624 CHAPTER 23. REPRESENTATION THEOREMS

Since ε is arbitrary, this shows |µ|(∪n

i=1Ei)≤ ∑

ni=1 |µ|(Ei) . Thus |µ| is finitely additive.

■In the case that λ is a complex measure, it is always the case that |λ |(Ω)< ∞. First is

a lemma.

Lemma 23.2.5 Suppose λ is a real valued measure called a signed measure (Definition10.13.2). Then |λ | is a finite measure.

Proof: Suppose λ : F →R is a vector measure (signed measure by Definition 10.13.2).By the Hahn decomposition, Theorem 10.13.5 on Page 301, Ω=P∪N where P is a positiveset and N is a negative one. Then on N, −λ acts like a measure in the sense that if A ⊆ Band A,B measurable subsets of N, then −λ (A)≤−λ (B). Similarly λ is a measure on P.

∑F∈π(Ω)

|λ (F)| ≤ ∑F∈π(Ω)

(|λ (F ∩P)|+ |λ (F ∩N)|)

= ∑F∈π(Ω)

λ (F ∩P)+ ∑F∈π(Ω)

−λ (F ∩N)

= λ((∪F∈π(Ω)F

)∩P)+−λ

((∪F∈π(Ω)F

)∩N)≤ λ (P)+ |λ (N)|

It follows that |λ |(Ω)< λ (P)+ |λ (N)| and so |λ | has finite total variation. ■

Theorem 23.2.6 Suppose λ is a complex measure on (Ω,S ) where S is a σ al-gebra of subsets of Ω. Then |λ |(Ω)< ∞.

Proof: If λ is a vector measure with values in C, Reλ and Imλ have values in R. Then

∑F∈π(Ω)

|λ (F)| ≤ ∑F∈π(Ω)

|Reλ (F)|+ |Imλ (F)|

= ∑F∈π(Ω)

|Reλ (F)|+ ∑F∈π(Ω)

|Imλ (F)|

≤ |Reλ |(Ω)+ |Imλ |(Ω)< ∞

thanks to Lemma 23.2.5. ■The following corollary is about representing a complex measure λ in terms of its total

variation |λ |. It is like representing a complex number in the form reiθ . In particular, Iwant to show that in the Radon Nikodym theorem

∣∣∣ dλ

∣∣∣ = 1 a.e. First is a lemma which is

interesting for its own sake and shows∣∣∣ dλ

∣∣∣≤ 1.

Lemma 23.2.7 Suppose (Ω,S ,µ) is a measure space and f is a function in L1(Ω,µ)with the property that ∣∣∣∣∫E

f dµ

∣∣∣∣≤ µ(E)

for all E ∈S . Then | f | ≤ 1 a.e.

Proof of the lemma: Consider the following picture where B(p,r)∩B(0,1) = /0.

624 CHAPTER 23. REPRESENTATION THEOREMSSince € is arbitrary, this shows |u| (U7_,£;) < Di, |u| (Ei). Thus || is finitely additive.aIn the case that A is a complex measure, it is always the case that |A| (Q) < ce. First isa lemma.Lemma 23.2.5 Suppose A is a real valued measure called a signed measure (Definition10.13.2). Then |A| is a finite measure.Proof: Suppose 2 : ¥ — R is a vector measure (signed measure by Definition 10.13.2).By the Hahn decomposition, Theorem 10.13.5 on Page 301, €2 = PUN where P is a positiveset and N is a negative one. Then on N, —A acts like a measure in the sense that if A C Band A,B measurable subsets of N, then —A (A) < —A (B). Similarly 4 is a measure on P.yy a F)|< d. (A (FOP)|+|A (FON)|)Fen(Q) Fen(Q= ) A(FnP)+ Yo -A(FNN)Fen(Q) Fen(Q)=A ((UremayF) OP) +—A ((UrenayF) AN) <A (P)+|A (N)|It follows that |A|(Q) <A (P)+ A (N)| and so |A| has finite total variation.Theorem 23.2.6 suppose A is a complex measure on (Q,.7) where S is ao al-gebra of subsets of Q. Then |A| (Q) < ee.Proof: If A is a vector measure with values in C, ReA and ImA have values in R. ThenLW) < YP [Rea (F)|+|Ima (F)|Fen(Q) Fen(Q)YY |ReA(F)|+ Y |ImAa(F)|Fen(Q) Fen(Q)|ReA| (Q) + |ImA| (Q) <0lAthanks to Lemma 23.2.5. HiThe following corollary is about representing a complex measure A in terms of its totalvariation |A|. It is like representing a complex number in the form re’®. In particular, Iwant to show that in the Radon Nikodym theorem Fa A| — = 1 ae. First is a lemma which isinteresting for its own sake and shows | | <1.Lemma 23.2.7 Suppose (Q,.%,\) is a measure space and f is a function in L'(Q, 1)with the property that[an sueforall E € SY. Then |\f| <lae.Proof of the lemma: Consider the following picture where B(p,r) 1 B(0,1) = 9.