Some obscure points about the proof of Hahn Decomposition Theorem 
The Hahn Decomposition Theorem. If $\nu$ is a signed measure on $(X,\mathcal{A})$, there exist a >positive set $P$ and a negative set $N$ for $\nu$ such that $P\cup N=X$ and $P\cap N=\emptyset.$

Proof Without loss of generality, we assume that $\nu$ does not assume the value $+\infty$. Let $$m = \sup\{\nu(E):E\in M, E \ \text{postive}\}$$
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, $$m\ge \nu(P)=\nu(P_j)+\nu(P\setminus P_j)\ge \nu(P_j)$$ for all j, since $P\setminus P_j\subseteq P$ and P is positive $\nu(P\setminus P_j)\ge 0$. Then $m=\nu(P)$  from this we obtain that $m<+\infty.$
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 $. 

Question 1. $\nu(A)>0?$

If yes, I can said that: since $\nu(A)>0$ and $A\subseteq N$ exists $B\subseteq A$ for $2$  such that $$\nu(B)>\nu(A)\Rightarrow \nu(B)>\nu(A)+\frac{1}{n}\quad\text{for same}\;n\in\mathbb{N}.$$
I don't understand why Folland's book ends like this: for $j$ sufficiently large we have $n<n_j$, and $B\subseteq A_{j-1}$, which contradicts the construction of $n_j$ and $A_j.$ 
Could you explain this to me in detail?
 A: If $A_{1} \subseteq N$ has positive measure but is not a positive set, then it must contain a subset with strictly greater measure. 
Let $n_{1}$ be the smallest integer such that there exists $A_{2} \subseteq A_{1}$ with $0 < m(A_{1}) + \frac{1}{n_{1}} \leq m(A_{2})$. If $A_{2}$ is a positive set, we're done. Otherwise, repeat and let $n_{2}$ be the smallest integer such that there exists $A_{3} \subseteq A_{2}$ with $m(A_{2}) + \frac{1}{n_{2}} \leq m(A_{3})$. Combining this with the previous inequality, $$0 < m(A_{1}) + \tfrac{1}{n_{1}} + \tfrac{1}{n_{2}} \leq m(A_{2}) + \tfrac{1}{n_{2}} \leq m(A_{3})$$ 
If we continue this process and never find a positive set, then $(A_{n})_{n = 1}^{\infty}$ is a decreasing set sequence since $A_{n} \subseteq A_{n + 1}$ whose limit is then $\bigcap_{n = 1}^{\infty} \: A_{n}$. For each $n$, we have $0 < \sum_{k = 1}^{n} \: \frac{1}{n_{k}} \leq m(A_{n})$. Taking limits, $$0 < \sum_{k = 1}^{\infty} \: \frac{1}{n_{k}} \: \leq \: m(\bigcap_{n = 1}^{\infty} \: A_{n})$$
However, the signed measure does not take on $+\infty$ as a value, which means that $m(\bigcap_{n = 1}^{\infty} \: A_{n})$ and therefore, $\sum_{k = 1}^{\infty} \: \frac{1}{n_{k}}$, is finite. Hence the sequence $(\frac{1}{n_{k}})$ converges to zero, which implies that $n_{k} \to \infty$. 
Now the set $A = \bigcap_{n = 1}^{\infty} \: A_{n}$ has positive measure so either it is a positive set and we are done. Or we can find a subset $B \subseteq A$ with greater measure. Let $N \in \mathbb{N}$ such that $m(A) + \frac{1}{N} \leq m(B)$. However, there exists $n_{k}$ such that $N < n_{k}$ and $B \subseteq A \subseteq A_{k}$. 
This is a contradiction because $N$ is smaller than $n_{k}$ and we choose $n_{k}$ as the smallest integer such that $B \subseteq A_{k}$ with $m(A_{k}) + \frac{1}{n_{k}} \leq m(B)$. Therefore, we conclude that $A$ is our positive set. 
