# 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!

• Am I correct in thinking that you’re considering only finite products (since otherwise your description of the product topology isn’t correct)? – Brian M. Scott Nov 30 '12 at 13:38
• @BrianM.Scott: I almost forgot the distinction between box topology and product topology. I meant the most popular one, which I think is product topology? Is it similar for sigma algebras to have box sigma algebra and product sigma algebra? Still I meant the most popular one. But if I can know the answers in both cases, that will be even greater. – Tim Nov 30 '12 at 13:50
• "[...] and then generate the smallest topology". This is sort of correct, but it would more precise to add "such that each projection onto each factor is continuous", i.e. the product topology: the initial topology w.r.t. the projections. – kahen Nov 30 '12 at 13:51
• Thanks, @user49437! That is a very good post! – Tim Nov 30 '12 at 18:57
• @kahen: Thanks! I was not very clear and yes, I meant product topology and product sigma algebra, instead of box ones. – Tim Nov 30 '12 at 18:58

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.

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 $A\in\mathcal{C}$. 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.

• Thanks, Michael! In the definition of strong mixing for a stochastic process, I wonder why the sigma algebra on the set of sample paths is the Borel sigma algebra of the product topology, but not the product Borel sigma algebra? See the first paragraph of strong mixing in the Wikipedia article here en.wikipedia.org/wiki/… – Tim Dec 2 '12 at 20:25
• Maybe the author thinks that the product topology is more elementary than the product $\sigma$-algebra? Either way, in that situation, it doesn't matter. – Michael Greinecker Dec 3 '12 at 8:02
• "elementary" in what sense? Why it doesn't matter? – Tim Dec 3 '12 at 8:03
• Elementary in the sense that more people know it. It doesn't matter because the two notions coincide in this specific case with countably many real-valued random variables. – Michael Greinecker Dec 3 '12 at 8:05

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).

• I just saw this as well mathoverflow.net/questions/39882/… – Learner Dec 1 '12 at 5:16
• +1. Thanks! The link was also mentioned in a link given by user49437. – Tim Dec 1 '12 at 14:04