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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 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.

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

Composition with a Borel measurable outer map preserves measurability

Statement

Let (X,A), (Y,B), and (Z,C) be measurable spaces. If f:(X,A)(Y,B) is measurable and g:(Y,B)(Z,C) is measurable, then gf:(X,A)(Z,C) is measurable.

In particular, if f is measurable and g is a Borel measurable function on its codomain, then gf is measurable.

Facts & Assumptions

Given: Measurable spaces (X,A), (Y,B), (Z,C), a measurable map f:XY, and a measurable map g:YZ.

[L1]

Measurability means that preimages of measurable sets are measurable. (A measurable function between measurable spaces)

Proof

technique · direct
1.1

Let CC. Since g is measurable, [L1] gives [given, L1] g1(C)B.

givenL1
2.1

Since f is measurable, [L1] applied again gives

step 1.1L1

(gf)1(C)=f1(g1(C))A.

So gf is measurable. [step 1.1, L1] ∎

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