I am struggling to understand the following theorem's proof on Folland's Real Analysis, page 89, lemma 3.7.
Suppose that $\nu$ and $\mu$ are finite measures on $(X, \mathcal M)$. Either $\nu \perp \mu$, or there exists $\epsilon > 0$ and $E \in \mathcal M$ such that $\mu(E) > 0$ and $\nu \geq \epsilon \mu$ on $E$ (that is, $E$ is a positive set for $\nu - \epsilon \mu$.
Proof: Let $P_n \cup N_n$ be a Hahn decomposition for $\nu - n^{-1}\mu$, and let $P = \displaystyle\bigcup_1^\infty P_n$ and $N = \displaystyle\bigcap_1^\infty N_n = P^c$. Then $N$ is a negative set of $\nu - n^{-1}\mu$ for all $n$, i.e. $\mathbf{0 \leq \nu(N)} $ $\leq n^{-1}\mu(N)$ for all $n$, so $\nu(N) = 0$. If $\mu(P) = 0$, then $\nu \perp \mu$. If $\mu(P) > 0$, then $\mu(P_n) > 0$ for some $n$, and $P_n$ is a positive set for $\nu - n^{-1}\mu$.
The boldfaced words are the parts that I do not understand. Why is it that $\nu(N)$ must be a nonnegative number? Furthurmore, wouldn't the definition of a negative set require any subset $A \subset N$ be a nonnegative set also? I proved that for any such subset we must have $\nu(A) \leq 0$ but I am not sure how it must be equal to zero. I also do not get why $\mu(P) = 0$ implies $\nu \perp \mu$, as well as the last statement. I feel like I am missing something obvious, I would be grateful if anyone could explain.