1102 CHAPTER 32. FOURIER TRANSFORMS

Therefore,mn (Vm \Km)≤ 2−m.

Letφ m ∈Cc (Vm) , Km ≺ φ m ≺Vm.

The statement Km ≺ φ m ≺ Vm means that φ m equals 1 on Km, has compact support in Vm,maps into [0,1] , and is continuous. Then φ m (x)→XER (x) a.e. because the set whereφ m (x) fails to converge to this set is contained in the set of all x which are in infinitelymany of the sets Vm \Km. This set has measure zero because

∞

∑m=1

mn (Vm \Km)< ∞

Thus φ m converges pointwise a.e to XER and so, by the dominated convergence theorem,

0 = limm→∞

∫Rn

f φ mdx = limm→∞

∫V1

f φ mdx =∫

ER

f dx≥ rm(ER).

Thus, mn (ER) = 0 and therefore mn (E) = limR→∞ mn (ER) = 0. Since r > 0 is arbitrary, itfollows

mn ([ f > 0]) = ∪∞k=1mn

([f > k−1])

= ∪∞k=1mn

([f+ > k−1])= mn

([f+ > 0

])= 0.

Hence f+ = 0 a.e. It follows that∫

f−φdx = 0 for all φ ∈Cc (Rn) because∫f−φdx =

∫f+φ −

∫f φ = 0.

Thus from what was just shown, with f− taking the place of f , it follows | f−|+ f−

2 = 0 andso f− = 0 a.e. also.

Corollary 32.3.6 Let f ∈ L1 (Rn) and suppose∫Rn

f (x)φ (x)dx = 0

for all φ ∈ G . Then f = 0 a.e.

Proof: Let ψ ∈Cc (Rn) . Then by the Stone Weierstrass approximation theorem, thereexists a sequence of functions, {φ k} ⊆ G such that φ k→ ψ uniformly. Then by the domi-nated convergence theorem, ∫

f ψdx = limk→∞

∫f φ kdx = 0.

By Lemma 32.3.5 f = 0.The next theorem is the main result of this sort.