TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck pass

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.

Composition with a Borel measurable outer map preserves measurability

Statement

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

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

Facts & Assumptions

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

[L1]

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

Proof

technique · direct

1.1givenL1

Let $C \in C$ . Since $g$ is measurable, [L1] gives $g^{- 1} (C) \in B$ .

2.1step 1.1L1∎

Since $f$ is measurable, [L1] applied again gives $(g \circ f)^{- 1} (C) = f^{- 1} (g^{- 1} (C)) \in A .$ So $g \circ f$ is measurable.

Depends on

A measurable function between measurable spaces

Dependency tree · 2 levels

Sources

John K. Hunter, Measure Theory (2011), Definition 3.1, p. 33, and composition discussion after Definition 3.3, p. 34