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 $$x we have that $$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?

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

$$0 < f(q) - y_{\alpha^{-1}(q)} - f(x) = \sum_{k \in \alpha^{-1}((x,q))} y_k \leqslant \sum_{k = n_0}^\infty y_k < \varepsilon.$$

• Ah, I see. Thank you Daniel... I suspected that in my prior analysis something was wrong but I didnt knew exactly how to formalize it. Nov 23, 2016 at 14:38

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.