The Jordan Decomposition Theorem, Folland 
The Jordan Decomposition Theorem - If $\nu$ is a signed measure, there exists unique positive measures $\nu^+$ and $\nu^-$ such that $\nu = \nu^+ - \nu^-$ and $\nu^+\perp \nu^-$.

Attempted proof - Let $X = P\cup N$ be a Hahn decomposition for $\nu$ where $P$ and $N$ are positive and negative sets respectively.. Then let us define the positive and negative measure as such $$\nu^{+}(E) := \nu(E\cap P) \ \ \ \nu^{-}(E) := - \nu(E\cap N)$$ then, 
\begin{align*}
\nu^+(E) - \nu^-(E) &= \nu(E\cap P) + \nu(E\cap N)\\
&= \nu(E)
\end{align*}
So we have $\nu = \nu^+ - \nu^-$ and $\nu^+\perp \nu^-$. 
Now I believe to complete this proof we need to show uniqueness. As in assuming we have $\nu = \tilde{\nu^{+}} - \tilde{\nu^{-}}$ is another such pair and we have $\tilde{\nu^{+}}\perp\tilde{\nu^{-}}$ we can find $\tilde{N}$ and $\tilde{P}$ such that $X = \tilde{P}\cup\tilde{N}$ and we need to check that $\tilde{P}$ is positive and $\tilde{N}$ is negative (not sure how to do that yet).
Then we need to show that $\tilde{\nu^{+}} = \nu^+$ and $\tilde{\nu^{-}} = \nu^-$ again I am not sure how to do that either yet.
Any suggestions is greatly appreciated.
 A: 
The Jordan Decomposition Theorem - If $\nu$ is a signed measure, there exists unique positive measures $\nu^+$ and $\nu^-$ such that $\nu = \nu^+ - \nu^-$ and $\nu^+\perp \nu^-$.

Proof - Let $X = P\cup N$ be a Hahn decomposition for $\nu$ where $P$ and $N$ are positive and negative sets respectively.. Then let us define the positive and negative measure as such $$\nu^{+}(E) := \nu(E\cap P) \ \ \ \nu^{-}(E) := - \nu(E\cap N)$$ then, 
\begin{align*}
\nu^+(E) - \nu^-(E) &= \nu(E\cap P) + \nu(E\cap N)\\
&= \nu(E)
\end{align*}
So we have $\nu = \nu^+ - \nu^-$ and $\nu^+\perp \nu^-$. 
To complete this proof, let us to show uniqueness. Suppose that $\nu = \tilde{\nu^{+}} - \tilde{\nu^{-}}$ is another such pair and we have $\tilde{\nu^{+}}\perp\tilde{\nu^{-}}$ we can find $\tilde{P}$ and $\tilde{N}$  such that $X = \tilde{P}\cup\tilde{N}$,  $\tilde{P}$ is null for $\tilde{\nu^{-}}$ and $\tilde{N}$ is null for $\tilde{\nu^{+}}$. So, for any measurable set $E\subset \tilde{P}$, 
$$ \nu(E)= \tilde{\nu^{+}}(E) - \tilde{\nu^{-}}(E)=\tilde{\nu^{+}}(E)-0 = \tilde{\nu^{+}}(E) \geq 0$$
So $\tilde{P}$ is a positive set for $\nu$. 
In a similar way, So, for any measurable set $F\subset \tilde{N}$, 
$$ \nu(F)= \tilde{\nu^{+}}(F) - \tilde{\nu^{-}}(F)=0-\tilde{\nu^{-}}(F)= -\tilde{\nu^{-}}(F) \leq 0$$
So $\tilde{N}$ is a negative set for $\nu$. 
So  $\tilde{P}$,  $\tilde{N}$ is another Hahn decomposition for $\nu$. So, from 3.3 Hahn Decomposition Theorem, we have that 
$P\Delta \tilde{P} = N \Delta \tilde{N}$ is a null set for $\nu$.
So, $P\setminus \tilde{P}$, $\tilde{P}\setminus P$, $N\setminus \tilde{N}$ and $\tilde{N} \setminus N$ are null sets for $\nu$
For any measurable set $H\subset X$, we have 
\begin{align*} 
\nu^+(H)&= \nu(H\cap P)= \nu(H\cap(P\cap\tilde{P} ))+\nu(H\cap(P\setminus \tilde{P}))=\nu(H\cap(P\cap\tilde{P}) )=\\&= \nu(H\cap(P\cap\tilde{P}) )+\nu(H\cap(\tilde{P} \setminus P))= \nu(H\cap \tilde{P})= \tilde{\nu^{+}}(H\cap \tilde{P}) - \tilde{\nu^{-}}(H\cap \tilde{P})=\\&=  \tilde{\nu^{+}}(H\cap \tilde{P})=  \tilde{\nu^{+}}(H\cap \tilde{P}) +  \tilde{\nu^{+}}(H\cap \tilde{N})=  \tilde{\nu^{+}}(H)
\end{align*}
So we have that $\nu^+=\tilde{\nu^{+}}$.
In a similar way, we can prove that $\nu^-=\tilde{\nu^{-}}$. 
(Or, if we had assumed, without loss of generality, that $\nu$ does not assume the value +\infty, we have 
$\nu^-=\nu - \nu^+ = \nu - \tilde{\nu^{+}}= \tilde{\nu^{-}}$).
A: If $\tilde\nu^+\perp\tilde\nu^-$ then by definition of singularity of measures, there exists a partition $(\tilde P, \tilde N)$ of $X$ such that $\tilde P$ is null for $\tilde\nu^-$ and $\tilde N$ is null for $\tilde\nu^+$. But this means $(\tilde P,\tilde N)$ is a Hahn decomposition for $\nu$, since $\tilde\nu^+$ and $\tilde\nu^-$ are (positive) measures. Recall that the Hahn decomposition theorem implies then that $P\bigtriangleup\tilde P$ is $\nu$-null. From here it is straightforward to show that $\nu^+=\tilde\nu^+$ and $\nu^-=\tilde\nu^-$.
