The Cantor function has derivative 0 almost everywhere, is not differentiable on the Cantor set, and still rises from 0 to 1

By The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, every point of $[0,1]\setminus C$ lies in an open interval on which $c$ is constant. Hence $c^{\mathrm{\prime }}(x)=0$ for all $x\mathrm{\notin }C$. Since $C$ is Lebesgue null by The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero, this proves $c^{\mathrm{\prime }}=0$ almost everywhere.

Fix $x\in C$. Write the ternary expansion of $x$ using only digits $0$ and $2$, and let $u_n\le x\le v_n$ be the two points of $C$ obtained by freezing the first $n$ ternary digits of $x$ and filling the remaining digits with all $0$'s and all $2$'s. Then $v_n-u_n=3^{-n}$ and $c(v_n)-c(u_n)=2^{-n}$ by the digit description of The Cantor set is exactly the set of ${\sum }_{k\ge 1}{a}_k{3}^{-k}$ with every $a_k\in \{0,2\}$, and this gives a bijection with $\{0,1{\}}^{\mathbb{N}}$ and the definition of the Cantor function. At least one of the two numerator differences $c(v_n)-c(x)$ and $c(x)-c(u_n)$ is at least $2^{-n-1}$. For that choice, the corresponding denominator is positive and at most $v_n-u_n=3^{-n}$, so one of the two secant slopes is at least $2^{-n-1}\mathrm{/}{3}^{-n}=\frac{1}{2}(3\mathrm{/}2{)}^n$. These lower bounds are unbounded, so $c$ cannot have a finite derivative at $x$.

The endpoint values $c(0)=0$ and $c(1)=1$ are part of The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, so the function still rises by one.

Steps 1.1, 2.1, and 2.2 prove the example.