False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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FALSE: dominated convergence holds without a dominating function

Statement

If $f_{n} \to f$ almost everywhere and each $f_{n}$ is integrable, then $\int f_{n} d μ \to \int f d μ .$

Facts & Assumptions

Given: The statement above.

[L1]

Dominated convergence spends one integrable majorant for the whole sequence (Dominated convergence).

Refutation

technique · direct

1.1

On $(0, 1)$ with Lebesgue measure, let [given, construct] $f_{n} : = (n + 1) χ_{(0, 1 / (n + 1))} .$ Then $f_{n} \to 0$ almost everywhere and every $f_{n}$ is integrable.

2.1

However $\int f_{n} d μ = 1$ for every $n$ , while $\int 0 d μ = 0$ . So the conclusion fails, and [L1] identifies the missing dominating function as the lost hypothesis.

step 1.1L1algebra∎

Depends on

Dominated convergence

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

John K. Hunter, Measure Theory Notes, Example 4.18 (standard reference, not scraped)