TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$

Statement

Assume the Axiom of Countable Choice.

Let $1 \leq p < \infty$ . Then $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, $1 \leq p < \infty$ , $ε > 0$ , and $f \in L^{p} (R^{n})$ .

[L1]

Box-step functions are dense in $L^{p} (R^{n})$ (Finite linear combinations of box indicators are dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L2]

Finite-measure measurable sets admit compact cores with bounded open neighbourhoods of arbitrarily small excess (A finite-measure measurable set in $R^{n}$ 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 $L^{p}$ (Minkowski's inequality for integrals, including $p = \infty$ ).

Proof

technique · direct

1.1L1L2givenchoose

By [L1], choose a box-step function [L1, L2, given, choose] $s = \sum_{j = 1}^{m} a_{j} 1_{E_{j}}$ with $∥ f - s ∥_{p} < ε / 2$ . For each $j$ , apply [L2] to choose $K_{j} \subseteq E_{j}$ and a bounded open set $O_{j} \supseteq K_{j}$ with $λ_{n} (E_{j} ∖ K_{j}), λ_{n} (O_{j} ∖ K_{j}) < {(\frac{ε}{2^{1 + 1 / p} m (1 + \sum_{j = 1}^{m} ∣ a_{j} ∣)})}^{p} .$

2.1L3L4step 1.1constructalgebra

For each $j$ , [L3] gives $η_{j} \in C_{c} (R^{n})$ with [L3, L4, step 1.1, construct, algebra] $0 \leq η_{j} \leq 1$ , $η_{j} = 1$ on $K_{j}$ , and $supp (η_{j}) \subseteq O_{j}$ . Put $g : = \sum_{j = 1}^{m} a_{j} η_{j} \in C_{c} (R^{n}) .$ Since $η_{j} - 1_{E_{j}}$ vanishes off $(E_{j} ∖ K_{j}) \cup (O_{j} ∖ K_{j})$ and has absolute value at most $1$ there, $∥ g - s ∥_{p} \leq \sum_{j = 1}^{m} ∣ a_{j} ∣ {(λ_{n} (E_{j} ∖ K_{j}) + λ_{n} (O_{j} ∖ K_{j}))}^{1 / p} < ε / 2$ by [L4].

3.1step 1.1step 2.1algebra∎

Therefore [step 1.1, step 2.1, algebra] $∥ f - g ∥_{p} \leq ∥ f - s ∥_{p} + ∥ s - g ∥_{p} < ε .$ Since $g \in C_{c} (R^{n})$ , the space $C_{c} (R^{n})$ is dense in $L^{p} (R^{n})$ .

Depends on

Dependency tree · 2 levels

Sources

Walter Rudin, Real and Complex Analysis, 3rd ed.

Cc(Rn)Cc​(Rn) is dense in Lp(Rn)Lp(Rn) for 1≤p<∞1≤p<∞