Showing that a function is not continuous, but integrable I am trying to prove that the function $$
F(x)=\sum_{n=1}^\infty \frac{1}{3^n}g(x-r_n)
$$
is not continuous in every point $r_n$, but still is integrable, Here $\{r_n\}$ is a dense sequence in $[0,1]$ and $g(x)=\sin\left(\frac{1}{x}\right)$ if $x\neq 0$, and $g(0)=0$.
For the part of the function being not continuous I've tried to get by the definition, showing that $|F(r_{n_0})-F(x)|> \epsilon>0  $ for every $x$ in a neighborhood of $r_{n_0}\in\{r_n\}$. 
I get to $$|F(r_{n_0})-F(x)|=\left|2\sum_{n=1}^\infty \frac{1}{3^n}
\sin\left(\frac{1}{2(r_{n_0}-x)}\right)\cos\left(\frac{1}{2(r_{n_0}+x-2r_n)}\right) \right|,$$ but I don't know how to go on.
To the part of being integrable I have no idea how to proceed.
Thanks in advance.

By the Weierstrass M test, the series converges uniformly on $\mathbb R.$ So at any point where all summands are continuous, $F$ is continuous. Since each summand is continuous at each irrational, the only possible points of discontinuity are in the set $\{r_1, r_2,\dots \}.$ That is a set of measure $0.$ Since the sum is uniformly bounded everywhere (by $\sum_n 1/3^n$), it is Riemann integrable on any closed bounded interval.
To show $F$ is discontinuous at each $r_n,$ note that
$$F(x) = \frac{\sin(1/(x-r_n))}{3^n}+ \sum_{k\ne n}\frac{\sin(1/(x-r_k))}{3^k}.$$
The first function on the right is discontinuous at $r_n,$ while the series on the right is continuous at $r_n.$
