Continuity of shifted measure function I originally wanted to show that if $E \subset \mathbb{R}$ is a Lebesgue measurable set of finite measure and we define $E_t = \{ x+t : x \in E \}$, then the function $f(t) = m(E \cap E_t)$ where $m$ is Lebesgue measure is continuous. I have reduced this to the case where $E$ is open using regularity and a uniform convergence argument. I have also proved that it is true when $E$ is an interval. However I have a problem with generalizing from intervals to open sets. Specifically, when multiple intervals intersect simultaneously, the behavior becomes hard to manage. My best idea so far has been to try to apply Arzela-Ascoli, but again if I get too many sets intersecting at once I have trouble bounding the derivative. Do you have any suggestions? 
Edit: I have reduced the problem further to specifically showing that $f$ is continuous at $0$, by saying that if it is continuous at 0 then you may apply the same argument with $E$ replaced by $E_t$ to show that the original $f$ is continuous at $t$. I still have some problems with having countably many intervals, however. For example, consider $E=\bigcup_{n=1}^\infty (n,n+1/n^p)$ for $p>1$. Then for $1/(n+1)^p < t < 1/n^p$ we have $f(t) = m(U)-nt$, which is continuous (since for $t < 1/n^p$ we have $|f(t)-f(0)|<1/n^{p-1}$) but clearly not differentiable. This means my Arzela-Ascoli theorem idea will be hard at best.
 A: I worked this out. First we note that we merely need to show that $f$ is continuous at $0$ for any measurable set of finite measure, because elsewhere we can reset $E=E_t$ and the same argument will apply. (Edit: I've talked to some classmates, and this previous step may not be quite right.) Now to do this, let $\varepsilon>0$. We obtain a finite union $U=\bigcup_{n=1}^N I_n$ of open intervals such that $m(E \Delta U)<\varepsilon/3$, where $\Delta$ is symmetric difference. We may assume because the union is finite that $\alpha = \min \, d(I_m,I_n)>0$ for $m \neq n$ (otherwise we may merge intervals without affecting the measure, in effect decomposing the interior of the closure of $U$ into intervals.)
Then $(\forall \delta > 0) \quad m(E \cap E_0)-m(E \cap E_\delta) = m(E - E_\delta)$. Now $E - E_\delta = (U - U_\delta) \cup (E - U) \cup (U_\delta - E_\delta)$. So $m(E - E_\delta) \leq m(U-U_\delta) + 2\varepsilon/3$, using containment, subadditivity and translation invariance. Then for $\delta < \min \{ \varepsilon/3N , \alpha\}$ we get $m(U - U_\delta) < \varepsilon/3$; this is essentially because if we're less than $\alpha$, the overlap is only between an interval and its own shifted copy, in which case the lost measure will be less than $\varepsilon/3N$ on each interval, hence adding up to less than $\varepsilon/3$. Then $m(E - E_\delta) < \varepsilon$, so the function is continuous at zero independent on $E$, consequently it is continuous everywhere. 
A: Lemma: Let $O\subset \mathbb{R}$ be an open set, then $O$ is a disjoint union of a countable collection of open intervals; this representation is unique.
Pf: for $x\in O$ let $I_x$ be the largest open interval about $x$ that is contained in $O$. Main observation is that for $x,y\in O$ either $I_x = I_y$ of $I_x\cap I_y =\emptyset$. Countability follows by observing that any open interval contains a rational number.
