23.3. REPRESENTATION FOR THE DUAL SPACE OF Lp 627

If {Ei}∞i=1 is a sequence of disjoint sets of S , let Fn = ∪n

i=1Ei, F = ∪∞i=1Ei. 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. Since a similar theorem will be proved in which λ hasvalues in an infinite dimensional space, I will prove this directly without using Theorem23.2.6. Let A1, · · · ,An be a partition of Ω. |ΛXAi | = wi(ΛXAi) = Λ(wiXAi) for somewi ∈ C, |wi|= 1. Thus

n

∑i=1|λ (Ai)| =

n

∑i=1|Λ(XAi)|= Λ(

n

∑i=1

wiXAi)

≤ ∥Λ∥(∫|

n

∑i=1

wiXAi |pdµ)

1p = ∥Λ∥(

∫Ω

dµ)1p = ∥Λ∥µ(Ω)

1p.

This is because if x ∈Ω, x is contained in exactly one of the Ai and so the absolute value ofthe sum in the first integral above is equal to 1. Therefore |λ |(Ω)< ∞ because this was anarbitrary partition. with |λ | finite.

It is also clear from the definition of λ that λ ≪ µ . Therefore, by the Radon Nikodymtheorem, there exists h ∈ L1(Ω) with λ (E) =

∫E hdµ = Λ(XE). Actually h ∈ Lq and satis-

fies the other conditions above. This is shown next.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 . (23.2)

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) , |sn| ≤ | f |. This follows fromTheorem 9.1.6 on Page 239 upon breaking f up into positive and negative parts of real andcomplex parts. 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 23.2 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))