Problem statement
I am reading Royden's Real Analysis (4th ed), and am struggling with Problem 51 in Ch 18 in pp. 385. This problem is related to the Lesbegue decomposition of functions of bounded variations. The exact problem statement is as follows.
Let $[a, b]$ be a closed, bounded interval and the function $f$ be of bounded variation on $[a, b]$. Show that there is an absolutely continuous (AC) function $g$ on $[a, b]$, and a function $h$ on [a, b] that is of bounded variation (BV) and has $h' = 0$ a.e. on $[a, b]$, for which $f = g + h$ on $[a, b]$. Then show that this decomposition is unique except for the addition of constants.
Background
Much earlier in the text it has been shown that if $f \in \textrm{BV}[a, b]$, then
$$ f(x) = \int ^x _a f' dm + \left[ f(x) - \int ^x _a f' dm \right] \, $$
where $m$ is the Lebesgue measure, the integral is AC and the function in the bracket is BV. This exercise is listed in the section introducing the Lebesgue Decomposition Theorem for measure and the Radon-Nikodym Theorem, so I would like to seek a solution using measures.
My attempt
We can assume $f$ is increasing on $[a, b]$, since if $f \in \textrm{BV}[a, b]$, then it is a difference of two increasing functions, and both BV and AC functions are closed under linear combination.
Let $f^*$ be the right-continuous version of $f$ by taking the right-hand-limit at each point in $[a, b]$, which is properly defined because $f$ is increasing. Moreover, $f^*$ is still increasing. Define the function $\phi = f^* - f$, then $\phi = 0$ a.e. on $[a, b]$, and $\phi \in \textrm{BV}[a, b]$ as the difference of two increaisng functions.
Let $X = [a, b]$, $\mathcal{M}$ be the $\sigma$-algebra of Borel sets in $[a, b]$ and $\mu$ the Lebesgue measure. Next we construct the Lebesgue-Stieltjes measure on $[a, b]$. Define $S$ to be a collection of set, consisting of $\emptyset$, $\{ a \}$ and all the intervals in the form $(c, d] \subset [a, b]$. Then $S$ is a semiring. Further define $\nu$ on $S$ by
$$ \nu (\{ a \}) = f^*(a) \quad \text{and} \quad \nu ((c, d]) = f^*(d) - f^*(c) \text{ for } (c, d] \subset [a, b]. $$
I skip the steps but $\nu$ is a premeasure on $S$ as in the usual definition of Lebesgue-Stieltjes measure. By the Carathéodory Extension Theorem, $\nu$ can be extended to a measure to $\sigma(S)$, which is clearly larger than $\mathcal{M}$ the Borel $\sigma$-algebra.
Note that $\mu([a, b]) = b - a < \infty$ and $\nu([a, b]) = \nu((a, b]) + \nu( \{ a \}) = f^*(b) < \infty$, so both $\mu$ and $\nu$ are finite. We can apply Lebesgue Decomposition Theorem to obtain two measures $\nu_0$ and $\nu_1$ on $([a, b], \mathcal{M})$, such that $\nu_0 \ll \mu$, $\nu_1 \perp \mu$ and $\nu = \nu_0 + \nu_1$. Observe that
$$ f^*(x) = (f^*(x) - f^*(a)) + f^*(a) = \nu ([a, x]) + \nu( \{ a \}) = \nu ([a, x]) \quad \text{ for all } x \in [a, b] \, , $$
so by the decompositon
$$ f^*(x) = \nu_0 ([a, x]) + \nu_1 ([a, x]) \implies f(x) = \nu_0 ([a, x]) + \nu_1 ([a, x]) + \phi(x) \, . $$
If we define $$ g(x) = \nu_0 ([a, x]) \quad \text{and} \quad h(x) = \nu_1 ([a, x]) + \phi(x) \, , $$ then $g$ inherits absolute continuity from $\nu_0$. Also $h \in \textrm{BV}[a, b]$ since $\phi \in \textrm{BV}[a, b]$ and $\nu_1 ([a, x])$ is increasing.
My problems
I have no idea on how to show $h' = 0$ a.e. on $[a, b]$. I tried using the mutual singularity of $\nu_1$, say $\nu_1(A) = \mu ([a, b] - A) = 0$, then
$$ \int ^x _a d \nu_1 = \int ^x _a \chi _{A} \, d \nu_1 \, $$
and if we $\color{red}{\textrm{take derivative on the integral}}$, then the derivative is $\chi _A$ which is $0$ a.e. [$\mu$]. However, taking the derivative of $\nu_1$ is not defined at this stage. Another method I tried is to use the Radon-Nikodym Theorem on $\nu_0$ to obtain a function $\zeta$ such that
$$ \nu_0 (E) = \int _E \zeta d \mu \, . $$
Then
$$ h(x) = f(x) - \int ^x _a g d\mu + \phi (x) \implies h'(x) = f'(x) - g(x) \, , $$
because $\phi' = 0$ a.e. on $[a, b]$ as a constant a.e. However, I would need to show $\color{red}{g = f'}$ which I struggle doing based solely on measures.
Any help is appreciated.