Denote the limit from the left by

If ${\displaystyle f\left(x^{+}\right)}$  and ${\displaystyle f\left(x^{-}\right)}$  exist and are finite then the difference ${\displaystyle f\left(x^{+}\right)-f\left(x^{-}\right)}$  is called the jump^[3] of ${\displaystyle f}$  at ${\displaystyle x.}$ 

Consider a real-valued function ${\displaystyle f}$  of real variable ${\displaystyle x}$  defined in a neighborhood of a point ${\displaystyle x.}$  If ${\displaystyle f}$  is discontinuous at the point ${\displaystyle x}$  then the discontinuity will be a removable discontinuity, or an essential discontinuity, or a jump discontinuity (also called a discontinuity of the first kind).^[4] If the function is continuous at ${\displaystyle x}$  then the jump at ${\displaystyle x}$  is zero. Moreover, if ${\displaystyle f}$  is not continuous at ${\displaystyle x,}$  the jump can be zero at ${\displaystyle x}$  if ${\displaystyle f\left(x^{+}\right)=f\left(x^{-}\right)\neq f(x).}$ 

Let ${\displaystyle f}$  be a real-valued monotone function defined on an interval ${\displaystyle I.}$  Then the set of discontinuities of the first kind is at most countable.

One can prove^[5][3] that all points of discontinuity of a monotone real-valued function defined on an interval are jump discontinuities and hence, by our definition, of the first kind. With this remark the theorem takes the stronger form:

Let ${\displaystyle f}$  be a monotone function defined on an interval ${\displaystyle I.}$  Then the set of discontinuities is at most countable.

This proof starts by proving the special case where the function's domain is a closed and bounded interval ${\displaystyle [a,b].}$ ^[6][7] The proof of the general case follows from this special case.

Proof when the domain is closed and bounded edit

Two proofs of this special case are given.

Proof 1 edit

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 ${\displaystyle a<x<b,}$ 

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,

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:

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. ${\displaystyle \blacksquare }$ 

Proof 2 edit

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 ${\displaystyle D.}$  ${\displaystyle \blacksquare }$ 

Proof of general case edit

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^[a]). 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. ${\displaystyle \blacksquare }$ 

Examples. Let x₁ < x₂ < x₃ < ⋅⋅⋅ be a countable subset of the compact interval [a,b] and let μ₁, μ₂, μ₃, ... be a positive sequence with finite sum. Set

${\displaystyle f(x)=\sum _{n=1}^{\infty }\mu _{n}\chi _{[x_{n},b]}(x)}$ 

where χ_A denotes the characteristic function of a compact interval A. Then f is a non-decreasing function on [a,b], which is continuous except for jump discontinuities at x_n for n ≥ 1. In the case of finitely many jump discontinuities, f is a step function. The examples above are generalised step functions; they are very special cases of what are called jump functions or saltus-functions.^[8][9]

More generally, the analysis of monotone functions has been studied by many mathematicians, starting from Abel, Jordan and Darboux. Following Riesz & Sz.-Nagy (1990), replacing a function by its negative if necessary, only the case of non-negative non-decreasing functions has to be considered. The domain [a,b] can be finite or have ∞ or −∞ as endpoints.

The main task is to construct monotone functions — generalising step functions — with discontinuities at a given denumerable set of points and with prescribed left and right discontinuities at each of these points. Let x_n (n ≥ 1) lie in (a, b) and take λ₁, λ₂, λ₃, ... and μ₁, μ₂, μ₃, ... non-negative with finite sum and with λ_n + μ_n > 0 for each n. Define

${\displaystyle f_{n}(x)=0\,\,}$  for ${\displaystyle \,\,x<x_{n},\,\,f_{n}(x_{n})=\lambda _{n},\,\,f_{n}(x)=\lambda _{n}+\mu _{n}\,\,}$  for ${\displaystyle \,\,x>x_{n}.}$ 

Then the jump function, or saltus-function, defined by

${\displaystyle f(x)=\,\,\sum _{n=1}^{\infty }f_{n}(x)=\,\,\sum _{x_{n}\leq x}\lambda _{n}+\sum _{x_{n}<x}\mu _{n},}$ 

is non-decreasing on [a, b] and is continuous except for jump discontinuities at x_n for n ≥ 1.^{[10][11][12][13]}

To prove this, note that sup |f_n| = λ_n + μ_n, so that Σ f_n converges uniformly to f. Passing to the limit, it follows that

${\displaystyle f(x_{n})-f(x_{n}-0)=\lambda _{n},\,\,\,f(x_{n}+0)-f(x_{n})=\mu _{n},\,\,\,}$  and ${\displaystyle \,\,f(x\pm 0)=f(x)}$ 

if x is not one of the x_n's.^[10]

Conversely, by a differentiation theorem of Lebesgue, the jump function f is uniquely determined by the properties:^[14] (1) being non-decreasing and non-positive; (2) having given jump data at its points of discontinuity x_n; (3) satisfying the boundary condition f(a) = 0; and (4) having zero derivative almost everywhere.

As explained in Riesz & Sz.-Nagy (1990), every non-decreasing non-negative function F can be decomposed uniquely as a sum of a jump function f and a continuous monotone function g: the jump function f is constructed by using the jump data of the original monotone function F and it is easy to check that g = F − f is continuous and monotone.^[10]

• Continuous function – Mathematical function with no sudden changes
• Bounded variation – Real function with finite total variation
• Monotone function

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