Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29
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The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}

Statement

Let m,n≥1. Under the identification Rm+n=Rm×Rn, one has B(Rm)⊗B(Rn)=B(Rm+n).

Facts & Assumptions

Given: Positive integers m,n and the identification Rm+n=Rm×Rn.

[L1]

The Borel sigma-algebra on a topological space is the sigma-algebra generated by its open sets. (The Borel sigma-algebra of a topological space)

[L2]

Continuous maps pull Borel sets back to Borel sets. (A continuous map has Borel preimages of Borel sets)

[A1]

The coordinate projections π1:Rm+n→Rm and π2:Rm+n→Rn are continuous.

[A2]

Every open set in Rm+n is a countable union of open rectangles U×V with U⊆Rm and V⊆Rn open.

Proof

technique · direct
1.1A1L2

If A⊆Rm and B⊆Rn are Borel, then A×B=π1−1(A)∩π2−1(B), so [L2] makes A×B Borel in Rm+n. Therefore every measurable rectangle for B(Rm)⊗B(Rn) belongs to B(Rm+n), and hence B(Rm)⊗B(Rn)⊆B(Rm+n).

2.1step 1.1A2L1∎

By [A2], every open set in Rm+n is a countable union of open rectangles, hence belongs to B(Rm)⊗B(Rn). Since [L1] says a Borel sigma-algebra is generated by the open sets, this gives the reverse inclusion. Combining with step 1.1 proves the equality.

Depends on

Used by

Dependency tree · two levels

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Sources