Set of discontinuity of monotone function is countable

Question

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.

Answer 1

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.

Answer 2

The usual definition of a series of nonnegative terms is as the supremum of the sums over finite subsets of the index set, $$\sum_{i\in I} x_i=\sup\biggl\{\sum_{j\in J}x_j:J\subseteq I\mbox{ is finite}\biggr\}.$$ (Note this definition does not quite work in general for series of positive and negative terms.)

The point then is that is $a< x<y< b$ and $F\!:[a,b]\to\mathbb R$ is increasing, then $$ F(a)\le F(x^-)\le F(x^+)\le F(y^-)\le F(b). $$ It follows from this that if $J\subset (a,b)$ is a finite set of discontinuities of $F$, then $$ \sum_{z\in J}F(z^+)-F(z^-)\le F(b)-F(a) $$ and therefore the same holds for the sum $S $ over all discontinuity points of $F$ in $(a,b)$. One way to think about this is to note that the (finitely many) intervals $(F (z^-),F (z^+)) $, $z\in J $, are pairwise disjoint and are all contained in $(F (a), F (b))$, so the sum of their lengths is at most the length of the whole interval.

Now, if $x$ as above is actually a discontinuity point of $F$, then the inequality $F(x^-)<F(x^+)$ is strict. It follows from this that the sum I called $S $ above is over a countable index set, since any series of uncountably many positive terms diverges.

Note that I did not include $a$ or $b$ in the above, but of course this does not change the conclusion that the set of discontinuities of $F$ in $[a,b]$ is countable. If we include them, then we need to use the convention that $F(a^-):=F(a)$ and $F(b^+):=F(b)$. With this convention the same conclusion (with the same bound) holds: If $D$ is the set of discontinuity points of $F$ in $[a,b]$, then $$ S=\sum_{x\in D}F(x^+)-F(x^-)\le F(b)-F(a). $$

Naturally, the same conclusion (that $D$ is actually countable) follows if $F$ is defined not just in a finite interval but on an unbounded one (even on $\mathbb R$), by noting that $\mathbb R=\bigcup_{N\in\mathbb N}[-N,N]$.

Answer 3

You're right, every discontinuity corresponds to a jump $F(x^-)\to F(x^+)$, and because $(F(x^-),F(x^+))\subset\mathbb{R}$ is an open interval, there is some rational in this interval. Also for the case of an unbounded monotone function, we can assume to the contrary there are uncountable many of these jumps, and find uncountable many rationals (by monotonicity the intervals $(F(x^-),F(x^+))\subset\mathbb{R}$ are disjoint), which is the desired contradiction. Indeed the proofs are really similar and your proof for the bounded case is easily generalized.