Let $f$ be a continuous function on $\mathbb R$ with compact support and $\mu$ a finitely additive measure which is in the dual space of $L^\infty(\mathbb R)$. Is the convolution $f\ast \mu(x)=\int_{\mathbb R} f(xy)d\mu(y)$ a continuous function in $x$? This is really an update of a question I asked, where I took $f$ to be only $L^\infty$ and I received the answer that in that case $f\ast \mu$ may not be continuous.

Since $f$ has compact support, it is uniformly continuous. Let $h$ be a uniform modulus of continuity. If $x'x < \varepsilon$, then $f(xy)f(x'y) < h(\varepsilon)$ for all $y$, hence $f*\mu(x)f*\mu(x') < h(\varepsilon)\\mu\$ (where $\\cdot\$ is variational norm) and $f*\mu$ is uniformly continuous. Edit: As Mateusz pointed out, it becomes more interesting if $f$ does not need to vanish at $\infty$. For uniformly continuous $f$, the above still works and for $\sigma$additive $\mu$ we can use dominated convergence as suggested by Davide. For arbitrary, bounded continuous $f$ and non$\sigma$additive $\mu$, $f*\mu$ need not be continuous: Let $\mu$ be defined by $\mu(f) = \lim_{n\to\infty, n\in\mathbb{N}} f(n)$ if the limit exists and extend $\mu$ by Hahn Banach to a positive linear functional (a BanachMazur limit). Let $f$ be a continuous function which is zero on $[n\frac1{n}, n+\frac1{n}]$ for every $n\in\mathbb{Z}$ and 1 for points which are more than $\frac2{n}$ away from every $n\in \mathbb{Z}$. Then $f*\mu(0) = 0 $ but $f*\mu(x) = 1$ for $x$ close to but unequal zero. 

