Let $\mathbb R^{\mathbb N}=\mathbb R\times\mathbb R\times\ldots$ be the space of all real sequences and endow it with product topology. Is the product $\sigma$-algebra generated by Borel subsets of $\mathbb R$ the same as the Borel $\sigma$-algebra generated by the product topology: $$\mathscr B(\mathbb R)\otimes\mathscr B(\mathbb R)\otimes\ldots=\mathscr B(\mathbb R\times\mathbb R\times\ldots)?$$ More generally, it is true if $\mathbb R$ is replaced by a second-countable topological space? If not, does at least $\subset$ or $\supset$ hold? What about uncountable products?

It is quite well-known that the claim is true for finitely many products (even for general second-countable topological space), but I can't seem to find a proof or disproof for the (un)countably infinite case. Any input is appreciated.

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    $\begingroup$ What exactly is an infinite product of borel algebras? I mean, is it the set of all $(U_i)_i:\,\,U_i$ open in $\mathbb{R}$, or just the set of such products with finitely many $U_i\neq\mathbb{R}$? $\endgroup$ – Peter Franek Apr 5 '15 at 7:49
  • $\begingroup$ The right hand side equals to the $\sigma$-algebra generated by cylinder sets. You might find p. 9-I, Remarks, A Probabilities and Potential By C. Dellacherie, P.-A. Meyer interesting. $\endgroup$ – shall.i.am Oct 3 '16 at 5:43

Yes, this holds for countable products of second-countable topological spaces. See 2.1.3 in these notes: https://web.ma.utexas.edu/mp_arc/c/02/02-156.pdf


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