I got stuck reading the proof of the Hanhn Decomposition Theorem in Folland's Real Analysis. The part that I don't understand is selected with red. Where did $\nu(A_{j-1})$ come from? Don't we only have $\nu(B)>n_j^{-1}$? (But I understand that's not enough to get the series of the next line)
He labelled $B$ as $A_{1}$ in order to begin the construction (see "$A_{1}$ is such a set").
First, if $N$ is not negative, then we can find a set $A_{1}$ with positive measure. Let $n_{1}$ be the smallest integer such that $0 < \frac{1}{n_{1}} < m(A_{1})$.
Next, we want to achieve a contradiction by showing that $A_{1} \subseteq N$ contains a positive set.
If $A_{1}$ has positive measure but is not a positive set, then we can find a subset $A_{2} \subseteq A_{1}$ such that $m(A_{1}) < m(A_{2})$. Let $n_{2}$ be the smallest integer such that $$m(A_{1}) + \tfrac{1}{n_{2}} < m(A_{2})$$ Continuing in this way, we have the sequence of subsets $A_{j - 1} \subseteq A_{j - 2} ... \subseteq ... A_{1}$ for some $j \in \mathbb{N}$. Now either $A_{j - 1}$ is a positive set and we are done. Or we can find a set $A_{j} \subseteq A_{j - 1}$ such that $m(A_{j - 1}) < m(A_{j})$. Let $n_{j}$ be the least integer satisfying $$(A_{j - 1}) + \tfrac{1}{n_{j}} < m(A_{j})$$ The process then continues and if we never find a positive set, then we consider the intersection $\bigcap_{j = 1}^{\infty} \: A_{j}$.
