Let $\varphi(x)=1_{[a,b]}(x)$ be the indicator function of the interval $[a,b]$. Then $f*\varphi$ is continuous by the fundamental theorem of calculus (in Königsberger it is proven for integrands which are regulated functions).
If $\varphi$ is a step function, i.e. a linear combination of indicator functions of singletons and intervals as above, then $f*\varphi$ is still continuous.
Any regulated function on a compact interval is bounded. So $\|f\|_{[-\pi,\pi]}$ exists.
Let $x_0 \in \mathbb{R}$ and $\varepsilon>0$.
Since $g$ is regulated and periodic, there exists a step function $\varphi$ (actually it is the periodic extension of a step function) such that $\|g-\varphi\|_{\mathbb{R}} < \varepsilon$.
Choose $\delta>0$ such that $|(f*\varphi)(x) - (f*\varphi)(x_0)|<\varepsilon$ for all $x \in (x_0-\delta,x_0+\delta)$. Then
$$
\begin{align*}
|(f*g)(x) - (f*g)(x_0)| &\leq |(f*g)(x) - (f*\varphi)(x)| + |(f*\varphi)(x) - (f*\varphi)(x_0)| \\
&\quad + |(f*\varphi)(x_0) - (f*g)(x_0)| \\
&\leq 2\varepsilon\|f\|_{[-\pi,\pi]} + \varepsilon
\end{align*}
$$
for all $x \in (x_0-\delta,x_0+\delta)$. This shows that $f*g$ is continuous on $\mathbb{R}$. Of course, any continuous function is regulated.
I am a little bit surprised that I ended up showing that $f*g$ is continuous, but in light of the answer given by nullUser it might not be out of this world because the regulated function $g$ equals an (everywhere) left-continuous or (everywhere) right-continuous function in all except countably many points.
Best Answer
If you have a continuous function, then you can approximate it by step functions, this should be clear. but if you fix a point, then you can just let this point be in the interior of a step and you can fix it all time to be the value of the function.
be careful: here I'm changing the length of the steps.