Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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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.

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Uniform convergence does not force convergence of integrals on an infinite-measure space

Statement refuted

Uniform convergence of integrable functions always implies convergence of their integrals.

Facts & Assumptions

Given: The functions fn:=(n+1)−1χ[0,n+1] on R.

[L1]

Bounded convergence is a finite-measure-space theorem (Bounded convergence on a finite measure space).

Counterexample

technique · direct
1.1

Since 0≤fn≤1/(n+1), the sequence (fn) converges uniformly to 0 on R.

given
2.1

Nevertheless, ∫Rfn dλ=(n+1)−1λ([0,n+1])=1 for every n, so the integrals do not converge to 0. This refutes the Statement and shows why [L1] needs finite total measure.

step 1.1L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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