TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Convergence in measure determines the limit almost everywhere

Statement

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

Facts & Assumptions

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

[L1]

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

[L2]

A property holds $μ$ -almost everywhere when its exceptional set is contained in a measurable $μ$ -null set. (Measure-null sets and almost-everywhere statements relative to a measure)

[L3]

For measurable $(E_{k})$ one has $μ (⋃_{k} E_{k}) \leq \sum_{k = 0}^{\infty} μ (E_{k})$ , and in particular $μ (E_{0} \cup E_{1}) \leq μ (E_{0}) + μ (E_{1})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1

For $m \geq 1$ put $E_{m} : = {∣ f - g ∣ > 1 / m}$ , and for $n \in N$ put $A_{n, m} : = {∣ f_{n} - f ∣ > 1 / (2 m)}$ and $B_{n, m} : = {∣ f_{n} - g ∣ > 1 / (2 m)}$ . If $x \in E_{m}$ and $x \notin A_{n, m} \cup B_{n, m}$ , then $∣ f (x) - g (x) ∣ \leq ∣ f (x) - f_{n} (x) ∣ + ∣ f_{n} (x) - g (x) ∣ \leq 1 / m$ , a contradiction. So $E_{m} \subseteq A_{n, m} \cup B_{n, m}$ for every $n, m$ . [given, L1, algebra] 2.1 Fix $m \geq 1$ and let $η > 0$ . By [L1] choose $n$ so large that $μ (A_{n, m}) < η / 2$ and $μ (B_{n, m}) < η / 2$ . Then step 1.1 and [L3] give $μ (E_{m}) \leq μ (A_{n, m}) + μ (B_{n, m}) < η$ . Since $η$ was arbitrary, $μ (E_{m}) = 0$ . [step 1.1, L1, L3] 3.1 If $f (x) \neq g (x)$ , then $∣ f (x) - g (x) ∣ > 1 / m$ for some $m \geq 1$ , so ${f \neq g} = ⋃_{m = 1}^{\infty} E_{m}$ . Step 2.1 makes every $E_{m}$ null, hence [L3] gives $μ ({f \neq g}) = 0$ . By [L2], $f = g$ $μ$ -almost everywhere. ∎

step 2.1L2L3

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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., Theorem 2.30 (standard reference, not scraped)