Is "product" of Borel sigma algebras the Borel sigma algebra of the "product" of underlying topologies? A Borel sigma algebra is the smallest sigma algebra generated by a topology.
The "product" of a family of Borel sigma algebras is to first take the Cartesian of the Borel sigma algebras, and then generate the smallest sigma algebra.
Similarly, the "product" of a family of topologies is to first take the Cartesian of the topologies, and then generate the smallest topology.
Is the "product" of some Borel sigma algebras the Borel sigma algebra for the "product" of their underlying topologies?
Thanks!
 A: It is asserted (with a short proof) in Billingsley's Convergence of Probability measures (second edition) on page 244, that this holds if the underlying spaces are separable. (He only considers the product of two spaces).
A: The Borel $\sigma$-algebra of the countable product of second countable topological spaces is the product of the Borel $\sigma$-algebras. it is always true that the Borel $\sigma$-algebra of the topological product space is at least as large as the product of the $\sigma$-algebras. Proofs of theis can be found for example in Kallenberg's book (see Lemma 1.2).
The Borel $\sigma$-algebra of the uncountable product of nontrivial (at least two points) Hasudorff spaces is always larger than the product of the $\sigma$-algebras. To see this, note that every point is closed in the product topology and therefore a Borel set. But by the construction of the product $\sigma$-algebra, a set can depend only on countably many coordinates. More precisely, there is a general result that if $A\in\sigma(\mathcal{F})$ then there is a countable family $\mathcal{C}\subseteq\mathcal{F}$ such that $A\in \sigma(\mathcal{C})$. To prove this, just check that the sets generated by a countable subfamily of $\mathcal{F}$ give you a $\sigma$-algebra containing $\mathcal{F}$. In particular, every set in the product $\sigma$-algebra is generated by countably many measurable rectangles.
