Hahn Decomposition Theorem In Folland

Question

I was reading the proof of Hahn Decomposition theorem from the textbook of Folland: precisely I was looking at the following text

I have the following question:

1. As Highlighted in the text above, why $m$ is finite? It may be infinite as there is no restriction on $X$. Why does the author consider it finite?
2. Again why $\nu(A)<\infty$ ? I do not understand also this.
3. I understand that even for $A$ also we get some $B$ with the property that $\nu(B)>\nu(A)+1/n$ but I do not understand how this leads to a contradiction.

I would be really thankful if someone could help me. Any help will be appreciated.

Answer 1

Perhaps if you consult Folland's definition of "signed measure" you will find that all values are finite. $\nu : \mathcal X \to (-\infty,+\infty)$.

According to Wikipedia:

There are two slightly different concepts of a signed measure, depending on whether or not one allows it to take infinite values. In research papers and advanced books signed measures are usually only allowed to take finite values, while undergraduate textbooks often allow them to take infinite values.

Suppose we allow infinite values. Then Lebesgue measure on $\mathbb R$ would be a finite measure, $E = \mathbb R$ would be a positive set, and $\nu(E)=\infty$ on that case.

Answer 2

I'm fairly confident that the assumption in the very first line is a typo. It should say $\nu < \infty$ rather than $\nu > -\infty$. It appears like $\nu > -\infty$ is not used anywhere in the proof whereas as you've noted $\nu < \infty$ is used multiple times.

($\dagger$) As for 3. the point is that for any set, $C$, contained in $N$ with $\nu(C) > 0$ we can find another set $C' \subset C$ with $\nu(C') > \nu(C)$.

Furthermore, by contradiction it is assumed that such a $C$ exists as otherwise $N$ would be negative. Then given this initial $C$ there exists a sequence $N \supset C \supset A_1 \ldots \supset A_k \supset$ of sets $A_i$ constructed using the result ($\dagger$) about positive sets of $N$. Additionally the $A_i$ are constructed so that $A_{i+1}$ approximates $A_i$ poorly so that the difference $\nu(A_{i+1}) - \nu(A_i)$ is large. Now as $N \supset \cap_{i = 1}^{\infty}A_i$ and $\nu(\cap_{i = 1}^{\infty}A_i) > 0$ by ($\dagger$) again there exists a $D \subset \cap_{i = 1}^{\infty}A_i$ with $\nu(\cap_{i = 1}^{\infty}A_i) > \nu(D) + n^{-1}$ for some $n$. It is shown that $n < n_j$ for some $j$ as well so that $\nu(D) > \nu(A_{j-1}) + n^{-1} > \nu(A_{j-1}) + n_j^{-1}$. This is the desired contradiction because $n_j$ was chosen so that for any $H \subset A_{j-1}$ we have that $\nu(A_{j-1}) + (n_j - 1)^{-1} > \nu(H)$ so take $H = D$ and you are done.