Lecture Notes on Measure Theory and Integration 6 — Signed Measures

Joel Feldman

University of British Columbia

November 22, 2019

1 6 Signed Measures

Definition 6.1 (Signed Measures). Let X be a nonempty set and M ⊂ P(X)bea σ-algebra. (a) A signed measure on (X, M) is a function ν : M→ [−∞, ∞] such that (i) ν(∅)=0 (ii) ν assumes at most one1 of the values ±∞.

(iii) If En n∈N ⊂M is disjoint, then  ∞ ∞ ν n=1En = ν(En) n=1 S  X 2 ∞ with the sum converging absolutely if ν n=1En is finite. (b) A set E ∈M is said to be positive (negative,S null)for the signed measure ν if F ∈M, F ⊂ E =⇒ ν(F ) ≥ 0 (≤ 0, = 0) Example 6.2. (a) If µ is a measure on (X, M), then µ is a signed measure and X is a positive set for µ. (b) If µ and ν are measures on (X, M), with at least one of them finite, then µ − ν is a signed measure. (c) If µ is a measure on (X, M) and f : X → R is measurable with at least one of max{f, 0} dµ and max{−f, 0} dµ finite, then R R ν(E)= f(x) dµ(x)= max{f, 0} dµ − max{−f, 0} dµ ZE ZE ZE is a signed measure. We call f an extended µ-integrable function and we write dν(x) = f(x) dµ(x). A set E ∈ M is positive (negative, null) for ν if and only if f(x) ≥ 0 (≤ 0, = 0) almost everywhere on E. Lemma 6.3 (Continuity). Let ν be a signed measure on (X, M).

(a) If En n∈N ⊂M with E1 ⊂ E2 ⊂ E3 ⊂··· , then

 ∞ ν =1En = lim ν(En) n n→∞ 1so that we don’t have to deal with∞ − ∞  2 S ∞ so that we don’t have to worry about the order of the terms in n=1 ν(En) P Signed Measures 2 November 22, 2019 (b) If En n∈N ⊂M with E1 ⊃ E2 ⊃ E3 ⊃··· and ν(E1) is finite, then

 ∞ ν =1En = lim ν(En) n n→∞   Proof. Problem Set 10, #1. T Lemma 6.4 (Positive sets). Let ν be a signed measure on (X, M). (a) If E is a positive set for ν and F ∈M with F ⊂ E, then F is a positive set for ν. N ∞ (b) If, for each n ∈ , En is a positive set for ν, then n=1 En is a positive set for ν. S Proof. Problem Set 10, #2. Theorem 6.5 (The Hahn Decomposition Theorem). If ν is a signed measure on (X, M), then there is a positive set P ∈ M and a negative set N ∈ M for ν such that P ∪ N = X and P ∩ N = ∅. If P ′, N ′ is any other such pair of sets, then P ∆P ′ = N∆N ′ is null. Proof. Assume, without loss of generality, that ν never takes the value +∞. (Oth- erwise consider −ν.) Step 1: Choice of P : Set

m = sup ν(E) E positive ˜  Then there exists a sequence Pn n∈N of positive sets such that  lim ν P˜n = m n→∞ n ˜
Let Pn = m=1 Pn. Then P1 ⊂ P2 ⊂ P3 ⊂··· , and each Pn is positive, and since ≥0 S ν(Pn)= ν(P˜n)+ ν(Pn \ P˜n), z }| { ν(P˜n) ≤ ν(Pn) ≤ m

∞ Set P = n=1 Pn. Then P is positive, and by continuity from below,

S ν(P ) = lim ν(Pn)= m =⇒ m< ∞ n→∞ since, by hypothesis, ν never takes the value +∞.

Signed Measures 3 November 22, 2019 Step 2: Proof that N ≡ X \ P is negative: N cannot contain a positive, nonnull subset A, because, if it did, P ∪A would be a positive subset with ν(P ∪A)= ν(P )+ν(A) > ν(P ), contradicting the maximality of ν(P ). On the other hand, if N fails to be negative, there must exist a subset B ⊂ N with B ∈ M and ν(B) > 0. Now it suffices to apply the lemma (proven below) that B ∈M, 0 < ν(B) < ∞ =⇒ ∃ A ∈M with A ⊂ B and A positive, nonnull Step 3: Uniqueness up to null sets: If P ′ is positive, N ′ is negative and P ′ ∩ N ′ = ∅, P ′ ∪ N ′ = X, then P \ P ′ is positive (since P \ P ′ ⊂ P , positive) is negative (since P \ P ′ ⊂ X \ P ′ = N ′, negative) =⇒ P \ P ′ is null Similarly P ′ \ P is null. So P ∆P ′ = P ∩ P ′c ∪ P c ∩ P ′ = N c ∩ N ′ ∪ N ∩ N ′c = N∆N ′ are null.

