TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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Convergence in L^1(mu) implies convergence in measure

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 $f_{n} \to f$ in measure.

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 $L^{1} (μ)$ means $\int ∣ f_{n} - f ∣ d μ \to 0$ . (Convergence in L^1(mu))

[L2]

Convergence in measure means that for every real $ε > 0$ , $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ . (Convergence in measure)

[L3]

For a nonnegative measurable function $h$ and a real $t > 0$ , $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

Proof

technique · direct

1.1L1L3algebra

Fix $ε > 0$ . Applying [L3] to $h : = ∣ f_{n} - f ∣$ and $t : = ε$ gives $μ ({∣ f_{n} - f ∣ \geq ε}) \leq ε^{- 1} \int ∣ f_{n} - f ∣ d μ$ for every $n$ .

2.1step 1.1L1L2∎

By [L1], the right-hand side in step 1.1 tends to $0$ as $n \to \infty$ . Hence $μ ({∣ f_{n} - f ∣ > ε}) \to 0$ for every $ε > 0$ , which is exactly [L2].

Depends on

Used by

Dependency tree · two levels

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Sources

Gerald B. Folland, Real Analysis, 2nd ed., Proposition 2.29 (standard reference, not scraped)