Is a finite product of Borel algebras also a Borel algebra? Let $X$ be a topological space. We denote the $\sigma$-algebra generated by the set of open subsets of $X$ by $\mathcal{B}(X)$.
Let $X$, $Y$ be topological spaces. We denote by $\mathcal{B}(X) \times \mathcal {B}(Y)$ the $\sigma$-algebra generated by sets of the form $M\times N$, $M\in \mathcal{B}(X)$, $N\in\mathcal{B}(Y)$.
My question is :

$\mathcal{B}(X)\times \mathcal{B}(Y) = \mathcal{B}(X\times Y)$?

The motivation came from my desire to explicitly construct the Haar measure on the idele group of $\mathbb{Q}$
 A: It's true that $\mathcal{B}(X) \times \mathcal{B}(Y) \subset \mathcal{B}(X \times Y)$.
To prove this it suffices to show that $C \times D \in \mathcal{B}(X \times Y)$ whenever $C \in \mathcal{B}(X)$ and $D \in \mathcal{B}(Y)$, since $\mathcal{B}(X) \times \mathcal{B}(Y)$ is generated by such sets.
In turn, it suffices to prove that $C \times Y \in \mathcal{B}(X \times Y)$ for all $C \in \mathcal{B}(X)$, and $X \times D \in \mathcal{B}(X \times Y)$ for all $D \in \mathcal{B}(Y)$. This would imply the preceding statement because $C \times D = (C \times Y) \cap (X \times D)$.
Define $\mathcal{M}:= \{ C \in \mathcal{B}(X) : C \times Y \in \mathcal{B}(X \times Y)\}$. Easily enough, $\mathcal{M}$ is a sigma algebra which contains all of the open sets of $X$, and thus $\mathcal{M}=\mathcal{B}(X)$. It follows that $C \times Y \in \mathcal{B}(X \times Y)$ for all $C \in \mathcal{B}(X)$. The proof that $X \times D \in \mathcal{B}(X \times Y)$ for all $D \in \mathcal{B}(Y)$ is very similar.
I also claim that if $X$ and $Y$ are both second-countable, then $\mathcal{B}(X \times Y) \subset \mathcal{B}(X) \times \mathcal{B}(Y)$.
To prove this, let $X$ and $Y$ both be second-countable, and choose countable bases $\Gamma_X$ and $\Gamma_Y$ for $X$ and $Y$, respectively. Let $\Phi = \{ U \times V : U \in \Gamma_X$ and $V\in \Gamma_Y \}$. Then $\Phi$ is a countable basis for $X \times Y$.
I claim that $\Phi$ is a generating set for $\mathcal{B}(X \times Y)$. To prove this, it suffices to show that every open set in $X \times Y$ lies within the sigma algebra generated by $\Phi$. But this is true because every open set in $X \times Y$ can be written as the countable union of sets in $\Phi$, since it is a countable basis.
Since $\Phi$ generates $\mathcal{B}(X \times Y)$, and because $ \Phi \subset \mathcal{B}(X) \times \mathcal{B}(Y)$, it easily follows that $\mathcal{B}(X \times Y) \subset \mathcal{B}(X) \times \mathcal{B}(Y)$.
