Talk:Discontinuities of monotone functions

Totality Assumption
That the discontinuities are all jumps (a straight line with a hole in it doesn't count) means that this theorem only applies to total functions. — Preceding unsigned comment added by 136.159.16.20 (talk) 01:28, 26 October 2015 (UTC)

Not the actual theorem
According to mavropnevma @ Art of Problem Solving, $$f$$ need not be monotone, only real-valued. Here is a proof he sent me via PM:

Let $$f \colon \mathbb{R} \to \mathbb{R}$$ have $$S$$ as set of points of discontinuity of first species. That means that, for any $$x_0\in S$$, both $$\ell_{-}(x_0) = \lim_{x\to x_0-} f(x)$$ and $$\ell_{+}(x_0) = \lim_{x\to x_0+} f(x)$$ do exist and are finite, but either they are not equal, or they are equal, but not equal to $$f(x_0)$$. In turn, that means that $$\max\{\ell_{-}(x_0), \ell_{+}(x_0), f(x_0)\} - \min\{\ell_{-}(x_0), \ell_{+}(x_0), f(x_0)\}> 0$$.

Denote by $$S_n = \{ x_0 \in S \mid \max\{\ell_{-}(x_0), \ell_{+}(x_0), f(x_0)\} - \min\{\ell_{-}(x_0), \ell_{+}(x_0), f(x_0)\} $$ $$\in [1/(n+1), 1/n)\}$$, for $$n=0,1,2,\ldots$$ (where for $$n=0$$ we take $$[1/(n+1), 1/n) = [1,\infty)$$). It is easy to see that $$S_n$$ is either empty, or made by isolated points (since lateral limits exist, each point in $$S_n$$ has an open neighbourhood empty of any other point from $$S_n$$). Therefore $$S_n$$ is at most countable.

But the sets $$S_n$$ are pairwise disjoint, while $$S = \bigcup_{n\geq 0} S_n$$, therefore $$S$$ itself is at most countable.

Draco Meteor (talk) 18:20, 16 May 2011 (UTC)
 * I think you're right, I just came here with the same complaint and am sorry nobody has replied in the last nine months. I'll try and find that proof and look at it as I don't have something about it myself. Dmcq (talk) 12:14, 7 February 2012 (UTC)


 * I'm confused! Your talk section heading "Not the actual theorem" and first sentence "... f need not be monotone, only real-valued" imply either:


 * 1) that the article is incorrect, and Froda proved it for real-valued functions; ___ XOR ___
 * 2) that the article is correct, but that Froda's theorem for monotone functions can be extended to real-valued functions.


 * Which is true?  yoyo (talk) 05:33, 10 December 2013 (UTC)


 * The article is incorrect in many ways. If it should have a name, it should be Darboux--Froda. The result proved by Froda is not about Monotone Functions, or even functions of one variable, as you can see in his manuscript, it is about real-valued functions of 'n' variables. Wawawalter (talk)

Is this really Froda's Theorem?
I am not sure if this theorem is actually due to Froda, as the article states.

Froda himself appears to disclaim credit in his 1929 doctoral thesis (which the article cites as the first proof of the theorem):


 * On connaît, depuis longtemps, la proposition suivante:


 * 1. Si une fonction uniforme f(x) ne presente que des discontinuités de première espèce, il n'y en a jamais qu'une infinité dénombrable, au plus.


 * G. Darboux, dans son "Mémoire sur les fonctions discontinues", en s'occupant des fonctions susceptibles d'intégration, selon Riemann, se rapproche de ce résultat, sans le formuler toutefois d'une manère explicite. Nous allons reprendre, en le précisant un peu, sa démonstration qui est intéressante, malgré son cractère élémentaire.

which according to my imperfect French means:


 * The following proposition has been known for a long time:


 * 1. If a single-valued function f(x) has discontinuities of the first kind, there are never more than countably many of them.


 * G Darboux, in his Mémoire sur les fonctions discontinues, considering functions integrable in the Riemann sense, came close to this result without, however, giving an explicit formulation. We will pick up [his work], making it a little more precise, his argument is interesting, notwithstanding its elementary character.

