# Borel sigma algebra on uncountable product and product sigma-algebra

Let $$\Omega_{i}$$ be metric spaces for $$i \in I$$ (uncountable). Consider two $$\sigma$$-algebras on the product space $$\Omega=\prod_{i \in I}\Omega_{i}$$:

1. The Borel $$\sigma$$-algebra on $$\Omega$$ which is the smallest $$\sigma$$-algebra containing all the open sets (with respect to the product topology). Call this $$\mathcal{B}(\Omega)$$

2. The $$\sigma$$-algebra generated by finite-dimensional/cylinder sets. Denote this by $$\sigma(\mathcal{C}_{f})$$.

In these lecture notes, on p.66, Exercise 26 asks to prove these $$\sigma$$-algebras are the same. One direction is easy. $$\mathcal{B}(\Omega) \subset \sigma(\mathcal{C}_{f})$$, as the basic open sets in the product topology belong to the class of finite-dimensional sets. I have no idea how to prove the other direction. Any help would be appreciated.

• This is false. It is true if $I$ is at most countable. Apr 8, 2020 at 9:12
• FWIW I think the exercise was stated incorrectly in the notes. Judging from the preceding pages, it seems he is trying to claim that the product of Borel $\sigma$-algebras is the same as the cylindrical $\sigma$-algebra. (Leave the Borel of the product out of it)
– SBK
Jul 27 at 14:39

Loosely speaking any set in $$\sigma (\mathcal C)$$ depends only on a countable number of coordinates. But an open set in $$\Omega$$ need not depend only on a countable number of coordinates. Hence the result is false.