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Convergence in L^1(mu) has an almost-everywhere convergent subsequence

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable with $f_{n}, f \in L^{1} (μ)$ . If $f_{n} \to f$ in $L^{1} (μ)$ , then some subsequence of $(f_{n})$ converges to $f$ $μ$ -almost everywhere.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable real-valued integrable functions $f_{n}, f \in L^{1} (μ)$ , and convergence of $(f_{n})$ to $f$ in $L^{1} (μ)$ .

[L1]

Convergence in $L^{1} (μ)$ implies convergence in measure. (Convergence in L^1(mu) implies convergence in measure)

[L2]

Convergence in measure has a subsequence converging almost everywhere to the same limit. (Riesz's subsequence theorem for convergence in measure)

Proof

technique · direct

1.1

By [L1], the sequence $(f_{n})$ converges to $f$ in measure.

2.1

Apply [L2] to step 1.1. The resulting subsequence converges to $f$ almost everywhere.

step 1.1L2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Corollary 2.32 (standard reference, not scraped)