The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}

If $A\subseteq {\mathbb{R}}^m$ and $B\subseteq {\mathbb{R}}^n$ are Borel, then $A\times B={\pi }_1^{-1}(A)\cap {\pi }_2^{-1}(B),$ so [L2] makes $A\times B$ Borel in ${\mathbb{R}}^{m+n}$. Therefore every measurable rectangle for $\mathcal{B}({\mathbb{R}}^m)\otimes \mathcal{B}({\mathbb{R}}^n)$ belongs to $\mathcal{B}({\mathbb{R}}^{m+n})$, and hence $\mathcal{B}({\mathbb{R}}^m)\otimes \mathcal{B}({\mathbb{R}}^n)\subseteq \mathcal{B}({\mathbb{R}}^{m+n})\mathrm{.}$

Let $O$ be open in ${\mathbb{R}}^{m+n}$ and let ${\mathcal{R}}_O$ be the family of all rational open boxes contained in $O$. By [L3], every $x\in O$ lies in a member of ${\mathcal{R}}_O$, so $O=\bigcup {\mathcal{R}}_O$; the reverse containment follows from its definition, including when $O$ is empty. This is a subfamily of one fixed countable basis, hence is at most countable; no choice of a box for each point is made. Each box ${\prod }_{j=1}^{m+n}(a_j,b_j)$ splits into the open rectangle $U\times V$, with $U={\prod }_{j=1}^m(a_j,b_j)$ and $V={\prod }_{j=m+1}^{m+n}(a_j,b_j)$. Both factors are Borel, so each box belongs to the product sigma-algebra. Closure under countable unions puts $O$ there as well. Since [L1] says the open sets generate the Euclidean Borel sigma-algebra, this proves the reverse inclusion. Combining with step 1.1 proves the equality, without any choice assumption.