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Let $ A, B \subseteq \mathbb {R} $ be Lebesgue measurable sets such that at least one of them has finite measure. Let $ f $ be the function defined by $$f (x) = m ((x + A) \cap B)$$ for each $ x \in \mathbb{R} $. Show that $ f $ is continuous.

Hint: Suppose first that $ A $ and $ B $ are intervals and then generalized to arbitrary sets using the regularity of the Lebesgue measure.

Some help please. Thanks.

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Can you prove it when $A$ and $B$ are both intervals? – Arturo Magidin Apr 16 '12 at 2:48
  1. We can assume that both $A$ and $B$ are open sets of finite measure.

    Denote $f_{A,B}(x):=\lambda((x+A)\cap B)=\int_\mathbb R\chi_B(y)\chi_{x+A}(y)\lambda(\mathrm dy)$. Assume without loss of generality that $B$ has finite measure. By outer regularity, there is a sequence $(O_k)_{k\geqslant 1}$ of open sets such that $B\subset O_k$ and $\lambda(O_k\setminus B)\leqslant k^{-1}$. We thus have $$\sup_{x\in\mathbb R}|f_{A,B}(x)-f_{A,O_k}(x)|\leqslant k^{-1},$$ and it therefore suffice to show the result when $B$ is an open set of finite measure.

    Considering $A\cap [-N,N]$ instead of $A$, we can also assume that $A$ is of finite measure. Since $f_{B,A}(x)=f_{A,B}(-x)$, we can also assume that $A$ is an open set of finite measure.

  2. An open set is a countable union of disjoint intervals hence we have to prove the result when $A$ is a finite union of disjoint intervals, then use again an approximation argument.

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Is this also true for $\mathbb{R}^n$? – Kaa1el Mar 26 '14 at 7:40

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