I try to give a proof for Hahn-Decomposition Theorem,but I’m not sure whether it is right.
Hahn-Decomposition-Theorem
If $\nu$ is a signed measure on $(X,\mathscr{M})$,there exist a positive set $P$ and a negative set $N$ for $\nu$ such that $P\cup N=X$ and $ P\cap N=\emptyset$
This is my proof:
WLOG,we can assume that $\nu$ does not assume $+\infty$(Otherwise consider $-\nu$).
$1^0$.Let $$ M=sup\left\{ \nu \left( E \right) :E\,\,is\,\,a\,\,positive\,\,set \right\} $$Thus there exists a sequence $\left\lbrace E_j\right\rbrace$ of positive sets such that $$\nu (E_j)\to M\qquad j\to \infty$$Let $P=\bigcup_{j=1}^{\infty}$.Obviously, $P$ is positive and $$ \nu \left( P \right) =\nu \left( \bigcup_{j=1}^{\infty}{E_j} \right) =\underset{j\rightarrow \infty}{\lim}\nu \left( E_j \right) =M $$
$2^0$.We claim that $N=X\backslash P$ is negative.
(1).If $E\subset N$ is positive and $\nu (E)>0$,then $E\cup P$ is a positive set and "larger" than $P$.Hence $N$ cannot contain any nonnull positive sets.
(2).If $E\subset N$ and $\nu (E)>0$,there exists $F\subset E$ with $\nu (F)>\nu (E)$.Indeed,we know that $E$ is not positive by (1),so there exists $U\cup E$ with $\nu (U)<0$,thus if we take $F=E\backslash U$,we will have $\nu (F)>\nu (E)$.
(3).If $N$ is not negative, it must contain a set $A$ with $\nu (A)>0$.Let $$ S=sup\left\{ \nu \left( C \right) :C\subset A,\nu \left( C \right) >\nu \left( A \right) >0 \right\} $$By (2),we know that $S>0$.Now,there exists a decreasing sequence $\left\lbrace C_n \right\rbrace_{n=1}^{\infty}(C_n\subset A,\forall n)$ such that $$ \underset{n\rightarrow \infty}{\lim}\nu \left( C_n \right) =S $$ Let $Q=\bigcap_{n=1}^{\infty} C_n\subset A$,we have$$ \nu \left( Q \right) =\nu \left( \bigcap_{n=1}^{\infty}{C_n} \right) =\underset{n\rightarrow \infty}{\lim}\nu \left( C_n \right) =S $$ By our assumption at the beginning, $S<\infty $.However, we can find a set $T\subset Q$ with $\nu (T)>\nu (Q)=S$ by (2),which is contradicted with the definition of $S$.