Let $(X,\mathcal M)$ be a measurable space.
Defintion: We say two measures $\nu_1$ and $\nu_2$ are mutually singular if there exist disjoint measurable subsets $A$ and $B$ s.t. $X=A\cup B$ and $\nu_1 (A)=\nu_2 (B)=0$.
The Jordan Decomposition Theorem says that we can always uniquely decompose a signed measure into the form of the difference of two mutually singular measures, i.e. we can find $\nu^+$ and $\nu^-$ for any signed measure $\nu$ s.t. $\nu=\nu^+ - \nu^-$.
There is a theorem on my text saying that, for any signed measure $\mu$ (maybe with additional condition that it is finite), we can always find two unique positive measures $\mu^+$ and $\mu^-$ s.t. $\mu=\mu^+ - \mu^-$ and $\| \mu^+ \|+\| \mu^- \|$ is minimal.
The author didn't specify what the definition of $\| \mu^+ \|$ is, but I guess it is just the total variation norm, i.e. $\| \mu^+ \| :=\sup\{|\mu^+(E_1)|+\cdots+ |\mu^+(E_n)|:n\in \mathbb Z^+,E_1,\cdots, E_n \text{ are disjoint in }X \text{ s.t. }\cup_i E_i \subset X\}$.
My question is how to prove this theorem? Can we use Jordan Decomposition to deduce it? Thanks for help.