This should be a comment on Yulia Kuznetsova's answer, but I didn't have the points and it's gone now anyway:

Can't you take $\mu$ as corresponding to a functional $f \mapsto f(0)$ if $f$ is an $L^\infty$ function continuous at $0$, extended to all of $L^\infty(\mathbb{R})$ by Hahn-Banach? Then if you take e.g. a step function $g = \mathbb{1}_{(0,\infty)}$ which is $0$ to the left of $0$ and $1$ to the right, $g \star \mu$ should be $0$ a.e. on $(-\infty,0)$ and $1$ a.e. on $(0,\infty)$, and so wouldn't have a continuous representative. What am I missing?