Is $\mathcal B(\mathbb R^n)$ generated by $\{E_1\times\cdots\times E_n:E_1,\ldots,E_n\in\mathcal B(\mathbb R)\}$? I have a question which someone already asked but got no answer. Here is the link:
Question
In the question, $\mathbb R^n$ is given the standard topology $\mathcal T(\mathbb R^n)$ and $B(\mathbb R^n)$ is the smallest $\sigma$-algebra on $\mathbb R^n$ containing $\mathcal T(\mathbb R^n)$. A similar thing is true for $\mathbb R$.
I tried to prove it referring to Crostul's comment. Here's my attempt:
By definition, $\mathcal B(\mathbb R^n)=\sigma(\mathcal T(\mathbb R^n))$. So what we should prove is $\sigma(\mathcal A)=\sigma(\mathcal T(\mathbb R^n))$. Let $\mathcal C=\{(a_1,b_1)\times\cdots\times (a_n,b_n):a_1,\ldots,a_n,b_1,\ldots,b_n\in\mathbb Q,a_1<b_1,\ldots,a_n<b_n\}$. I proved that $C$ is a basis for $T(\mathbb R^n)$. Since $C\subseteq\mathcal T(\mathbb R^n)$, it follows that
$$\sigma(\mathcal C)\subseteq\sigma(\mathcal T(\mathbb R^n)).\qquad(1)$$
Since $\mathcal C$ is a countable basis for $\mathcal T(\mathbb R^n)$, every set in $\mathcal T(\mathbb R^n)$ is a countable union of sets in $\mathcal C$. Since $\sigma(\mathcal C)$ has all countable unions of sets in $C$, it follows that $\mathcal T(\mathbb R^n)\subseteq\sigma(\mathcal C)$. Since $\sigma(\mathcal T(\mathbb R^n))$ is a smallest $\sigma$-algebra containing $\mathcal T(\mathbb R^n)$, we have
$$\sigma(\mathcal T(\mathbb R^n))\subseteq\sigma(\mathcal C).\qquad (2)$$
By (1) and (2), we have $\sigma(\mathcal C)=\sigma(\mathcal T(\mathbb R^n))$. So if we prove $\sigma(\mathcal A)=\sigma(\mathcal C)$, then we are done. $\sigma(\mathcal C)\subseteq\sigma(\mathcal A)$ is obvious, since $\mathcal C\subseteq\mathcal A$. So it suffices to prove $\mathcal A\subseteq\sigma(\mathcal C)$.
I'm stuck with proving $\mathcal A\subseteq\sigma(\mathcal C)$. Please help me with completing my proof or show me another way.
 A: Let $\mathcal{C}=\{\prod\limits_{i=1}^n E_i: E_1,...,E_n\in \mathcal{B}(\mathbb{R})\}$. Since $\mathcal{C}$ in particular contains all sets of the form $\prod\limits_{i=1}^n (a_i, b_i)$ and all open sets of $\mathbb{R}^n$ are countable unions of such products, it follows that $\mathcal{B}(\mathbb{R}^n) \subset\sigma(\mathcal{C})$. Denoting the projection of $\mathbb{R}^n$ onto the $i$th factor by $p_i$, we see that $\{A\subset \mathbb{R}: p_i^{-1}(A)\in \mathcal{B}(\mathbb{R}^n)\}$ is a $\sigma$-algebra that contains the open sets, hence it contains $\mathcal{B}(\mathbb{R})$, i.e. $p_i^{-1}(E)\in \mathcal{B}(\mathbb{R}^n)$ for $E\in \mathcal{B}(\mathbb{R})$. But then, if $E_1, E_2,...$ up to $E_n$ are Borel-sets, we see that $\prod\limits_{i=1}^n E_i=\bigcap\limits_{i=1}^n p_i^{-1}(E_i)\in \mathcal{B}(\mathbb{R}^n)$ so that $\mathcal{C}\subset \mathcal{B}(\mathbb{R}^n)$. Since $\mathcal{B}(\mathbb{R}^n)$ is a $\sigma$-algebra, it holds that $\sigma(\mathcal{C})\subset \mathcal{B}(\mathbb{R}^n)$ and both desired inclusions are established.
