Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A theorem by Zygmunt Zahorski states that a necessary and sufficient condition for a subset of $\mathbb{R}$ to be the nondifferentiability set of a continuous real function is that it is the union of a $G_\delta$ set and a $G_{\delta \sigma}$ set of zero measure.

On the other hand it is not hard to see that the nondifferentiability set of an arbitrary real function is always a $G_{\delta \sigma}$ set.

My question is: why can't we leave the word 'continuous' out in Zahorski's theorem? In other words, what would be an example of a (necessarily discontinuous) real function whose nondifferentiability set is not a union of a $G_\delta$ set and a $G_{\delta \sigma}$ set of zero measure?

share|improve this question
add comment

1 Answer 1

up vote 6 down vote accepted

Apparently, continuity is not essential.

According to A. Brudno, Continuity and differentiability (Russian), Rec. Math [Mat. Sbornik], N.S. 13 (55), (1943), 119–134 (MathSciNet review here, online article here), the set of non-differentiability of any real function is the union of a $G_{\delta}$-set with a $G_{\delta\sigma}$-set of measure zero.

I quote from Brudno's English summary:

In the present paper we investigate the structure of the set of points, in which the function of one real variable is not differentiable. All the functions are only supposed to be finite in every point.

(...)

Theorem IV. In order that the set $Q$ should be the totality of points, in which a function f(x) does not possess a derivative, it is necessary and sufficient that $Q = G_{\delta} + G_{\delta\sigma}$ $(\operatorname{mes}G_{\delta\sigma} = 0)$.

Since I do not read Russian, I cannot tell you anything about the methods of proof apart from the fact that it involves an investigation of the sets where the Dini derivatives are infinite or distinct.

share|improve this answer
    
Mr. Buehler, thank you for your answer. Unfortunately I can't read Russian either but at least I can understand the interesting results from the summary. Indeed, in the article by Zahorski it is assumed that the function is continuous (see pages 176-177 of a French translation of the original, available here: numdam.org/item?id=BSMF_1946__74__147_0). I wonder how badly (if at all) dropping the assumption on continuity complicates his argument. –  LostInMath Dec 15 '10 at 16:07
    
Hi, LostInMath. It was a pleasure. I've been thinking about this a bit but it is still unclear to me where and exactly how continuity enters in Zahorski's argument - at the moment I don't have access to the works of Caratheodory, Saks and Hausdorff he refers to. I've found an exposition of the Zahorski-Brudno result in English (the paper looks rather scary, to be honest). But maybe it's giving you some clues and inspiration: Theorem 3.4.1 on page 114 in here projecteuclid.org/euclid.rae/1212763954 (Garg: A unified theory of bilateral derivates). PS: call me Theo, please :) –  Theo Buehler Dec 16 '10 at 6:27
    
Hi, Theo. I looked a bit more closely into the references mentioned in the article by Zahorski. The book by Saks is available at archive.org and the referred part of the book by Caratheodory is available at Google Books. You can see that Saks doesn't assume continuity in his theorems 4.2 & 4.4 but Caratheodory does in his Satz 2. Also, compare Caratheodory's Satz 2 with Brudno's Thm II and Garg's Corollary 2.4.2. where continuity is not assumed. BTW, that Garg's article does look scary. However, taking the earlier theorems for granted, the proof of 3.4.1. is quite elementary and not too hard. –  LostInMath Dec 17 '10 at 17:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.