I wish to prove that $\nu << \mu $ iff $(\nu^{+}<<\mu \operatorname{and} \nu^{-}<<\mu)$. Where $\nu^{+}$ and $\nu^{-}$ are referring to the positive measures which arise from the Hahn-Jordan decomposition theorem. The reverse implication is trivial but I am struggling to show the forward implication.
Note that I have seen this similar looking question
Prove that: $\nu \ll \mu$ iff $|\nu| \ll \mu$
but the answer uses a definition of $\vert \nu \vert$ which I am not familiar with and have not been taught which suggests to me that the question I have been given should be solvable without resorting to what the person who answered the above question used.
My attempt thus far has been:
$\nu <<\mu$ implies that of $\mu(A)=0$ then $\nu(A)=\nu^{+}(A)-\nu^{-}(A)=0$
Which would imply for all $A$ such that $\mu(A)=0$ we have $\nu^{+}(A)=\nu^{-}(A)$. From this it is not immediate that $\nu^{+}(A)$ and $\nu^{-}(A)$ should be zero. To go further I let $P$ and $N$ be the positive and negative sets (in the proof of the Hahn-Jordan decomposition theorem) such that $\nu^{+}(A)=\nu(A\cap P)$ and $\nu^{-}(A)=-\nu(A\cap N)$. We then have for all $A$ such that $\mu(A)=0$ that
$$\nu(A\cap P)=-\nu(A\cap N)$$
Although, even doing this, it is still unclear to me as to why $\nu(A\cap P)=-\nu(A\cap N)=0$
Where do I go from here...?
If you are using the definition of absolute continuity (for example, the one in Cohn's text), then $\mu$ is a positive measure and the argument goes as follows:
Suppose $\nu <<\mu $ and let $(P,N)$ be a Hahn-Jordan decomposition for $\nu$.
Now, if $\mu(A)=0,$ then $\mu(A\cap P)=0\ $ because $A\cap P\subseteq A,$ and so we have $\nu^+(A)=\nu(A\cap P)=0.$ Thus $\nu^+<<\mu.\ $
Similarly, $\nu^-(A)=\nu(A\cap N)=0\Rightarrow \nu^-<<\mu.$ Therefore $|\nu|=\nu^++\nu^-<<\mu.$