13.5. THE DUAL SPACE OF C0 (X) 331

It remains to turn the inequality around. Let π (Ω) be a partition. Then

∑A∈π(Ω)

|ν (A)|= ∑A∈π(Ω)

sgn(ν (A))ν (A)≡∫

f dν

where sgn(ν (A)) is defined to be a complex number of modulus 1 such that

sgn(ν (A))ν (A) = |ν (A)|

andf (ω) = ∑

A∈π(Ω)

sgn(ν (A))XA (ω) .

Therefore, choosing π (Ω) suitably, since ∥ f∥∞≤ 1,

∥ν∥− ε = |ν |(Ω)− ε ≤ ∑A∈π(Ω)

|ν (A)|= Tν ( f )

= |Tν ( f )|= |θ (ν)( f )| ≤ ∥θ (ν)∥ ≤ ∥ν∥

Thus θ preserves norms. Hence it is one to one also. Why is θ onto?Let Λ ∈ (L∞ (Ω; µ))′ . Then define

ν (E)≡ Λ(XE) (13.6)

This is obviously finitely additive because Λ is linear. Also, if µ (E) = 0, then XE = 0 inL∞ and so Λ(XE) = 0. If π (Ω) is any partition of Ω, then

∑A∈π(Ω)

|ν (A)| = ∑A∈π(Ω)

|Λ(XA)|= ∑A∈π(Ω)

sgn(Λ(XA))Λ(XA)

= Λ

(∑

A∈π(Ω)

sgn(Λ(XA))XA

)≤ ∥Λ∥

and so ∥ν∥ ≤ ∥Λ∥ showing that ν ∈ BV (Ω; µ). Also from 13.6, if s = ∑nk=1 ckXEk is a

simple function,∫sdν =

n

∑k=1

ckν (Ek) =n

∑k=1

ckΛ(XEk

)= Λ

(n

∑k=1

ckXEk

)= Λ(s)

Then letting f ∈ L∞ (Ω; µ) , there exists a sequence of simple functions converging to funiformly off a set of µ measure zero and so passing to a limit in the above with s replacedwith sn it follows that Λ( f ) =

∫f dν and so θ is onto. ■

13.5 The Dual Space of C0 (X)

Consider the dual space of C0(X) where X is a Polish space in which the balls have compactclosure. It will turn out to be a space of measures. To show this, the following lemma willbe convenient. Recall C0 (X) is defined as follows.

Definition 13.5.1 f ∈C0 (X) means that for every ε > 0 there exists a compact setK such that | f (x)|< ε whenever x /∈ K. Recall the norm on this space is

∥ f∥∞≡ ∥ f∥ ≡ sup{| f (x)| : x ∈ X}