Lemma 2.2.89F2BA6481FD8

Proof. Let $p_1$ and $p_2$ be the projections of $Z_1\times Z_2$ onto its factors. For $i=1,2$, the sets $E\subset Z_ i$ with $p_ i^{-1}(E)$ Borel form a $\sigma $-algebra which contains the open sets, because $p_ i$ is continuous; hence it contains all Borel sets. Thus $A_1\times A_2=p_1^{-1}(A_1)\cap p_2^{-1}(A_2)$ is Borel whenever $A_1$ and $A_2$ are Borel. Conversely, we may assume that $Z_1$ and $Z_2$ are nonempty, and we choose countable dense subsets $D_1\subset Z_1$ and $D_2\subset Z_2$. Let $W\subset Z_1\times Z_2$ be open and $(z_1,z_2)\in W$. There is $r>0$ with $B(z_1,r)\times B(z_2,r)\subset W$, where $B(z_ i,r)$ is the open ball in $Z_ i$. Choose $a_ i\in D_ i$ with $\operatorname {dist}(a_ i,z_ i)<r/3$ and a rational number $q\in (r/3,2r/3)$; then $z_ i\in B(a_ i,q)\subset B(z_ i,r)$ by the triangle inequality. Hence $W$ is the union of those of the countably many products $B(a_1,q_1)\times B(a_2,q_2)$, with $a_ i\in D_ i$ and positive rational $q_ i$, that are contained in $W$. Each of them is a product of open sets, so every open subset of $Z_1\times Z_2$, and therefore every Borel subset, belongs to the $\sigma $-algebra generated by the products of Borel sets. $\square$