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Let $A$ be bounded Borel subset of the real line with $\lambda(A)>0$ where $\lambda$ is the Lebesgue measure. How does one show that $f(x)=\lambda(A \cap (x+A)$ is continuous in $x$? My attempt:

Let $(x_n)$ be any sequence converging to $x$, then it suffices to show $f(x_n)$ converges to $f(x)$. In particular, if one shows $\mathbb{1}_{A \cap (x_n+A)}$ converges to $\mathbb{1}_{A \cap (x+A)}$ then by dominating convergence we are done. Unfortunately, if $x$ is an isolated point this does not seem work.

Is it possible to fix this argument? Much appreciated.

KCd
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Tanizaki
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    I think your argument can be fixed. Basically, you can show that when $A$ is an open set, we do have the convergence you need. Then you can argue about the fact that every Borel set can be approximated by an open set. – Izaak van Dongen Apr 16 '24 at 19:28
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    The method suggest by @Izaak is perhaps more useful to think through than my answer, as this is a very common way of proving statements in measure theory. – Dowdow Apr 16 '24 at 20:52

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Here's a different approach: Using the fact that for $f\in L^1(\mathbb{R})$ and $g\in L^\infty(\mathbb{R})$, the convolution $(f*g)(x)$ is a continuous function of $x$ (a proof is sketched here), we see that $$f(x)=\int_{\mathbb{R}}\chi_{A\cap(x+A)}(t)dt=\int_{\mathbb{R}}\chi_A(t)\chi_{-A}(x-t)dt=(\chi_A*\chi_{-A})(x).$$ Since $A$ is bounded and Borel, $\chi_A, \chi_{-A}\in L^1(\mathbb{R})\cap L^\infty(\mathbb{R}) $ and thus $f$ is continuous as desired.

Dowdow
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