convolution of characteristic functions Suppose $A$ and $B$ are measurable subsets of $\mathbb{R}$ of finite  positive measure. Show that the convolution $\chi_A*\chi_B$ is continuous and not identically $0$. Use this to prove that $A+B$ contains a segment.
We have $$(\chi_A*\chi_B)(x)=\int_{-\infty}^{\infty} \chi_A(x-y)\chi_B(y)~dm(y)$$
For $y\notin B$ we have $\chi_B(y)=0$. So,
$$(\chi_A*\chi_B)(x)=\int_B\chi_A(x-y)~dm(y)$$
For $x-y\notin A$ we have $\chi_A(x-y)=0$ 
$x-y\in A\Rightarrow y\in x-A$ so, we have $y\in B\cap (x-A)$. So, 
$$(\chi_A*\chi_B)(x)=\int_{B\cap(x-A)}~dm(y)=m(B\cap(x-A))$$
So, $$(\chi_A*\chi_B)(x)=m(B\cap(x-A))$$
EDIT 1 : Consider $x_n\rightarrow x $ ( converging sequence) then i see that
$$\cdots (B\cap(x_n-A))\supseteq (B\cap(x_{n+1}-A))\supseteq(B\cap(x_{n+2}-A))\supseteq\cdots (B\cap(x-A))$$ 
So, $m(B\cap (x_n-A))\rightarrow m(B\cap (x-A))$
(EDIT : i have said this before but then some one said that convergence may not imply that containment but now i tried and i strongly feel it holds if it converges)
I guess this would be sufficient to say that convolution is continuous....
$\epsilon-\delta$ way of proving continuity.. 
Given $\epsilon >0$ i have to choose $\delta>0$ such that $|x-y|<\delta$ implies $$\big|(\chi_A*\chi_B)(x)-(\chi_A*\chi_B)(y)\big|=\big|m(B\cap(x-A))-m(B\cap(y-A))\big|<\epsilon$$
I know $m(A)-m(B)=m(A\cap B^c)$ if $B\subset A$. I am not so sure if that holds in general or if it holds at least in this case...
Suppose every thing is super perfect then i would get
$$\big|m(B\cap(x-A))-m(B\cap(y-A))\big|=\big|m(B\cap(x-A))\cap (B\cap(y-A))^c\big|=\big| m(B\cap (x-A)\cap (y-A)^c)\big|$$  
I do not know where to go from here..
Even if i prove this is continuous i am not very sure how to use this and prove that $A+B$ contains a segment...
Please suggest some path...
 A: 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 $1 \le p < \infty$)
A: 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.
A: I like to think about the continuity of $f(x) = m(A \cap (-B + x))$ in this way: When $A, B$ are finite unions of open intervals, this is true and for every $\epsilon > 0$, there are finite unions of open intervals $U, V$ such that $m(A \Delta U), m(B \Delta V) < \epsilon$.
Also continuity implies that if $f(a) > 0$ then $f(x) > 0$ on some interval $I$ around $a$ so that $I \subseteq A + B$.
