Proof that a monotone function has countably many discontinuities

Proof

We assign a different rational number to each discontinuity that is not an endpoint of $I$.

1. First let $f$ be increasing: $x \le y$ implies $f(x) \le f(y)$. Let $D$ be the set of points of $I$ where $f$ is not continuous.
2. Call $c \in I$ an inner point if $I$ has a point $p < c$ and a point $q > c$. Then $(p, q) \subseteq I$, since $I$ is an interval.
3. Let $c$ be an inner point. Apply the existence of one-sided limits to $f$ restricted to $(p, q)$. Write $c^-$ and $c^+$ for its left and right limits at $c$.
4. That result gives $c^- = \sup_{p < x < c} f(x) \le f(c) \le \inf_{c < x < q} f(x) = c^+$.
5. Claim: $c^+ \le f(t)$ for each $t \in I$ with $t > c$. Choose $s$ with $c < s < \min(t, q)$. Then $c^+ \le f(s) \le f(t)$ by step 4 and monotonicity.
6. In the same way, $f(t) \le c^-$ for each $t \in I$ with $t < c$.
7. Claim: if $c$ is an inner point in $D$, then $c^- < c^+$. Suppose instead $c^- = c^+$. Then step 4 gives $c^- = f(c) = c^+$.
8. Let $\varepsilon > 0$. By the definition of one-sided limits, there are $\delta_1, \delta_2 > 0$ with these properties. We have $\abs{f(x) - f(c)} < \varepsilon$ for $x \in (p, q)$ with $c - \delta_1 < x < c$, and also for $x \in (p, q)$ with $c < x < c + \delta_2$.
9. Let $\delta = \min(\delta_1, \delta_2, c - p, q - c)$. Each $x \in I$ with $\abs{x - c} < \delta$ lies in $(p, q)$, so $\abs{f(x) - f(c)} < \varepsilon$ by step 8.
10. By the epsilon-delta form of continuity, $f$ is continuous at $c$. This contradicts $c \in D$, which proves the claim of step 7.
11. Let $D_0$ be the set of inner points in $D$. For $c \in D_0$, the density of the rationals gives a rational $r(c)$ with $c^- < r(c) < c^+$.
12. Let $c < e$ be two points of $D_0$. Put $t = \frac{c + e}{2}$, so $t \in I$ and $c < t < e$. Steps 5 and 6 give $c^+ \le f(t) \le e^-$.
13. So $r(c) < c^+ \le e^- < r(e)$. Hence $r \colon D_0 \to \Q$ is injective.
14. By the rationals are countable, there is a bijection $\N \to \Q$. Its inverse $j \colon \Q \to \N$ is injective. By composition of injections, $\varphi = j \circ r \colon D_0 \to \N$ is injective.
15. A point of $D$ that is not inner is the least element or the greatest element of $I$. So $D \setminus D_0$ has at most two points.
16. Define $\psi \colon D \to \N$ by $\psi(c) = \varphi(c) + 2$ for $c \in D_0$. Send the least element of $I$ to $0$, if it lies in $D$. Send the greatest element to $1$, if it lies in $D$ and differs from the least.
17. The map $\psi$ is injective, since $\varphi$ is injective and its shifted values are at least $2$. So $D$ is countable.
18. Now let $f$ be decreasing, and put $g = -f$. Then $g$ is increasing. Also $\abs{g(x) - g(c)} = \abs{f(x) - f(c)}$, so $f$ and $g$ have the same points of continuity.
19. By steps 1 to 17 applied to $g$, the set of discontinuities of $f$ is countable.