TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Almost uniform convergence implies almost-everywhere convergence and convergence in measure

Statement

Let $(X, A, μ)$ be a measure space and let $f_{n}, f : X \to R$ be measurable. If $f_{n} \to f$ almost uniformly, then $f_{n} \to f$ $μ$ -almost everywhere and $f_{n} \to f$ in measure.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , measurable functions $f_{n}, f : X \to R$ , and almost-uniform convergence of $(f_{n})$ to $f$ .

[L1]

Almost-uniform convergence means that for every $ε > 0$ there is a measurable $E$ with $μ (E) < ε$ such that $f_{n} \to f$ uniformly on $X ∖ E$ . (Almost uniform convergence)

[L2]

Almost-everywhere convergence means pointwise convergence off a measurable null set. (Convergence almost everywhere relative to a measure)

[L3]

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

[L4]

If $A \subseteq B$ are measurable, then $μ (A) \leq μ (B)$ . (Measures are monotone)

Proof

technique · direct

1.1

For each $m \geq 1$ , [L1] gives a measurable set $E_{m}$ with $μ (E_{m}) < 1 / m$ such that $f_{n} \to f$ uniformly on $X ∖ E_{m}$ . Let $N : = ⋂_{m = 1}^{\infty} E_{m}$ . Since $N \subseteq E_{m}$ for every $m$ , [L4] gives $μ (N) \leq 1 / m$ for every $m$ , hence $μ (N) = 0$ . If $x \in X ∖ N$ , then $x \notin E_{m}$ for some $m$ , and uniform convergence on $X ∖ E_{m}$ implies $f_{n} (x) \to f (x)$ . Therefore $f_{n} \to f$ almost everywhere by [L2].

L1L2L4

1.2

Fix $ε > 0$ and $η > 0$ . By [L1] choose a measurable set $E$ with $μ (E) < η$ such that $f_{n} \to f$ uniformly on $X ∖ E$ . Then there is $N$ such that for $n \geq N$ and $x \in X ∖ E$ one has $∣ f_{n} (x) - f (x) ∣ \leq ε$ , so ${∣ f_{n} - f ∣ > ε} \subseteq E$ . Hence $μ ({∣ f_{n} - f ∣ > ε}) \leq μ (E) < η$ for $n \geq N$ . Since $η$ was arbitrary, [L3] follows.

L1L3L4

2.1

Steps 1.1 and 1.2 prove the two asserted conclusions.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

8 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

Gerald B. Folland, Real Analysis, 2nd ed., Exercise 39 (standard reference, not scraped)