Question

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?

Answer 1

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 $Q = G_{\delta} + G_{\delta\sigma}$ $(\mathrm{mes}\,G_{\delta\sigma} = 0)$.

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.