Discontinuities of monotone functions

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{{efn|So for instance, these intervals need not be pairwise disjoint nor is it required that they intersect only at endpoints. It is even possible that \left[a_n, b_n\right] \subseteq \left[a_{n+1}, b_{n+1}\right] for all n}}). 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 1 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 . Then is a non-decreasing function on [,], which is continuous except for jump discontinuities at for ≥ 1. In the case of finitely many jump discontinuities, 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 [,] 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 ( ≥ 1) lie in (, ) and take λ1, λ2, λ3, ... and μ1, μ2, μ3, ... non-negative with finite sum and with λ + μ > 0 for each . 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 [, ] and is continuous except for jump discontinuities at for ≥ 1. (1) being non-decreasing and non-positive; (2) having given jump data at its points of discontinuity ; (3) satisfying the boundary condition () = 0; and (4) having zero derivative almost everywhere. Property (4) can be checked following , and . Without loss of generality, it can be assumed that is a non-negative jump function defined on the compact [,], with discontinuities only in (,). Note that an open set of (,) is canonically the disjoint union of at most countably many open intervals ; that allows the total length to be computed ℓ()= Σ ℓ(). Recall that a null set is a subset such that, for any arbitrarily small ε' > 0, there is an open containing with ℓ() 0 and a normalised non-negative jump function , let () be the set of points such that :{f(t)-f(s)\over t -s} > c for some , with () is open and has total length ℓ(()) ≤ 4 −1 (() – ()). Note that () consists the points where the slope of is greater that near . By definition () is an open subset of (, ), so can be written as a disjoint union of at most countably many open intervals = (, ). Let be an interval with closure in and ℓ() = ℓ()/2. By compactness, there are finitely many open intervals of the form (,) covering the closure of . 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 (,), (,), ... 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 . == See also ==