Cartesian Product of Borel Sets is Borel Again

Let $E$ and $F$ be Borel measurable subsets of $\mathbb R^{d_1}$ and $\mathbb R^{d_2}$, respectively. Then $E \times F$ is also Borel measurable in $\mathbb R^{d_1 + d_2}$.

I suppose it is necessary to show that elements of a generator of Borel $\sigma$-algebra on $\mathbb R^{d_1 + d_2}$ is Borel. That is, show that rectangles, i.e., Cartesian product of intervals, are Borel. It seems that I need to use Fubini' theorem. But I do not know how. Any help, please? Thank you!

Solution 1:

Hint: Let $A = \left\{ E \subseteq \mathbb{R}^{d_1} \mid \text{$E \times \mathbb{R}^{d_2}$ is Borel in $\mathbb{R}^{d_1 + d_2}$} \right\}$. Show that $A$ is a $\sigma$-algebra containing all open sets of $\mathbb{R}^{d_1}$, hence all Borel sets. Then make a similar argument for $F$.