CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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.

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences

Statement

Assume Countable Choice. Let $1 \leq p \leq \infty$ . If $u_{n} \to u$ in $L^{p} (μ)$ , then some subsequence of $(u_{n})$ admits measurable representatives converging almost everywhere to a measurable representative of $u$ .

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and a norm-convergent sequence $(u_{n})$ in $L^{p} (μ)$ .

[L1]

Under Countable Choice, Riesz-Fischer completeness states that every norm-convergent sequence in $L^{p} (μ)$ has an almost-everywhere convergent subsequence of representatives (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ). Applying that representative-selection clause in step 1.1 is the use of Countable Choice here.

Proof

technique · Choose a rapidly convergent subsequence from an $L^p$-convergent sequence and re-use the subsequence construction inside Riesz-Fischer

1.1L1given

The sequence $(u_{n})$ is Cauchy because it converges in norm. Applying [L1] to that Cauchy sequence gives an $L^{p}$ limit together with an almost-everywhere convergent subsequence of representatives. Because metric limits are unique, the limit supplied by [L1] must be the given $u$ .

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Assuming Countable Choice, LpLp-convergent sequences have almost-everywhere convergent subsequences

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences