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 
$f\ast \mu(x)=\int\_{\mathbb R} f(x-y)d\mu(y)$ 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. 



Convolution of a continuous function and a finitely additive measure
Since $f$ has compact support, it is uniformly continuous. Let $h$ be a uniform modulus of continuity. If $|x'-x| < \varepsilon$, then $|f(x-y)-f(x'-y)| < h(\varepsilon)$ for all $y$, hence $|f\*\mu(x)-f\*\mu(x')| < h(\varepsilon)\|\mu\|$ (where $\|\cdot\|$ is variational norm) and $f\*\mu$ is uniformly continuous.




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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 Banach-Mazur 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.


