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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 1p<

Statement

Assume the Axiom of Countable Choice.

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

Facts & Assumptions

Given: The Axiom of Countable Choice, 1p<, ε>0, and fLp(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
1.1

By [L1], choose a box-step function [L1, L2, given, choose] s=j=1maj1Ej with fsp<ε/2. For each j, apply [L2] to choose KjEj and a bounded open set OjKj with λn(EjKj), λn(OjKj)<(ε21+1/pm(1+j=1maj))p.

L1L2givenchoose
2.1

For each j, [L3] gives ηjCc(Rn) with [L3, L4, step 1.1, construct, algebra] 0ηj1, ηj=1 on Kj, and supp(ηj)Oj. Put g:=j=1majηjCc(Rn). Since ηj1Ej vanishes off (EjKj)(OjKj) and has absolute value at most 1 there, gspj=1maj(λn(EjKj)+λn(OjKj))1/p<ε/2 by [L4].

L3L4step 1.1constructalgebra
3.1

Therefore [step 1.1, step 2.1, algebra] fgpfsp+sgp<ε. Since gCc(Rn), the space Cc(Rn) is dense in Lp(Rn).

step 1.1step 2.1algebra

Depends on

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