Let $\nu$ be a signed measure which is absolutely continuous to a sigma-finite measure $\mu$. Show that $\frac{d|\nu|}{d\mu}=|\frac{d\nu}{d\mu}|$, where $|\nu|$ is the total variation measure of $\nu$.
Now by definition, for any $E\in F$ we have $|\nu|(E)=\nu^+(E)+\nu^-(E)=\nu(E\cap P)-\nu(E\cap P^c)$, where $\{P,P^c\}$ is a Hahn decomposition of $\nu$ and $\{\nu^+,\nu^-\}$ is the corresponding Jordan decomposition. Using these two expressions, I found that $\frac{d|\nu|}{d\mu}=\frac{d\nu^+}{d\mu}+\frac{d\nu^-}{d\mu}=(1_P-1_{P^c})\frac{d\nu}{d\mu}$, but I'm not sure how to equate either of these expressions with $|\frac{d\nu}{d\mu}|$.
Note that $|\nu|$ is a measure i.e. non-negative valued. Now if you go through the proof of Radon-Nikodym theorem you see that proof has two parts, one is for (non-negative) measure in which case Radon-Nikodym derivative is non-negative and another part is for signed measure which has been done by considering positive and negative parts of signed measure giving extended real valued Radon-Nikodym derivative. Therefore $\frac{d|\nu|}{d\mu}\geq 0$ and so taking modules in the equation $\frac{d|\nu|}{d\mu}=\frac{d\nu^+}{d\mu}+\frac{d\nu^-}{d\mu}=(1_P-1_{P^c})\frac{d\nu}{d\mu}$ we have $\frac{d|\nu|}{d\mu}=|(1_P-1_{P^c})||\frac{d\nu}{d\mu}|=1|\frac{d\nu}{d\mu}|=| \frac{d\nu}{d\mu}|$ , $\mu$ a.e.