(E ∩ P ) = µ(E ∩ N) = 0, so ν+(E) = ν(E ∩ P ) = 0 and ν−(E) = −ν(E ∩ N) = 0.

We can therefore apply Theorem 4.10 to ν+ and ν− to obtain the following extension of the Radon-Nikodym Theorem.

Theorem 8.7. Let µ be a positive ﬁnite measure and ν a ﬁnite signed measure on (X, M), such that ν << µ. Then there exists h ∈ L1(X, µ) such that

(8.14) F dν = F h dµ, Z Z X X for all F ∈ L1(X, |ν|).

Proof. Note that ν+(X) = ν(P ) < ∞ and ν−(X) = −ν(N) < ∞. Apply 1 Theorem 4.10 to ν, obtaining h ∈ L (X, µ), and let h = h+ − h−.

The only diﬀerence between (4.47) and (8.14) is that this time h need not be ≥ 0. 8. Signed Measures and 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 and

(8.17) Ej ∈ M disjoint sequence =⇒ ρ Ej = ρ(Ej). [j Xj

In this case, we do not allow ρ to assume any “inﬁnite” values. It is clear that we can set

(8.18) ν0(S) = Re ρ(S), ν1(S) = Im ρ(S), and νj are signed measures, which do not take on either +∞ or −∞ as a value. We have

(8.19) ρ(S) = ν0(S) + iν1(S), and we can deﬁne

(8.20) F dρ = F dν0 + i F dν1, Z Z Z where in turn F dνj are deﬁned as in (8.11). There is an obvious extension of the Radon-NikR odym Theorem: if µ is a ﬁnite positive measure and 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 and ν be ﬁnite positive measures on (X, F). For τ ∈ [0, ∞), set ϕτ = ν − τµ. For each such τ, ϕτ is a signed measure. 112 8. Signed Measures and Complex Measures

2. Show that there exists a family Pτ and a family Nτ of elements of F such that P0 = X, for each τ ∈ [0, ∞), Pτ , Nτ is a partition of X,

τ1 < τ2 =⇒ Pτ1 ⊃ Pτ2 , and ϕτ is Pτ -positive and Nτ -negative, ∀ τ ∈ [0, ∞). Hint. To get the nesting property, make a preliminary Hahn decompo- + sition Pα, Nα for α ∈ Q = Q ∩ [0, ∞), and set + e e Pτ = {Pα : α ∈ Q , α ≤ τ}. \ 3. For each k ∈ Z+ = {0, 1, 2, . .e. }, set −k + hk(x) = sup {τ ∈ 2 · Z : x ∈ Pτ }. + Show that hk % h ∈ M (X) and ν = hµ + ρ, ρ ⊥ µ, obtaining another proof of the Lebesgue-Radon-Nikodym Theorems 4.10– 4.11.

4. Suppose µ1 and µ2 are ﬁnite positive measures on (X, M), and form the signed measure ν = µ1 − µ2. Show that

|ν|(E) ≤ µ1(E) + µ2(E), ∀ E ∈ M. Hint. First look at E ⊂ P and E ⊂ N, with P and N as in Theorem 8.2.

5. Take ν as in Exercise 4. With f dν deﬁned by (8.11), show that for 1 f ∈ L (X, µ1 + µ2), R

f dν = f dµ1 − f dµ2. Z Z Z

Hint. ν+ + µ2 = ν− + µ1.

6. Let M(X, F) denote the set of ﬁnite signed measures on (X, F). Show that this is a linear space and, in fact, a Banach space, with norm (8.21) kνk = |ν|(X).

7. Suppose νj are finite signed measures on (Xj, Fj). Show that ν1 × ν2 is a well-defined, finite signed measure on (X1 × X2, F1 ⊗ F2), and

kν1 × ν2k = kν1k · kν2k. 8. Let E be a Banach space. Deﬁne the notion of an E-valued measure on (X, M), and develop some basic properties.