Proving the Hahn Decompostion Theorem from Folland 
The Hahn Decomposition Theorem - If $\nu$ is a signed measure on $(X,M)$, there exists a positive set $P$ and a negative set $N$ for $\nu$ such that $P\cup N = X$ and $P\cap N = \emptyset$. If $P',N'$ is another such pair, then $P \ \triangle \ P' = N \ \triangle \ N'$ is null for $\nu$.

Attempted proof - Without loss of generality assume $\nu$ does not attain $+\infty$. Let $$m = \sup\{\nu(E):E\in M, E \ \text{posistive}\}$$ then let us choose a positive sequence of sets $\{P_j\}$ such that $m = \lim_{j\rightarrow \infty}\nu(P_j)$. Let $P = \bigcup_{1}^{\infty}P_j$ then $P$ is positive and $m \geq \nu(P)\geq \nu(P_j)$.
Now set $N = X\setminus P$, we need to show that $N$ is negative. Note that $N$ contains no non-null positive sets, otherwise suppose $P'\subset N$ positive, then $P'\cup P$ would be positive and $\nu(P'\cup P) = \nu(P') + \nu(P) > m$. Also note that if $N$ contains a set $A$ such that $\nu(A) > 0$ then there is an $A'\subset A$ with $\nu(A') > \nu(A)$ (Since $A$ non-null, can't be positive so there is a $B\subset A$ with $\nu(B) < 0$ then $\nu(A\setminus B) = \nu(A) - \nu(B) > \nu(A)$).
Suppose $N$ is not negative, then let $n_1$ be the smallest natural number so there is a $B\subset N$ with $\nu(B) > 1/n$. Let $A_1$ be such a set $B$. Let $n_2$ be the smallest natural number such that there is a $B\subset A_1$ with $\nu(B) > \nu(A_1) + 1/n$. Let $A_2$ be such a set $B$. Continuing as so... then we have a sequence of natural numbers and a sequence of sets. In particular our sequence of sets is decreasing. Let $A = \bigcap_{1}^{\infty}A_j$, then $$\infty > \nu(A) = \lim_{j\rightarrow \infty}\nu(A_j)\geq \sum_{1}^{\infty}\frac{1}{n_j}$$ So since the sum converges $\lim_{j\rightarrow \infty}n_j = \infty$. Now suppose $A'\subset A$ such that $\nu(A') > \nu(A)$. We can find an $n$ such that $\nu(A_1) > \nu(A) + 1/n$. Then $n\geq n_j$ for all $j$ which is a contradiction since $n_j\rightarrow \infty$.
I am not sure if this is completely correct, sorry if it is sort of messy.
 A: @Wolfy , your proof is correct. I have copied it here and made minor adjustments (specially in the end) to make it clearer. I have also included the proof of "uniqueness".

The Hahn Decomposition Theorem - If $\nu$ is a signed measure on $(X,M)$, there exists a positive set $P$ and a negative set $N$ for $\nu$ such that $P\cup N = X$ and $P\cap N = \emptyset$. If $P',N'$ is another such pair, then $P \ \triangle \ P' = N \ \triangle \ N'$ is null for $\nu$.

Proof: - Without loss of generality assume $\nu$ does not attain $+\infty$. Let $$m = \sup\{\nu(E):E\in M, E \ \text{posistive}\}$$ then let us choose a positive sequence of sets $\{P_j\}$ such that $m = \lim_{j\rightarrow \infty}\nu(P_j)$. Let $P = \bigcup_{1}^{\infty}P_j$ then $P$ is positive and, for all $j$, $m \geq \nu(P)\geq \nu(P_j)$, so $\nu(P)=m$.
Now set $N = X\setminus P$.  Note that: 


*

*$N$ contains no non-null positive sets, otherwise suppose $P'\subset N$ non-null positive, then $P'\cup P$ would be positive and $\nu(P'\cup P) = \nu(P') + \nu(P) > m$. 

*if $N$ contains a set $A$ such that $\nu(A) > 0$ then there is an $A'\subset A$ with $\nu(A') > \nu(A)$ (Since $A$ non-null, can't be positive so there is a $B\subset A$ with $\nu(B) < 0$ then $\nu(A\setminus B) = \nu(A) - \nu(B) > \nu(A)$).
Now, let us prove by contradiction that $N$ is negative. 
Suppose $N$ is not negative, then let $n_1$ be the smallest natural number so there is a $B\subset N$ with $\nu(B) > 1/n_1$. Let $A_1$ be such a set $B$. Let $n_2$ be the smallest natural number such that there is a $B\subset A_1$ with $\nu(B) > \nu(A_1) + 1/n_2$. Let $A_2$ be such a set $B$. Continuing as so... then we have a sequence of natural numbers and a sequence of sets. In particular our sequence of sets is decreasing. Let $A = \bigcap_{1}^{\infty}A_j$, then, since $\nu(A_1)<\infty$, we have
$$\infty > \nu(A) = \lim_{j\rightarrow \infty}\nu(A_j)\geq \sum_{1}^{\infty}\frac{1}{n_j}$$ 
So since the sum converges, we have $ \lim_{j\rightarrow \infty}n_j = \infty $. 
On the other hand, the sequence $\{\mu(A_j)\}_j$ is an increasing set of real numbers.
Now suppose $A'\subset A$ such that $\nu(A') > \nu(A)$. Then we can find an $n$ such that $\nu(A') > \nu(A) + 1/n$. Then, for each $j$, $A\subset A_j$ and 
$$\nu(A') > \nu(A) +1/n >\nu(A_j)+1/n$$ 
By the definition of $n_{j+1}$, we have $n_{j+1}\leq n$.  So we have that for all $j$, $n_{j+1}\leq n$. Contradiction, since $\lim_{j\rightarrow \infty}n_j = \infty$. 
So we have proved that there exists a positive set $P$ and a negative set $N$ for $\nu$ such that $P\cup N = X$ and $P\cap N = \emptyset$. 
Now let us prove that If $P',N'$ is another such pair, then $P \ \triangle \ P' = N \ \triangle \ N'$ is null for $\nu$.
Suppose  $P',N'$ is another such pair. Then, for any $H\subset P'\setminus P$ we have $H\subset P'\setminus P\subset N$ , so $\nu(H)\leq 0$. On the other hand, we have $H\subset P'\setminus P\subset P'$, so $\nu(H)\geq 0$. So se have $\mu(H)=0$.  So $P'\setminus P$ is a null set. 
In a similar way, for any $H\subset P\setminus P'$ we have $H\subset P\setminus P'\subset N'$ , so $\nu(H)\leq 0$. On the other hand, we have $H\subset P\setminus P'\subset P$, so $\nu(H)\geq 0$. So se have $\mu(H)=0$.  So $P\setminus P'$ is a null set.
Since $P \Delta P'=(P\setminus P') \cup  (P'\setminus P)$,we have that $P \Delta P'$ is a null set 
To complete the proof, note that 
$$P \Delta P'=(P\setminus P') \cup  (P'\setminus P)= (P\cap N') \cup  (P'\cap N)=(N'\setminus N)\cup(N\setminus N') =N \Delta N'$$
