612 CHAPTER 20. REPRESENTATION THEOREMS

Then by the Dominated Convergence theorem,

||XFn −XF ||p→ 0.

Therefore, by continuity of Λ,

λ (F) = Λ(XF) = limn→∞

Λ(XFn) = limn→∞

n

∑k=1

Λ(XEk) =∞

∑k=1

λ (Ek).

This shows λ is a complex measure with |λ | finite.It is also clear from the definition of λ that λ ≪ µ . Therefore, by the Radon Nikodym

theorem, there exists h ∈ L1(Ω) with

λ (E) =∫

Ehdµ = Λ(XE).

Actually h ∈ Lq and satisfies the other conditions above. Let s = ∑mi=1 ciXEi be a simple

function. Then since Λ is linear,

Λ(s) =m

∑i=1

ciΛ(XEi) =m

∑i=1

ci

∫Ei

hdµ =∫

hsdµ . (20.3.9)

Claim: If f is uniformly bounded and measurable, then

Λ( f ) =∫

h f dµ.

Proof of claim: Since f is bounded and measurable, there exists a sequence of simplefunctions, {sn} which converges to f pointwise and in Lp (Ω). This follows from Theorem11.3.9 on Page 241 upon breaking f up into positive and negative parts of real and complexparts. In fact this theorem gives uniform convergence. Then

Λ( f ) = limn→∞

Λ(sn) = limn→∞

∫hsndµ =

∫h f dµ,

the first equality holding because of continuity of Λ, the second following from 20.3.9 andthe third holding by the dominated convergence theorem.

This is a very nice formula but it still has not been shown that h ∈ Lq (Ω).Let En = {x : |h(x)| ≤ n}. Thus |hXEn | ≤ n. Then

|hXEn |q−2(hXEn) ∈ Lp(Ω).

By the claim, it follows that

||hXEn ||qq =∫

h|hXEn |q−2(hXEn)dµ = Λ(|hXEn |q−2(hXEn))

≤ ||Λ||∣∣∣∣|hXEn |q−2(hXEn)

∣∣∣∣p = ||Λ|| ||hXEn ||

qpq ,