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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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Cc(Rn) is dense in Lp(Rn) for 1≤p<∞

Statement

Assume the Axiom of Countable Choice.

Let 1≤p<∞. Then Cc(Rn) is dense in Lp(Rn).

Facts & Assumptions

Given: The Axiom of Countable Choice, 1≤p<∞, ε>0, and f∈Lp(Rn).

[L2]

Finite-measure measurable sets admit compact cores with bounded open neighbourhoods of arbitrarily small excess (A finite-measure measurable set in Rn has a compact core and a bounded open neighbourhood of arbitrarily small excess).

[L3]

Compact sets inside bounded open sets admit explicit compactly supported continuous cutoffs (A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff).

[L4]

Minkowski's inequality holds in Lp (Minkowski's inequality for integrals, including p=∞).

Proof

technique · direct
1.1L1L2givenchoose

By [L1], choose a box-step function [L1, L2, given, choose] s=∑j=1maj 1Ej with ∥f−s∥p<ε/2. For each j, apply [L2] to choose Kj⊆Ej and a bounded open set Oj⊇Kj with λn(Ej∖Kj), λn(Oj∖Kj)<(ε21+1/pm(1+∑j=1m∣aj∣))p.

2.1L3L4step 1.1constructalgebra

For each j, [L3] gives ηj∈Cc(Rn) with [L3, L4, step 1.1, construct, algebra] 0≤ηj≤1, ηj=1 on Kj, and supp⁡(ηj)⊆Oj. Put g:=∑j=1maj ηj∈Cc(Rn). Since ηj−1Ej vanishes off (Ej∖Kj)∪(Oj∖Kj) and has absolute value at most 1 there, ∥g−s∥p≤∑j=1m∣aj∣(λn(Ej∖Kj)+λn(Oj∖Kj))1/p<ε/2 by [L4].

3.1step 1.1step 2.1algebra∎

Therefore [step 1.1, step 2.1, algebra] ∥f−g∥p≤∥f−s∥p+∥s−g∥p<ε. Since g∈Cc(Rn), the space Cc(Rn) is dense in Lp(Rn).

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Sources