CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)

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.

Null-set modifications defeat everywhere representative recovery

Statement refuted

Whenever $f, \hat{f} \in L^{1}$ , the inverse Fourier integral equals the chosen representative f at every point.

Facts & Assumptions

Given: The Axiom of Countable Choice ( ${A C}_{ω}$ ) and $f = 1_{{0}}$ on $R$ .

[F1]

A singleton on the line is Lebesgue null (Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null).

[F2]

The integral of a nonnegative function over a null set vanishes (A nonnegative integral over a null set vanishes).

[F3]

Null modifications do not change any transform value (The integral transform is representative independent).

Counterexample

1.1

F1 and F2 give $\int ∣ f ∣ = 0$ , so f represents the zero integrable class. F3 gives $\hat{f} (ξ) = 0$ for every frequency, and hence the transform is integrable too.

F1F2F3given

2.1

The inverse integral at zero is $\int \hat{f} (ξ) d ξ = 0$ , whereas $f (0) = 1$ . Thus the claimed every-point assertion fails. In fact the mean oscillation about the chosen value f(0) equals one on every centered interval, so zero is not a Lebesgue point with that value. This respects the actual almost-everywhere inversion theorem. Countable choice is inherited from F1.

F1F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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

Gerald Teschl, Topics in Real and Functional Analysis (2017) (standard reference, not scraped)