TITLE: Limit of Lebesgue measure of interesection of a set $E$ with its translation
QUESTION [6 upvotes]: Let $E$ be a Lebesgue measurable set in $\mathbb{R}$. Prove that
$$\lim_{x\rightarrow 0} m(E\cap (E+x))=m(E).$$

REPLY [4 votes]: First assume that $m(E)<\infty$. In this case you can write 
$$m(E\cap (E+x))=\int_{\mathbb R} \mathbf 1_{E}(u)\mathbf 1_E(u-x)=\mathbf 1_{E}*\mathbf 1_{-E} (x)\, .$$
Since $f=\mathbf 1_E\in L^1$ (because $m(E)<\infty$) and $g=\mathbf 1_{-E}\in L^\infty$, the convolution $f*g$ is an everyywhere defined $continuous$ function. (This is a nontrivial but quite well known fact). It follows that $\lim_{x\to 0} m(E\cap (E+x))=\mathbf 1_{E}*\mathbf 1_{-E}(0)=\int_{\mathbb R} \mathbf 1_E(u)\mathbf 1_E(u)\, du=m(E)$.
If $m(E)=\infty$, we have to show that $m(E\cap (E+x))\to \infty$ as $x\to 0$. Write $E$ as $E=\bigcup_n E_n$, where the sequence $(E_n)$ is increasing and $m(E_n)<\infty$ for all $n$. Then whe have $\liminf_{x\to 0} m(E\cap (E+x))\geq \liminf_{x\to 0} m(E_n\cap (E_n+x))=m(E_n)$ for all $n$, so $\liminf_{x\to 0} m(E\cap (E+x))\geq \sup_n m(E_n)=\infty$.