I am trying to show that given $E \subset \mathbb R^n$ measurable, then $\mathbb R^m \times E \subset \mathbb R^{m+n}$ is measurable but I don't have a clue how to show this. I could think of the following:
If $E$ is a measurable set, then $E=H \setminus N$, with $H$ a $G_{\delta}$ set and $N$ a null set, so $H= \bigcap_{n \in \mathbb N} G_n$, with $G_n$ is open.
So $$\mathbb R^m \times E= \mathbb R^m \times (H \setminus N)$$$$=(\mathbb R^m \times \bigcap_{n \in \mathbb N} G_n) \setminus (\mathbb R^m \times N)$$$$=\bigcap_{n \in \mathbb N} (\mathbb R^m \times G_n) \setminus (\mathbb R^m \times N)$$
So it is sufficient to prove the proposition for open and null sets. I would appreciate help to show the proposition for these two types of sets. Thanks in advance
For open sets, the product is also open.
For the case of null sets, use the definition of the Lebesgue outer measure to show that of $E $ is null, then so is $[-N,N]^m\times E $ for arbitrary $N $. Then take countable unions.