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Given two sets of finite measure in $\mathbb{R}$ say, $E$ and $F$, and their characteristic functions $\chi_E$ and $\chi_F$, can somebody show that $\chi_E\ast\chi_F(x)$ (the convolution) is a continuous function of $x$? This is a qual problem from an old qual that I'm studying, and I cannot figure it out. If we were dealing with continuous functions or mollifiers or something it would be straightforward, but what if the sets $E$ and $F$ are somehow pathological, like the Cantor set, or something like that?

Thanks!

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    @Jonas I am learning these basic analysis facts slowly, and I am not always clear on how to apply them, but I do know that.2011-03-09

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let $f(x)=\int\chi_E(y)\chi_F(x-y)dy$. Then $|f(s)-f(t)|=|\int\chi_E(y)(\chi_F(s-y)-\chi_F(t-y))dy|\leq\mu(E)|\int(\chi_F(s-y)-\chi_F(t-y))dy|$. approximate $F$ with finitely many open intervals (call this union of intervals $U$) with $\mu(F\Delta U)<\epsilon$. if $|s-t|$ is less than the distance between any two of the intervals making up $U$, then we have $\int(\chi_U(s-y)-\chi_U(t-y))dy=0$ (hopefully this is obvious). also note that $|\int(\chi_F(s-y)-\chi_F(t-y))dy-\int(\chi_U(s-y)-\chi_U(t-y))dy|\leq2\epsilon$. Hence for $|s-t|$ small with respect to the chosen $U$ we have $|f(s)-f(t)|\leq2\epsilon\mu(E)$.

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    I'm just going to go ahead an accept this one.2011-03-11