Lemma 6.6. Let E ∈ M with 0 < ν(E) < ∞. Then there exists an A ∈ M with A ⊂ E that is positive and nonnull. Proof. ◦ If E is positive, choose A = E. 1 ◦ Otherwise choose E1 ⊂ E with E1 ∈ M and ν(E1) < − n1 where n1 is the smallest natural number for which such an E1 exists. <0

· ν(E \ E1) > 0, since ν(E1)+ν(E \ E1)= ν(E) > 0. · If E \ E1 is positive,z choose}| { A = E \ E1, and we’re done. disjoint k−1 ◦ Otherwise, inductively choose Ek ⊂ E \ j=1 Ej with Ek ∈ M and with 1 ν(Ek) < − where nk is the smallest natural number for which such an Ek nk S z}|{ exists. <0 k k k · ν E \ j=1 Ej > 0, since ν E \ j=1 Ej + j=1 ν(Ej)= ν(E) > 0. k k · IfE \ Sj=1 Ej
is positive, choose SA = E \
jP=1 Ej,z and}| { we’re done. S S Signed Measures 4 November 22, 2019 ∞ ◦ If the induction never terminates, choose A = E \ j=1 Ej. Proof that A is nonnull: S ∞ Since all of the subsets in the union E = A ∪ j=1 Ej are disjoint   <−1/n

0 < ν(E)= ν(A)+ ν(Ej) =⇒ ν(A) > 0 j=1 z }| { X z }| { ∞ Furthermore, since ν(E) is assumed to be finite, j=1 ν(Ej) converges absolutely, which forces limj→∞ nj = ∞. P Proof that A is positive: It suffices to prove that, for each ε > 0, there is no B ⊂ A with B ∈ M and ν(B) < −ε. Let ε> 0. Choose k ∈ N with 1 <ε. Then nk−1

∞ k−1

A = E \ Ej ⊂ E \ Ej j=1 j=1 [ [ By construction, nk is the smallest natural number for which there there is a B ∈ M with B ⊂ E \ k−1 E and ν(B) < − 1 . So there is no B ∈ M with j=1 j nk B ⊂ A and ν(B) < − 1 . So there is no B ∈M with B ⊂ A and ν(B) < −ε. nk−S1

Definition 6.7. Two signed measure µ, ν on (X, M) are said to be mutually sin- gular, denoted µ ⊥ ν, if there exist sets E, F ∈M such that E ∪F = X, E ∩F = ∅, E is null for µ and F is null for ν. Example 6.8 (mutually singular measures). Let X = R and M = L, the σ-algebra of Lebesgue measurable sets. Then the measures µ = m = Lebesgue measure ν = point mass or Dirac measure at 0 That is (see Example 2.15),

1 if 0 ∈ E ν(E)= (0 if 0 ∈/ E are mutually singular, because µ {0} = 0 and ν R \{0} =0

Signed Measures 5 November 22, 2019 Theorem 6.9 (The Jordan Decomposition Theorem). If ν is a signed measure on (X, M), then there are unique positive measures ν+, ν− on (X, M) such that ν+ ⊥ ν− and ν = ν+ − ν−.

Proof. Existence: Let X = P ∪ N be the Hahn decomposition of Theorem 6.5. Then

ν+(A)= ν(A ∩ P ) ν−(A)= −ν(A ∩ N)

works. Uniqueness: ′ ′ ′ ′ ′ ′ ′ ′ ′ Let ν = ν+ − ν− also work. Let P , N be such that P ∪ N = X, P ∩ N = ∅, N ′ ′ ′ is null for ν+ and P is null for ν−. Then

′ ′ ′ ′ ′ P is positive for ν (since A ⊂ P =⇒ ν(A)= ν+(A) − ν−(A)= ν+(A) ≥ 0) N ′ is negative for ν =⇒ P ′ ∪ N ′ is another Hahn decomposition for ν =⇒ P ∆P ′ is null for ν

So, if A ∈M, ⊂N ′ ′ ′ ′ ′ ′ ′ ′ ν+(A)= ν+(A ∩ P )+ ν+(A ∩ N )= ν+(A ∩ P ) ′ z ⊂}|P { ′ ′ ′ ′ ′ = ν+(A ∩ P ) − ν−(A ∩ P ) ⊂P ∆P = ν(A ∩ P ′)= ν(A ∩z P}|∩{P ′)+ ν A ∩ (P ′ \ P ) ′ = ν(A ∩ P ∩ P ) ⊂Pz }| {
= ν(A ∩ P )= ν+(A ∩ P ) − ν−(A ∩ P )= ν+(A ∩ P ) = ν+(A) z }| {

′ and, similarly, ν−(A)= ν−(A).

Signed Measures 6 November 22, 2019