Q: Prove that a product of Borel algebras is a sigma algebra I was working on the following question and the very last step in my proof required the following:

If $X_1, \dots, X_n$ are separable metric spaces, then $\prod\limits_{i=1}^n \mathcal{B}(X_i)$ is a sigma algebra.

How do I show this? I'm asking a new question because this short claim is much simpler than what I originally asked. I no longer need people to answer the original complicated question, just this simpler and weaker statement. At the same time, I want to leave the other question available on its own to help people in the future who are looking for the full problem.

A: You cannot prove this. $\mathbb R^{2} \setminus [(0,1) \times (0,1)]$ cannot be expressed as  a product $A \times B$ for any sets $A$ and $B$. Hence the collection of sets $A \times B$ such that $A$ and $B$ are Borel sets in $\mathbb R$ is not  a sigma algbra.
