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!
 A: 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 $F$.
A: I don't see the meaning of constructing the $\sigma$-algebra $A$ in Mike's answer. Well, I think actually we can avoid that redundant construction.
What we need to prove is just: If $E$ is a Borel set in $\mathbb R^{d_1}$, then $E\times \mathbb R^{d_2}$ is a Borel set in $\mathbb R^{d_1+d_2}$.
Proof: This is simple. Because the set in the second position of the Cartesian product is the whole space $\mathbb R^{d_2}$, we can interchange the operations of taking Cartesian product and taking union/intersection. We can prove a similar result for $F$ in the same way. And therefore, if $E$ is a Borel set in $\mathbb R^{d_1}$ and $F$ is a Borel set in $\mathbb R^{d_2}$, then $E\times F(=[E\times \mathbb R^{d_2} ]\cap [\mathbb R^{d_1}\cap F]\,)$ is a Borel set in $\mathbb R^{R_1+R_2}$.
$\tag*{$\square$}$
