A jump function has derivative zero almost everywhere

By Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used, the discontinuity set of $F$ in $(a,b)$ is at most countable; enumerate it as $(s_n{)}_{n\ge 1}$. Define two discrete finite measures on $[a,b]$ by ${\mu }_L:={\sum }_{n\ge 1}(F(s_n)-F(s_n^{-})){\delta }_{s_n},{\mu }_R:={\beta }_a{\delta }_a+{\sum }_{n\ge 1}(F(s_n^{+})-F(s_n)){\delta }_{s_n},$ and put $\mu :={\mu }_L+{\mu }_R$. The masses are nonnegative, and for every $x>a$ the definition of The jump function of a nondecreasing function on a compact interval gives $J_F(x)={\mu }_L((a,x])+{\mu }_R([a,x))\mathrm{.}$ Because $\mu$ is concentrated on the countable set $\{a,s_1,s_2,\dots \}$, it is singular with respect to Lebesgue measure.

Fix $x\in (a,b)$ and $h>0$ small. From the representation in step 1.1 one gets $0\le J_F(x+h)-J_F(x)\le \mu ([x,x+h]),0\le J_F(x)-J_F(x-h)\le \mu ([x-h,x])\mathrm{.}$ Apply Differentiation of sigma-finite Borel measures finite on compact sets to $\mu$ and the interval families $[x,x+h]$ and $[x-h,x]$. Since the absolutely continuous part of $\mu$ is zero, the two interval ratios $\mu ([x,x+h])\mathrm{/}h$ and $\mu ([x-h,x])\mathrm{/}h$ tend to $0$ for almost every $x$. Therefore the right and left difference quotients of $J_F$ both tend to $0$ for almost every $x\in (a,b)$.

At every point where both one-sided difference quotients tend to $0$, the two-sided derivative exists and equals $0$. Hence $J_F^{\mathrm{\prime }}$ exists and is $0$ almost everywhere on $(a,b)$.