The Jordan decomposition theorem for signed measures If $\nu$ is a signed measure there exists unique positive measures $\nu^{+}$ and $\nu^{-}$ s.t. $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$
Would appreciate if someone can guide me through the proof.
Proof: let $X=P\cup N$ be Hahn decomposition for $\nu$, define $\nu^{+}(E):=\nu(E\cap P)$ and $\nu^{-}(E):=-\nu(E\cap N)$ then both $\nu^{-}$ and $\nu^{+}$ are positive. Also $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$
1) why is $\nu=\nu^{+}-\nu^{-}$ by the above definition we have 
$\nu(E)=\nu^{+}(E)-\nu^{-}(E)=\nu(E\cap P)+\nu(E\cap N)$  for some $E\in\mathcal{M}$ and $E\neq \emptyset$
2) Also why  $\nu^{+}\perp\nu^{-}$? Is it because $P$ is null for $\nu^{-}$ and $N$ is null for $\nu^{+}$? and $P\cup N=X$,$P\cap N=\emptyset$
As for the uniqueness part. Let $\nu=\mu^{+}-\mu^{-}$ be another decomposition.  Let $E,F$ be measurable s.t. $E\cup F=X$ and $F\cap E=\emptyset$ be another Hahn decompotion.  Also $\mu^{+}(F)=\mu^{-}(E)=0$. Then $P\triangle E$ is $\nu$-null. Why is it $\nu$-null?
We need to show that $\nu^{-}=\mu^{-}$ so we pick $A\in\mathcal{M}$ then 
$\mu^{+}(A)=\mu^{+}(A\cap E)=\nu(A\cap E)=\nu(A\cap P)=\nu^{+}(A)$
How do we get the second equality? Does the third equality follows from the fact that $P\triangle E$ is $\nu$-null?
 A: I think the idea of the proof is: Find a potential $\nu^+,\nu^-$ positive using Hahn dec; Via the construction it follows mutual singularity; (showing that is is unique is intuitively a consequence of $P\triangle E$) Get new pair of positive measures s.t. $\nu=\mu^+-\mu^-$, mutual singularity $\Rightarrow$ Hahn dec, but any difference in the two positive (negative) Hahn sets is a null set. 
1) I think you have to do the steps the other way around using a Hahn dec. to obtain your two measures (one positive and one negative) which are a candidate to be proved to be the unique pair:
$$
\nu(E)=\nu(E\cap (P\cup N))=\nu(E\cap P) +\nu(E\cap N) 
$$
2) Yes, $\forall A\subset N$
$$
\nu^+(A)=\nu(A\cap P) = 0
$$
Similar steps for $\nu^-$.
3.?1) $P\triangle E$ is $\nu$-null is a result from Hahn Decomposition Thm when you have two Hahn dec for the same signed measure.
3.?2) $$\nu(A\cap E)=\nu^+(A\cap E)-\nu^-(A\cap E)=\nu^+(A\cap E)+\nu(A\cap E\cap F)  $$ but $E\cap F=\emptyset$ because $E,F$ is a Hahn dec for $\nu$. 
3.?3)I do not see it the way I prove it, $$ \nu(A\cap E\cap (P\cup N))= \nu(A\cap E\cap P)+\nu(A\cap E\cap N) $$ 
but the $E$ is positive and $N$ negative, then 
$$
\nu(A\cap E\cap N)\geq 0 \text{ because }A\cap E\cap N\subset E
$$
$$
\nu(A\cap E\cap N)\leq 0 \text{ because }A\cap E\cap N\subset N
$$
$$
\Rightarrow \nu(A\cap E\cap N)=0.
$$
Similarly
$$
\nu(A\cap P\cap (E\cup F))=\nu(A\cap P\cap E)+\nu(A\cap P\cap F)=\nu(A\cap P\cap E)
$$ 
and so 
$$ 
\nu(A\cap  P)=\nu(A\cap E) 
$$
