LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Assuming countable choice, simple functions are continuous on a large closed core

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 $s : E \to R$ be a simple measurable function. Then for every $ε > 0$ there is a closed set $F \subseteq E$ such that $λ_{n} (E ∖ F) < ε$ and $s ∣_{F}$ is continuous.

Facts & Assumptions

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

[L1]

If the distinct values of $s$ are $c_{1}, \dots, c_{m}$ and $E_{j} : = s^{- 1} ({c_{j}})$ , then the sets $E_{j}$ are measurable, pairwise disjoint, their union is $E$ , and $s = \sum_{j = 1}^{m} c_{j} 1_{E_{j}}$ . (A simple function and its canonical representation)

[L2]

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)

[L3]

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

1.1

Let $c_{1}, \dots, c_{m}$ be the distinct values of $s$ , and let $E_{j} : = s^{- 1} ({c_{j}})$ . By [L1], the sets $E_{j}$ are measurable, pairwise disjoint, and cover $E$ . For each $j$ choose a compact set $K_{j} \subseteq E_{j}$ with $λ_{n} (E_{j} ∖ K_{j}) < ε / m .$ This is possible by [L2].

L1L2choose

2.1

Put $F : = ⋃_{j = 1}^{m} K_{j}$ . Each $K_{j}$ is closed in $R^{n}$ , so the finite union $F$ is closed and lies in $E$ . Also $E ∖ F = ⋃_{j = 1}^{m} (E_{j} ∖ K_{j}),$ so [L3] and step 1.1 give $λ_{n} (E ∖ F) \leq \sum_{j = 1}^{m} λ_{n} (E_{j} ∖ K_{j}) < ε .$

step 1.1L3algebra

3.1

Fix $x \in F$ . By step 1.1, $x$ lies in exactly one $K_{j}$ . In the subspace $F$ , the set $K_{j} = F ∖ ⋃_{i \neq j} K_{i}$ is open because the other $K_{i}$ are closed and finite in number. On that neighbourhood $K_{j}$ , the restriction $s ∣_{F}$ is constant with value $c_{j}$ . So $s ∣_{F}$ is locally constant at every point of $F$ , hence continuous.

step 1.1step 2.1

4.1

The closed set $F$ from step 2.1 has $λ_{n} (E ∖ F) < ε$ , and step 3.1 makes $s ∣_{F}$ continuous.

step 2.1step 3.1∎

Depends on

Used by

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

Dependency tree · two levels

25 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)