23.8. EXERCISES 643

by the Riesz representation theorem for the dual of L1 (µ) , there exists σ ∈ L∞ (µ) withL̃( f ) =

∫X f σdµ. In particular,

L( f ) =∫

Xf σdµ for f ∈C0 (X) .

It remains to verify that |σ |= 1.If E is measurable, the regularity of µ implies there exists a sequence of nonnegative

bounded functions fn ∈ Cc (X) such that fn (x)→XE (x) a.e. and in L1 (µ) . Then usingthe dominated convergence theorem,∫

Edµ = lim

n→∞

∫X

fndµ = limn→∞

Λ( fn)≥ limn→∞|L fn|= lim

n→∞

∣∣∣∣∫Xfnσdµ

∣∣∣∣= ∣∣∣∣∫Eσdµ

∣∣∣∣and so if µ (E)> 0, 1≥

∣∣∣ 1µ(E)

∫E σdµ

∣∣∣which shows from Lemma 23.2.7 that |σ | ≤ 1 a.e.

But also, choosing f1 appropriately, ∥ f1∥∞≤ 1, |L f1|+ ε > ∥L∥ = µ (X). Letting

ω (L f1) = |L f1| , |ω|= 1,

µ (X) = ∥L∥= sup∥ f∥∞≤1

|L f | ≤ |L f1|+ ε = ωL f1 + ε =∫

Xf1ωσdµ + ε

=∫

XRe( f1ωσ)dµ + ε ≤

∫X|σ |dµ + ε ≤ µ (X)+ ε

and since ε is arbitrary, µ (X)≤∫

X |σ |dµ ≤ µ (X) which requires |σ |= 1 a.e. since it wasshown to be no larger than 1 and if it is smaller than 1 on a set of positive measure, then theabove could not hold.

If ν (E)≡∫

E σdµ,by Corollary 23.2.9, |ν |(E) =∫

E |σ |dµ =∫

E 1dµ = µ (E) ■Sometimes people write L( f ) =

∫X f dν ≡

∫X f σd |ν | =

∫X f σdµ where σd |ν | is the

polar decomposition of the complex measure ν (E)≡∫

E σdµ . Then with this convention,the above representation is L( f ) =

∫X f 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, then 0 =∫

X f σd |ν1−ν2| , |σ |= 1, andso it will follow that |ν1−ν2|(X) = 0 because you could approximate σ̄ with a sequencefn and after using the dominated convergence theorem, you would get |ν1−ν2|(X) =0. Hence ν1 = ν2.

Corollary 23.7.7 Let L ∈L (C0 (X) ,C) . Then there exists a unique complex measureν such that for all f ∈C0 (X) ,L( f ) =

∫X f dν and |ν |(X) = ∥L∥.

23.8 Exercises1. Suppose µ is a vector measure having values in Rn or Cn. Can you show that |µ|

must be finite? Hint: You might define for each ei, one of the standard basis vectors,the real or complex measure, µei

given by µei(E) ≡ ei · µ (E) . Why would this

approach not yield anything for an infinite dimensional normed linear space in placeof Rn? Have a look at the proof of Theorem 23.1.3.

2. The Riesz representation theorem of the Lp spaces can be used to prove a very inter-esting inequality. Let r, p,q ∈ (1,∞) satisfy 1

r = 1p +

1q −1. Then 1

q = 1+ 1r −

1p > 1

r