Discontinuities of monotone functions

Definitions

Denote the limit from the left by 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 f\left(x^+\right) := \lim_{z \searrow x} f(z) = \lim_{\stackrel{h \to 0}{h 0}} f(x+h).

If f\left(x^+\right) and f\left(x^-\right) exist and are finite then the difference f\left(x^+\right) - f\left(x^-\right) is called the ** jump** of f at x.

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

Precise statement

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

One can prove 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 f be a monotone function defined on an interval I. Then the set of discontinuities is at most countable.

Proofs

This proof starts by proving the special case where the function's domain is a closed and bounded interval [a, b]. 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 I := [a, b] be an interval and let f : I \to \R be a non-decreasing function (such as an increasing function). Then for any a 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 \alpha 0 and let x_1 be n points inside I at which the jump of f is greater or equal to \alpha: f\left(x_i^+\right) - f\left(x_i^-\right) \geq \alpha,\ i=1,2,\ldots,n

For any i=1,2,\ldots,n, f\left(x_i^+\right) \leq f\left(x_{i+1}^-\right) so that f\left(x_{i+1}^-\right) - f\left(x_i^+\right) \geq 0. Consequently, \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} and hence n \leq \frac{f(b) - f(a)}{\alpha}.

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

Define the following sets: S_1: = \left{x : x \in I, f\left(x^+\right) - f\left(x^-\right) \geq 1\right}, S_n: = \left{x : x \in I, \frac{1}{n} \leq f\left(x^+\right) - f\left(x^-\right)

Each set S_n is finite or the empty set. The union 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 S_i,\ i=1,2,\ldots is at most countable, their union S is also at most countable.

If 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. \blacksquare

Proof 2

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

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

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

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

Proof of general case

Suppose that the domain of f (a monotone real-valued function) is equal to a union of countably many closed and bounded intervals; say its domain is \bigcup_{n} \left[a_n, b_n\right] (no requirements are placed on these closed and bounded intervals). It follows from the special case proved above that for every index n, the restriction f\big\vert_{\left[a_n, b_n\right]} : \left[a_n, b_n\right] \to \R of f to the interval \left[a_n, b_n\right] has at most countably many discontinuities; denote this (countable) set of discontinuities by D_n. If f has a discontinuity at a point x_0 \in \bigcup_{n} \left[a_n, b_n\right] in its domain then either x_0 is equal to an endpoint of one of these intervals (that is, x_0 \in \left{a_1, b_1, a_2, b_2, \ldots\right}) or else there exists some index n such that a_n in which case x_0 must be a point of discontinuity for f\big\vert_{\left[a_n, b_n\right]} (that is, x_0 \in D_n). Thus the set D of all points of at which f is discontinuous is a subset of \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 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 f is an interval 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 I_n with the property that any two consecutive intervals have an endpoint in common: I = \cup_{n=1}^\infty I_n. If I = (a,b] \text{ with } a \geq -\infty then 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 \left(\alpha_n\right)_{n=1}^{\infty} is a strictly decreasing sequence such that \alpha_n \rightarrow a. In a similar way if I = [a,b), \text{ with } b \leq +\infty or if I = (a,b) \text{ with } -\infty \leq a In any interval 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. \blacksquare

Jump functions

Examples. Let x1 2 3 1, μ2, μ3, ... be a positive sequence with finite sum. Set

: f(x) = \sum_{n=1}^{\infty} \mu_n \chi_{[x_n,b]} (x)

where χA 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.

More generally, the analysis of monotone functions has been studied by many mathematicians, starting from Abel, Jordan and Darboux. Following , 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 λn + μn 0 for each n. Define

:f_n(x)=0,, for ,, x for ,, x x_n.

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

: f(x)=,,\sum_{n=1}^\infty f_n(x) =,, \sum_{x_n\le x} \lambda_n + \sum_{x_n

is non-decreasing on [a, b] and is continuous except for jump discontinuities at x**n for n ≥ 1.

To prove this, note that sup |f**n| = λn + μn, so that Σ f**n converges uniformly to f. Passing to the limit, it follows that

:f(x_n)-f(x_n-0)=\lambda_n,,,, f(x_n+0)-f(x_n)=\mu_n,,,, and ,, f(x\pm 0)=f(x)

if x is not one of the x**n's.

Conversely, by a differentiation theorem of Lebesgue, the jump function f is uniquely determined by the properties:For more details, see

(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.

Property (4) can be checked following , and . Without loss of generality, it can be assumed that f is a non-negative jump function defined on the compact [a,b], with discontinuities only in (a,b).

Note that an open set U of (a,b) is canonically the disjoint union of at most countably many open intervals I**m; that allows the total length to be computed ℓ(U)= Σ ℓ(I**m). Recall that a null set A is a subset such that, for any arbitrarily small ε' 0, there is an open U containing A with ℓ(U)

Proposition 1. For c 0 and a normalised non-negative jump function f, let Uc(f) be the set of points x such that

:{f(t)-f(s)\over t -s} c

for some s, t with s U**c(f) is open and has total length ℓ(U**c(f)) ≤ 4 c−1 (f(b) – f(a)).

Note that Uc(f) consists the points x where the slope of h is greater that c near x. By definition Uc(f) is an open subset of (a, b), so can be written as a disjoint union of at most countably many open intervals I**k = (a**k, b**k). Let J**k be an interval with closure in I**k and ℓ(J**k) = ℓ(I**k)/2. By compactness, there are finitely many open intervals of the form (s,t) covering the closure of J**k. On the other hand, it is elementary that, if three fixed bounded open intervals have a common point of intersection, then their union contains one of the three intervals: indeed just take the supremum and infimum points to identify the endpoints. As a result, the finite cover can be taken as adjacent open intervals (s**k,1,t**k,1), (s**k,2,t**k,2), ... only intersecting at consecutive intervals. Hence :\ell(J_k) \le \sum_m (t_{k,m} - s_{k,m}) \le \sum_m c^{-1}(f(t_{k,m})-f(s_{k,m})) \le 2 c^{-1}(f(b_k)-f(a_k)). Finally sum both sides over k.

Proposition 2. If f is a jump function, then f '(x) = 0 almost everywhere.

To prove this, define

:Df(x)= \limsup_{s,t\rightarrow x,,, s

a variant of the Dini derivative of f. It will suffice to prove that for any fixed c 0, the Dini derivative satisfies D**f(x) ≤ c almost everywhere, i.e. on a null set.

Choose ε 0, arbitrarily small. Starting from the definition of the jump function f = Σ f**n, write f = g + h with g = Σn≤N f**n and h = Σn**N f**n where N ≥ 1. Thus g is a step function having only finitely many discontinuities at x**n for n ≤ N and h is a non-negative jump function. It follows that D**f = g' +D**h = D**h except at the N points of discontinuity of g. Choosing N sufficiently large so that Σn**N λn + μn

By construction Df ≤ c off an open set with length less than 4ε/c. Now set ε' = 4ε/c — then ε' and c are arbitrarily small and Df ≤ c off an open set of length less than ε'. Thus Df ≤ c almost everywhere. Since c could be taken arbitrarily small, Df and hence also f ' must vanish almost everywhere. As explained in , 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.

Notes

References

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References

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5. This is a simple example of how [[Lebesgue covering dimension]] applies in one real dimension; see for example {{harvtxt. Edgar. 2008.
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