The result I am trying to prove is the following:
Let $F:\Bbb R\to\Bbb R$ be increasing. Then the set of points, at which it is discontinuous, is countable.
I have been reading this form Folland's Real Analysis and he proves this by considering the sum $\sum_{|x|<N}[F(x^+)-F(x^-)]$ which has to be finite by some sort of telescoping sum. Now, I am not sure how he defines this sum. I am assuming he does this by using nets (let me know if there is another way to look at this), but he doesn't introduce the concept of nets until the next chapter. So I was looking for some other ways to prove this result and I found the following result (from Stein-Shakarchi) which is quite similar but assumes $F$ to be bounded.
A bounded increasing function $F$ on $[a,b]$ has at most countably many discontinuities.
This is shown by using the density of rationals in $\Bbb R$, i.e. $\exists r_x\in\Bbb Q$ s. t. $F(x^-)<r_x<F(x^+)$ corresponding to each point of discontinuity $x$.
Now I feel like there shouldn't be a problem in doing the exact same thing for unbounded increasing function. Is this right? Secondly how does this proof relate to the first one? I feel like there isn't much of a difference between the two proofs. To be precise I think that the previous sum is well defined/finite because of this countability of discontinuous points. What do you think about this way of thinking?
Edit. Here is the Folland's proof in more detail:
Since $F$ is increasing, the intervals $(F(x^-),F(x^+))$ are disjoint for all $\forall x\in\Bbb R$. Moreover, $\forall|x| < N$ such intervals are contained in $(F(-N),F(N^+))$ and so
$$\sum_{|x|<N}[F(x^+)-F(x^-)] \leqslant F(N) – F(-N) < \infty. $$
Hence the set $\{ x \in (-N,N) : F(x^+) \neq F(x^-)\}$ is countable.
Best Answer
An essentially equivalent question was recently asked, which I answered before realizing it was a duplicate of this one. My solution is essentially the same as those posted here, but phrased somewhat differently and may be of interest.
Let $D$ denote the set of discontinuity points. For each $x\in D$, the left and right limits differ, and are therefore the endpoints of a non-empty open interval. In this manner we obtain a collection of such intervals $$ \{I_d\colon d\in D\}. $$ By monotonicity the intervals are disjoint. Choosing one rational number from each interval therefore yields an injection from $D$ into $\mathbb Q$. Hence $D$ is countable.