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Borel sigma-algebra of a product of second-countable spaces

Statement

Let X,Y be second-countable topological spaces and give X×Y the product topology. Then B(X×Y)=B(X)⊗B(Y).

Facts & Assumptions

Given: Second-countable topological spaces X,Y with their product topology and the product sigma-algebra.

[F1]

A second-countable space has an at most countable basis, and every open set is a union of basis elements. (Second countability: an at most countable basis for the topology)

[F2]

Finite products of at most countable sets are at most countable. (A product of two at most countable sets is at most countable)

[F4]

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

[F5]

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

Proof

Given: Second-countable spaces X,Y.

1.1givenF1F2F3F4

Choose at most countable bases U for X and V for Y using [F1]. Their product U×V is at most countable by [F2], and its members U×V form a basis for X×Y by [F3]. Each such rectangle lies in B(X)⊗B(Y); every open subset of X×Y is a union of a subfamily of this countable basis, hence a countable union of those rectangles. By [F4], this proves B(X×Y)⊆B(X)⊗B(Y).

2.1givenF3F4F5algebra∎

The coordinate projections πX:X×Y→X and πY:X×Y→Y are continuous by the definition of the product topology [F3]. For Borel A⊆X and C⊆Y, A×C=πX−1(A)∩πY−1(C) is Borel in X×Y. Thus every generator rectangle of the product sigma-algebra [F5] belongs to B(X×Y), giving the reverse inclusion.

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Sources