Cantor function

Counterexample

Let $C \subseteq [0, 1]$ be the Cantor set. For $x \in [0, 1]$, write $x$ in base $3$ as $x = 0.a_1 a_2 a_3 \ldots$ with $a_k \in \set{0, 1, 2}$. If some digit equals $1$, delete all digits after the first $1$. Then replace each digit $2$ by $1$, and read the result $0.b_1 b_2 b_3 \ldots$ in base $2$. The value is $c(x) = \sum_k b_k 2^{-k}$, and it does not depend on the choice of the base-$3$ expansion.

The function $c \colon [0, 1] \to [0, 1]$ is continuous and nondecreasing, with $c(0) = 0$ and $c(1) = 1$. It is constant on each open interval removed in the construction of $C$, for example $c(x) = 1/2$ on $(1/3, 2/3)$. So $c'(x) = 0$ at every point of $[0, 1] \setminus C$. The Cantor set is a null set, so $c' = 0$ almost everywhere. But $c$ is not constant.