968 CHAPTER 26. INTEGRALS AND DERIVATIVES

By the Riesz representation theorem for positive linear functionals of this sort, there existsa unique Radon measure, µ1 such that for all f ∈Cc (Rn) ,∫

f dµ1 = L f =∫

f dµ.

Now let V be an open set and let Kk ≡{

x ∈V : dist(x,VC

)≤ 1/k

}∩B(0,k). Then {Kk} is

an incresing sequence of compact sets whose union is V. Let Kk ≺ fk ≺ V. Then fk (x)→XV (x) for every x. Therefore,

µ1 (V ) = limk→∞

∫fkdµ1 = lim

k→∞

∫fkdµ = µ (V )

and so µ = µ1 on open sets. Now if K is a compact set, let

Vk ≡ {x ∈ Rn : dist(x,K)< 1/k} .

Then Vk is an open set and ∩kVk = K. Letting K ≺ fk ≺Vk, it follows that fk (x)→XK (x)for all x ∈ Rn. Therefore, by the dominated convergence theorem with a dominating func-tion, XRn

µ1 (K) = limk→∞

∫fkdµ1 = lim

k→∞

∫fkdµ = µ (K)

and so µ and µ1 are equal on all compact sets. It follows µ = µ1 on all countable unionsof compact sets and countable intersections of open sets.

Now let E be a Borel set. By regularity of µ1, there exist sets, H and G such thatH is the countable union of an increasing sequence of compact sets, G is the countableintersection of a decreasing sequence of open sets, H ⊆ E ⊆ G, and µ1 (H) = µ1 (G) =µ1 (E) . Therefore,

µ1 (H) = µ (H)≤ µ (E)≤ µ (G) = µ1 (G) = µ1 (E) = µ1 (H) .

therefore, µ (E) = µ1 (E) and this proves the lemma.

Corollary 26.7.9 Suppose µ is a complex Borel measure defined on Rn for which thereexists a µ measurable set, N such that for all Borel sets, E, µ (E) = µ (E ∩N) wherem(N) = 0. Then

dµ

dm(x) = 0 m a.e.

Proof: Each of Re µ+,Re µ−, Im µ+, and Im µ− are real measures having values in[0,∞) and so by Lemma 26.7.8 each is a Radon measure having the same property that µ

has in terms of being supported on a set of m measure zero. Therefore, for ν equal to anyof these, dν

dm (x) = 0 m a.e. This proves the corollary.

26.8 Exercises1. Suppose A and B are sets of positive Lebesgue measure in Rn. Show that A−B must

contain B(c,ε) for some c ∈ Rn and ε > 0.

A−B≡ {a−b : a ∈ A and b ∈ B} .