False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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FALSE: Egorov's theorem holds on every measure space

Statement refuted

Egorov's theorem holds on every measure space.

Facts & Assumptions

Given: Lebesgue measure on $R$ and the sequence $f_{n} : = χ_{[n, n + 1]}$ .

[L2]

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

Refutation

technique · direct

1.1

For each fixed $x \in R$ one has $f_{n} (x) = 0$ for all $n > x + 1$ , so $f_{n} (x) \to 0$ pointwise on $R$ .

given

2.1

Let $E \subseteq R$ be measurable with $λ (E) < + \infty$ . If $[n, n + 1] \subseteq E$ for infinitely many $n$ , then $λ (E) \geq \sum 1 = + \infty$ , impossible. Hence infinitely many indices $n$ satisfy $[n, n + 1] \subseteq̸ E$ , so for each such $n$ there is $x_{n} \in [n, n + 1] ∖ E$ with $f_{n} (x_{n}) = 1$ . Therefore $\sup_{x \in R ∖ E} ∣ f_{n} (x) - 0 ∣ \geq 1$ for infinitely many $n$ , and the convergence cannot be uniform on $R ∖ E$ . Thus [L2] fails.

step 1.1L2algebra

3.1

So pointwise almost-everywhere convergence does not force almost-uniform convergence on this infinite-measure space. The claim is false.

step 1.1step 2.1∎

Depends on

Almost uniform convergence

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Terence Tao, 245A Notes 4: Modes of convergence, Example 4 (standard reference, not scraped)
Gerald B. Folland, Real Analysis, 2nd ed., Section 2.4, Example (ii) (standard reference, not scraped)