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Continuous functions are dense in Lp for compactly supported finite measures

Statement

Let μ be a finite regular Borel measure on Rd concentrated on a compact set K⊆Rd, and let 1≤p<∞. Form the real Lp(μ) space using the Borel σ-algebra. Then restrictions to K of real-valued compactly supported continuous functions on Rd are dense in Lp(μ;R).

Facts & Assumptions

Given: A finite regular Borel measure μ concentrated on compact K, an exponent 1≤p<∞, a real-valued Borel class [f]∈Lp(μ;R), and ε>0.

[F1]

Lp(μ) consists of measurable functions with finite (∫∣f∣p dμ)1/p, with functions identified when equal μ-almost everywhere (The function space Lp(μ) for 0<p<∞, The space Lp(μ) as the quotient by null functions).

[F2]

A measure is a nonnegative countably additive set function on a sigma-algebra, and regularity here means Borel sets can be approximated from inside by compact sets and outside by open sets with arbitrarily small measure excess (Measures on sigma-algebras); regularity is the hypothesis on μ.

[F3]

If F is compact and O is open with F⊂O, there is a smooth compactly supported cutoff ψ with 0≤ψ≤1, ψ=1 near F, and supp⁡ψ⊂O (Test function cutoffs and euclidean localization).

[F4]

Dominated convergence applies to ∣f−fM∣p≤∣f∣p and fM→f almost everywhere (Dominated convergence).

[F5]

For 1≤p<∞, the quotient Lp(μ) has a norm whose triangle inequality is Minkowski's inequality (The Lp norm descends to the quotient and makes Lp a normed space for 1≤p≤∞).

Proof

1.1givenF1F4F5choose

Choose a measurable representative f of [f]. Truncate it to fM=max⁡(−M,min⁡(f,M)). Since ∣f−fM∣p≤∣f∣p and fM→f pointwise wherever f is finite, [F4] makes ∥f−fM∥Lp(μ)<ε/2 for some large M. Partition [−M,M] into finitely many intervals of length at most δ and replace each value of fM by a fixed endpoint of its interval. If μ(K)>0, choose δ with δμ(K)1/p<ε/2; if μ(K)=0, any δ>0 suffices. The resulting finite-valued Borel simple function s satisfies ∣s−fM∣≤δ pointwise, so ∥s−fM∥Lp(μ)≤δ μ(K)1/p. By [F5] and the triangle inequality, ∥s−f∥Lp(μ)<ε. Since μ is concentrated on K, replace s by 1Ks: this finite Borel simple function represents the same class and admits a representation with every set Ej⊆K. Thus such supported simple functions are dense in Lp(μ).

2.1step 1.1F2F3choose

Write such a simple function as s=∑j=1Ncj1Ej with Borel Ej⊆K. For any η>0, regularity [F2] gives compact Fj⊆Ej and open Oj⊇Ej with μ(Oj∖Fj)<η. By [F3] choose ψj∈Cc∞(Oj) with 0≤ψj≤1 and ψj=1 on Fj. Outside Oj∖Fj the functions ψj and 1Ej agree; everywhere their difference has magnitude at most one. Therefore ∥ψj−1Ej∥Lp(μ)p≤μ(Oj∖Fj)<η.

3.1step 1.1step 2.1F5choose∎

The triangle inequality [F5] gives ∥∑j=1Ncjψj−s∥Lp(μ)≤∑j=1N∣cj∣η1/p. Choose η small enough that this is below ε/2, and choose s from step 1.1 within ε/2 of f. Then the compactly supported continuous function ∑jcjψj is within ε of f in Lp(μ).

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