Discontinuities of monotone functions

In the mathematical field of analysis, a well-known theorem describes the set of discontinuities of a monotone real-valued function of a real variable; all discontinuities of such a (monotone) function are necessarily jump discontinuities and there are at most countably many of them.

Usually, this theorem appears in literature without a name. It is called Froda's theorem in some recent works; in his 1929 dissertation, Alexandru Froda stated that the result was previously well-known and had provided his own elementary proof for the sake of convenience.^[1] Prior work on discontinuities had already been discussed in the 1875 memoir of the French mathematician Jean Gaston Darboux.^[2]

Denote the limit from the left by $${\displaystyle f\left(x^{-}\right):=\lim _{z\nearrow x}f(z)=\lim _{\stackrel {h\to 0}{h>0}}f(x-h)}$${\displaystyle f\left(x^{-}\right):=\lim _{z\nearrow x}f(z)=\lim _{\stackrel {h\to 0}{h>0}}f(x-h)} and denote the limit from the right by $${\displaystyle f\left(x^{+}\right):=\lim _{z\searrow x}f(z)=\lim _{\stackrel {h\to 0}{h>0}}f(x+h).}$${\displaystyle f\left(x^{+}\right):=\lim _{z\searrow x}f(z)=\lim _{\stackrel {h\to 0}{h>0}}f(x+h).}

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

Consider a real-valued function ${\displaystyle f}${\displaystyle f} of real variable ${\displaystyle x}${\displaystyle x} defined in a neighborhood of a point ${\displaystyle x.}${\displaystyle x.} If ${\displaystyle f}${\displaystyle f} is discontinuous at the point ${\displaystyle x}${\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 ${\displaystyle x}${\displaystyle x} then the jump at ${\displaystyle x}${\displaystyle x} is zero. Moreover, if ${\displaystyle f}${\displaystyle f} is not continuous at ${\displaystyle x,}${\displaystyle x,} the jump can be zero at ${\displaystyle x}${\displaystyle x} if ${\displaystyle f\left(x^{+}\right)=f\left(x^{-}\right)\neq f(x).}${\displaystyle f\left(x^{+}\right)=f\left(x^{-}\right)\neq f(x).}

Let ${\displaystyle f}${\displaystyle f} be a real-valued monotone function defined on an interval ${\displaystyle I.}${\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 ${\displaystyle f}${\displaystyle f} be a monotone function defined on an interval ${\displaystyle I.}${\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 ${\displaystyle [a,b].}${\displaystyle [a,b].}^{[6] [7]} The proof of the general case follows from this special case.

Two proofs of this special case are given.

Let ${\displaystyle I:=[a,b]}${\displaystyle I:=[a,b]} be an interval and let ${\displaystyle f:I\to \mathbb {R} }${\displaystyle f:I\to \mathbb {R} } be a non-decreasing function (such as an increasing function). Then for any ${\displaystyle a<x<b,}${\displaystyle a<x<b,} $${\displaystyle f(a)~\leq ~f\left(a^{+}\right)~\leq ~f\left(x^{-}\right)~\leq ~f\left(x^{+}\right)~\leq ~f\left(b^{-}\right)~\leq ~f(b).}$${\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 ${\displaystyle \alpha >0}${\displaystyle \alpha >0} and let ${\displaystyle x_{1}<x_{2}<\cdots <x_{n}}${\displaystyle x_{1}<x_{2}<\cdots <x_{n}} be ${\displaystyle n}${\displaystyle n} points inside ${\displaystyle I}${\displaystyle I} at which the jump of ${\displaystyle f}${\displaystyle f} is greater or equal to ${\displaystyle \alpha }${\displaystyle \alpha }: $${\displaystyle f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\geq \alpha ,\ i=1,2,\ldots ,n}$${\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 i=1,2,\ldots ,n,} ${\displaystyle f\left(x_{i}^{+}\right)\leq f\left(x_{i+1}^{-}\right)}${\displaystyle f\left(x_{i}^{+}\right)\leq f\left(x_{i+1}^{-}\right)} so that ${\displaystyle f\left(x_{i+1}^{-}\right)-f\left(x_{i}^{+}\right)\geq 0.}${\displaystyle f\left(x_{i+1}^{-}\right)-f\left(x_{i}^{+}\right)\geq 0.} Consequently, $${\displaystyle {\begin{alignedat}{9}f(b)-f(a)&\geq f\left(x_{n}^{+}\right)-f\left(x_{1}^{-}\right)\\&=\sum _{i=1}^{n}\left[f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\right]+\sum _{i=1}^{n-1}\left[f\left(x_{i+1}^{-}\right)-f\left(x_{i}^{+}\right)\right]\\&\geq \sum _{i=1}^{n}\left[f\left(x_{i}^{+}\right)-f\left(x_{i}^{-}\right)\right]\\&\geq n\alpha \end{alignedat}}}$${\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 ${\displaystyle n\leq {\frac {f(b)-f(a)}{\alpha }}.}${\displaystyle n\leq {\frac {f(b)-f(a)}{\alpha }}.}

