# Can an right-continuous increasing function have uncountable many jump points?

I am presenty studying Darboux-Stieljes Integral thorough the book

Elementary Analysis: The theory of Calculus

By Kenneth A. Ross (Edition 2)

35.14 Theorem

Let $F$ be a right-continuous increasing function on $[a, b]$. Then we have $F = F_c + F_d$, where $F_c$ is a continuous increasing function on $[a, b]$, and $F_d=\sum c_j J_{u_j}$ where {$u_1,u_2,...$} are the jump points of $F$ and $c_j$ is the jump at $u_j$ , i.e., $c_j = F(u_j^+)−F(u_j^-)$ for each $j$. (If there are no jump points, then $F$ is continuous, $F = F_c$ and $F_d = 0$.) In general, there might be finitely many or infinitely many such $u_j$ , so we will not decorate the sums until we need to.

Below is the proof.

Where I get lost is in the statement of the proof. I couldn't understand why $F$, a right continuous increasing function on $[a,b]$ must have a countable (or at most countable) number of jump points, as the author write these:

$F_d=\sum c_j J_{u_j}$

and

{$u_1,u_2,...$}

I don't believe the author will include the case while the set of jump points is uncountable infinite. I can't find any help in the proof above.

I believe the jump points should be countable, just an unbelievable belief. It is because I think that uncountablity, often occurs on $\mathbb R$ with treatments of rationals and irrationals, but I don't think (just don't believe) an function related to rationals and irrationals will be increasing or (or at least and) be right-continuous. Actually I don't really have any ideas.

I would like an answer to explain how the statement make sense. If the statement indeed does not make sense, then I want an explanation about why it doesn't make sense. I am pretty sure that if it does not make sense, it wouldn't affect the text after the theorem.

Any help will be appreciate.

NOTES: I don't know which tag(s) does countability relates to, maybe someone may help me to add an tag.

Note that any jump between $f(x)_- = \alpha$ and $f(x)_+ = \beta$, say, spans a rational between $\alpha$ and $\beta$. So there are at most as many jumps as there are rationals.
• Ah! I get it. This maps jump into $\mathbb Q$. Thx a lot! Feb 23, 2018 at 10:45