Prove that $f$ have a jump discontinuity at any rational point 
Let $\alpha:\Bbb N\to \Bbb Q$ a bijection, and a function
$$f:\Bbb R\to\Bbb R,\quad x\mapsto \sum_{k\in N_x}y_k$$
where $\sum_{k=0}^\infty y_k<\infty$ and $y_k>0$ for all $k$, and
$$N_x:=\{k\in\Bbb N:\alpha(k)\le x\}$$
Prove that $f$ have a jump discontinuity at every rational $q$, where the size of the discontinuity is $y_n$ where $n=\alpha^{-1}(q)$.

I dont have a clear picture of the jump discontinuities of $f$, in concrete I dont know if my proof fit to the meaning "$y_n$ is the size of the discontinuity at $q$", so Im not totally sure that the proof is fine. A confirmation will be welcome.
First of all: observe that $f$ is an strictly increasing function and then injective. Proof: for $x<y$ we have that $N_x\subsetneq N_y$ because exists infinitely many rationals between any two real points.
Because $y_k>0$ for all $k$ then $f$ is strictly increasing.$\Box$
Now observe that if $q\in\Bbb Q$ then
$$\{k\in\Bbb N:\alpha(k)< q\}\subsetneq \{k\in\Bbb N:\alpha(k)< q\}\cup\{\alpha^{-1}(q)\}=N_q$$
what means that approaching from below to $q$ no matter how close we are there is a distance from the left limit to $f(q)$ of $y_{\alpha(q)}$, i.e.
$$\lim_{x\to q^-}f(x)=L^-=f(q)-y_{\alpha(q)}$$
In a similar way no matter how close we approach from above to $q$ there are rational points between our sequence and $q$, then
$$\lim_{x\to q^+}f(x)=L^+=f(q)+y_{\alpha(q)}$$
But from this analysis, not totally formalized, we have then two consecutive jump discontinuities of size $y_{\alpha(q)}$: one from $L^-$ to $q$, and another from $q$ to $L^+$.
It is this analysis correct? There is something wrong?
 A: The part showing that $f$ is strictly increasing is good. The part concerning the left-hand limit at $q$ is not quite complete, from
$$\{ k \in \mathbb{N} : \alpha(k) < q\} = N_q \setminus \{ \alpha^{-1}(q)\}$$
alone, you can only conclude that
$$L^- \leqslant f(q) - y_{\alpha^{-1}(q)},$$
to have the equality, you also need that the set on the left is exhausted by the $N_x$ for $x < q$,
$$\{ k \in \mathbb{N} : \alpha(k) < q\} = \bigcup_{x < q} N_x.$$
Then how you conclude $L^- = f(q) - y_{\alpha^{-1}(q)}$ depends on what you have to work with, and which way you like most. Let's postpone the discussion of that a little.
In the part about the right-hand limit, you have an error. Since
$$N_q = \bigcap_{x > q} N_x,$$
we do in fact have
$$\lim_{x\to q^+} f(x) = L^+ = f(q).$$
Concerning the arguments that $L^- = f(q) - y_{\alpha^{-1}(q)}$ and $L^+ = f(q)$, if we can use some measure theory, these follow directly from continuity properties of general measures, since
$$\mu \colon A \mapsto \sum_{y_k \in A} y_k$$
defines a finite positive measure on $\mathscr{P}(\mathbb{R})$, and $f(x) = \mu((-\infty, x])$.
If we want to throw less theory at the problem, we note that
$$r_n := \sum_{k = n}^\infty y_k$$
defines a monotonically decreasing sequence converging to $0$. Now let $\varepsilon > 0$ be given, and choose $n_0$ with $r_{n_0} < \varepsilon$. Then let
$$\delta = \min \bigl\{ \lvert y_k - q\rvert : k \in \mathbb{N}, k < n_0, y_k \neq q\bigr\}.$$
If $q < x < q+\delta$, then
$$0 < f(x) - f(q) = \sum_{k \in \alpha^{-1}((q,x])} y_k \leqslant \sum_{k = n_0}^\infty y_k < \varepsilon,$$
since $\alpha^{-1}((q,x]) \subset \{ k \in \mathbb{N} : k \geqslant n_0\}$. By the analogous argument, for $q - \delta < x < q$, we have
$$0 < f(q) - y_{\alpha^{-1}(q)} - f(x) = \sum_{k \in \alpha^{-1}((x,q))} y_k \leqslant \sum_{k = n_0}^\infty y_k < \varepsilon.$$
A: Just for the record, I will add an answer that definitively cleared my mind about this question.
The point is that I can understand a formal solution but, for this case, I dont "see" where comes the asymmetry for the different lateral continuity when $r\in\Bbb Q$. But for the sketched proof below the asymmetry is very clear.
Because this is about continuity lets see what happen with the distance between images. For $x<r$ we have that
$$d(f(x),f(r))=\sum_{k\in N_r}y_k-\sum_{k\in N_x}y_k=\sum_{k\in N_r\setminus N_x}y_k$$ 
and for $x>r$
$$d(f(x),f(r))=\sum_{k\in N_x}y_k-\sum_{k\in N_r}y_k=\sum_{k\in N_x\setminus N_r}y_k$$ 
because $f$ is strictly increasing. Then the distance is determined by the set of the index of the summation. Now observe that
$$N_r\setminus N_x=\{k\in\Bbb N: x<\alpha(k)\le r\}\implies \alpha(N_r\setminus N_x)=\{q\in\Bbb Q:x<q\le r\}$$
and
$$N_x\setminus N_r=\{k\in\Bbb N:r<\alpha(k)\le x\}\implies \alpha(N_x\setminus N_r)=\{q\in\Bbb Q:r<q\le x\}$$
Now we have that
$$\bigcap_{x\in\Bbb R\\x<r}\{q\in\Bbb Q:x<q\le r\}=\begin{cases}\{r\},&\text{if }r\in\Bbb Q\\\emptyset,&\text{if }r\notin\Bbb Q\end{cases}$$
and
$$\bigcap_{x\in\Bbb R\\x>r}\{q\in\Bbb Q:r<q\le x\}=\emptyset,\;\forall r\in\Bbb R$$
From here we can show easily that any lateral limit of $f$ it is equal to it image (definition of continuity) except the case for the left limit of $r$ being rational. This is the jump discontinuity.
