convolution of characteristic functions

Solution 1:

Here is another proof for the continuity of $\chi_A*\chi_B$. Let us in fact show, more generally, that if $f$ is integrable on $\mathbb R$ and $g$ is measurable and bounded, then $f*g$ is a continuous function. (Then take $f:=\chi_A$ and $g:=\chi_B$, of course).

The function $f*g(x)=\int_{\mathbb R} f(x-y)g(y)\, dy$ is clearly well-defined at every point $x\in\mathbb R$. Moreover, one may write $$f*g(x)=\int_{\mathbb R} \tau_x f(y) g(y)\, dy\, ,$$ where $\tau_xf(y)=f(x-y)$. In an even more formal way, $$f*g(x)=\Phi_g (\tau_x f)\, ,$$ where $\Phi_g :L^1(\mathbb R)\to \mathbb R$ is the linear functional defined by $\Phi_g(u)=\int_{\mathbb R} u(y)g(y)\, dy$. This makes sense, and $\Phi_g$ is a continuous linear functional on $L^1(\mathbb R)$, because $g$ is bounded.

So, to prove that $f*g$ is a continuous function, you only have to check that the map $x\mapsto \tau_x f$ is continuous from $\mathbb R$ into $L^1(\mathbb R)$.

But now, there is no miracle: you need to use some approximation argument, just like in the other answers already given to your question. For example, you can first check that the map $x\mapsto \tau_x f$ is indeed continuous if $f$ is continuous and compactly supported (this should cause no difficulty); and then, taking a sequence $(f_n)$ of continuous and compactly supported functions tending to $f$ with respect to the $L^1$ norm, observe that $\tau_xf_n\to \tau_x f$ uniformly on $\mathbb R$ as $n\to\infty$, because $\Vert \tau_x f_n-\tau_x f\Vert_{L^1}=\Vert f_n-f\Vert_{L^1}$.

(A proof can be found on Real and Complex Analysis, 3rd Edition, theorem 9.5. In fact, this function is uniformly continuous from $\mathbb{R}$ into $L^p(\mathbb{R})$ where $1 \le p < \infty$)

Solution 2:

For the part about the open interval notice that if $m(B\cap(x-A))>0$ then $x\in A+B$, so it's enough to show that $m(B\cap(x-A))>0$ on some interval. But this function is continuous and not identically zero.

Added: Proof of continuity.

So pick a sequence of continuous functions $f_k$ with the followign properties: $\sup_\mathbb{R} f_k \leq 1$, $f_k(x)\to \chi_B(x)$ for a.e. $x\in \mathbb{R}$, and $\int_\mathbb{R} | f_k - \chi_B| dx \to 0$.

First notice that for any integrable function $h$ and bounded $g$ we have, for every $x\in \mathbb{R}$, $$ |h*g(x)|=\left| \int_\mathbb{R} h(x-y)g(y)dy \right| \leq \sup_\mathbb{R} |g| \int_ \mathbb{R} |h(x-y)|dy =\sup_\mathbb{R} |g| \int_\mathbb{R} |h(z)|dz. $$ This is just a change of variables in the last equality.

Now putting $h= \chi_A$ and $g=f_k-\chi_B$ we arrive at $$ |\chi_A*\chi_B(x)-\chi_A*f_k(x)| \leq \int_\mathbb{R} | f_k-\chi_B| dy\to 0, \qquad \text{as } k\to \infty. $$
Therefore $\chi_A*f_k$ converges uniformly on $\mathbb{R}$ to $\chi_A*\chi_B$. Therefore it's enough to prove that $\chi_A*f_k$ is continuous. To see this fix $k$ and take a sequence $x_n\to x$, and define $F_n(y)= f_k(x_n-y)$, and $F(y)=f_k(x-y)$, then since $f_k$ is continuous we get that $F_n(y)\to F(y)$ everywhere. Moreover $|F_n(y)|\leq 1$ (since $|f_k|\leq 1$), and the constant function $1$ is integrable over $A$. Therefore the DCT gives $$ \lim_{n\to \infty}\chi_A *f_k(x_n)=\lim_{n\to \infty}\int_A F_n(y)dy = \int_A F(y)dy = \chi_A*f_k(x). $$ This finishes the proof.