# Two finite positive measures are mutually singular if and only if their minimum is zero.

measure-theorysigned-measures

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