CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-27

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Fatou can be strict and domination can fail simultaneously

Statement refuted

Whenever $f_{n} \to f$ almost everywhere and each $f_{n}$ is integrable, Fatou's lemma is an equality and dominated convergence is automatic.

Facts & Assumptions

Given: The spike sequence $f_{n} : = (n + 1) χ_{(0, 1 / (n + 1))}$ on $(0, 1)$ .

[L1]

Fatou's lemma is only a one-sided inequality (Fatou's lemma).

[L2]

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

Counterexample

technique · direct

1.1givenalgebra

The sequence $f_{n}$ converges pointwise almost everywhere to $0$ , but $\int_{0}^{1} f_{n} d λ = (n + 1) λ ((0, 1 / (n + 1))) = 1$ for every $n$ .

2.1

Therefore [step 1.1, L1, L2] ∎ $\int {lim inf}_{n} f_{n} d λ = 0 < 1 = {lim inf}_{n} \int f_{n} d λ,$ so Fatou is strict, and the unchanged integral also shows that no dominated convergence conclusion can hold. This is exactly the hypothesis loss recorded in [L1] and [L2].

Depends on

Used by

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Dependency tree · two levels

12 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)