False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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FALSE: equality almost everywhere with a measurable function implies measurability

Statement

False claim. If $g$ equals a measurable function almost everywhere, then $g$ is measurable. This fails on incomplete measure spaces.

Facts & Assumptions

Given: The Axiom of Choice and the Borel measure space $(R, B (R), λ ∣_{B (R)})$ , the Cantor set $C$ , and the homeomorphism $ψ (x) = x + c (x)$ from $[0, 1]$ onto $[0, 2]$ .

[L1]

The set $K = ψ [C]$ is compact and has Lebesgue measure $1$ , so it has positive outer measure; every positive-outer-measure subset of $R$ contains a nonmeasurable subset. (The homeomorphism $x \mapsto x + c (x)$ sends the Cantor set onto a compact set of Lebesgue measure $1$ , Every subset of $R$ of positive Lebesgue outer measure contains a nonmeasurable subset)

[L2]

The Cantor set has Lebesgue measure zero, and Lebesgue measure is complete. (The Cantor set is an uncountable subset of $R$ of Lebesgue measure zero, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume)

[L3]

The map $ψ$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ . (The map $x \mapsto x + c (x)$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ )

Refutation

technique · direct

1.1

By [L1], choose a nonmeasurable subset $N \subseteq K$ , and put [L1, L2, L3, choose] $E : = ψ^{- 1} [N] \subseteq C$ . Since $E \subseteq C$ and $C$ is Lebesgue null, [L2] makes $E$ Lebesgue measurable. If $E$ were Borel, then continuity of $ψ^{- 1} ∣_{K}$ from [L3] would make $N$ Borel in the subspace $K$ , hence Lebesgue measurable, which contradicts the choice of $N$ . So $E$ is Lebesgue measurable but not Borel.

L1L2L3choose

2.1

Let $f = 0$ and $g = 1_{E}$ on $R$ . The function $f$ is Borel [step 1.1, L2] measurable, and $g = f$ on $R ∖ C$ , so $g = f$ almost everywhere with respect to $λ ∣_{B (R)}$ because $C$ is a measurable null set by [L2]. But

$g^{- 1} ((1 / 2, \infty)) = E,$

and $E$ is not Borel by step 1.1, so $g$ is not measurable on this incomplete measure space. [step 1.1, L2] ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

John K. Hunter, Measure Theory, Example 2.22 (standard reference, not scraped)