In his Analysis 1, Konrad Königsberger defines a convolution of two $2\pi$ periodic regulated functions (that have left and right limits at every point ) $f$ and $g$ as:
$$
(f*g)(x):=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)g(x-t)dt
$$
where $\mathbb{T}$ is an arbitrary $2\pi$ long interval. Right after this, he states that $f*g$ is clearly $2\pi$ periodic and regulated. To me, it is not obvious that it's regulated.
I've managed to prove that if the functions are continuous, then their convolution is also continuous, but I can't use the same argument to prove that it's regulated.
What seemed as a promising attempt was to prove that the convolution of two step (piecewise-continuous) functions is a piecewise linear function. The problem is that I cant find a nice way to show this. Knowing that this holds, one could approximate $f(t)g(x-t)$ with $\varphi(t)\psi(x-t)$ such that $|f(t)g(x-t)-\varphi(t)\psi(x-t)|<\frac{\varepsilon}{2(b-a)}$ for all $t$, and $\varphi*\psi$ with a step function $\rho$ such that $||\varphi*\psi-\rho|| < \frac{\varepsilon}{2}$.
Edit:
I managed to prove it using piecewise linear and step functions: First we must prove that the convolution of two step functions with at most one point of discontinuities is a piecewise linear function. To do this, we can let:
\begin{equation*}
\begin{split}
&\varphi(t)=
\begin{cases}
\alpha & x\in[0,a)\\
\beta & x\in[a,2\pi)
\end{cases}\\
&\psi(t)=
\begin{cases}
\gamma & x\in[0,b)\\
\delta & x\in[b,2\pi)
\end{cases}
\end{split}
\end{equation*}
They're both extended $2\pi$ periodically. We can assume that $a\leq b$. Let $x\in[0,2\pi)$
\begin{equation*}
\begin{split}
&2\pi(\varphi*\psi)(x) = \int_0^{2\pi}\varphi(t)\psi(x-t)dt = \alpha\int_0^a\psi(x-t)dt + \beta\int_a^{2\pi}\psi(x-t)dt
&
\end{split}
\end{equation*}
Now you simply look at cases $x\in[0,a);x\in[a,b);x\in[b,2\pi)$ and notice, after a few substitutions, that the function is linear on all of them.
When $\varphi$ and $\psi$ are arbitrary step functions they can be written as sums of step functions with one point of discontinuity. Using the bilinearity of the convolution and the fact that the sum of piecewise linear functions is piecewise linear, we get that convolution of any two step functions is piecewise linear, thus obviously regulated.
Finally, if we let $f,g$ be $2\pi$ periodic regulated functions, we pick step functions $\varphi,\psi,\rho$ such that:
\begin{equation*}
\begin{split}
&1.||g-\psi|| < \frac{\varepsilon}{3||f||}\\
&2.||f-\varphi|| < \frac{\varepsilon}{3||\psi||}\\
&3.||\varphi*\psi-\rho||<\frac{\varepsilon}{3}
\end{split}
\end{equation*}
From here, we get that for all $x\in[0,2\pi)$
\begin{equation*}
\begin{split}
|(f*g)(x)-\rho(x)|\leq|(f*g)(x)-(f*\psi)(x)|+|(f*\psi)(x)-(\varphi*\psi)(x)| +
|(\varphi*\psi(x)-\rho(x))|<\varepsilon
\end{split}
\end{equation*}
Best Answer
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.