Two finite positive measures are singular iff a condition is satisfied Q) Let $\mu,\nu$ be two finite positive measures on $X$. Prove that $\mu\perp \nu \iff \mu=(\mu-\nu)^+$
To prove that $\mu\perp \nu\implies \mu=(\mu-\nu)^+$, I can consider the signed measure $\lambda=\mu-\nu$, by uniqueness in Jordan's decomposition theorem, $(\mu-\nu)^+=\lambda^+=\mu$ but here I am not using the fact that $\mu\perp \nu$?
To prove that $\mu=(\mu-\nu)^+\implies \mu\perp \nu$, I know that I need to find disjoint $E_1,E_2\subset X$ such that $E_1\cup E_2=X$ and $\mu(A\cap E_2)=\nu(A\cap E_1)=0$ for any measurable set $A$ and am not sure how to find these $E_1,E_2$?
Any suggestions?
 A: It's helpful to keep in mind the uniqueness condition for the Jordan decomposition theorem: If $\lambda$ is a signed measure and there are positive measures $\lambda_1$, $\lambda_2$ with $\lambda = \lambda_1-\lambda_2$ and $\lambda_1\perp\lambda_2$, then $\lambda_1=\lambda^+$ and $\lambda_2 = \lambda^-$. This is where the assumption $\mu\perp\nu$ comes into play; setting $\lambda = \mu-\nu$ then implies $\mu = \lambda^+ = (\mu-\nu)^+$.
For the other direction one can use the Hahn Decomposition Theorem (which is a precursor of Jordan decomposition). Since $\lambda = \mu-\nu$ is a signed measure, this theorem provides a $\lambda$-positive set $P$ and $\lambda$-negative $N$ such that $X = P\cup N$ and $P\cap N=\emptyset$. These sets further have the property $\lambda^+(A)=\lambda(A\cap P)$ and similarly for $\lambda^-$ in the Jordan decomposition $\lambda = \lambda^+-\lambda^-$.
Taking $P$ and $N$ as above, I claim $E_1=P$ and $E_2=N$ work in showing $\mu\perp\nu$. Indeed, $\mu = \lambda^+$ implies $\mu(A\cap N) = \lambda^+(A\cap N) = \lambda((A\cap N)\cap P)= 0$ for any measurable set $A$. Since $\mu = \lambda^+$ implies $\nu = \lambda^-$, we also have $\nu(A\cap P) = \lambda^-(A\cap P) = \lambda((A\cap P)\cap N) = 0$. This completes the proof.
