Let $\nu$ be a signed measure, then E is $\nu$-null iff $|\nu|(E)=0$ I am having trouble with understanding the proof of the forward direction. In all the proofs I have seen it's always the case that the following fact is implied.
Taking a Hahn Decomposition $X=P\cup N$.
$\nu(E)=0 \Rightarrow \nu(E\cap P)=0$ since $E\cap P\subset E$. However why is this true? An example, proving $E$ is $\nu$-null iff $|\nu| (E)=0$.
I can only deduce $0=\nu(E)=\nu(E\cap P)+\nu (E\cap N)$ and $\nu(E\cap P)\geq 0$ since it is a subset of P, but $\nu (E\cap N)$ could still be negative.
Alternatively, are there any other proofs?
 A: Your reasoining is right. Your problem seems to be with the definition of null set for signed measures.
Let $\nu$ be a signed measure on the $\sigma$-algebra $\Sigma$, then the definition of  $\nu$-null set is:

Definition: $E$ is $\nu$-null set if for all $A\in \Sigma$ and $A\subseteq E$ we have $\nu(A)=0 \phantom{i}$.  

Note that if $E$ is a $\nu$-null set then $\nu(E) =0$. But the implication does not work in the reverse direction. That is to say: we may have  $\nu(E) =0$ without $E$ being a $\nu$-null set. 
Let us prove that if $E$ is a  $\nu$-null then $|\nu|(E)=0$. 
Proof: Taking a Hahn Decomposition $X=P\cup N$.
Since $E$ is a  $\nu$-null and $E\cap P\subset E$, we have $\nu(E\cap P)=0$. 
Since $E$ is a  $\nu$-null and $E\cap N\subset E$, we have $\nu(E\cap N)=0$. 
So we have 
$$|\nu|(E) = \nu(E\cap P)+ \nu(E\cap P)=0$$
Remark: The implication $$\nu(E)=0 \Rightarrow \nu(E\cap P)=0 \textrm{ since } E\cap P\subset E$$ is actually wrong. The correct statement is 
$$E \textrm{ is a } \nu\textrm{-null set }   \Rightarrow \nu(E\cap P)=0 \textrm{ since } E\cap P\subset E$$
