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$m$ is Lebesgue measure and $E$ is any Lebesgue measurable set. It is shown here that $f(x)=m(E\cap(E+x))$ is continuous at $0$ (even if $m(E)=\infty$, $\lim_{x\to 0} f(x)=f(0)$).

This is Exercise 3.4.16(ii) in Srivastava's "A Course on Borel Sets." The hint given is to use the monotone class theorem.

If $E=\bigcup_0^{\infty}[2n,2n+1]$ or $E=[0,2]\cup\bigcup_1^{\infty}[2n+1,2n+2]$, then the range of $f$ is $\{0,\infty\}$, $\{1,\infty\}$, respectively.

Assuming additionally that $m(E)<\infty$, how does one prove, using the monotone class theorem, that $f$ is continuous?

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This is my guess (and should be taken as a comment, and not a solution) :

  1. Prove it when $E = (a,b)$ is a bounded open interval. In that case, $f(x) = b-a-|x|$.

  2. If $E_1\subset E_2\subset \ldots$ and $E = \cup E_n$, then the corresponding sequence of functions $\{f_i\}$ is increasing, and each $f_i$ is continuous. You want to conclude that limit function is also continuous.

    By intersecting with $[-n,n]$ for large enough $n$, perhaps you can assume that the convergence is happening inside a compact set. Then by Dini's theorem, the convergence is uniform.

  3. Do the same thing for $E_1\supset E_2\supset \ldots$.

Now you see that the collection of sets $E$ for which the corresponding function is continuous forms a monotone class. By the monotone class theorem, it must be a $\sigma$-algebra, and so must contain all Borel sets.

From here, perhaps you can use regularity of the measure to conclude this for all measurable sets $E$.

There are probably a thousand mistakes in what I have just said, but hopefully it is a decent starting point.

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