A pure jump function can have dense discontinuities and derivative 0 almost everywhere

Every summand $x\mapsto {\mathbf{1}}_{\{q_n\le x\}}{2}^{-n}$ is nondecreasing, so $J$ is nondecreasing. If $x<y$, choose $q_m\in \mathbb{Q}\cap (x,y]$; then the $m$th summand contributes $0$ at $x$ and $2^{-m}$ at $y$, so $J(y)-J(x)\ge 2^{-m}>0$. Hence $J$ is increasing. At a rational point $q_m$, the value of the $m$th summand jumps by $2^{-m}$, so $J$ is discontinuous at $q_m$. Thus the discontinuity set contains $\mathbb{Q}\cap (0,1]$, hence is dense in $[0,1]$.

Let $x\in [0,1)$ be irrational. Given $\epsilon >0$, choose $N$ so large that ${\sum }_{n>N}{2}^{-n}<\epsilon \mathrm{/}3$. Because $x\mathrm{\ne }{q}_n$ for $n\le N$, there is a neighborhood of $x$ containing none of the finitely many rationals $q_1,\dots ,q_N$, so the first $N$ partial sums are constant on that neighborhood. The tail contributes less than $\epsilon \mathrm{/}3$ on either side, so $J$ is continuous at $x$. Also $J(0)=0$ because every $q_n$ is positive. Therefore the discontinuity set is exactly $\mathbb{Q}\cap (0,1]$.

The function $J$ has no endpoint defect at $0$, and because the enumeration has no repetitions, at each $q_n$ its jump size is exactly $2^{-n}$. Therefore claim 2 of A nondecreasing function splits uniquely into a jump part and a continuous part identifies the jump function of $J$ with $J$ itself. Countable Choice is assumed, so the jump-function theorem A jump function has derivative zero almost everywhere gives $J^{\mathrm{\prime }}=0$ almost everywhere.

Steps 1.1 through 3.1 prove the example.