59.14. CONVOLUTION AND SUMS 1901

Theorem 59.14.3 Let µ be a finite measure on B (E) where E is a separable Banach spaceand let f ∈ Lp (E; µ) . Then for any ε > 0, there exists a uniformly continuous, bounded gdefined on E such that

|| f −g||Lp(E) < ε.

Proof: As usual in such situations, it suffices to consider only f ≥ 0. Then by Theorem11.3.9 on Page 241 and an application of the monotone convergence theorem, there existsa simple measurable function,

s(x)≡m

∑k=1

ckXAk (x)

such that || f − s||Lp(E) < ε/2. Now by regularity of µ there exist compact sets, Kk and

open sets, Vk such that 2∑mk=1 |ck|µ (Vk \K)1/p < ε/2 and by Corollary 59.14.2 there exist

uniformly continuous functions gk having values in [0,1] such that gk = 1 on Kk and 0 onVC

k . Then consider

g(x) =m

∑k=1

ckgk (x) .

This function is bounded and uniformly continuous. Furthermore,

||s−g||Lp(E) ≤

(∫E

∣∣∣∣∣ m

∑k=1

ckXAk (x)−m

∑k=1

ckgk (x)

∣∣∣∣∣p

dµ

)1/p

≤

(∫E

(m

∑k=1|ck|∣∣XAk (x)−gk (x)

∣∣)p)1/p

≤m

∑k=1|ck|(∫

E

∣∣XAk (x)−gk (x)∣∣p dµ

)1/p

≤m

∑k=1|ck|(∫

Vk\Kk

2pdµ

)1/p

= 2m

∑k=1|ck|µ (Vk \K)1/p < ε/2.

Therefore,|| f −g||Lp ≤ || f − s||Lp + ||s−g||Lp < ε/2+ ε/2.

This proves the theorem.

Lemma 59.14.4 Let A ∈B (E) where µ is a finite measure on B (E) for E a separableBanach space. Also let xi ∈ E for i = 1,2, · · · ,m. Then for x ∈ Em,

x→ µ

(A+

m

∑i=1

xi

), x→ µ

(A−

m

∑i=1

xi

)