My course in measure theory provided a lengthy proof of Hahn decomposition theorem, in the form of a supporting lemma and a proper proof, both proved hard-analytically (a lot of $\varepsilon$s, $\delta$s and iterative constructions). I thought of an alternative, very concise, proof of both the results, leaning of Zorn's lemma.
The results are the following:
Lemma (Fundamental lemma on signed measures). Let $(X, \Sigma)$ be a measurable space, and let $\nu$ be a (finite) signed measure on it. Let also $A \in \Sigma$ be a set of positive measure ($\nu(A) \geq 0$). Then there exists a set $A^+$ such that:
- $A^+ \subset A$,
- $A^+$ is a positive set,
- $\nu(A^+) \geq \nu(A)$.
We called positive a set of positive measure such that any of its measurable subsets is itself of positive measure. Similarly we defined a negative set.
Theorem (Hahn decomposition theorem) Under the same hypotheses of the preceding lemma, there exists two measurable disjoint subsets $X^+$ and $X^-$ of $X$ such that:
- $X^+$ is positive, $X^-$ is negative,
- $X^+ \cup X^- = X$,
- If $Y^+$ and $Y^-$ satisfies the same conditions above, then $\nu(X^+ \Delta Y^+) = \nu(X^- \Delta Y^-) = 0$, that is, the decomposition is essentially unique.
My proof of the lemma goes like this:
Let $P = \{ F \subseteq A \,:\, \text{$F$ is positive}\}$. $P$ is non-empty ($\varnothing \in P$) and partially ordered by inclusion, infact it is a $\sigma$-ideal of $\Sigma$ (this was also proven by my teacher, and it is indeed trivial to check). Then let $T$ be a chain in $P$: clearly $\bigcup T$ is an upper bound for $T$ and a positive set. Then it follows by Zorn's lemma that $P$ admits a maximal element $A^+$. Let's show that $\nu(A^+) \geq \nu(A)$. We proceed by contradiction: suppose $\nu(A^+) < \nu(A)$. Then, by additivity, $F = A \setminus A^+$ is a set of positive measure, using the same argument which gave us $A^+$ we select a positive subset $F^+$ of $F$. But then $A^+ \cup F^+$ would contradict the maximality of $A^+$, hence the inequality.
The Hahn decomposition follows trivially.
The claim that $\bigcup T$ is an upper bound for $T$ is obviously false, because the chain might easily be uncountable and thus we have no guarantee on the ``positivity'' of the union. So the proof, stated like this, it's not correct. Yet I wonder if there might be a more delicate way to contsruct the upper bound of $T$ such that we can preserve positivity.
Seems to me a more serious problem with your proof of the lemma is this: How do you know that $\nu(F^+)>0$? If $\nu(F^+)=0$ there's no contradiction. So you have to show that every set of positive measure contains a poitive set of positive measure. My impression is that that's the hard part of the lemma; once you know that the lemma follows easily.
You can fix your concerns about measurability of $T$ in various ways. The simplest is probably to forget Zorn's lemma; instead of looking for a maximal positive set look for a positive set of maximal measure.