Let $X=X\_{1}\times X\_{2}$ is locally compact space, and define $$E=\{E\_{1}\times E\_{2}\mid E\_{i}\text{ is a Borel set in }X\_{i}\;,\text{ for}\; i=1,2\}$$ Now why the Baire sets of $X$ are in the $\sigma$-algebra generated by $E$? Of course that every Baire set is Borel too so all Baires of $X\_{i}$ is a Borel of it too.
Locally Compact: every point has a neighborhood with compact closure.


Baires: The smallest σ -algebra of subsets of $X$ such that each function in $C\_{C}(X)$ is measurable with respect to that.



Baire sets of $X$ possess the required Cartesian product property
The idea of the proof is to make use of compactness to show that sets of the form $f^{-1}(0)^{c}$ are countable unions of boxes in $E$.


For a proof, assume that $f:X^{2}\rightarrow[0,1]$ is a continuous function with compact support. Let $U=f^{-1}(0,1]$. Then since $f$ has compact support, the sets $f^{-1}[a,1]$ are compact for $a>0$, so the set $U$ is Lindelof. If $x\in U$, then by the definition of the product topology there are open sets $A\_{1}\subseteq X\_{1},A\_{2}\subseteq X\_{2}$ with $x\in A\_{1}\times A\_{2}\subseteq U$. If we set $A\_{x}=A\_{1}\times A\_{2}$, then $A\_{x}\in E$. Furthermore, by the Lindelof property, there is some $\{x\_{n}|n\in\mathbb{N}\}\subseteq U$ with $U=\bigcup\_{n}A\_{x\_{n}}$. Therefore, we have $U$ be in the $\sigma$-algebra generated by $E$ as well. Therefore since each the $\sigma$-algebra generated by $E$ contains each cozero set of compact support, the $\sigma$-algebra generated by $E$ contains each Baire set.


