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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.

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    $\begingroup$ This is false. It is true if $I$ is at most countable. $\endgroup$
    – Kavi Rama Murthy
    Apr 8, 2020 at 9:12
  • $\begingroup$ 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) $\endgroup$
    – SBK
    Jul 27 at 14:39

1 Answer 1

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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.

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