If $E$ open set. Why $\left\{(x,y):x-y\in E\right\}$ is Lebesgue measurable?

Question

Without using that $f(x,y)=x-y$ is continuous. If $E$ is open set in $\mathbb{R}$. Why $\left\{(x,y)\in\mathbb{R}^2:x-y\in E\right\}$ is a Lebesgue measurable set? $\left\{(x,y)\in\mathbb{R}^2:x-y\in E\right\}$ is open set?

Answer 1

Let $S=\{(x,y)\mid x-y\in E\}$. Given a point $(a,b)\in S$. Then $a-b\in E$, and hence, there is a neighbourhood of $a-b$, say $]a-b-r, a-b+r[\subseteq E$ for some $r>0$. Then the open disc with radius $r/2$ aroud the center $(a,b)$ is contained in $S$: if $(x,y)$ is in this disc, then $a-r/2<x<a+r/2$ and $b-r/2<y<b+r/2$, thus $a-b-r<x-y<a-b+r$, hence $x-y\in E$.