230 CHAPTER 9. BASIC FUNCTION SPACES

9.4 Approximation TheoremsFirst is a significant result on approximating with simple functions in Lp.

Theorem 9.4.1 Let f ∈ Lp (Ω) for p ≥ 1. Then for each ε > 0 there is a simplefunction s such that ∥ f − s∥p ≤ ε .

Proof: It suffices to consider the case where f ≥ 0 because you can then apply whatis shown to the positive and negative parts of the real and imaginary parts of f to getthe general case. Thus, suppose f ≥ 0 and in Lp (Ω) . By Theorem 6.1.10, there ex-ists a sequence of simple functions increasing to f . Then | f (ω)− sn (ω)|p ≤ | f (ω)|p.This is a suitable dominating function. Then by the dominated convergence theorem,0 = limn→∞

∫Ω| f (ω)− sn (ω)|p dµ which establishes the desired conclusion unless p = ∞.

Use Proposition 9.3.9 to get a set of measure zero N such that off this set, | f (ω)| ≤∥ f∥

∞. Then consider f XNC . It is a measurable and bounded function so by Theorem

6.1.10, there is an increasing sequence of simple functions {sn} converging uniformly tothis function. Hence, for n large enough, ∥ f − sn∥∞

< ε . ■

Theorem 9.4.2 Let µ be a regular Borel measure on Rn and f ∈ Lp (Rn). Then foreach p≥ 1, p ̸= ∞, there exists g a continuous function which is zero off a compact set suchthat ∥ f −g∥p < ε .

Proof: Without loss of generality, assume f ≥ 0. First suppose that f is 0 off some ballB(0,R). There exists a simple function 0 ≤ s ≤ f such that

∫| f − s|p dµ < (ε/2)p . Thus

it suffices to show the existence of a continuous function h which is zero off a compact setwhich satisfies (

∫|h− s|p dµ)1/p < ε/2. Let

s(x) =m

∑i=1

ciXEi (x) , Ei ⊆ B(0,R)

where Ei is in Fp. Thus each Ei is bounded. By regularity, there exist compact sets Ki and

open sets Vi with Ki ⊆ Ei ⊆Vi ⊆ B(0,R) and ∑mi=1(cp

i µ (Vi \Ki))1/p

< ε/2.

Now define hi (x) ≡dist(x,VC)

dist(x,K)+dist(x,VC). Thus hi equals zero off a compact set and it

equals 1 on Ki and 0 off Vi. Let h≡ ∑mi=1 cihi. Thus 0≤ h≤max{ci, i = 1, · · · ,m} . Then∫

|cihi− ciXEi |p dµ ≤ cp

i XVi−Ki ≤ cpi µ (Vi \Ki)

It follows that, from the Minkowski inequality,(∫ ∣∣∣∣∣ f −∑i

cihi

∣∣∣∣∣p

dµ

)1/p

≤(∫| f − s|p dµ

)1/p

+

(∫ ∣∣∣∣∣s−∑i

cihi

∣∣∣∣∣p

dµ

)1/p

≤ ε

2+

(∫ (∑

i|ciXEi − cihi|

)p

dµ

)1/p

≤ ε

2+∑

i

(∫|ciXEi − cihi|p dµ

)1/p

≤ ε

2+∑

i

(cp

i µ (Vi \Ki))1/p

<ε

2+

ε

2= ε