I am reading this proof of the Hahn decomposition from Billingsley Probability and Measure.
The Hahn Decomposition Theorem 32.1. For any additive set function $\varphi$, there exist disjoint sets $A^{+}$ and $A^{-}$such that $A^{+} \cup A^{-}=\Omega, \varphi(E) \geq 0$ for all $E$ in $A^{+}$, and $\varphi(E) \leq 0$ for all $E$ in $A^{-}$.
A set $A$ is positive if $\varphi(E) \geq 0$ for $E \subset A$ and negative if $\varphi(E) \leq 0$ for $E \subset A$. The $A^{+}$and $A^{-}$in the theorem decompose $\Omega$ into a positive and a negative set. This is the Hahn decomposition.
Proof. Let $\alpha=\sup [\varphi(A): A \in \mathscr{F}]$. Suppose that there exists a set $A^{+}$ satisfying $\varphi\left(A^{+}\right)=\alpha$ (which implies that $\alpha$ is finite). Let $A^{-}=\Omega-A^{+}$. If $A \subset A^{+}$and $\varphi(A)<0$, then $\varphi\left(A^{+}-A\right)>\alpha$, an impossibility; hence $A^{+}$is a positive set. If $A \subset A^{-}$and $\varphi(A)>0$, then $\varphi\left(A^{+} \cup A\right)>\alpha$, an impossibility; hence $A^{-}$is a negative set.
It is therefore only necessary to construct a set $A^{+}$for which $\varphi\left(A^{+}\right)=\alpha$. Choose sets $A_{n}$ such that $\varphi\left(A_{n}\right) \rightarrow \alpha$, and let $A=\cup_{n} A_{n}$. For each $n$ consider the $2^{n}$ sets $B_{n i}$ (some perhaps empty) that are intersections of the $B_{n}=\left[B_{k=1}: 1 \leq i \leq 2^{n}\right]$ of these sets partitions $A$. Clearly, $A_{k}$. The collection $\mathscr{B}_{n}=\left[B_{n i}: 1 \leq i \leq 2^{n}\right]$ of these sets partitions $A$. Clearly, $\mathscr{B}_{n}$ refines $\mathscr{B}_{n-1}$ : Let $B_{n j}$ is contained in exactly one of the $B_{n-1, i}$. Let $C_{n}$ be the union of those $B_{n i}$ in $\mathscr{B}_{n}$ for which $\varphi\left(B_{n i}\right)>0$.
At this point of the proof, I have a question. Why couldn't I conclude the theorem by constructing the set
$A=\bigcup_n C_n$ becuase we have $\alpha = \lim_{n\to \infty}\varphi(A_n)\leq \lim_{n\to \infty}\varphi(C_n)$?
The proof goes on to say
Since $A_{n}$ is the union of certain of the $B_{n i}$, it follows that $\varphi\left(A_{n}\right) \leq \varphi\left(C_{n}\right)$. Since the partitions $\mathscr{B}_{1}, \mathscr{B}_{2}, \ldots$ are successively finer, $m<n$ implies that $\left(C_{m} \cup \cdots \cup\right.$ $\left.C_{n-1} \cup C_{n}\right)-\left(C_{m} \cup \cdots \cup C_{n-1}\right)$ is the union (perhaps empty) of certain of the sets $B_{n i}$; the $B_{n i}$ in this union must satisfy $\varphi\left(B_{n i}\right)>0$ because they are contained in $C_{n}$. Therefore, $\varphi\left(C_{m} \cup \cdots \cup C_{n-1}\right) \leq \varphi\left(C_{m} \cup \cdots \cup C_{n}\right)$, so that by induction $\varphi\left(A_{m}\right) \leq \varphi\left(C_{m}\right) \leq \varphi\left(C_{m} \cup \cdots \cup C_{n}\right)$. If $D_{m}=\cup_{n=m}^{\infty} C_{n}$, then by Lemma 1 (take $\left.E_{v}=C_{m} \cup \cdots \cup C_{m+v}\right) \varphi\left(A_{m}\right) \leq \varphi\left(D_{m}\right)$. Let $A^{+}=\cap_{m=1}^{\infty} D_{m}$ (note that $\left.A^{+}=\lim \sup _{n} C_{n}\right)$, so that $D_{m} \downarrow A^{+}$. By Lemma $1, \alpha=\lim _{m} \varphi\left(A_{m}\right)$ $\leq \lim _{m} \varphi\left(D_{m}\right)=\varphi\left(A^{+}\right)$. Thus $A^{+}$does have maximal $\varphi$-value.
but frankly, I am confused as to what the motivation is for going further as they have done. We want intuitively, the biggest possible set with positive signed measure. Why couldn't I just find this by unioning all the little partitions which have positive measure i.e. unioning the $C_n$ together?
$(X, \mathcal{B}, \varphi) $ be a signed measure space.
Let $S=\{\varphi(A): A> 0\}$
Let $\alpha=\sup S$
$\varphi(\emptyset) =0\in S$ implies $0<\alpha<\infty$ ( as $\varphi$ can attained atmost one of $\infty$ or $-\infty$ ,assume $-\infty$ is attained)
Since $\alpha=\sup S, \exists (A_n)$ sequence of positive sets such that $\varphi(A_n)\to \alpha$
Let $A=\bigcup_{n} A_n$ .
Claim $1$: $A$ is a positive set.
It follows from the fact the countable union of positive sets is positive (see here).
Claim $2$ :$\varphi(A) =\alpha$
$A>0$ implies $\varphi(A) \le \alpha$.
Now $A=A\setminus A_i\cup A_i$ implies
$\begin{align}\varphi(A) &=\varphi(A\setminus A_i) +\varphi(A_i)\\&\ge \varphi(A_i)\\&\ge \alpha\end{align}$
Hence $\varphi(A) =\alpha$
Calim $3$: $B=A^c$ is a negative set.
Let $E\subset B$ such that $E>0$ .Then $A\cup E>0$ and $A\cap E=\emptyset$ implies $\begin{align}\alpha\ge \varphi(A\cup E) &=\varphi(A)+\varphi(E)\\&=\alpha+\varphi(E)\end{align}$
Hence $\varphi(E) =0$.
Hence every non null subsets of $B$ are negative. Hence $B<0$.
Hence $(A, B) $ is the required Hahn decomposition of $X$ w.r.to $\varphi$.