Since ${\displaystyle f(b)-f(a)<\infty }${\displaystyle f(b)-f(a)<\infty } we have that the number of points at which the jump is greater than ${\displaystyle \alpha }${\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_{1}:=\left\{x:x\in I,f\left(x^{+}\right)-f\left(x^{-}\right)\geq 1\right\},} $${\displaystyle S_{n}:=\left\{x:x\in I,{\frac {1}{n}}\leq f\left(x^{+}\right)-f\left(x^{-}\right)<{\frac {1}{n-1}}\right\},\ n\geq 2.}$${\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 ${\displaystyle S_{n}}${\displaystyle S_{n}} is finite or the empty set. The union ${\displaystyle S=\bigcup _{n=1}^{\infty }S_{n}}${\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 ${\displaystyle S_{i},\ i=1,2,\ldots }${\displaystyle S_{i},\ i=1,2,\ldots } is at most countable, their union ${\displaystyle S}${\displaystyle S} is also at most countable.

If ${\displaystyle f}${\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. ${\displaystyle \blacksquare }${\displaystyle \blacksquare }

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

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

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

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

Suppose that the domain of ${\displaystyle f}${\displaystyle f} (a monotone real-valued function) is equal to a union of countably many closed and bounded intervals; say its domain is ${\displaystyle \bigcup _{n}\left[a_{n},b_{n}\right]}${\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 ${\displaystyle n,}${\displaystyle n,} the restriction ${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}:\left[a_{n},b_{n}\right]\to \mathbb {R} }${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}:\left[a_{n},b_{n}\right]\to \mathbb {R} } of ${\displaystyle f}${\displaystyle f} to the interval ${\displaystyle \left[a_{n},b_{n}\right]}${\displaystyle \left[a_{n},b_{n}\right]} has at most countably many discontinuities; denote this (countable) set of discontinuities by ${\displaystyle D_{n}.}${\displaystyle D_{n}.} If ${\displaystyle f}${\displaystyle f} has a discontinuity at a point ${\displaystyle x_{0}\in \bigcup _{n}\left[a_{n},b_{n}\right]}${\displaystyle x_{0}\in \bigcup _{n}\left[a_{n},b_{n}\right]} in its domain then either ${\displaystyle x_{0}}${\displaystyle x_{0}} is equal to an endpoint of one of these intervals (that is, ${\displaystyle x_{0}\in \left\{a_{1},b_{1},a_{2},b_{2},\ldots \right\}}${\displaystyle x_{0}\in \left\{a_{1},b_{1},a_{2},b_{2},\ldots \right\}}) or else there exists some index ${\displaystyle n}${\displaystyle n} such that ${\displaystyle a_{n}<x_{0}<b_{n},}${\displaystyle a_{n}<x_{0}<b_{n},} in which case ${\displaystyle x_{0}}${\displaystyle x_{0}} must be a point of discontinuity for ${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}}${\displaystyle f{\big \vert }_{\left[a_{n},b_{n}\right]}} (that is, ${\displaystyle x_{0}\in D_{n}}${\displaystyle x_{0}\in D_{n}}). Thus the set ${\displaystyle D}${\displaystyle D} of all points of at which ${\displaystyle f}${\displaystyle f} is discontinuous is a subset of ${\displaystyle \left\{a_{1},b_{1},a_{2},b_{2},\ldots \right\}\cup \bigcup _{n}D_{n},}${\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 ${\displaystyle D}${\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 ${\displaystyle f}${\displaystyle f} is an interval ${\displaystyle I}${\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 ${\displaystyle I_{n}}${\displaystyle I_{n}} with the property that any two consecutive intervals have an endpoint in common: ${\displaystyle I=\cup _{n=1}^{\infty }I_{n}.}${\displaystyle I=\cup _{n=1}^{\infty }I_{n}.} If ${\displaystyle I=(a,b]{\text{ with }}a\geq -\infty }${\displaystyle I=(a,b]{\text{ with }}a\geq -\infty } then ${\displaystyle I_{1}=\left[\alpha _{1},b\right],\ I_{2}=\left[\alpha _{2},\alpha _{1}\right],\ldots ,I_{n}=\left[\alpha _{n},\alpha _{n-1}\right],\ldots }${\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 ${\displaystyle \left(\alpha _{n}\right)_{n=1}^{\infty }}${\displaystyle \left(\alpha _{n}\right)_{n=1}^{\infty }} is a strictly decreasing sequence such that ${\displaystyle \alpha _{n}\rightarrow a.}${\displaystyle \alpha _{n}\rightarrow a.} In a similar way if ${\displaystyle I=[a,b),{\text{ with }}b\leq +\infty }${\displaystyle I=[a,b),{\text{ with }}b\leq +\infty } or if ${\displaystyle I=(a,b){\text{ with }}-\infty \leq a<b\leq \infty .}${\displaystyle I=(a,b){\text{ with }}-\infty \leq a<b\leq \infty .} In any interval ${\displaystyle I_{n},}${\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. ${\displaystyle \blacksquare }${\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

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

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

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

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

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