Prove that $f\ast g$ is continuous if $f\in C(\mathbb{T})$ and $g\in R(\mathbb{T})$ 
Prove that $f\ast g$ is continuous if $f\in C(\mathbb{T})$ and $g\in R(\mathbb{T})$ (Meaning $f$ is continuous and periodic and $g$ is Riemann integrable and periodic).

So basically, if we define $$C(x) = f\ast g(x) = \frac{1}{2\pi} \int_0^{2\pi} f(x-t)g(t)dt$$
We want to prove that $C(x)$ is continuous.
Let's look at:$$ \lim_{x\to x_0} C(x) = \lim_{x\to x_0} \frac{1}{2\pi} \int_0^{2\pi} f(x-t)g(t)dt$$
Now, If I could insert the limit inside the integral I would finished the proof, right?
If so, how do I explain it is a "legal" move?
 A: To answer the question you asked (more or less, "Can I swap the limit and the integral?"), I have no idea. There are some cases where you can, and I can never remember the rules. I had a student who called such swaps "engineer's prerogative," but I think this is a little unkind to engineers. I'm sure others can point you to theorems. But often it's easier just to do things by hand. 
If you instead look at
\begin{align}
  C(x) - C(x_0) 
&= \frac{1}{2\pi} \int_0^{2\pi} f(x-t)g(t)~dt - \frac{1}{2\pi} \int_0^{2\pi} f(x_0-t)g(t)~dt\\
&= \frac{1}{2\pi} \int_0^{2\pi} [f(x-t) - f(x_0 - t)]\cdot g(t)~dt\\
&=  \frac{1}{2\pi} \int_0^{2\pi} [f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)~ds
\end{align}
you can estimate the integrand: Since $f$ is continuous on the compact set $[0,2\pi]$, it's uniformly continuous. So for $\epsilon > 0$, there's a $\delta$ such that $|p - q| < \delta$ implies $|f(p) - f(q)| < \epsilon$. Furthermore, because $g$ is integrable, so is $|g|$ (explanation.) That means that for $|x - x_0| < \delta$, we have
\begin{align}
 | C(x) - C(x_0) | 
&=  \frac{1}{2\pi} |\int_0^{2\pi} [f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)ds| \\
&\le  \frac{1}{2\pi} \int_0^{2\pi} |[f((x-x_0) - s) - f(-s)]\cdot g(x_0 + s)|ds \\
&=  \frac{1}{2\pi} \int_0^{2\pi} \epsilon \cdot |g(x_0 + s)|ds \\
&=  \epsilon \frac{1}{2\pi} \int_0^{2\pi} |g(x_0 + s)| ds \\
&=  \epsilon \frac{1}{2\pi} \int_0^{2\pi} |g(s)| ds \text{, by periodicity}\\
&= \epsilon M
\end{align}
where $M$ is the integral of $|g|$. 
So as $\epsilon$ goes to zero, the difference goes to $0$ and $C$ is continuous at $x_0$. 
