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Finite disjoint unions of measurable rectangles form an algebra generating the product sigma-algebra

Statement

Let (X,A) and (Y,B) be measurable spaces. The family of finite disjoint unions of measurable rectangles in X×Y is an algebra of subsets of X×Y, and it generates AB.

Facts & Assumptions

Given: Measurable spaces (X,A) and (Y,B).

[L1]

A measurable rectangle has the form A×B with AA and BB. (Measurable rectangles in a product of measurable spaces)

[L2]

The product sigma-algebra AB is the sigma-algebra generated by the measurable rectangles. (The product sigma-algebra and its finite iterates)

[A1]

For rectangles, (A1×B1)(A2×B2)=(A1A2)×(B1B2), and (X×Y)(A×B)=(Ac×Y)(A×Bc).

[A2]

If Rj=Aj×Bj for 1jn, take the nonempty Boolean atoms generated by A1,,An in X and by B1,,Bn in Y. The products of an X-atom and a Y-atom are finitely many pairwise disjoint measurable rectangles partitioning X×Y, and each Rj, hence also j=1nRj, is the union of a subfamily of these product atoms.

Proof

technique · direct
1.1

By [L1] and [A1], the intersection of two measurable rectangles is again a measurable rectangle, and the complement of a measurable rectangle is a finite union of measurable rectangles.

L1A1
2.1

Let R be the family of finite disjoint unions of measurable rectangles. It contains and X×Y. If E,FR, then [A2] disjointifies the finite union EF into finitely many pairwise disjoint measurable rectangles, so EFR. Likewise EF=EFc belongs to R by step 1.1. Thus R is an algebra.

step 1.1A2
3.1

Every measurable rectangle belongs to R, so [L2] gives AB=σ(measurable rectangles)σ(R). The reverse inclusion holds because every member of R is a finite union of measurable rectangles and hence lies in AB. Therefore σ(R)=AB, and R is an algebra generating the product sigma-algebra.

L2step 2.1

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Dependency tree · two levels

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