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

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Assuming countable choice, simple approximants to a measurable function can be made uniformly convergent on a large closed set

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 bounded and measurable. Then for every $ε > 0$ there are a closed set $F \subseteq E$ and simple functions $s_{m} : E \to R$ such that

$λ_{n} (E ∖ F) < ε$ ;
each restriction $s_{m} ∣_{F}$ is continuous;
$s_{m} \to f$ uniformly on $F$ .

Facts & Assumptions

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

[L1]

Every measurable function admits simple approximations $s_{m}$ with $∣ s_{m} ∣ \leq ∣ f ∣$ for every $m$ and $s_{m} (x) \to f (x)$ for every $x$ . (Every measurable function admits simple approximations dominated by its absolute value)

[L2]

On a finite measure space, almost-everywhere convergence implies almost-uniform convergence. (Egorov's theorem)

[L3]

Assuming countable choice, every Lebesgue measurable subset of $R^{n}$ has compact subsets of arbitrarily close measure from inside. (Assuming countable choice, the Lebesgue measure of a measurable set is the supremum of the measures of its compact subsets)

[L4]

Every simple measurable function on a finite-measure Lebesgue set is continuous on a large closed core. (Assuming countable choice, simple functions are continuous on a large closed core)

[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

By [L1], choose simple functions $s_{m} : E \to R$ with $∣ s_{m} ∣ \leq ∣ f ∣$ and $s_{m} (x) \to f (x)$ for every $x \in E$ . Since $λ_{n} (E) < + \infty$ , [L2] gives a measurable set $G \subseteq E$ with $λ_{n} (E ∖ G) < ε / 2$ such that $s_{m} \to f$ uniformly on $G$ .

1.2L3L4choose

By [L3], choose a compact set $K \subseteq G$ with $λ_{n} (G ∖ K) < ε / 4$ . For each $m \geq 1$ , apply [L4] to $s_{m}$ with tolerance $ε 2^{- m - 3}$ , obtaining a closed set $F_{m} \subseteq E$ such that $λ_{n} (E ∖ F_{m}) < ε 2^{- m - 3}$ and $s_{m} ∣_{F_{m}}$ is continuous.

2.1step 1.1step 1.2L5algebra

Put $F : = K \cap ⋂_{m = 1}^{\infty} F_{m}$ . Then $F$ is closed, $F \subseteq K \subseteq G \subseteq E$ , and $E ∖ F \subseteq (E ∖ G) \cup (G ∖ K) \cup ⋃_{m = 1}^{\infty} (E ∖ F_{m}) .$ So [L5] together with steps 1.1 and 1.2 gives $λ_{n} (E ∖ F) < \frac{ε}{2} + \frac{ε}{4} + \sum_{m = 1}^{\infty} ε 2^{- m - 3} < ε .$ Because $F \subseteq G$ , the convergence $s_{m} \to f$ remains uniform on $F$ . And because $F \subseteq F_{m}$ , each $s_{m} ∣_{F}$ is continuous as a restriction of $s_{m} ∣_{F_{m}}$ .

3.1step 2.1∎

The closed set $F$ and the simple approximants $s_{m}$ satisfy all three assertions.

Depends on

Used by

Assuming countable choice, Lusin's theorem on finite-measure subsets of Rⁿ Theorem

Dependency tree · two levels

33 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

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