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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!

Martin Sleziak
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Tim
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    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
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    "[...] 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
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    +Brian I don't mean to nitpick, but for clarity for later readers, the standard product $\sigma$-algebra (generated by cylinders) and the "box" $\sigma$-algebra (generated by rectangles) are still the same for countably infinite products (because you can intersect countably many cylinders to get the rectangles). If you're indexing uncountably many dimensions, every cylinder is an "infinite" rectangle, but you can not intersect uncountably many cylinders to get the smaller rectangles anymore, so the product $\sigma$-algebra is properly contained in the "box" $\sigma$-algebra. – Travis Bemrose Dec 02 '15 at 12:18

2 Answers2

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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.

ABIM
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Michael Greinecker
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  • 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 http://en.wikipedia.org/wiki/Mixing_%28mathematics%29#Mixing_in_stochastic_processes – Tim Dec 02 '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 03 '12 at 08:02
  • "elementary" in what sense? Why it doesn't matter? – Tim Dec 03 '12 at 08:03
  • 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 03 '12 at 08:05
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    Is there a counterexample for finite products of non-2nd-countable spaces? – Cronus Oct 20 '20 at 16:46
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    @Cronus Yes, just take a Hausdorff space with cardinality larger than the continuum and take the product with itself. The diagonal is closed and hence a Borel set in the product but is not in the product $\sigma$-algebra. – Michael Greinecker Oct 20 '20 at 22:07
  • Regarding your first point, would that be Lemma 1.2 in Kallenberg's book? I'm asking because here he assumes that the spaces are separable metric spaces, but as far as I can tell he only uses second countability and not the metrics at all. – Danny Feb 02 '21 at 21:36
  • @DannyHansen Yes. I think Kallenberg simply has no need for nonmetrizable topological spaces. – Michael Greinecker Feb 02 '21 at 21:42
  • @Michael Greinecker The link is not working, can you please give us a new one? – Carlo Mantegazza Oct 27 '23 at 05:36
  • @CarloMantegazza It works when I click on it. Anyway, it is the book "Foundations of Modern Probability." – Michael Greinecker Oct 27 '23 at 07:20
  • @Michael Greinecker Thanks – Carlo Mantegazza Oct 28 '23 at 09:43
  • ``it is always true that the Borel -algebra of the topological product space is at least as large as the product of the -algebras." Where can I find a proof of this fact? Kallenberg only gives the proof for countably many second countable spaces – Atom Feb 06 '24 at 23:46
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    @Atom A function between two topological spaces is Borel-to-Borel measurable if and only if the preimage of every open set is measurable. So, if every projection is continuous (as in the product topology), it is certainly Borel measurable. And the product $\sigma$-algebra is the smallest one making every projection measurable. – Michael Greinecker Feb 06 '24 at 23:52
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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).

Learner
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