CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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$L^{p}$ -convergent sequences have almost-everywhere convergent subsequences

Statement

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: A norm-convergent sequence $(u_{n})$ in $L^{p} (μ)$ .

[L1]

Riesz-Fischer completeness already 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 $1 \leq p \leq \infty$ ).

Proof

Proof technique: Choose a rapidly convergent subsequence from an $L^{p}$ -convergent sequence and re-use the subsequence construction inside Riesz-Fischer.

1.1

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 $u$ .

L1given

2.1

That subsequence is the required one.

step 1.1∎

Depends on

Riesz-Fischer completeness of $L^p$ for $1 \le p \le \infty$

Used by

The published typewriter sequence shows why Riesz-Fischer only promises a subsequence Remark

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Measure, Integration & Real Analysis, Proposition 7.23 (standard reference, not scraped)
John K. Hunter, Measure Theory, Corollary 7.11 (standard reference, not scraped)

LpLp-convergent sequences have almost-everywhere convergent subsequences