A monotone function is differentiable almost everywhere by the Lebesgue-Stieltjes route

Replacing $F$ by $-F$ if necessary, we may assume that $F$ is nondecreasing. By A nondecreasing function splits uniquely into a jump part and a continuous part, write $F=J_F+C_F$ where $J_F$ is the jump part and $C_F$ is continuous and nondecreasing. By A jump function has derivative zero almost everywhere, $J_F^{\mathrm{\prime }}=0$ almost everywhere.

Let $\mu$ be the Lebesgue-Stieltjes measure of $C_F$. Since $C_F$ is continuous, Interval formulas and atoms for a Lebesgue-Stieltjes measure shows that $\mu$ has no atoms. The differentiation theorem for measures Differentiation of sigma-finite Borel measures finite on compact sets applied to the shrinking interval families $(x-h,x]$ and $(x,x+h]$ therefore gives a full-measure set on which the left and right interval ratios of $\mu$ both converge to the same finite density. By the interval formulas for Lebesgue-Stieltjes measures, those interval ratios are exactly the left and right difference quotients of $C_F$. Hence all four Dini derivatives of $C_F$ agree finitely almost everywhere, and The four Dini derivatives always exist in the extended reals, satisfy the one-sided order inequalities, and detect finite differentiability implies that $C_F^{\mathrm{\prime }}$ exists almost everywhere.

On the common full-measure set where $J_F^{\mathrm{\prime }}$ and $C_F^{\mathrm{\prime }}$ exist, one has $F^{\mathrm{\prime }}=J_F^{\mathrm{\prime }}+C_F^{\mathrm{\prime }}=C_F^{\mathrm{\prime }}$. Therefore $F^{\mathrm{\prime }}$ exists almost everywhere on $(a,b)$. Using A subset of $\mathbb{R}$ has Lebesgue outer measure zero if and only if it has measure zero in the sense of countable closed-interval covers, this is exactly the claimed almost-everywhere differentiability statement.