$\mu,\nu$ be two complex measures on measurable space $(X,\mathfrak{M})$. Prove that, $\mu\perp\nu\iff |a\mu+b\nu|=|a||\mu|+|b||\nu|$ for all $a,b\in\Bbb{C}$
I have proved that if $\mu\perp\nu$ then $|\mu+\nu|=|\mu|+|\nu|$ and $a\mu\perp b\nu$ for $a,b\in\Bbb{C}$ and we know that $|a\mu|=|a||\mu|$. This proves one direction of the problem.
For the other direction, I tried to prove that $\mu\perp |\nu|$. If I amable to prove this then $\nu<<|\nu|$ and $\mu\perp|\nu|$ imply that $\nu\perp\mu$.
We let $\mu=\mu_a+\mu_s$ be the Lebesgue decomposition of $\mu$ with respect to $|\nu|$. So I have to show $\mu_a=0$.
By Radon Nikodym theorem, there is $f\in L^1(|\nu|)$ such that $d\mu_a=f\ d|\nu|$.
I cannot proceed from here. Can anyone help finish the proof? Thanks for your help in advance.