Continuous function differentiable on $[0,1]\setminus\mathbb{Q}$, but nondifferentiable on all of $\mathbb{Q}\cap[0,1]$? I'm trying to work out an example of a continuous function which is differentiable at all irrationals but nondifferentiable at all rationals in $[0,1]$.
Since $\mathbb{Q}$ is countable, list it as $\mathbb{Q}=\{q_0,q_1,q_2,\dots\}$. Define a map $g\colon\mathbb{Q}\to\mathbb{R}$ by $q_n\mapsto 2^{-n}$. Since $\sum_{n=0}^\infty\frac{1}{2^n}$ is absolutely convergent, $\sum_{r\in\mathbb{Q}}g(r)$ is too. Then define $f\colon [0,1]\to\mathbb{R}$ by
$$
f(x)=\sum_{r\in\mathbb{Q};r<x}g(r)
$$
which is well defined.
It is not hard to see that $f$ is monotonically increasing on $[0,1]$, and thus Riemann integrable on $[0,1]$. I've been able to show that $f$ is continuous at all irrationals, but discontinuous at all rationals. I can add this if needed.
By the fundamental theorem of calculus, the function $F\colon [0,1]\to\mathbb{R}$ defined by
$$
F(x)=\int_0^x f
$$is continuous, and differentiable at all irrationals since $f$ is continuous at all irrationals.
The example on page 7 of these notes, Math 131AH Winter 2003, Prof. Terry Tao remarks that $F$ is actually nondifferentiable at every rational, by use of the mean value theorem.
I'm confused because I don't see how I can apply the mean value theorem. I don't think I'm intended to apply it to $f$, since $f$ is not differentiable on any nondegenerate interval. Also, although $F$ is continuous, I don't think I can apply the mean value theorem to it without assuming that $F$ is differentiable on all the rationals in some interval, which seems like a large assumption. 
How can the mean value theorem (if necessary), show that $F$ is discontinuous at all rationals? Thanks.
 A: Your example of a function continuous at irrationals but discontinuous at rationals can be rephrased as follows: writing $\mathbb Q\cap[0,1] = \{q_0,q_1,q_2,\dots\}$ as before, define
$$
f_n(x) = \begin{cases} 0, &\text{if } x\le q_n, \\ 2^{-n}, &\text{if } x>q_n.\end{cases}
$$
Then your example $f(x)$ is simply $f(x) = \sum_{n=0}^\infty f_n(x)$. In this formulation, it's easy to see where the discontinuity at some $q_k$ comes from: all the functions being added are continuous there except for $f_k$.
So to construct a function differentiable at irrationals but nondifferentiable at rationals, I suggest trying $F(x) = \sum_{n=0}^\infty F_n(x)$, where
$$
F_n(x) = \begin{cases} 0, &\text{if } x\le q_n, \\ 2^{-n}(x-q_n), &\text{if } x>q_n.\end{cases}
$$
(And yes, this $F$ really is $\int f$, although it's not necessary to know that: one can deduce the differentiability properties of $F$ directly from the above definition.)
A: Suppose not i.e., $F$ is differentiable at some rational point $x_0 \in [0,1]$ and 
therefore $f$ is continuous at this $x_0$. Let $x_0 \in [0,x]$ with $0\le x\le 1$.
Since $0\le x_0 \le x$ then $x-x_0\ge 0$ so by Archimedean property there exists a positive integer $n \ge 1$ such that $n(x-x_0)>1$ i.e., $x-x_0 > \frac{1}{n}$. Also, since $f$ is increasing we have $f(x-x_0)>f(\frac{1}{n})$     ...........(*).
Since $F$ is cts at arbitrary $x_0 \in [0,1]$ the for $\epsilon >0$ and choose a suitable $\delta = \frac{\epsilon}{M}$ with $M>0$ such that $|F(x)-F(x_0)| < \epsilon$, whenever $0<|x-x_0|<\delta$.
\begin{align}
 \left| {F\left( x \right) - F\left( {x_0 } \right)} \right| &= \left| {\int_0^x {f\left( t \right)dt}  - \int_0^{x_0 } {f\left( t \right)dt} } \right| 
\\ 
  &= \left| {\int_0^x {f\left( t \right)dt}  + \int_{x_0 }^0 {f\left( t \right)dt} } \right| = \left| {\int_{x_0 }^x {f\left( t \right)dt} } \right| 
\\
&\le \int_{x_0 }^x {\left| {f\left( t \right)} \right|dt}  \le \frac{M}{{\delta \left( {x - x_0 } \right)}}\left( {x - x_0 } \right) = \epsilon .....(**) 
 \end{align}
(Similary, can be done by MVT). On the other hand, since $f$ increasing using (*) we have
\begin{align}
 \left| {F\left( x \right) - F\left( {x_0 } \right)} \right| = \left| {\int_{x_0 }^x {f\left( t \right)dt} } \right| 
= \left| {\int_0^{x - x_0 } {f\left( {t + x_0 } \right)dt} } \right| &\ge \left| {\int_0^{{\textstyle{1 \over n}}} {f\left( {t + x_0 } \right)dt} } \right| 
\\
&= \left| {\int_{x_0 }^{{\textstyle{1 \over n}} + x_0 } {f\left( t \right)dt} } \right| \\ 
  &= \left| {F\left( {x_0  + \frac{1}{n}} \right) - F\left( {x_0 } \right)} \right| (< \epsilon \,\,\text{by (**)})  \\ 
 \end{align}
which is a contradiction. Since $x_0$ is an arbitrary rational in $[0,1]$, so that $F$ is not cts on [0,1] therefore $F$ is not differentiable at any rational $x_0 \in [0,1]$.