128.250.94.230 (talk) 07:11, 3 September 2012 (UTC)


 * I would read that as saying that the result was sitting around in the background for a long time, assumed to be true, without anyone having formulated it explicitly and actually proved it. --Dfeuer (talk) 22:23, 6 June 2013 (UTC)


 * I don't agree, if you read his proof, he clearly explains that the core of the result is well done in Darboux, and that it is the essential piece. What is missing is the proper handling of countable union of countable sets, which was muddled in Darboux's Memoire. I would not call it Froda's Theorem. And it certainly is not about monotone functions. This entry should not be named this way. Wawawalter — Preceding undated comment added 14:30, 11 April 2014 (UTC)

The stronger form
Who was responsible for the first published proof of "the stronger form" of Froda's theorem, viz.: Let f be a monotone function defined on an interval I. Then the set of discontinuities is at most countable. ?

I'm making the good-faith assumption that this is not original research by a Wikipedian editor.

yoyo (talk) 05:43, 10 December 2013 (UTC)
 * According to the links I posted below, this was at least done by Young around 1907 (not by Froda in 1929). However, this is a special case of Young's more general results, and it seems that special case was likely known decades earlier. Gumshoe2 (talk) 05:35, 3 February 2022 (UTC)

Name change
This is following up on what looks like a discussion here from ten years ago. There are clarifying remarks/references in these stackexchange/mathoverflow posts: It seems to be completely illegitimate (and unsourced) to say that this theorem is "Froda's theorem". Even if he had been the first to prove it (which seems untrue anyway), it is much more commonly known by no name. (This is putting it rather mildly.) Gumshoe2 (talk) 05:24, 3 February 2022 (UTC)
 * https://math.stackexchange.com/questions/1500155/proof-of-frodas-theorem
 * https://hsm.stackexchange.com/questions/13937/history-of-the-darboux-froda-theorem
 * https://mathoverflow.net/questions/231360/a-search-for-theorems-which-appear-to-have-very-few-if-any-hypotheses/231462#231462
 * I have made corresponding changes, and changed the title of the page. Gumshoe2 (talk) 05:33, 3 February 2022 (UTC)

Why weaken the result on countable sets?
I added a section "Negative" to show that every countable subset of the real line is the set of discontinuities of a monotone function. replaced it with a much weaker result. See my version here. JRSpriggs (talk) 19:21, 4 February 2022 (UTC)
 * I don't know why the header "Negative" was used. It's well known that a non-decreasing function on [a,b] is equivalent to a positive measure, i.e. a positive bounded functional on C([a,b]). That's the modern point of view in WP:RSs like Apostol's Mathematical Anaylsis, Kolmagorov & Fomin, Loomis, Riesz & Nagy, Rudin, etc. In those sources a dictionary is set up between complex and real measures on [a,b] versus the terminology of functions of bounded variation, the Jordan decomposition (a real fn of BV is canonically a difference of non-negative monotone functions), total variation (its norm) and the Stieltjes integral. These examples of step functions are easy to understand in this context. A positive measure of norm one is called a probability measure or state. A convex combination of states is also a state; so a convex combination of normalised non-decreasing functions gives another non-decreasing function. Although straightforward, a little bit of work is required to check where the atoms of the state occur, i.e. checking rigorously where the jump discontinuities occur (see Apostol's section on "Step functions as integrators", etc). In those terms, discrete random variables can have accumulation points at jump discontinuities in [a,b] which need not be easy to understand. Mathsci (talk) 20:57, 4 February 2022 (UTC)
 * Most general statement (with citation to proof in Riesz & Sz-Nagy) is now in the article. Mathsci (talk) 22:41, 4 February 2022 (UTC)


 * I meant to say the inversion rather than the negation. Instead of saying "every uncountable set of reals are not the discontinuities of a monotone function", I am saying "every countable set of reals are the discontinuities of a monotone function". Instead of ~P→~Q, P→Q.
 * I was not talking about measures, I do not see their relevance here.
 * What I left out was saying that if q is not in S, then one can use the definition of continuity (with epsilon and delta) to show that q is a point where fS is continuous. This is true even if q is a limit point of S as it would be if S were the rational numbers. JRSpriggs (talk) 02:05, 5 February 2022 (UTC)
 * At points of discontinuity, you have not taken notice of discontinuities from both the left and the right, which is a failing. Mathsci (talk) 03:05, 6 February 2022 (UTC)
 * Discontinuities includes both left and right discontinuities. Depending on which version of the Heaviside step function one uses, one could choose to make fS either: left-continuous and right-discontinuous, left-discontinuous and right-continuous, or both left-discontinuous and right-discontinuous. Why is not mentioning that a failure? JRSpriggs (talk) 18:41, 6 February 2022 (UTC)

"Inversion"?
The book of Riesz and Sz.-Nagy is a classic and its first pages are very easy to read:



