If $\nu$ and $\mu$ are finite and positive measures then $\nu \perp \mu$ if and only if $\|\nu - \mu \|= \|\nu\| + \|\mu \|$
Here, $\nu \perp \mu$ means that $\nu$ and $\mu$ are singular measures, and the total variation $\| \nu \|$ is defined as $|\nu |(E) = \sup \{ |\nu(A)| +|\nu (B)|: A, B \quad \text{disjoint}, A \cup B \subset E \}$.
My work: Suppose $\nu \perp \mu$, then there exist $A, \; B$ such that $A\cup B=X$ and $A\cap B =\phi$ with $\nu(E)=\nu(E\cap A)$ and $\mu(E)=\mu(E\cap B)$ for every $E$. So,
\begin{align} \|\nu - \mu\| & = |\nu -\mu|(X) = |\nu -\mu |(A \cup B) \\[8pt] & = \sup \{ |(\nu -\mu)(A)|+|(\nu -\mu)(B)| \} \\[8pt] &= \sup \{ \nu(A)+\mu(B) \} \\[8pt] & = \sup \{ |\nu(A)|\} + \sup \{|\mu(B)| \} \\[8pt] & = \sup \{ |\nu(A)|+|\nu(B)|\} + \sup \{|\mu(A)|+|\mu(B)| \} \\[8pt] & =|\nu|(X)+|\mu|(X)=\|\nu\|+\|\mu\|. \end{align}
Is that correct and complete argument for the direct direction? and how can I prove the converse direction??