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

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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.

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Finite products of standard Borel spaces are standard Borel

Statement

Every finite product of standard Borel spaces, with its finite product sigma-algebra, is standard Borel.

Facts & Assumptions

Given: Standard-Borel spaces $(E_{j}, E_{j})$ for $j < n$ .

[F1]

Each $E_{j}$ is measurably isomorphic to the Borel space of a Polish space. (Standard Borel spaces)

Proof

1.1

Choose Polish presentations $h_{j} : E_{j} \to P_{j}$ from [F1]. The product map $h = \prod_{j < n} h_{j}$ is a bijection from $\prod E_{j}$ to $\prod P_{j}$ .

2.1

A finite product of Polish spaces is Polish, and inverse images under $h$ of its Borel rectangles are exactly the finite product measurable rectangles. Therefore $h$ transports its Borel sigma-algebra to $⨂_{j < n} E_{j}$ , proving the claim.

step 1.1∎

Depends on

Used by

Assuming the Axiom of Choice, Kolmogorov extension for arbitrary families of standard Borel coordinate spaces Theorem

Dependency tree · two levels

7 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

Biskup, MATH 275D notes, Lemma 2.8 (standard reference, not scraped)