Discontinuities of monotone functions

Two proofs of this special case are given.

Let ${\displaystyle I:=[a,b]}$ be an interval and let ${\displaystyle f:I\to \mathbb {R} }$ be a non-decreasing function (such as an increasing function). Then for any ${\displaystyle a<x<b,}$

$${\displaystyle f(a)~\leq ~f\left(a^{+}\right)~\leq ~f\left(x^{-}\right)~\leq ~f\left(x^{+}\right)~\leq ~f\left(b^{-}\right)~\leq ~f(b).}$$

Let ${\displaystyle \alpha >0}$ and let ${\displaystyle x_{1}<x_{2}<\cdots <x_{n}}$ be ${\displaystyle n}$ points inside ${\displaystyle I}$ at which the jump of ${\displaystyle f}$ is greater or equal to ${\displaystyle \alpha }$:

$${\displaystyle f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\geq \alpha ,\ i=1,2,\ldots ,n}$$

For any ${\displaystyle i=1,2,\ldots ,n,}$ ${\displaystyle f\left(x_{i}^{+}\right)\leq f\left(x_{i+1}^{-}\right)}$ so that ${\displaystyle f\left(x_{i+1}^{-}\right)-f\left(x_{i}^{+}\right)\geq 0.}$ Consequently,

$${\displaystyle {\begin{alignedat}{9}f(b)-f(a)&\geq f\left(x_{n}^{+}\right)-f\left(x_{1}^{-}\right)\\&=\sum _{i=1}^{n}\left[f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\right]+\sum _{i=1}^{n-1}\left[f\left(x_{i+1}^{-}\right)-f\left(x_{i}^{+}\right)\right]\\&\geq \sum _{i=1}^{n}\left[f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\right]\\&\geq n\alpha \end{alignedat}}}$$

and hence ${\displaystyle n\leq {\frac {f(b)-f(a)}{\alpha }}.}$

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

Define the following sets:

$${\displaystyle S_{1}:=\left\{x:x\in I,f\left(x^{+}\right)-f\left(x^{-}\right)\geq 1\right\},}$$

$${\displaystyle S_{n}:=\left\{x:x\in I,{\frac {1}{n}}\leq f\left(x^{+}\right)-f\left(x^{-}\right)<{\frac {1}{n-1}}\right\},\ n\geq 2.}$$

Each set ${\displaystyle S_{n}}$ is finite or the empty set. The union ${\displaystyle S=\bigcup _{n=1}^{\infty }S_{n}}$ contains all points at which the jump is positive and hence contains all points of discontinuity. Since every ${\displaystyle S_{i},\ i=1,2,\ldots }$ is at most countable, their union ${\displaystyle S}$ is also at most countable.

If ${\displaystyle f}$ is non-increasing (or 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. ${\displaystyle \blacksquare }$

So let ${\displaystyle f:[a,b]\to \mathbb {R} }$ is a monotone function and let ${\displaystyle D}$ denote the set of all points ${\displaystyle d\in [a,b]}$ in the domain of ${\displaystyle f}$ at which ${\displaystyle f}$ is discontinuous (which is necessarily a jump discontinuity).

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

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

Thus every ${\displaystyle d\in D}$ is associated with a unique rational number (said differently, the map ${\displaystyle D\to \mathbb {Q} }$ defined by ${\displaystyle d\mapsto y_{d}}$ is injective). Since ${\displaystyle \mathbb {Q} }$ is countable, the same must be true of ${\displaystyle D.}$ ${\displaystyle \blacksquare }$

Suppose that the domain of ${\displaystyle f}$ (a monotone real-valued function) is equal to a union of countably many closed and bounded intervals; say its domain is ${\displaystyle \bigcup _{n}\left[a_{n},b_{n}\right]}$ (no requirements are placed on these closed and bounded intervals[lower-alpha 1]). It follows from the special case proved above that for every index ${\displaystyle n,}$ the restriction ${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}:\left[a_{n},b_{n}\right]\to \mathbb {R} }$ of ${\displaystyle f}$ to the interval ${\displaystyle \left[a_{n},b_{n}\right]}$ has at most countably many discontinuities; denote this (countable) set of discontinuities by ${\displaystyle D_{n}.}$ If ${\displaystyle f}$ has a discontinuity at a point ${\displaystyle x_{0}\in \bigcup _{n}\left[a_{n},b_{n}\right]}$ in its domain then either ${\displaystyle x_{0}}$ is equal to an endpoint of one of these intervals (that is, ${\displaystyle x_{0}\in \left\{a_{1},b_{1},a_{2},b_{2},\ldots \right\}}$) or else there exists some index ${\displaystyle n}$ such that ${\displaystyle a_{n}<x_{0}<b_{n},}$ in which case ${\displaystyle x_{0}}$ must be a point of discontinuity for ${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}}$ (that is, ${\displaystyle x_{0}\in D_{n}}$). Thus the set ${\displaystyle D}$ of all points of at which ${\displaystyle f}$ is discontinuous is a subset of ${\displaystyle \left\{a_{1},b_{1},a_{2},b_{2},\ldots \right\}\cup \bigcup _{n}D_{n},}$ which is a countable set (because it is a union of countably many countable sets) so that its subset ${\displaystyle D}$ 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 ${\displaystyle f}$ is an interval ${\displaystyle I}$ that is not closed and bounded (and hence by Heine–Borel theorem not compact). Then the interval can be written as a countable union of closed and bounded intervals ${\displaystyle I_{n}}$ with the property that any two consecutive intervals have an endpoint in common: ${\displaystyle I=\cup _{n=1}^{\infty }I_{n}.}$ If ${\displaystyle I=(a,b]{\text{ with }}a\geq -\infty }$ then ${\displaystyle I_{1}=\left[\alpha _{1},b\right],\ I_{2}=\left[\alpha _{2},\alpha _{1}\right],\ldots ,I_{n}=\left[\alpha _{n},\alpha _{n-1}\right],\ldots }$ where ${\displaystyle \left(\alpha _{n}\right)_{n=1}^{\infty }}$ is a strictly decreasing sequence such that ${\displaystyle \alpha _{n}\rightarrow a.}$ In a similar way if ${\displaystyle I=[a,b),{\text{ with }}b\leq +\infty }$ or if ${\displaystyle I=(a,b){\text{ with }}-\infty \leq a<b\leq \infty .}$ In any interval ${\displaystyle I_{n},}$ 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. ${\displaystyle \blacksquare }$