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== Proofs ==

This proof starts by proving the special case where the function's domain is a closed and bounded interval <math>[a, b].</math>{{sfn|Apostol|1957|pp=162–3}}{{sfn|Hobson|1907|p=245}} The proof of the general case follows from this special case.

=== Proof when the domain is closed and bounded ===

Two proofs of this special case are given.

==== Proof 1 ====

Let <math>I := [a, b]</math> be an interval and let <math>f : I \to \R</math> be a non-decreasing function (such as an [[Monotonic function|increasing]] function).
Then for any <math>a < x < b,</math>
<math display=block>f(a) ~\leq~ f\left(a^+\right) ~\leq~ f\left(x^-\right) ~\leq~ f\left(x^+\right) ~\leq~ f\left(b^-\right) ~\leq~ f(b).</math>
Let <math>\alpha > 0</math> and let <math>x_1 < x_2 < \cdots < x_n</math> be <math>n</math> points inside <math>I</math> at which the jump of <math>f</math> is greater or equal to <math>\alpha</math>:
<math display=block>f\left(x_i^+\right) - f\left(x_i^-\right) \geq \alpha,\ i=1,2,\ldots,n</math>

For any <math>i=1,2,\ldots,n,</math> <math>f\left(x_i^+\right) \leq f\left(x_{i+1}^-\right)</math> so that <math>f\left(x_{i+1}^-\right) - f\left(x_i^+\right) \geq 0.</math>
Consequently,
<math display=block>\begin{alignat}{9}
f(b) - f(a)
&\geq f\left(x_n^+\right) - f\left(x_1^-\right) \\
&= \sum_{i=1}^n \left[f\left(x_i^+\right) - f\left(x_i^-\right)\right] + \sum_{i=1}^{n-1} \left[f\left(x_{i+1}^-\right) - f\left(x_i^+\right)\right] \\
&\geq \sum_{i=1}^n \left[f\left(x_i^+\right) - f\left(x_i^-\right)\right] \\
&\geq n \alpha
\end{alignat}</math>
and hence <math>n \leq \frac{f(b) - f(a)}{\alpha}.</math>

Since <math>f(b) - f(a) < \infty</math> we have that the number of points at which the jump is greater than <math>\alpha</math> is finite (possibly even zero).

Define the following sets:
<math display=block>S_1: = \left\{x : x \in I, f\left(x^+\right) - f\left(x^-\right) \geq 1\right\},</math>
<math display=block>S_n: = \left\{x : x \in I, \frac{1}{n} \leq f\left(x^+\right) - f\left(x^-\right) < \frac{1}{n-1}\right\},\ n\geq 2.</math>

Each set <math>S_n</math> is finite or the [[empty set]]. The union
<math>S = \bigcup_{n=1}^\infty S_n</math> contains all points at which the jump is positive and hence contains all points of discontinuity. Since every <math>S_i,\ i=1,2,\ldots</math> is at most countable, their union <math>S</math> is also at most countable.

If <math>f</math> is non-increasing (or [[Monotonic function|decreasing]]) then the proof is similar. This completes the proof of the special case where the function's domain is a closed and bounded interval.
<math>\blacksquare</math>

==== Proof 2 ====

For a monotone function <math>f</math>, let <math>f\nearrow</math> mean that <math>f</math> is monotonically non-decreasing and let <math>f\searrow</math> mean that <math>f</math> is monotonically non-increasing. Let <math>f : [a, b] \to \R</math> is a monotone function and let <math>D</math> denote the set of all points <math>d \in [a, b]</math> in the domain of <math>f</math> at which <math>f</math> is discontinuous (which is necessarily a jump discontinuity).

Because <math>f</math> has a jump discontinuity at <math>d \in D,</math> <math>f\left(d^-\right) \neq f\left(d^+\right)</math> so there exists some [[rational number]] <math>y_d \in \Q</math> that lies strictly in between <math>f\left(d^-\right) \text{ and } f\left(d^+\right)</math> (specifically, if <math>f \nearrow</math> then pick <math>y_d \in \Q</math> so that <math>f\left(d^-\right) < y_d < f\left(d^+\right)</math> while if <math>f \searrow</math> then pick <math>y_d \in \Q</math> so that <math>f\left(d^-\right) > y_d > f\left(d^+\right)</math> holds).

