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

Let $X=X_{1}\times X_{2}$ is locally compact space, and define $$E=\{E_{1}\times E_{2}\;|\; E_{i}\; \text{is a Borel set in}\; X_{i}\; ,\; \text{for}\; i=1,2\}$$ 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.

• I deleted a straightforward answer based on the Stone-Weierstrass theorem because apparently AmirHosein doesn't assume the Hausdorff condition. Please specify what you mean by local compactness: a compact neighborhood for every point? a base of compact neighborhoods for every point? a base of closed compact neighborhoods for every point? What is your definition of the Baire $\sigma$-algebra? The one generated by the zero sets of continuous functions or the one generated by (closed?) compact $G_\delta$-sets? – Martin Jan 2 '13 at 22:49
• – Martin Jan 5 '13 at 15:05

Original MO link: Baire sets of $$X$$ possess the required Cartesian product property (as mentioned in the comment)
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.