Limit of Lebesgue measure of interesection of a set $E$ with its translation Let $E$ be a Lebesgue measurable set in $\mathbb{R}$. Prove that
$$\lim_{x\rightarrow 0} m(E\cap (E+x))=m(E).$$ 
 A: 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 $\liminf_{x\to 0} m(E\cap (E+x))\geq \sup_n m(E_n)=\infty$.
A: Recall that a Lebesgue measurable set $E$ can be approximated arbitrarily well by a finite union of disjoint closed intervals $S=I_1\cup\ldots\cup I_n$, i.e. $\epsilon=m(E\Delta S)$ can be made arbitrarily small. Thus we have that
$$\begin{align}
|m(E\cap (E+x))-m(S\cap (S+x))|&\leq m((E\cap (E+x))\Delta(S\cap (S+x)))\\
&\leq m(E\Delta S)+m((E+x)\Delta(S+x))\\
&\leq 2m(E\Delta S) = 2\epsilon
\end{align}$$
can be made arbitrarily small, so it suffices to show that $\lim\limits_{x\to 0} m(S\cap (S+x))=m(S)$.
Note that for sufficiently small $x$, the intervals $I_i$ and $I_j$ for $i\ne j$ are separated by more than $x$, so $I_i\cap (I_j+x)=\emptyset$. Thus for sufficiently small $x$ we have 
$$S\cap (S+x)=(I_1\cap (I_1+x))\cup \cdots \cup (I_n\cap (I_n+x))$$
and each set in the union is disjoint. Observing that the desired result clearly holds for intervals, we have
$$\begin{align}
\lim\limits_{x\to 0}m(S\cap (S+x))&=\lim\limits_{x\to 0}\sum\limits_{i=1}^nm(I_i\cap (I_i+x))\\
&= \sum\limits_{i=1}^n\lim\limits_{x\to 0} m(I_i\cap (I_i+x))\\
&= \sum\limits_{i=1}^nm(I_i)=m(S).
\end{align}$$  
