Decomposition of the variation of a signed measure as $|\mu|(A) = \int_A |\frac{d\mu_{1a}}{d\mu_2}-1|d\mu_2 + \mu_{1s}(A)$, where $\mu=\mu_1-\mu_2$ Let $\mu_1$ and $\mu_2$ be two finite measures on $(\Omega, \mathcal{F})$. Let $\mu_1 = \mu_{1a}+\mu_{1s}$ be the Lebesgue decomposition of $\mu_1$ w.r.t. $\mu_2$, that is, $\mu_{1a} \ll \mu_2$ and $\mu_{1s}\perp \mu_2$. Let $\mu = \mu_1 - \mu_2$. I'd like to show that for all $A\in \mathcal{F}$,
$$ |\mu|(A) = \int_A |h-1|d\mu_2 + \mu_{1s}(A) $$
where $h = \frac{d\mu_{1a}}{d\mu_2}$ is the Radon-Nikodym derivative of $\mu_{1a}$ w.r.t. $\mu_2$.
I know that we have the following decomposition: $$|\mu|=\mu_+ + \mu_- $$
Here, $\mu_+(A) = \mu(A \cap \Omega_+)$ and $\mu_-(A)=\mu_-(A \cap \Omega_-)$, where $\Omega = \Omega_+ \cup \Omega_-$ is the Hahn decomposition of $\Omega$ w.r.t $\mu$. Also, there exists a finite measure $\lambda$ such that $\mu_1 = \mu_+ + \lambda$ and $\mu_2 = \mu_-+\lambda$ with $\lambda = 0$ iff $\mu_1 \perp \mu_2$. By the definition of Radon-Nikodym derivative, we have $\mu_{1a}(A) = \int_A h d\mu_2$ for all $A\in \mathcal{F}$.
I am unable to use these facts to prove the desired result. Any hint as to how I should proceed would be highly appreciated.
Edit: This problem is exercise 4.13 from the book "Measure theory and Probability Theory" by Krishna B. Athreya and Soumendra N. Lahiri.
 A: While you know the definitions, it is always good to realize that upon experimentation, we can actually stumble upon what the Hahn-decomposition, and the Jordan decomposition of $|\mu|$ are.
How? Let us start with $\mu = \mu_1 - \mu_2$. We substitute $\mu_1 = \mu_{1a} + \mu_{1s}$ to get $\mu = \mu_{1a} + \mu_{1s}-\mu_2$.. Let us rearrange to get $\mu = (\mu_{1a} - \mu_2) + \mu_{1s}$. Let $h = \frac{d \mu_{1a}}{d \mu_2}$.Finally, evaluated at any set $A$, the equality reads :
$$
\mu(A) = \mu_{1a}(A) - \mu_2(A) + \mu_{1s}(A) = \int_{A} h d \mu_2 - \int_A 1 d \mu_2 + \mu_{1s}(A) \\ = \int_{A} (h-1)d \mu_2 + \mu_{1s}(A) \\ = \color{green}{\int_{A \cap \{h \leq 1\}} (h-1) d \mu_2} + \color{blue}{\int_{A\cap\{h > 1\}} (h-1) d \mu_2  + \mu_{1s}(A)}
$$
Recognize that the green part is negative for all $A$ and the blue part is positive for all $A$ (note that $\mu_{1s}$ is a non-negative measure, why?). This immediately hints at the following Jordan decomposition : $-\mu^-(A)$ is the green part and $\mu^+(A)$ is the blue part.
To show this, since the negative-positive sign part is clear, it's enough to show that the green and blue part are mutually singular as measures (functions of $A$, if you like).
That's not too difficult : let $\mu_{2}$ concentrate on $B$ and $\mu_{1s}$ concentrate on $B^c$. The green part then concentrates on $B \cap \{h \leq 1\}$, while the blue concentrates on the rest. To see this, if $C \subset B \cap \{h \leq 1\}$ then the blue part is zero. On the other hand, if either $C \subset B^c$ or if $C \subset \{h > 1\}$ the green part is zero.
Hence, the Hahn decomposition is also established, and the green part is $-\mu^-$ with the blue part being $\mu^+$. The result follows from $\mu = \mu^- + \mu^+$.
