Discontinuous derivative, positive on a dense set Whether there exists a continuos monotone function $f: [0,1]\rightarrow [0,1]$ with the following properties:
(1) $f$ strictly increase, $f(0)=0$ and $f(1)=1$;
(2) there is no interval $A\subset [0,1]$, where the derivative $f'$  (a) exists, (b) is continuous and (c) is finite;
(3) there exists a set $B\subset [0,1]$ dense in $[0,1]$, where the derivative $f'$ (a) exists, (b) is positive and (c) is finite?
I can construct an example of function $f:[0,1]\rightarrow [0,1]$ with condition (1), whose derivative is either $0$, or $+\infty$ (at points, where the dewrivative exists). It follows from Lebesgues theorem that $f'$ is zero on some set $A\subset[0,1]$, which has Lebesgue measure 1 and, whence, is dence in $[0,1]$. Clearly, since the derivative is either $0$, or $+\infty$ (in the example, which I can construct), and the function is invertible, then $f'\neq 0$ at any interval.
In other words, I can answer "yes" to my question if ommit the word "positive" in the condition (3).
I hope that my question is "natural", but neither have found "in the internet" neither its proof, nor counter example, nor this question as it is (for example as open problem).
 A: The answer is yes, see Pompeiu derivative. Short explanation : denote $(q_j)_{j \in \mathbb{N}}$ an enumeration of all rationals in $[0,1]$, and denote $$g : x \mapsto \sum \limits_{j \in \mathbb{N}} 2^{-j} \sqrt[3]{x-q_j}, \ \mbox{ and }\ f : x \mapsto \frac{g(x)-g(0)}{g(1)-g(0)}$$
It is easy to see that $g$ is well defined and strictly increasing, so $f$ is well-defined, strictly increasing and $f(0)=0$, $f(1)=1$. It is clear that if we want $f$ satisfying $(2)$ and $(3)$, it is enough to check that $g$ statisfies $(2)$ and $(3)$.


*

*$(2)$ is obvious : for every $j \in \mathbb{N}$, $g'(q_j) = +\infty$, and $\{q_j, j\in \mathbb{N}\}$ is dense in $[0,1]$.

*Let us prove $(3)$. We consider some open interval $I = ]x_0-\varepsilon,x_0+\varepsilon[ \subset [0,1]$, and prove that that $B \cap I \neq \varnothing$. It is easy to show that $g'(x) = \frac{1}{3}\sum \limits_{j \in \mathbb{N}} \frac{2^{-j}}{\sqrt[3]{(x-q_j)^2}}$ is finite at every point where this sum is finite. The measure of $\tilde{B} := [0,1] \backslash \bigcup \limits_{j \in \mathbb{N}} \big[ q_j-\frac{\varepsilon}{2^{j+2}}, q_j+\frac{\varepsilon}{2^{j+2}}\big]$ is $|\tilde{B}| \ge 1 - \sum \limits_{j=0}^{\infty} \frac{2\varepsilon}{2^{j+2}} = 1- \varepsilon$. Thus $\tilde{B}$ and $I$ cannot be disjoint. For $x \in \tilde{B} \cap I$, $\sum \limits_{j \in \mathbb{N}} \frac{2^{-j}}{\sqrt[3]{(x-q_j)^2}} \le \sum \limits_{j \in \mathbb{N}} \frac{1}{\sqrt[3]{\varepsilon}} \frac{2}{\sqrt[3]{2}}\big)^{-j} < +\infty$ because $\frac{2}{\sqrt[3]{2^2}} < 1$, so $B \cap I \neq \varnothing$

I think there is even a function $f$ continuous, satisfying $(1)$, $(2)$, $(3)$ and such that $f$ has a finite derivative everywhere (but unbounded on every interval because of 2). It would require more effort to build than what is above.
