Let $(\Omega_n, \tau_n)_n$ be a sequence of metrizable topological spaces. Let $\sigma (\tau_n)$ be the Borel $\sigma$-algebra on $\Omega_n$. Let $\Omega :=\prod_{n =1}^\infty \Omega_n$ and $\pi_n: \Omega \to \Omega_n$ be the canonical projection map. Let
- $\bigotimes_n \sigma (\tau_n)$ be the smallest $\sigma$-algebra on $\Omega$ such that all maps $\pi_n$'s are measurable.
- $\sigma (\bigotimes_n \tau_n)$ be the Borel $\sigma$ algebra on $\Omega$ that is generated by the product topology $\bigotimes_n \tau_n$ on $\Omega$.
It follows from this answer that below theorem is false without the assumption that each $\tau_n$ is second-countable. However, I manage to "prove it" without this assumption.
False Theorem: $\bigotimes_n \sigma (\tau_n) = \sigma (\bigotimes_n \tau_n)$.
Could you elaborate on where I made a mistake? Thank you so much!
My attempt: We have $\bigotimes_n \tau_n$ is metrizable and thus perfectly normal. Hence $\sigma (\bigotimes_n \tau_n)$ is the smallest $\sigma$-algebra on $\Omega$ such that every continuous function whose domain is $\Omega$ is measurable. We have $\pi_n$ is continuous w.r.t. $\bigotimes_n \tau_n$. Hence $\pi_n$ is measurable w.r.t. $\sigma (\bigotimes_n \tau_n)$. It follows that
$$
\bigotimes_n \sigma (\tau_n) \subset \sigma (\bigotimes_n \tau_n).
$$
Let's prove the reverse, i.e.,
$$
\sigma (\bigotimes_n \tau_n) \subset \bigotimes_n \sigma (\tau_n).
$$
Let
$$
\mathcal C := \{\pi^{-1}_n(A_n) \mid n \in \mathbb N, A_n \in \sigma (\tau_n)\}.
$$
Then $\bigotimes_n \sigma (\tau_n) = \sigma (\mathcal C)$. So we want to prove
$$
\sigma (\bigotimes_n \tau_n) \subset \sigma (\mathcal C).
$$
Let
$$
\mathcal D := \{\pi^{-1}_n(A_n) \mid n \in \mathbb N, A_n \in \tau_n\}.
$$
Then $\bigotimes_n \tau_n$ is the topology on $\Omega$ generated by $\mathcal D$. This also means that $\mathcal D$ is a subbase of $\bigotimes_n \tau_n$. This implies
$$
\sigma (\bigotimes_n \tau_n) = \sigma (\mathcal D) .
$$
The claim then follows trivially by the fact that $\mathcal D \subset \mathcal C$.
Best Answer
The problem is the last step. It is true that $\mathcal{D}$ is a subbase of $\bigotimes_n \tau_n$, but this does not imply they generate the same $\sigma$-algebra. Indeed, every element of $\bigotimes_n \tau_n$ is a union of finite intersections of elements of $\mathcal{D}$. These unions may not be countable, and so may not be in $\sigma(\mathcal{D})$. (This is fixed by the second-countable assumption, which can be used to show that every open set is actually a countable union of finite intersections of elements of $\mathcal{D}$.)