This is a part of my answer here, but it should completely answer your questions too.
I use the notation $$\lim_{y\to x^{+}}f(y)=f(x^+)$$ $$\lim_{y\to x^{-}}f(y)=f(x^-)$$
There is a very nice way of constructing, given a sequence $\{x_n\}$ of real numbers, a function which is continuous everywhere except the elements of $\{x_n\}$ [That is, discontinuous on a countable set $A\in\Bbb R$]. Let $\{c_n\}$ by any nonnegative summable sequence [that is $\sum\limits_{n\geq 0} c_n$ exists finitely], and let $$s(x)=\sum_{x_n<x} c_n$$
What we do is sum through the indices that satisfy the said inequality. Because of absolute convergence, order is irrelevant. The function is monotone increasing because the terms are nonnegative, and $s$ is discontinuous at each $x_n$ because $$s(x_n^+)-s(x_n^-)=c_n$$
However, it is continuous at any other $x$: see xzyzyz's proof with the particular case $c_n=n^{-2}$. In fact, this function is lower continous, in the sense $\lim\limits_{y\to x^{-}}f(y)=f(x^-)=f(x)$ for any value of $x$. If we had used $x_n\leq x$, it would be upper continuous, but still discontinuous at the $x_n$.
To see the function has the said jumps, note that for $h>0$, we have $$\begin{align} s(x_n^+)-s(x_n^-)&=\\ \lim_{h\to 0^+} s(x_k+h)-s(x_k-h)&=\lim_{h\to 0^+}\sum_{x_n<x_k+h} c_n-\sum_{x_n<x_k-h}c_n\\&=\lim_{h\to 0^+}\sum_{x_k-h\leq x_n<x_k+h} c_n\end{align}$$
and we can take $\delta$ so small that whenever $0<h<\delta$, for any given $x_m\neq x_k$, $x_m\notin [x_k-\delta,x_k+\delta)$, so the only term that will remain will be $c_k$, as desired.
Let $P_n$ be the partition including $0,1$ and $q_1,...,q_n$. We denote this partition by $P_n=\{t_0=0,t_1,...t_n,t_{n+1}=1\}$. Moreover, let $s_i$ be the term in $\sum p_k$ corresponding to $t_i$ for $1\leq i\leq n$.
Observe that
$h(t_i)\geq \sum_{j=1}^is_j$ and
$h(t_i)+\sum_{j=i}^ns_j\leq 1$
$\implies h(t_i) \leq 1-\sum_{j=i}^ns_j$
We can now bound the upper and lower Riemann sums.
$$L(f,P_n)\geq \sum_{i=1}^nh(t_i)(t_{i+1}-t_i) $$$$\geq \sum_{i=1}^n\left((t_{i+1}-t_i)\sum_{j=1}^i s_j\right)=\sum_{i=1}^n(s_i-t_is_i)$$
This last sum is just a rearrangement of $\sum_{k=1}^np_k-p_kq_k$.
$$U(f,P_n)\leq \sum_{i=0}^nh(t_{i+1})(t_{i+1}-t_i)$$$$\leq \sum_{i=0}^n(t_{i+1}-t_i)(1-\sum_{j=i+1}^ns_j)$$$$=1- \sum_{i=0}^n\left((t_{i+1}-t_i)\sum_{j=i+1}^ns_j\right)$$$$ = 1-\sum_{i=1}^ns_it_i=1-\sum_{k=1}^np_kq_k.$$
Taking limits as $n\to\infty$, we find the desired integral to be $$\fbox{1-$\sum p_kq_k$}.$$
Best Answer
You can find an answer in this thread. Although the situation is more special, because there $a_k = 2^{-k}$ is chosen, the proof doesn't change. Let $x \in \mathbb{R}$ be irrational, and let $N \in \mathbb{N}$ with $\sum_{k=N1}^\infty a_n < \varepsilon/2$. Now define $$\delta:= \min_{i=1,\ldots,N}|q_i-x|.$$ For any $|y-x| < \delta$ we have $q_k \in [0,x)$ if and only if $q_k \in [0,y)$ for all $k=1,\ldots,N$. Thus $$S(x) \cap [1,N] = S(y) \cap [1,N]$$ and therefore $$|f(x)-f(y)| \le 2 \sum_{k=N+1}^\infty a_k<\varepsilon.$$