TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 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.

A generating family on the codomain suffices to test measurability

Let $(X, A)$ and $(Y, B)$ be measurable spaces, let $G \subseteq P (Y)$ generate $B$ , and let $f : X \to Y$ be a function. If

$f^{- 1} (G) \in A for every G \in G,$

then $f$ is measurable as a map $(X, A) \to (Y, B)$ .

Given: Measurable spaces $(X, A)$ and $(Y, B)$ , a function $f : X \to Y$ , and a generating family $G$ for $B$ .

[L1]

A function is measurable exactly when the preimage of every measurable set in the codomain is measurable in the domain. (A measurable function between measurable spaces)

technique · direct

1.1L1algebra

Let

$C : = {B \subseteq Y : f^{- 1} (B) \in A} .$

Preimages preserve complements and countable unions, so $C$ is a sigma-algebra on $Y$ . [L1, algebra]

2.1givenstep 1.1

By hypothesis, every member of $G$ lies in $C$ . Since [given, step 1.1] $B$ is the sigma-algebra generated by $G$ , one has $B \subseteq C$ .

3.1step 2.1L1∎

Therefore $f^{- 1} (B) \in A$ for every $B \in B$ , and [step 2.1, L1] [L1] says exactly that $f$ is measurable.

Depends on