19.7. MEASURES AND POSITIVE LINEAR FUNCTIONALS 519

Now note that |t0|+ ti +ε ≥ 0 and so from the definition of µ and Lemma 19.7.5, 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 )).

The first and last terms cancel. Therefore this is no larger than

(2|t0|+ |b|+µ(spt( f ))+ ε)ε +

≤∫

f dµ

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 for all f ∈ Cc(Ω), f real valued. Hence equalityholds because L(− f ) ≤ −

∫f dµ so L( 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 gives theexistence part of the Riesz representation theorem.

It only remains to prove uniqueness. Suppose both µ1 and µ2 are measures on Ssatisfying 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 ).

Thus, taking the inf for all V ⊇K, µ1(K)≤ µ2(K) for all K. Similarly, the inequality can bereversed and so it follows the two measures are equal on compact sets. By the assumptionof inner regularity on open sets, the two measures are also equal on all open sets. By outerregularity, they are equal on all sets of S . ■

Example 19.7.9 Let L( f ) =∫

∞

−∞f (t)dt for all f ∈Cc (R) where this is just the ordinary

Riemann integral. Then the resulting measure is known as one dimensional Lebesguemeasure.

Example 19.7.10 Let L( f ) =∫

∞

−∞· · ·∫

∞

−∞f (x)dx1 · · ·dxn for f ∈Cc (Rn). Then the result-

ing measure is mn, n dimensional Lebesgue measure.

Here is a nice observation.

Proposition 19.7.11 In Example 19.7.10 the order of integration is not important. Thesame functional is obtained in any order.