Measure of intersection of a set with its translate I'm trying to prove that if $A \subset \mathbb{R}^d$ is a Lebesgue mensurable set with $\mu(A) \neq 0$ and $\mu(A) < \infty$, then $$\lim_{\| x \| \to 0} \mu(A \cap (A+x)) = \mu(A).$$
I think I can prove this for parallelepipeds, but I'm unable to make it for other sets. Can anyone help me?
 A: First consider the case of an open set $U$.
Because $U$ is open, for all $t\in U$, $\lim\limits_{\|x\|\to 0}1_{U+x}(t)=1_U(t)=1$.  It follows that $$\lim\limits_{\|x\|\to 0}1_U(t)1_{U+x}(t)=1_U(t)$$ for all $t\in\mathbb R^d$. By the dominated convergence theorem1, $$\lim\limits_{\|x\|\to 0}\int_{\mathbb R^d}1_U(t)1_{U+x}(t)\,d\mu(t)=\int_{\mathbb R^d}1_U(t)\,d\mu(t).$$ In other words, 
$$\lim\limits_{\|x\|\to 0}\mu(U\cap(U+x))=\mu(U).$$

To reduce the general case to this, Let $\varepsilon>0$ and let $U$ be an open set such that $\mu(U\setminus A)<\varepsilon$.  For all $x$, $$[U\cap(U+x)]\setminus[A\cap(A+x)]= [(U\setminus A)\cap(U+x)]\cup[U\cap((U+x)\setminus(A+x)],$$ so $\mu([U\cap(U+x)]\setminus[A\cap(A+x)])<2\varepsilon$.  It follows that if $\mu((U\cap(U+x))>\mu(U)-\varepsilon$ for sufficiently small $\|x\|$, then $\mu((A\cap(A+x))>\mu(A)-3\varepsilon$ for sufficiently small $\|x\|$. Hence it suffices to show that the result is true for the open set $U$, as done above.

1 In case the use of dominated convergence for a continuous family of functions rather than a sequence as usual seems troubling, note that we can apply it with an arbitrary sequence $(x_n)$ in $\mathbb R^d$ converging to $0$, and having the equality hold for all such sequences is equivalent to equality for the continuous limit.
