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While studying old real analysis qualifying exams, I've gotten stuck trying to solve the following problem. Any hints would be appreciated.

If $E$ is a Lebesgue measurable subset of $\mathbb R$, show that $$ \lim_{t \to 0} \mu( E \cap (E+t) ) = \mu(E). $$ Here $\mu$ denotes Lebesgue measure.

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Certainly the limit is at most $\mu(E)$. Let $(a_1,b_1),\ldots,(a_n,b_n)$ be intervals with union $I$ such that $\mu(E\Delta I)<\epsilon$ where $\Delta$ denotes symmetric difference. Then $$\begin{align} \mu(E\cap (E+t)) &\ge \mu(I\cap (I+t))-\mu(I\setminus E)-\mu((I+t)\setminus (E+t))\\ &>\sum_{i=1}^n (b_i-a_i-t) -2\epsilon\\ &= \mu(I)-nt-2\epsilon>\mu(E)-nt-3\epsilon \end{align}$$ and letting $\epsilon,t\to 0$ gives the desired result.

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  • $\begingroup$ Should the last term on the first line be $\mu((I+t)\setminus(E+t))$? $\endgroup$ – Cardboard Box Dec 31 '13 at 16:30
  • $\begingroup$ @Dane Yes, thanks. $\endgroup$ – Alex Becker Dec 31 '13 at 19:07
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If $\mu E < \infty$, apply dominated convergence to $$ \int \chi_{E}(x)\chi_{E}(x-t)d\mu(x) = \mu (E\cap(E+t)). $$

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    $\begingroup$ But now we have to show that $\chi_E(x-t) \to \chi_E(x)$ for almost every $x$. That's not so easy to do from scratch. $\endgroup$ – Nate Eldredge Dec 31 '13 at 1:53
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    $\begingroup$ Outer measure is from scratch, and you can show that $\phi(x-t)$ converges pointwise a.e. to $\phi(x)$ as $t\rightarrow 0$ if $\phi$ is a finite sum of characteristic functions of intervals, even if the intervals overlap. So, it's not too difficult with a standard 3-epsilon type of argument. $\endgroup$ – Disintegrating By Parts Dec 31 '13 at 3:24

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