Lebesgue decomposition of an increasing function This problem asserts that any increasing function $f$ on $[a,b]$ can be decomposed as $$F=F_A+F_C+F_J,$$ with $F_A$ is absolutely continuous, $F_J$ is a jump function, and $F_C$ is a singular function, i.e., it is continuous and  $F'_C=0$, a.e.. Both $F_A$, $F_J$, $F_C$ are increasing on $[a,b]$. Moreover, the decomposition is unique upto a constant.

From Stein's book, I know that $F=F-J+J$ where $J$ is the jump function associated to $F$, and $F-J$ is increasing and continuous. But how can I proceed from here to prove the claim above?

Any hints would be greatly appreciated!
My suggestion is as follows: because $f$ is increasing on $I=[a,b]$, then it has bounded variation and hence $u\in BV(I)$. Thus, we have $Du$, as a Radon measure, is well defined. Now, by Lebesgue decomposition theorem, we could decompose $Du$ into two measures, namely the one $\mu_1&lt;&lt;\mathcal{L}^1$ and $\mu_2\bot \mathcal{L}^1$ such that $Du=\mu_1+\mu_2$. We future write $\mu_j:=\mu_2\lfloor J$ where $J$ is the jump set and $\mu_s:=\mu_2\lfloor(I-J)$. Now, $\mu_1$ is actually your $F_A$, $\mu_j$ is $F_J$ and $\mu_s$ is $F_C$.
My suggestion is as follows: because $f$ is increasing on $I=[a,b]$, then it has bounded variation and hence $u\in BV(I)$. Thus, we have $Du$, as a Radon measure, is well defined. Now, by Lebesgue decomposition theorem, we could decompose $Du$ into two measures, namely the one $\mu_1&lt;&lt;\mathcal{L}^1$ and $\mu_2\bot \mathcal{L}^1$ such that $Du=\mu_1+\mu_2$. We future write $\mu_j:=\mu_2\lfloor J$ where $J$ is the jump set and $\mu_s:=\mu_2\lfloor(I-J)$. Now, $\mu_1$ is actually your $F_A$, $\mu_j$ is $F_J$ and $\mu_s$ is $F_C$.