292 CHAPTER 12. THE CONSTRUCTION OF MEASURES

(If x ∈Vi, this follows from 12.3.14. If x /∈Vi both sides equal 0.) Therefore,

L f = L(n

∑i=1

f hi)≤ L(n

∑i=1

hi(ti + ε))

=n

∑i=1

(ti + ε)L(hi)

=n

∑i=1

(|t0|+ ti + ε)L(hi)−|t0|L

(n

∑i=1

hi

).

Now note that |t0|+ ti +ε ≥ 0 and so from the definition of µ and Lemma 12.3.5, this is nolarger than

n

∑i=1

(|t0|+ ti + ε)µ(Vi)−|t0|µ(spt( f ))

≤n

∑i=1

(|t0|+ ti + ε)(µ(Ei)+ ε/n)−|t0|µ(spt( f ))

≤ |t0|

µ(spt( f ))︷ ︸︸ ︷n

∑i=1

µ(Ei)+ |t0|ε +n

∑i=1

tiµ(Ei)+ ε(|t0|+ |b|)

n

∑i=1

tiε

n+ ε

n

∑i=1

µ(Ei)+ ε2−|t0|µ(spt( f )).

From 12.3.13 and 12.3.12, the first and last terms cancel. Therefore this is no larger than

(2|t0|+ |b|+µ(spt( f ))+ ε)ε

+n

∑i=1

ti−1µ(Ei)+ εµ(spt( f ))+n

∑i=1

(|t0|+ |b|)ε

n

≤∫

f dµ +(2|t0|+ |b|+2µ(spt( f ))+ ε)ε +(|t0|+ |b|)ε

Since ε > 0 is arbitrary,

L f ≤∫

f dµ (12.3.15)

for all f ∈ Cc(Ω), f real. Hence equality holds in 12.3.15 because L(− f ) ≤ −∫

f dµ soL( f )≥

∫f dµ . Thus L f =

∫f dµ for all f ∈Cc(Ω). Just apply the result for real functions

to the real and imaginary parts of f . This proves the Lemma.This gives the existence part of the Riesz representation theorem.It only remains to prove uniqueness. Suppose both µ1 and µ2 are measures on S

satisfying the conclusions of the theorem. Then if K is compact and V ⊇ K, let K ≺ f ≺V .Then

µ1(K)≤∫

f dµ1 = L f =∫

f dµ2 ≤ µ2(V ).