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The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}

Statement

Let $m, n \geq 1$ . Under the identification $R^{m + n} = R^{m} \times R^{n},$ one has $B (R^{m}) \otimes B (R^{n}) = B (R^{m + n}) .$

Facts & Assumptions

Given: Positive integers $m, n$ and the identification $R^{m + n} = R^{m} \times R^{n}$ .

[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 $π_{1} : R^{m + n} \to R^{m}$ and $π_{2} : R^{m + n} \to R^{n}$ are continuous.

[A2]

Every open set in $R^{m + n}$ is a countable union of open rectangles $U \times V$ with $U \subseteq R^{m}$ and $V \subseteq R^{n}$ open.

Proof

technique · direct

1.1

If $A \subseteq R^{m}$ and $B \subseteq R^{n}$ are Borel, then $A \times B = π_{1}^{- 1} (A) \cap π_{2}^{- 1} (B),$ so [L2] makes $A \times B$ Borel in $R^{m + n}$ . Therefore every measurable rectangle for $B (R^{m}) \otimes B (R^{n})$ belongs to $B (R^{m + n})$ , and hence $B (R^{m}) \otimes B (R^{n}) \subseteq B (R^{m + n}) .$

A1L2

2.1

By [A2], every open set in $R^{m + n}$ is a countable union of open rectangles, hence belongs to $B (R^{m}) \otimes B (R^{n})$ . 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.

step 1.1A2L1∎

Depends on

Used by

Dependency tree · two levels

25 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

John K. Hunter, Measure Theory, Proposition 5.3 (standard reference, not scraped)