For brevity monotone functions will mean monotone functions on an interval that are non-negative and non-decreasing. A monotone jump (or saltus) function $$H$$ is one with countably many jump discontinuities, enumerated as $$a_n$$. At points of continuity $$H(x) = H(x\pm 0)$$; otherwise the jumps are given by $$u_n=H(a_n)-H(a_n -0)$$ and $$v_n=H(a_n +0) -H(a_n)$$. In this case $$H(x)$$ can be written in the explicit form
 * $$H(x) = \sum_{a_n \le x} u_n + \sum_{a_n < x} v_n.$$

Given an arbitrary monotone function $$F$$ with discontinuities at ($$a_n$$), its jump data is given by $$u_n=F(a_n)-F(a_n-0)$$ and $$v_n=F(a_n+0) -F(a_n)$$. From these a jump function $$H$$ can be defined by the formula above, and $$ G(x)=F(x) -H(x)$$ is then continuous and monotone. In this way any monotone function can be written in a unique way as the sum of a continuous monotone function and a jump function.

Since the formula for $$H(x)$$ is a positive combination of characteristic functions, it is a uniformly convergent sum, so the analysis of is particularly simple. From the perspective of Hardy's "A Course in Pure Mathematics", only first or second year undergraduate material occurs. No ideas from measure theory, evenly thinly disguised.

's new section on "inversion" is unsourced (no WP:RS of any form) and there is no motivation or explanation for the header "inversion" or "contrapositive" (the header "negative" was previously used). That proposed content attempts to repeat, only partially, the elementary and clean-cut results of Riesz and Sz.-Nagy above on "jump functions". I am puzzled why there might be some reluctance to use or find WP:RSs (in French, German, Russian, English, Hungarian, ...). F. Riesz produced several definitive representation theorems, all discussed in the monograph: the section on "Linear functionals on the space C" directly involves monotone functions (positive functionals) and has applications to the moment problem. Mathsci (talk) 02:22, 6 February 2022 (UTC)


 * Inversion is not the same thing as contrapositive. I just linked the article on contrapositive because it contains a definition of inversion. JRSpriggs (talk) 18:41, 6 February 2022 (UTC)


 * I wrote the section in response to "Should the article include a construction of a monotone function having any given countable set as its points of discontinuity? —David Eppstein" in the discussion on the Project talkpage. I do not see where you have answered that question in all the stuff you put into the article. JRSpriggs (talk) 18:52, 6 February 2022 (UTC)


 * The problem with your section "Jump functions" lies at the very beginning. You say "Let $x$1 < $x$2 < $x$3 < ⋅⋅⋅ be a countable subset of the compact interval [$a$,$b$] ..." as if every countable set of reals can be put in this form. That is not so. Although the rational numbers is countable, it cannot be enumerated in a strictly increasing sequence. JRSpriggs (talk) 02:32, 9 February 2022 (UTC)
 * There wasn't an error, despite your edit summary. Please read the section and the references more carefully.


 * The section starts with a particular class of Examples, specifically [generalised] step functions which are very special cases of jump functions. Standard step functions correspond to those with finitely many discontinuities $a$ < $x$1 < ... < $x$$M$ < $b$. More generalised step functions are just those which arise from subdividing [$a$,$b$] with $x$1 < $x$2 < $x$3 < ⋅⋅⋅ or even ⋅⋅⋅ < $x$–2 < $x$–1 < $x$0 <  $x$1 < $x$2 < ⋅⋅⋅


 * But these generalised functions are extremely special — they were only given as examples. General saltus-functions or jump functions are discussed in the paragraph starting "More generally, ..." in the article; they summarise the material in the WP:RS . Because it might not have been clear, I have now distinguished between [generalised] step functions and the far more complicated jump functions. The section on jump functions describes them explicitly: with some minor changes of notation, the definitions are exactly what can be found in F. Riesz. (von Neumann also gave a similar treatment in his 1950 book, "Functional Operators".) Since Riesz & Sz.-Nagy, many authors have provided more elementary proofs of the main result of Lebesgue on differentiation of monotone functions — that the derivative of a jump function vanishes a.e. These proofs appear in Colloq. Math., American Mathematical Monthly, etc; and the maths forums Stack Exchange/Overflow or blogs, such as Terence Tao's. Mathsci (talk) 04:07, 9 February 2022 (UTC)

