Let $E \subseteq \mathbb R^n$ be Lebesgue measurable. Consider the set $E^- \subseteq \mathbb R^n \times \mathbb R^n$ defined by $$E^- : = \left \{(x,y) \in \mathbb R^n \times \mathbb R^n\ |\ x - y \in E \right \}.$$ Show that $E^-$ is Lebesgue measurable.
My Attempt $:$ I tried by writing $E^-$ as $$\begin{align*} E^- & = \bigcup\limits_{y \in \mathbb R^n} (E + y) \times \{y\}. \end{align*}$$
Now, any translate of a Lebesgue measurable set is also Lebesgue measurable and has the same Lebesgue measure. Singletons are clearly Lebesgue measurable with zero Lebesgue measure and cartesian product of Lebesgue measurable sets is again Lebesgue measurable with measure equals to the product of the respective measures. So for each $y \in \mathbb R^n$ the cartesian product $(E + y) \times \{y\}$ is Lebesgue measurable. However, the arbitrary union of Lebesgue measurable sets might not be Lebesgue measurable failing to which it's not possible to conclude that $E^-$ is Lebesgue measurable from the argument that I proposed. Is there any nicer way to conclude the same thing for which we don't run into trouble?
Thanks a lot for your kind attention.
EDIT $:$ Another idea is to take the Lebesgue measurable function $(x,y) \mapsto \chi_{E} (x - y)$ and then take inverse image of $\{1\}$ under this function which is precisely $E^-.$