It will now be shown that if <math>d, e \in D</math> are distinct, say with <math>d < e,</math> then <math>y_d \neq y_e.</math>
If <math>f \nearrow</math> then <math>d < e</math> implies <math>f\left(d^+\right) \leq f\left(e^-\right)</math> so that <math>y_d < f\left(d^+\right) \leq f\left(e^-\right) < y_e.</math>
If on the other hand <math>f \searrow</math> then <math>d < e</math> implies <math>f\left(d^+\right) \geq f\left(e^-\right)</math> so that <math>y_d > f\left(d^+\right) \geq f\left(e^-\right) > y_e.</math>
Either way, <math>y_d \neq y_e.</math>

Thus every <math>d \in D</math> is associated with a unique rational number (said differently, the map <math>D \to \Q</math> defined by <math>d \mapsto y_d</math> is [[Injective function|injective]]).
Since <math>\Q</math> is countable, the same must be true of <math>D.</math>
<math>\blacksquare</math>

=== Proof of general case ===

Suppose that the domain of <math>f</math> (a monotone real-valued function) is equal to a union of countably many closed and bounded intervals; say its domain is <math>\bigcup_{n} \left[a_n, b_n\right]</math> (no requirements are placed on these closed and bounded intervals{{efn|So for instance, these intervals need not be [[Disjoint sets|pairwise disjoint]] nor is it required that they intersect only at endpoints. It is even possible that <math>\left[a_n, b_n\right] \subseteq \left[a_{n+1}, b_{n+1}\right]</math> for all <math>n</math>}}).
It follows from the special case proved above that for every index <math>n,</math> the [[Restriction of a map|restriction]] <math>f\big\vert_{\left[a_n, b_n\right]} : \left[a_n, b_n\right] \to \R</math> of <math>f</math> to the interval <math>\left[a_n, b_n\right]</math> has at most countably many discontinuities; denote this (countable) set of discontinuities by <math>D_n.</math>
If <math>f</math> has a discontinuity at a point <math>x_0 \in \bigcup_{n} \left[a_n, b_n\right]</math> in its domain then either <math>x_0</math> is equal to an endpoint of one of these intervals (that is, <math>x_0 \in \left\{a_1, b_1, a_2, b_2, \ldots\right\}</math>) or else there exists some index <math>n</math> such that <math>a_n < x_0 < b_n,</math> in which case <math>x_0</math> must be a point of discontinuity for <math>f\big\vert_{\left[a_n, b_n\right]}</math> (that is, <math>x_0 \in D_n</math>).
Thus the set <math>D</math> of all points of at which <math>f</math> is discontinuous is a subset of <math>\left\{a_1, b_1, a_2, b_2, \ldots\right\} \cup \bigcup_{n} D_n,</math> which is a countable set (because it is a union of countably many countable sets) so that its subset <math>D</math> must also be countable (because every subset of a countable set is countable).

In particular, because every interval (including open intervals and half open/closed intervals) of real numbers can be written as a countable union of closed and bounded intervals, it follows that any monotone real-valued function defined on an interval has at most countable many discontinuities.

To make this argument more concrete, suppose that the domain of <math>f</math> is an interval <math>I</math> that is not [[Closed set|closed]] and [[Bounded set|bounded]] (and hence by [[Heine–Borel theorem]] not [[Compact set|compact]]).
Then the interval can be written as a countable union of closed and bounded intervals <math>I_n</math> with the property that any two consecutive intervals have an [[Interval (mathematics)#Notations for intervals|endpoint]] in common: <math>I = \cup_{n=1}^\infty I_n.</math>
If <math>I = (a,b] \text{ with } a \geq -\infty</math> then <math>I_1 = \left[\alpha_1, b\right],\ I_2 = \left[\alpha_2, \alpha_1\right], \ldots, I_n = \left[\alpha_n, \alpha_{n-1}\right], \ldots</math> where <math>\left(\alpha_n\right)_{n=1}^{\infty}</math> is a strictly decreasing [[sequence]] such that <math>\alpha_n \rightarrow a.</math> In a similar way if <math>I = [a,b), \text{ with } b \leq +\infty</math> or if <math>I = (a,b) \text{ with } -\infty \leq a < b \leq \infty.</math>
In any interval <math>I_n,</math> there are at most countable many points of discontinuity, and since a countable union of at most countable sets is at most countable, it follows that the set of all discontinuities is at most countable.
<math>\blacksquare</math>