334 CHAPTER 13. REPRESENTATION THEOREMS

What follows is the Riesz representation theorem for C0(X)′.

Theorem 13.5.5 Let L ∈ (C0(X))′. Then there exists a finite Radon measure µ anda function σ ∈ L∞(X ,µ) such that for all f ∈C0 (X) ,

L( f ) =∫

Xf σdµ.

Furthermore, µ (X) = ∥L∥ , |σ |= 1 a.e. and if ν (E)≡∫

E σdµ then µ = |ν | .

Proof: From the above there exists a unique Radon measure µ such that for all f ∈Cc (X) ,Λ f =

∫X f dµ . Then for f ∈Cc (X) ,

|L f | ≤ λ (| f |) = Λ(| f |) =∫

X| f |dµ = ∥ f∥L1(µ).

Since µ is both inner and outer regular, Cc(X) is dense in L1(X ,µ). (See Theorem 9.4.2)Therefore L extends uniquely to an element of (L1(X ,µ))′, L̃. By the Riesz representationtheorem for L1 for finite measure spaces, there exists a unique σ ∈ L∞(X ,µ) such that forall f ∈ L1 (X ,µ) , L̃ f =

∫X f σdµ. In particular, for all f ∈ C0 (X) ,L f =

∫X f σdµ and it

follows from Lemma 13.5.4, µ (X) = ∥L∥.It remains to verify |σ | = 1 a.e. For any continuous f ≥ 0,Λ f ≡

∫X f dµ ≥ |L f | =

|∫

X f σdµ| . Now if E is measurable, the regularity of µ implies that there exists a sequenceof nonnegative bounded functions fn ∈Cc (X) such that fn (x)→XE (x) a.e. and in L1 (µ) .Then using the dominated convergence theorem in the above,∫

Edµ = lim

n→∞

∫X

fndµ = limn→∞

Λ( fn)≥ limn→∞|L fn|

= limn→∞

∣∣∣∣∫Xfnσdµ

∣∣∣∣= ∣∣∣∣∫Eσdµ

∣∣∣∣and so if µ (E) > 0, 1 ≥

∣∣∣ 1µ(E)

∫E σdµ

∣∣∣which shows from Theorem 13.2.8 that |σ | ≤ 1 µ

a.e. But also, from Theorem 13.2.8, if ∥ f∥∞≤ 1,

|µ|(X) = µ (X) = ∥L∥= sup∥ f∥∞≤1

∣∣∣∣∫Xf σdµ

∣∣∣∣≤ ∫X| f | |σ |dµ

≤∫

X|σ |dµ ≤

∫X

dµ = µ (X)

and so |σ | = 1 a.e. since µ (X) =∫

X |σ |dµ = µ (X) and it is known that |σ | ≤ 1. If |σ |were less than 1 on a set of positive measure, this could not hold.

It only remains to verify µ = |ν |. Recall ν (E)≡∫

E σdµ. By Theorem 13.2.8, |ν |(E) =∫E |σ |dµ =

∫E 1dµ = µ (E) and so µ = |ν | . ■

Sometimes people write∫

X f dν ≡∫

X f σd |ν | where σd |ν | is the polar decompositionof the complex measure ν . Then with this convention, the above representation is

L( f ) =∫

Xf dν , |ν |(X) = ∥L∥ .

Also note that at most one ν can represent L. If there were two of them ν i, i = 1,2, thenν1−ν2 would represent 0 and so |ν1−ν2|(X) = 0. Hence ν1 = ν2.