CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Assuming countable choice, on a bounded measurable set, Lusin's closed core can be chosen compact

Statement

Assume the Axiom of Countable Choice.

Let $n \geq 1$ , let $E \subseteq R^{n}$ be Lebesgue measurable with $λ_{n} (E) < + \infty$ , and suppose $E$ is bounded. Let $f : E \to R$ be measurable. Then for every $ε > 0$ there is a compact set $K \subseteq E$ such that $λ_{n} (E ∖ K) < ε$ and $f ∣_{K}$ is continuous.

Facts & Assumptions

Given: The Axiom of Countable Choice, a bounded Lebesgue measurable set $E \subseteq R^{n}$ of finite measure, a measurable function $f : E \to R$ , and a real $ε > 0$ .

[L1]

Assuming countable choice, Lusin's theorem gives a closed set $F \subseteq E$ with $λ_{n} (E ∖ F) < ε$ such that $f ∣_{F}$ is continuous. (Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n)

[L2]

In $R^{n}$ , a subset is compact if and only if it is closed and bounded. (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line)

Proof

technique · direct

1.1L1choose

By [L1], choose a closed set $F \subseteq E$ with $λ_{n} (E ∖ F) < ε$ such that $f ∣_{F}$ is continuous.

2.1step 1.1L2

Because $F \subseteq E$ and $E$ is bounded, the set $F$ is bounded. Since $F$ is also closed in $R^{n}$ , [L2] makes $F$ compact.

3.1step 2.1∎

Taking $K : = F$ proves the claim.

Depends on

Used by

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Theorem 5.15 (standard reference, not scraped)