CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

Equality almost everywhere with a measurable function can fail on an incomplete space

Statement refuted

That equality almost everywhere with a measurable function forces measurability even on an incomplete measure space.

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 positive-measure compact set $K = ψ [C]$ contains a nonmeasurable subset $N$ , and $E = ψ^{- 1} [N]$ is a Lebesgue measurable subset of the Cantor set that is not Borel. (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 $[0, 2]$ )

Counterexample

technique · direct

1.1

Let $f = 0$ and $g = 1_{E}$ . The set $C$ is Borel null, so $g = f$ on [L1] $R ∖ C$ and hence $g = f$ almost everywhere.

2.1

But [step 1.1, L1] $g^{- 1} ((1 / 2, \infty)) = E$ , and [L1] says $E$ is not Borel. Thus $g$ is not measurable for the Borel sigma-algebra, even though it agrees almost everywhere with the measurable function $f$ .

step 1.1L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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, Example 2.22 (standard reference, not scraped)