Prove or disprove:
Let $G$ be a locally compact Hausdorff topological group and $\mu$ be a left haar measure on $G$. $f\in L^1(G,\mu)$, $g\in L^{\infty}(G,\mu)$. Then $f*g$ is measurable with respect of $(G,\mu)$.
(I can prove it when $(G,\mu)$ is $\sigma$-finite. But I don't know if it is true when $(G,\mu)$ isn't $\sigma$-finite.)
I will give a proof by myself.
Proposition: Let $G$ be a locally compact group and $\mu$ be a left haar measure on $G$. $f\in\mathscr{L}^1(G), g\in\mathscr{L}^\infty(G)$. Then $f*g$ is continuous. Obviously it is measurable.
Proof: Firstly, we suppose $f\in\mathscr{K}^\mathbb{C}(G)$ (Complex Compact support continuous function on $G$). Then similar to proposition 9.1.6 in Donald Cohn-Measure Theory-Second Edition, we get $\forall t_0\in G, \forall \epsilon >0, \exists A$ is the open neighborhood of $t_0$ such that $\forall t\in A$, \begin{equation*} \int |f(ts)-f(t_0s)|\,\mathrm{d}\mu(s)<\frac{\epsilon}{\|g\|_{L^\infty}+1}. \end{equation*} Then \begin{equation*} \begin{split} |f*g(t)-f*g(t_0)|&\le\int |f(ts)-f(t_0s)||g(s^{-1})|\,\mathrm{d}\mu(s)\\ &\le\|g\|_{L^\infty}\int |f(ts)-f(t_0s)|\,\mathrm{d}\mu(s)<\epsilon. \end{split} \end{equation*} Then $f*g$ is continuous.
Secondly, we prove the theorem. There exists $f_n\in\mathscr{K}^\mathbb{C}(G)$ such that $\|f_n-f\|_{L^1}\rightarrow 0$. Then $\forall t\in G$, \begin{equation*} |f_n*g(t)-f*g(t)|\le \|f_n-f\|_{L^1}\|g\|_{L^\infty}. \end{equation*} Then $f_n*g$ converges to $f*g$ uniformly. Then $f*g$ is continuous.