(1) The article is hard to read because it is cluttered with a lot of unnecessary information and repetition. (2) "More generally" does not appear in the article. (3} The formulas "Passing to the limit, it follows that f(xn + 0) − f(xn) = λn and f(xn) − f(xn − 0) = μn." has mu and lambda switched. JRSpriggs (talk) 03:27, 10 February 2022 (UTC)

Please use use, not mathsci. Mathsci (talk) 07:11, 10 February 2022 (UTC)
 * It is strange that mathsci replaced JRSpriggs' material with a weaker version, and it is also strange that he also added a generalized version; I see no reason not to replace the three given constructions with the (one) generalized construction added by mathsci, as all constructions and proofs only differ in pretty minor contextual details. The general one given by mathsci is the best of the three. Two additional comments:
 * it seems mathsci made an error in saying that f(a)=0 for the constructed function
 * I have no idea why mathsci added a lengthy & detailed proof that jump functions have derivative 0 almost everywhere. I am skeptical that it is good content for any wiki page, but it has only marginal interest on this one. It is especially strange for inclusion when the proof of the construction of $f$ (which is the main point) is basically contained in the single phrase "Passing to the limit, it follows that" which is very much undetailed.
 * Gumshoe2 (talk) 05:36, 10 February 2022 (UTC)
 * These comments are comments are not accurate. The most general results starts in the second paragraph on "jump functions"; the simple examples about (generalised) step functions are described in the first paragraph. The later paragraphs summarise the 1932 work of F. Riesz on the most general type of jump functions on a compact interval (this generalises easily to any interval including the whole reals). The WP:RSs are in, Rubel's short note and Komornik's lecture notes. Mathsci (talk) 07:30, 10 February 2022 (UTC)
 * By definition the jump functions are defined to vanish at $a$. That applies for the $f$$n$ also. For reasons best known to themselves, User:Gumshoe2 decided to write an article on "Discontinuities of monotone functions". WP:RSs show that the main result in the topic concerns "Lebesgue's differentiation theorem, which can be found in the classic text of Riesz & Sz.-Nagy: one of the main and famous consequences is that jump functions have zero derivative zero. There are numerous accounts of that; it's material that can be found in American Mathematical Monthly, so extremely accessible and attractive on undergraduates. Mathsci (talk) 07:56, 10 February 2022 (UTC)
 * f(a)=0 if and only if none of the x_i are equal to a. (Also, I'm not sure why you think I wrote this article. I didn't.) Gumshoe2 (talk) 21:44, 10 February 2022 (UTC)
 * Looking more carefully, I see that the detailed proof is also not very well written. It uses epsilon in the first paragraph without defining it; it introduces the language of "null set", never to be revisited; the phrase "it is elementary that, if three fixed bounded open intervals have a non-empty intersection, then their union contains one of the three intervals" is baffling and probably misstated; it does not define the notation Uc. Gumshoe2 (talk) 05:47, 10 February 2022 (UTC)
 * Some more problems: c is also not defined in the fourth sentence; the trick in the third sentence is completely unnecessary since f is assumed monotone. To be clear- I think this proof should not be included, but if it will be included, then it must be written more clearly and correctly. Gumshoe2 (talk) 05:57, 10 February 2022 (UTC)

