A right-continuous nondecreasing function splits uniquely as absolutely continuous plus jump plus singular continuous

Let ${\mu }_F$ be the Lebesgue-Stieltjes measure of $F$. By Every finite Borel measure on R has a unique absolutely continuous, discrete, and singular-continuous decomposition, ${\mu }_F=f\lambda +{\mu }_d+{\mu }_s$ with $f\in L^1([a,b])$ chosen nonnegative almost everywhere, ${\mu }_d$ discrete, and ${\mu }_s$ singular and atomless. Let $A(x):=F(a)+{\int }_{[a,x]}fd\lambda ,$ let $J$ be the distribution function of ${\mu }_d$ normalized by $J(a)=0$, and let $S$ be the distribution function of ${\mu }_s$ normalized by $S(a)=0$. By the Lebesgue-Stieltjes correspondence Assuming countable choice, finite-on-compacts Borel measures on $\mathbb{R}$ correspond to nondecreasing right-continuous functions modulo constants, the measure of $A+J+S$ is exactly ${\mu }_F$, and at $x=a$ the three functions satisfy $A(a)+J(a)+S(a)=F(a)$. Hence $F=A+J+S$.

The integral form of $A-F(a)$ is absolutely continuous by Absolute continuity of the integral, so $A$ is absolutely continuous as well, and because $f\ge 0$ almost everywhere the function $A$ is increasing. Also The indefinite integral of an $L^1$ function is differentiable almost everywhere gives $A^{\mathrm{\prime }}=f$ almost everywhere. The discrete measure ${\mu }_d$ is carried by its atoms, so the corresponding distribution function is exactly the jump function of $F$ by Interval formulas and atoms for a Lebesgue-Stieltjes measure. The atomless singular measure ${\mu }_s$ has a continuous distribution function $S$, again by the interval formulas.

Uniqueness follows from the uniqueness of the measure decomposition of ${\mu }_F$ and the uniqueness part of the Lebesgue-Stieltjes correspondence: the absolutely continuous, discrete, and singular-continuous parts of the measure are unique, so the three normalized distribution functions are unique as well.

Apply Differentiation of sigma-finite Borel measures finite on compact sets to ${\mu }_s$. Its absolutely continuous part is zero, so the differentiation density is $0$ almost everywhere. The interval formulas therefore give $S^{\mathrm{\prime }}=0$ almost everywhere. If $S$ is nonconstant, then it is continuous, nondecreasing, and has derivative $0$ almost everywhere, so $S$ is a singular function by A singular function on a compact interval.

Steps 1.1 through 3.1 prove the theorem.