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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?


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If you take $F=\{0\}$ then $\chi_E \ast \chi_F = \chi_E$, but this function is not continuous usually (unless $E = \emptyset$). – Yuval Filmus Mar 9 '11 at 17:19
For the convolution $f \ast g $ to be well-defined, $f$ and $g$ have to be Lebesgue integrable. This means they have to be measurable which in your case means that $E$ and $F$ have to be measurable sets and have to have finite measure. If you integrate the characteristic function of the Cantor set (which is measurable with measure $0$) you get $\int_{\mathbb{R}} \chi_{Cantor} dt = 0$, i.e. the characteristic function of the Cantor set is Lebesgue integrable. As for proving the continuity of $\int \chi_E \cdot \chi_F dt$: maybe you could use that $\chi_E \cdot \chi_F = \chi_{E \cap F}$. – Rudy the Reindeer Mar 9 '11 at 17:31
@JBeardz: Do you know translations are continuous in $L^p$? – Jonas Teuwen Mar 9 '11 at 17:40
@Yuval: isn't $\chi_E * \chi_F =0$ in this case? – Fabian Mar 9 '11 at 17:41
@Yuval: Isn't $\chi_{\lbrace 0\rbrace}=0$ Lebesgue almost surely, so that $\chi_E * \chi_F$ is the zero function? Oh, Fabian beat me to it! – Byron Schmuland Mar 9 '11 at 17:46
up vote 0 down vote accepted

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 think absolute values are missing for this to be correct, and with absolute values your "hopefully obvious" equality is not true anymore (but can be corrected). – Plop Mar 9 '11 at 19:18
I'm just going to go ahead an accept this one. – Jon Beardsley Mar 11 '11 at 3:07

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