This question is related to Prove that Borel sigma field on R(d) is the smallest sigma-field that makes all continuous functions f:R(d)->R measurable. I'm trying to prove this exercise in Durrett's book myself, and wonder if the following elementary arguments are valid:
First of all, for every continuous $f:\mathbb R^d\to \mathbb R$, $f^{-1}(E)$ is open, for any open set $E\in \mathbb R$. So clearly the Borel $\sigma$-field $\mathcal B^d$ makes all continuous function measurable.
To show the $\mathcal B^d$ is the smallest such $\sigma$-field $\mathcal F$, let $d=2$ for simplicity of arguments. Consider two continuous functions: $g(x,y)=x$ and $h(x,y)=y$. Note that $g^{-1}((a_1, b_1))=(a_1, b_1)\times(-\infty, \infty)$, and $h^{-1}((a_2, b_2))=(-\infty, \infty)\times(a_2, b_2)$. So $\mathcal F$ must contain $\{(a_1, b_1)\times(-\infty, \infty):a_1, b_1\in \mathbb R\}$ and $\{(-\infty, \infty)\times(a_2, b_2):a_2, b_2\in \mathbb R\}$, and hence $\{(a_1, b_1)\times(a_2, b_2):a_1, b_1, a_2, b_2\in \mathbb R\}$. This implies $\mathcal F\supset \mathcal B^2$ and hence $\mathcal F = \mathcal B^2$. Extension to $d>2$ or $d=1$ is obvious.
Is this proof correct? If there're flaws, I'd appreciate it if someone can point them out. Thanks a lot!
