LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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.

Assuming countable choice, cylinder-measurable events depend on only countably many coordinates

Statement

Assume countable choice. For every $A \in C_{I}$ there is a countable $J \subseteq I$ such that $x ∣_{J} = y ∣_{J}$ implies $x \in A ⟺ y \in A$ .

Facts & Assumptions

Given: Countable choice, the arbitrary product $E$ , and its cylinder sigma-algebra.

[F1]

The cylinder sigma-algebra is generated by finite-coordinate cylinders. (Coordinate maps, finite-coordinate cylinders, and the cylinder $σ$ -algebra)

[F2]

Under countable choice, a countable union of countable sets is countable. (Countable unions of at most countable sets, assuming ${A C}_{ω}$ )

Proof

1.1

Let $D$ be the events determined by some countable coordinate set. Complements preserve the same support. For a countable family in $D$ , countable choice selects one countable support for each event, and [F2] makes their union countable; that union determines the union of the events. Thus $D$ is a sigma-algebra.

2.1

Every finite-coordinate cylinder belongs to $D$ , using its finite support. By generated-sigma minimality and [F1], $C_{I} \subseteq D$ .

F1∎

Depends on

Used by

A noncylinder path functional may fail to be measurable Counterexample
The cylinder sigma-algebra need not be the full path-space power set Remark

Dependency tree · two levels

15 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

Shalizi, Building Processes, proof of Theorem 29 (standard reference, not scraped)