Folland Exercise 3.8 Let $\nu$ is a signed measure and $\mu$ is a positive measure then $\nu \ll \mu$ iff ${\nu}^{+} \ll \mu$ and ${\nu}^{-} \ll \mu$.
My try:
Converse part is easy. 
For forward implication, let $\nu \ll \mu$ and $E \in \mathcal{M}$ such that $\mu(E)=0$
$\Rightarrow \nu(E)=0$
$\Rightarrow {\nu}^{+}(E)={\nu}^{-}(E)$
Since ${\nu}^{+}\perp {\nu}^{-}$ $\exists P,N \in \mathcal{M}$ such that P is ${\nu}^{-}$ null and N is $\nu^{+}$ null.
$\Rightarrow$ $\nu^+(E)=\nu(E \cap P)$
But I am not able to proceed and show ${\nu}^{+}(E)=0$
Thanks for help!
 A: Let $X = P \cup N$ be the Hahn decomposition of the measure space, and $\nu = \nu^+ - \nu^-$.
Assume $\nu \ll \mu$. Let's check that $\nu^+(E)=\nu^-(E) = 0$. We know that since $\mu(E) = 0$ and $\mu$ is positive, $P \cap E \subseteq E$ gives $\mu(P \cap E) = 0$ and so $\nu(P \cap E) = 0$. But this is $\nu^+(E)$. Similarly $\nu^-(E) = 0$.
The converse is easy: if $\nu^+ \ll \mu$ and $\nu^- \ll \mu$, then $\mu(E) = 0$ implies $\nu^+(E) = \nu^-(E) = 0$, whence $\nu(E) = \nu^+(E) - \nu^-(E) = 0-0 = 0$, as wanted.
A: Exercise 8 - If $\nu\ll \mu$ if and only if $|\nu|\ll \mu$ if and only if $\nu^+\ll\mu$ and $\nu^-\ll\mu$.
Proof:
Suppose $\nu\ll \mu$. If $E\in M$ and $\mu(E) = 0$ then, for any $F\in M$ such that $F\subseteq E$, we have $\mu(F) = 0$ so we have
$$\nu(F) = 0 $$
So $E$ is $\nu$-null. By exercise 2, $|\nu|(E) = 0$. Therefore, $|\nu|\ll \mu$.
Now suppose $|\nu|\ll \mu$. If $E\in M$ and $\mu(E) = 0$ then $$\nu^+(E) +\nu^-(E)=|\nu|(E) = 0$$ and hence $$\nu^+(E) = \nu^-(E) = 0$$ This shows that $\nu^+\ll \mu$ and $\nu^-\ll \mu$.
Finally, suppose that $\nu^+\ll\mu$ and $\nu^-\ll\mu$. If $E\in M$ and $\mu(E) = 0$, then $$\nu(E) = \nu^+(E) - \nu^-(E) = 0$$ and hence $\nu\ll\mu$
