LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Laws commute with measurable maps

Statement

Let $X : (Ω, F, P) \to (S, Σ)$ be a random element, and let $g : (S, Σ) \to (T, T)$ be measurable. Then $g \circ X$ is a random element and for every $B \in T$ , $P_{g \circ X} (B) = P_{X} (g^{- 1} (B)) .$

Facts & Assumptions

Given: A random element $X$ and a measurable map $g$ as in the Statement.

[L1]

Composition of measurable maps is measurable (Composition with a Borel measurable outer map preserves measurability).

[L2]

The law of a random element is defined by pullback of measurable target sets (Law or distribution of a random element).

[L3]

The law of any random element is a probability measure (The law of a random element is a probability measure).

Proof

technique · direct

1.1

By [L1], the composite $g \circ X$ is measurable, hence a random element.

givenL1

1.2

For every $B \in T$ , $(g \circ X)^{- 1} (B) = X^{- 1} (g^{- 1} (B)) .$ Therefore [L2] gives $P_{g \circ X} (B) = P ((g \circ X)^{- 1} (B)) = P (X^{- 1} (g^{- 1} (B))) = P_{X} (g^{- 1} (B)) .$ This right-hand side is defined because [L3] makes $P_{X}$ a probability measure on $(S, Σ)$ .

L2L3

2.1

Steps 1.1 and 1.2 prove the measurable-map compatibility of laws.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

S. R. S. Varadhan, Probability Theory, Section 1.4 (standard reference, not scraped)
J. R. Norris, Probability and Measure, Section 3.3 (standard reference, not scraped)