A strictly increasing singular function from a dense series of scaled Cantor functions

Enumerate all closed rational intervals $I_n=[u_n,v_n]$ with $0\le u_n<v_n\le 1$. For each $n$, let $c_n$ be the function that is $0$ on $[0,u_n]$, is $1$ on $[v_n,1]$, and on $[u_n,v_n]$ is the affine rescaling of the Cantor function. The Cantor function is continuous by The Cantor function is continuous on $[0,1]$, so each $c_n$ is continuous, nondecreasing, and takes values in $[0,1]$. Define $S(x):={\sum }_{n\ge 1}{2}^{-n-1}{c}_n(x)$. The series converges uniformly because each summand is bounded by $2^{-n-1}$, so $S$ is continuous and nondecreasing.

If $x<y$, choose a rational interval $I_n$ with $x<u_n<v_n<y$. Then $c_n(x)=0$ and $c_n(y)=1$, so $S(y)-S(x)\ge 2^{-n-1}>0$. Hence $S$ is strictly increasing. For each $n$, the derivative of $c_n$ is $0$ almost everywhere because off the scaled Cantor set inside $I_n$ the function is locally constant 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, and that scaled Cantor set is null by 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. Since each $c_n$ is nondecreasing and Countable Choice is assumed, the term-by-term differentiation theorem Fubini's theorem on term-by-term differentiation for pointwise sums of nondecreasing functions applies and gives $S^{\mathrm{\prime }}(x)={\sum }_{n\ge 1}{2}^{-n-1}{c}_n^{\mathrm{\prime }}(x)=0$ almost everywhere.

The function $S$ is continuous, nondecreasing, strictly increasing, and has derivative $0$ almost everywhere, so it is a singular function by A singular function on a compact interval.

Steps 1.1 through 3.1 prove the example.