Two finite positive measures are mutually singular if and only if their minimum is zero. Let $\mu$ and $v$ be two finite positive measures on a measurable space $(X, \mathcal{B})$. Prove that $\mu \perp v \iff \mu + v = |\mu - v|$
For the forward direction, I think I need to use the Jordan decomposition theorem, particularly the uniqueness. For the backwards, I am not sure.
Let $\rho = \mu - v$ be a signed measure. Then by uniqueness, as we have $\mu \perp v$, it follows that $\rho_{+} = \mu$ and $\rho_{-} = v$. How do I proceed?
Here is a similar question: Two finite positive measures are singular iff a condition is satisfied

By Radon-Nykodim there exists $h$ such that
$d(\mu-\nu)=hd(|\mu-\nu|)=hd(\mu+\nu)$.
This implies $(1-h)d\mu=(h+1)d\nu$. Since $|h|=1$ $\mu+\nu$-a.e.  wlog we can suppose that $|h|=1$ everywhere.   Now let $A=h^{-1}(1),B=h^{-1}(-1)$. Then $\nu_{|A}\equiv 0, \mu_{|B}\equiv 0$ and $X=A\cup B$, proving that the two measures are mutually singular.
For the other direction, let $A,B$ be complementary sets on which $\mu,\nu$ are respectively supported and let $h:=\chi_{A}-\chi_B
$
.  Now,$d|\mu-\nu|\ge hd(\mu-\nu)=d(\mu+\nu)\ge d|\mu-\nu$ and the result follows
