Let $A,B \subset \Bbb{R}$ be Borel measurable with $m(A),m(B) \in (0,\infty)$, then prove $\chi_A \ast \chi_B$ is continuous.
Attempt:
Note
$$f(x)=\chi_A \ast \chi_B (x)=\int \chi_A(x-y)\chi_B(y)dy.$$
As $C_c(\Bbb{R}) \subset L^1(\Bbb{R})$ is dense, there exists $g_n \in C_c(\Bbb{R}$) such that
$$\Vert g_n - \chi_A \Vert_1 \to 0.$$ Then for all $x \in \Bbb{R}$, one has $$ \begin{align} \vert g_n \ast \chi_B(x) - \chi_A \ast \chi_B(x) \vert&\leq\int \chi_B(x-y)\vert g_n(y)-\chi_A(y) \vert dy\\ &\leq \Vert g_n - \chi_A \Vert_1\to 0. \end{align} $$ Thus $g_n \ast \chi_B \to \chi_A \ast \chi_B$ thus it suffices to show $g_n \ast \chi_B$ is continuous. As $g_n$ is uniformly continuous, for all $\epsilon >0$, there is a $\delta>0$ such that if $\vert x - y \vert<\delta$, then $\vert g_n(x) - g_n(y) \vert<\epsilon$. So for $x_1,x_2$ such that $\vert x_1 - x_2 \vert<\delta$, we have $$ \begin{align} \vert g_n \ast \chi_B (x_1) - g_n \ast \chi_B(x_2) \vert&\leq \int \vert g_n(x_1-y)-g_n(x_2-y) \vert \chi_B(y)dy\\ &\leq \epsilon m(B). \end{align} $$ as $\epsilon>0$ was arbitrary, one has $g_n \ast \chi_B$ is uniformly continuous thus $\chi_A \ast \chi_B$ is the uniform limit of continuous functions and thus continuous.
Does this proof work? Can someone, anyone help me out?! Also why would someone vote to close this it’s perfectly valid!!