Does differentiability almost everywhere imply continuity on an interval?

Question

I suppose there are differentiable almost everywhere functions whose sets of discontinuities are dense. How to prove or disprove it?

Additionally, is Thomae's function $T(x)$ raised to some power greater than 2 an example? (With 2 it isn't differentiable anywhere by Hurwitz's theorem.) Or maybe $\begin{cases} e^{-\frac 1 {T(x)}} & \textrm{if $x\in\mathbb Q$} \\ 0 & \textrm{otherwise} \end{cases}$?

Answer 1

By Khinchin's theorem, for almost every real $x$ there are at most finitely many rationals $p/q$ (where $p,q$ are integers with $q > 0$) with $|x - p/q| < 1/q^3$. Consider the function $f(x)$ such that $f(p/q) = 1/q^4$ for rational $p/q$ in lowest terms, $f(x) = 0$ otherwise. Note that if $x$ is irrational and $|x - p/q| \ge 1/q^3$, $$ \left|\frac{f(x) - f(p/q)}{x - p/q}\right| \le \frac{1/q^4}{1/q^3} = \frac{1}{q} $$ and as a result $f'(x)$ exists and is $0$ for any $x$ in Khinchin's set. But $f$ is discontinuous at every rational.

Answer 2

You may consider the function $f(x) = \text{sign}(x) e^{-|x|}$ and an enumeration $q_1,q_2,\ldots$ of the elements of $\mathbb{Q}$. If you manage to prove that $$ F(x)\stackrel{\text{def}}{=}\sum_{n\geq 1}\frac{f(n(x-q_n))}{2^n} $$ is almost everywhere differentiable, you have your counter-example.
For short: condensation of singularities.