606 CHAPTER 20. REPRESENTATION THEOREMS

Proof: First here is a claim.Claim: Suppose |µ|(E) = ∞. Then there are disjoint subsets of E, A and B such that

E = A∪B, |µ (A)| , |µ (B)|> 1 and |µ|(B) = ∞.Proof of the claim: From the definition of |µ| , there exists a partition of E,π (E) such

that∑

F∈π(E)|µ (F)|> 20(1+ |µ (E)|) . (20.2.6)

Here 20 is just a nice sized number. No effort is made to be delicate in this argument. Alsonote that µ (E)∈C because it is given that µ is a complex measure. Consider the followingpicture consisting of two lines in the complex plane having slopes 1 and -1 which intersectat the origin, dividing the complex plane into four closed sets, R1,R2,R3, and R4 as shown.

R1

R2

R3

R4

Let π i consist of those sets, A of π (E) for which µ (A)∈Ri. Thus, some sets, A of π (E)could be in two of the π i if µ (A) is on one of the intersecting lines. This is not important.The thing which is important is that if µ (A) ∈ R1 or R3, then

√2

2 |µ (A)| ≤ |Re(µ (A))| and

if µ (A) ∈ R2 or R4 then√

22 |µ (A)| ≤ |Im(µ (A))| and Re(z) has the same sign for z in R1

and R3 while Im(z) has the same sign for z in R2 or R4. Then by 20.2.6, it follows that forsome i,

∑F∈π i

|µ (F)|> 5(1+ |µ (E)|) . (20.2.7)

Suppose i equals 1 or 3. A similar argument using the imaginary part applies if i equals 2or 4. Then, ∣∣∣∣∣ ∑

F∈π i

µ (F)

∣∣∣∣∣ ≥∣∣∣∣∣ ∑F∈π i

Re(µ (F))

∣∣∣∣∣= ∑F∈π i

|Re(µ (F))|

≥√

22 ∑

F∈π i

|µ (F)|> 5

√2

2(1+ |µ (E)|) .

Now letting C be the union of the sets in π i,

|µ (C)|=

∣∣∣∣∣ ∑F∈π i

µ (F)

∣∣∣∣∣> 52(1+ |µ (E)|)> 1. (20.2.8)

Define D≡ E \C.