False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: composing a Lebesgue measurable function with a continuous map preserves measurability

Statement

False claim. Let $X, Y \subseteq R$ carry their trace Lebesgue sigma-algebras. If $f : X \to Y$ is continuous and $g : Y \to R$ is Lebesgue measurable, then $g \circ f$ is Lebesgue measurable.

Facts & Assumptions

Given: The Axiom of Choice, the Cantor set $C$ , the homeomorphism $ψ (x) = x + c (x)$ from $[0, 1]$ onto $[0, 2]$ , and $K : = ψ [C]$ .

[L1]

The compact set $K$ has positive Lebesgue measure and therefore contains a nonmeasurable subset $N$ . The set $C$ is Lebesgue null, Lebesgue measure is complete, and $ψ$ is a homeomorphism. (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, 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, The map $x \mapsto x + c (x)$ is a homeomorphism from $[0, 1]$ onto $[0, 2]$ )

[L2]

The indicator of a measurable set is measurable. (An indicator function is measurable exactly when its set is measurable)

Refutation

technique · direct

1.1

Choose a nonmeasurable $N \subseteq K$ by [L1], put [L1, L2, choose] $E : = ψ^{- 1} [N] \subseteq C$ , let $g : = 1_{E} : [0, 1] \to R$ , and let $f : = ψ^{- 1} : [0, 2] \to [0, 1]$ . Completeness makes $E$ Lebesgue measurable in $R$ , hence measurable in the trace sigma-algebra on $[0, 1]$ , so [L2] makes $g$ measurable, while [L1] makes $f$ continuous.

2.1step 1.1L1

But

$(g \circ f)^{- 1} ((1 / 2, \infty)) = f^{- 1} [E] = N,$

which does not lie in the trace sigma-algebra $L (R) ∣_{[0, 2]}$ , even though $g$ is measurable and $f$ is continuous. [step 1.1, L1] ∎

Depends on

Used by

FALSE: the Borel-representative discipline in convolution is unnecessary because continuous precomposition always preserves Lebesgue measurability False statement

Dependency tree · two levels

38 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

John K. Hunter, Measure Theory, Definition 3.3 and Example 2.22 (standard reference, not scraped)