Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck 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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A function can have measurable level sets without being measurable

Statement refuted

That measurability of every level set {x:f(x)=a} is enough to make f measurable.

Facts & Assumptions

Given: The Axiom of Choice, the Lebesgue measurable space ([0,1],L(R)∣[0,1]), a Vitali set V⊆[0,1], and the function f:[0,1]→R defined by f(x)=x for x∉V, f(x)=x+2 for x∈V.

Counterexample

technique · direct
1.1given

Every level set of f is empty, a singleton, or a two-point set, so every [given] level set is measurable.

2.1step 1.1L1∎

Yet f−1([2,3])=V, and [L1] says that V is not measurable. Therefore [step 1.1, L1] f is not measurable.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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