598 CHAPTER 20. REPRESENTATION THEOREMS

The plan is to show h is real and nonnegative at least a.e. Therefore, consider the set whereImh is positive.

E = {x ∈Ω : Imh(x)> 0} ,

Now let g = XE and use 20.1.1 to get

λ (E) =∫

E(Reh+ i Imh)d(µ +λ ). (20.1.2)

Since the left side of 20.1.2 is real, this shows

0 =∫

E(Imh)d(µ +λ )

≥∫

En

(Imh)d(µ +λ )

≥ 1n(µ +λ )(En)

where

En ≡{

x : Imh(x)≥ 1n

}Thus (µ +λ )(En) = 0 and since E = ∪∞

n=1En, it follows (µ +λ )(E) = 0. A similar argu-ment shows that for

E = {x ∈Ω : Imh(x)< 0},

(µ +λ )(E) = 0. Thus there is no loss of generality in assuming h is real-valued.The next task is to show h is nonnegative. This is done in the same manner as above.

Define the set where it is negative and then show this set has measure zero.Let E ≡ {x : h(x) < 0} and let En ≡ {x : h(x) < − 1

n}. Then let g = XEn . Since E =∪nEn, it follows that if (µ +λ )(E)> 0 then this is also true for (µ +λ )(En) for all n largeenough. Then from 20.1.2

λ (En) =∫

En

h d(µ +λ )≤−(1/n)(µ +λ )(En)< 0,

a contradiction. Thus it can be assumed h≥ 0.At this point the argument splits into two cases.Case Where λ ≪ µ. In this case, h < 1.Let E = [h≥ 1] and let g = XE . Then

λ (E) =∫

Eh d(µ +λ )≥ µ(E)+λ (E).

Therefore µ(E) = 0. Since λ ≪ µ , it follows that λ (E) = 0 also. Thus it can be assumed

0≤ h(x)< 1

for all x.