Borel sigma algebra on uncountable product and product sigma-algebra

Question

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.

Answer 1

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.