# Baire sets of $X$ possess the required Cartesian product property

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.

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math.stackexchange.com/q/268533 –  Martin Jan 3 '13 at 8:24

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.

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