TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n

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 let $f : E \to R$ be measurable. Then for every $ε > 0$ there exist a real $M > 0$ and a closed set $F \subseteq E \cap f^{- 1} ([- M, M])$ such that $λ_{n} (E ∖ F) < ε$ and $f ∣_{F}$ is continuous. In particular, $f ∣_{F}$ is bounded.

Facts & Assumptions

Given: The Axiom of Countable Choice, a measurable function $f : E \to R$ on a Lebesgue measurable set $E \subseteq R^{n}$ with $λ_{n} (E) < + \infty$ , and a real $ε > 0$ .

[L1]

Measurability means preimages of Borel sets are measurable. (A measurable function between measurable spaces)

[L2]

Measures are continuous from below on increasing measurable sets. (Continuity from below for measures)

[L3]

Assuming countable choice, for bounded measurable functions on finite-measure Lebesgue sets there is a large closed set on which simple approximants are continuous and converge uniformly to the function. (Assuming countable choice, simple approximants to a measurable function can be made uniformly convergent on a large closed set)

[L4]

A uniform limit of continuous functions is continuous. (A uniform limit of continuous functions is continuous, so $C (X, Y)$ is closed in $Y^{X}$ under the uniform metric)

[L5]

For measurable $(A_{k})$ one has $μ (⋃_{k} A_{k}) \leq \sum_{k = 0}^{\infty} μ (A_{k})$ . (Finite and countable subadditivity of measures)

Proof

technique · direct

1.1L1L2choose

For $m \geq 1$ , put $E_{m} : = E \cap f^{- 1} ([- m, m])$ . Because $[- m, m]$ is a Borel subset of $R$ , [L1] makes every $E_{m}$ measurable. The sets $E_{m}$ increase with $m$ , and their union is all of $E$ because $f$ is real-valued. So [L2] gives $λ_{n} (E) = \sup_{m \geq 1} λ_{n} (E_{m}) .$ Choose $M \geq 1$ with $λ_{n} (E ∖ E_{M}) < ε / 2$ .

2.1step 1.1L3L4

On $E_{M}$ one has $∣ f ∣ \leq M$ , so $f ∣_{E_{M}}$ is bounded. Apply [L3] to the bounded measurable function $f ∣_{E_{M}}$ with tolerance $ε / 2$ . This gives a closed set $F \subseteq E_{M}$ and simple functions $s_{m} : E_{M} \to R$ such that $λ_{n} (E_{M} ∖ F) < ε / 2,$ each $s_{m} ∣_{F}$ is continuous, and $s_{m} \to f ∣_{E_{M}}$ uniformly on $F$ . By [L4], $f ∣_{F}$ is continuous.

3.1step 1.1step 2.1L5algebra

Because $F \subseteq E_{M} \subseteq E$ , one has $E ∖ F = (E ∖ E_{M}) \cup (E_{M} ∖ F) .$ Hence [L5] together with steps 1.1 and 2.1 gives $λ_{n} (E ∖ F) < ε / 2 + ε / 2 = ε .$ The set $F$ also lies in $E \cap f^{- 1} ([- M, M])$ by step 1.1.

4.1step 2.1step 3.1∎

The set $F$ and the bound $M$ satisfy the theorem.

Depends on

Used by

Dependency tree · two levels

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Sources

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