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.
Best Answer
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 $\mu = \mu^- + \mu^+$.