The Cantor function is continuous and nondecreasing, climbs from $0$ to $1$, and is constant on every interval removed in the construction of the Cantor set, so all of its increase happens on a set of measure zero

Collect what is now known about the Cantor function $c : [0,1] \to \mathbb{R}$ (The Cantor function on $[0,1]$, defined on the Cantor set through ternary digits and extended constantly across each removed interval) and the Cantor set $C$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds).

• $c$ is continuous on $[0,1]$ (The Cantor function is continuous on $[0,1]$).
• $c$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), and $c(0) = 0$, $c(1) = 1$: it climbs the whole way from $0$ to $1$ (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, claims 2 and 3).
• $c$ is constant on every interval removed in the construction of $C$: if $u < v$ lie in $C$ and $(u,v) \cap C = \varnothing$, then $c$ is constant on $[u,v]$; and every point of $[0,1] \setminus C$ lies in the open interval $(u,v)$ of such a pair (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, claim 4).
• $C$ has content zero, and therefore measure zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, claim 2, Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover)).

All of the increase happens on $C$, in the following exact sense. Let $x < y$ in $[0,1]$ with $c(x) < c(y)$. Then $(x,y) \cap C \ne \varnothing$. Indeed, suppose $(x,y) \cap C = \varnothing$ and pick any $t$ with $x < t < y$. Then $t \notin C$, so $t$ lies in the open interval $(u,v)$ of a pair $u < v$ of points of $C$ with $(u,v) \cap C = \varnothing$, and $c$ is constant on $[u,v]$. Now $u < t < y$, and $u \in C$, so $u \notin (x,y)$ and therefore $u \le x$; symmetrically $x < t < v$ and $v \in C$ give $v \ge y$. Hence $[x,y] \subseteq [u,v]$ and $c(x) = c(y)$, contrary to assumption. So a nondegenerate interval on which $c$ actually rises must meet $C$, a set of measure zero, while on the complement of $C$ the function is locally constant.

What is not claimed here. Nothing above says that $c$ is differentiable anywhere, that its derivative vanishes anywhere, or that $c$ is singular: no notion of derivative is available at this point in the reading order, and no notion of Lebesgue measure is developed in the library as it stands. Measure zero here is exactly Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover), a condition on covers by intervals, and every statement above is a statement about $c$, about $C$, and about that covering condition, and about nothing else.

Why this is worth recording at all. A continuous nondecreasing function that climbs from $0$ to $1$ might be expected to do its climbing on a set that is large in some sense; $c$ does all of it on a set that is null and, being nowhere dense (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points, claim 5), small in category as well. The companion page pushes the same observation one step further: $c$ maps the null set $C$ onto the whole of $[0,1]$.