Comments. The wikipedia article almost everywhere as written cannot be used to describe subsets of the reals. On the other hand, the article null set is not so bad. For subsets of a compact interval, a null set is anything contained in an open set of arbitrarily small length. Such an open set is a disjoint union of countably many intervals — an enumeration of the rationals allows the connected component to be labelled. The rest of the proof is just taken from Lee Rubel's 1963 article: it's modelled on F. Riesz's 1932 proof; and it's reproduced in Komornik's 2016 account (originally in French). Here's is a snippet in English: "Since $$[a_n^\prime, b_n^\prime]$$ is compact, there exists a finite subcover $$(s_1,t_1), \ldots, (s_N,t_N)$$. Choose a finite subcover with $$N$$ as small as possible. Then no point of $$(s_k, t_k)$$ is covered more than twice, because if three intervals have a common point, then one of them belongs to the union of the other two." This direct quote from Rubel's note is more clearly explained by Lebesgue covering dimension (LCT). If three open intervals $$ I, J, K$$ have $$ I \cap J \cap K \ne \emptyset$$, then one of the three $$K$$, say, must be contained in $$ I \cup J$$. This is explained in Gerard Edgar's book "Measure, Topology of Fractals" — the fact is founds in many WP:RSs, including Hurewicz & Wallman, Munkres, etc. In this simple case, it's enough to take the infimum and supremum to identify the endpoints. For the LCT, there are coloured pictures describing the 2-D case. In 1-D, the open cover can also described by pictures with only consecutive intervals intersecting. The Coll. Math. article of Lee Rubel is available online in Poland but is not in OCR format; but anybody can read the small portion with the proof. Mathsci (talk) 07:05, 10 February 2022 (UTC)
 * I don't have any issue with the sources, I have issue with what you have written on the wiki page. Gumshoe2 (talk) 07:15, 10 February 2022 (UTC)
 * Your haste to make edits shows that you can't really have had time to read the sources or understood the content particularly carefully, since there are so many of them (20 or so?). For example did you read Edgar's 2008 book, one of the most accessible WP:RSs on topological or covering dimension? The word "almost everywhere" was originally rendered as almost everywhere because the article almost everywhere was not sufficiently explicit for subsets of the reals. Quibbling about a.e. vs null set or overlapping intervals that cover a compact set seems pointless. The summary of what I've written is extremely close to the original, but the words and order has been changed. That's the way wikipedia is written (compare the recent edits to Richard Lynn and E. O. Wilson, which are covered by extremely rigid discretionary sanctions). This "elementary" topic has been discussed many times, e.g. this StackExchange post |about the topic. As explained several times on this page, it's been done many times (see the citations in the article). Mathsci (talk) 09:40, 10 February 2022 (UTC)
 * I don't care whether or not you will acknowledge that you have made some errors and problems in your editing, or whether you will only contribute these kinds of bizarre remarks to the talk page. I'm glad you have now fixed some of the problems I pointed out. Please fix the others as well. Gumshoe2 (talk) 21:49, 10 February 2022 (UTC)
 * Funny tone to use. Have you ever read WP:dickery, as in Hickory Dickory Dock? Mathsci (talk) 02:32, 11 February 2022 (UTC)
 * Following the excellent advice "Ensure contributions to discussions are clear, concise, and address the matter at hand", I will point out again the problems which have not been addressed:
 * f(a)=0 if and only if none of the x_i are equal to a, i.e. it is not always true as written
 * the phrase "it is elementary that, if three fixed bounded open intervals have a non-empty intersection, then their union contains one of the three intervals" is baffling and probably misstated
 * it does not define the notation Uc
 * the trick in the fourth sentence is completely unnecessary since f is assumed monotone.
 * Gumshoe2 (talk) 02:45, 11 February 2022 (UTC)
 * Since you seem to be having difficulties with Lebesgue covering dimension, I'll explain it to you again. If $$A, B, C$$ are open intervals with $$A \cap B \cap C \ne \emptyset$$, then one of the intervals, $$C$$ say, must satisfy $$ C\subseteq A \cup B$$. That's all. The trick in the article was used by Riesz, not by me; it was also used by Komornik. The definition of $$U_c$$ is clear enough and is the definition given by Rubel: it is the set of points $$x$$ in $$(a,b)$$ for which there are $$s < x  c.$$ That's what's in the article. Mathsci (talk) 05:38, 11 February 2022 (UTC)

Arbitrary break

 * Who are you trying to fool? It is only in the last couple of hours that you fixed problem #1 and then fixed problem #3 . Anyway, I am glad you fixed them. #2 and #4 remain:
 * Your new version "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" of #2 is still clearly a misstatement. You have not cited sources very carefully, since the relevant parts of those books have slightly different aims than what you intend (they seem to not be immediately dealing with the case of zero derivative, like you are). In the proof you have added, f is monotone non-decreasing so D−f is nonnegative everywhere, and automatically less than or equal to D+f. So if D+f is zero a.e. then the same is true of D−f. No trick needed, the only argument needed is the fact that the only number between 0 and 0 is 0. Gumshoe2 (talk) 06:06, 11 February 2022 (UTC)


 * $$\limsup D^+f$$ could be infinite, hence the need for care: see Riesz & Sz.-Nagy, page 7; von Neumann, pages 71–73; Komornik, page 159. Also, Rubel writes: "By the Heine-Borel theorem, there is a finite subcovering, say $$(s_k,t_k)$$ where $$k=1,2, \ldots, N;$$ and if we choose a subcovering with $$N$$ as small as possible, then each point of $$\bigcup (s_k,t_k)$$ lies in at most two of the intervals $$(s_k,t_k),$$ because given any three intervals with a common point, some one is contained in the union of the other two." Note the phrase starting "because given ..." Mathsci (talk) 14:35, 11 February 2022 (UTC)