Let $A,B$ be Borel subsets of $\mathbb{R}$ with positive measure. Want to show that $\textbf{1}_A*\textbf{1}_B$ is continuous and not identically zero.
I have showed that $\textbf{1}_A*\textbf{1}_B$ is continuous, as $\textbf{1}_A,\textbf{1}_B\in L^2(\mathbb{R})$. And $$\textbf{1}_A*\textbf{1}_B(x) = \int_{\mathbb{R}}\textbf{1}_A(x-t)\textbf{1}_B(t)dt = \int_B \textbf{1}_A(x-t)dt = m(B\cap x-A)$$ where $x-A$ is translation of the set A. I'm having trouble on how to prove that the set $B\cap x-A$ has positive measure. I only know that $B\cap x-A$ is not empty. Can someone help me with this? Or is there another way to solve this problem? Thank you
$$\int\int1_A(x-t)1_B(t)dtdx=\int\int1_A(x-t)1_B(t)dxdt=\int 1_B(t)\int1_A(x-t)dxdt=$$$$\int 1_B(t)\lambda(t+A)dt=\int1_B(t)\lambda(A)dt=\lambda(A)\int1_B(t)dt=\lambda(A)\lambda(B)>0$$
This cannot be true if $\int1_A(x-t)1_B(t)dt$ is the zero-function.