Let $\nu$ be a signed measure over measurable space $(X,\mathcal{M})$. Denote
$\tilde{N}=\left\{ u\in \mathcal{M} \; :\; \mbox{$u$ is $\nu$-negative}
\right\}$ and $\tilde{P}=\left\{ u\in \mathcal{M} \; :\; \mbox{$u$ is
$\nu$-positive} \right\}$
- $\exists N\in \tilde{N}\Big(\exists P \in \tilde{P}\Big( N\cap P=\emptyset \; \mbox{ and }\;
N\cup P = X\Big)\Big)$
- $\forall N,N^{'}\in \tilde{N}\Big(\forall P,P^{'}\in \tilde{P}
\Big(N\sqcup P =N^{'}\sqcup P^{'}=X \; \Longrightarrow
\nu( P\triangle \tilde{P})=\nu( N\triangle \tilde{N})=0\Big)\Big) $
Proof:
Part 1
Without loss of generality, let $\infty$ is not in the range of $\nu$
(otherwise substituting $\nu$ with $-\nu$ in the theorem yields the
same result). Define $\tilde{E}=\left\{ \nu(F)\;:\; F\in
\tilde{P} \right\} $ and denote $m=\sup \tilde{E}$. The
definition of $\tilde{E}$ and the supremum implies the existence of a
sequence $ \left\{ P_{j} \right\}_{j=1}^{\infty}\subset
\tilde{P} $ such that $\lim \nu\left( P_{j} \right)=m$. Denote
$P=\bigcup_{j=1}^{\infty}P_{j}$, then $P$ is measurable. Denote
$P_{0}=\emptyset$ and let $A\subset P$ be any set, then we get
\[ A\cap \left( P_{j}\setminus \bigcup_{k=0}^{j-1}P_{k} \right)\subset
P_{j}, \]
and since $P_{j} $ is $\nu$-positive, every subset of $P_{j}$ must be
$\nu$-positive, we thus get $\nu\left( A\cap P_{j} \setminus
\bigcup\limits_{k=0}^{j-1}P_{k}\right)\ge 0$ and from this we can see that
\[
\begin{array}{lll}
\nu(A) &=&\nu(A\cap P)=\nu\left( A\cap \left(
\bigcup_{j=1}^{\infty}P_{j}\setminus
\bigcup_{k=0}^{j-1}P_{k} \right) \right) = \nu\left(
\bigcup_{j=1}^{\infty}A\cap \left( P_{j}\setminus
\bigcup_{k=0}^{j-1}P_{k} \right) \right) \\ &=&
\sum\limits_{j=1}^{\infty} \nu\left( A \cap \left( P_{j}\setminus
\bigcup_{k=0}^{j-1}P_{k} \right) \right) \ge 0.
\end{array}\]
Therefore $P$ must be $\nu$-positive. Denote $N=X\setminus P$. Suppose
there exists $A\in 2^{N}\cap \mathcal{M}$(that is a measurable
$A$ that is subset to $N$) such that $A$ is $\nu$-positive and
$\nu(A)>0$, and $A\cap P=\emptyset$ and $A\cup P$ is $\ne$-positive
(since for any $K\subset (A\cup P)$ and measurable we get
$\nu(K)=\nu\left( K\cap (A\cup P) \right)=\nu(K\cap A)+\nu(K\cap P)\ge 0$ )
and $\nu(A\cup P)=\nu(A)+\nu(P)=\nu(A)+m>m$, but this contradict our
definition of $m$, hence
\begin{equation}
N \; \mbox{ does not contain a $\nu$-positive set $A$ such that
$\nu(A)>0$ }.
\tag{1}
\end{equation}
If $F\in 2^{N}\cap \mathcal{M}$ and $\nu(F)>0$, then
we using property (1) we may find a set
$C\in 2^{F}\cap \mathcal{M}$ such that $\nu(C)<0$ (otherwise $F$
would be $\nu$-positive). Denote $K=F\setminus C$, then
$\nu(K)=\nu(F)-\nu(C)>\nu(F)$ and we get
\begin{equation}
\Big( F\in 2^{N}\cap \mathcal{M} \; \mbox{and}\; \nu(F)>0
\Big) \Longrightarrow \exists K\in 2^{F}\cap \mathcal{M} \Big(
\nu(F)<\nu(K)\Big).
\tag{2}
\end{equation}
Now suppose that $N$ is not $\nu$-negative, then we get $A_{1}\in
2^{N}\cap \mathcal{M}$ such that $\nu(A_{1})>0$ and thus, using
property (2), there exists at list one measurable set $B\subsetneq
A_{1}$ such that $0<\nu(A_{1})<\nu(B)$. Therefore we may define
\[ n_{1}=\min\left\{ n\in \mathbb{N}\setminus 0 \; : \; \exists B\in
2^{A_{1}}\cap
\mathcal{M}\Big(\nu(B)-\nu(A_1)>\frac{1}{n}\right\}. \]
Hence, we may find $A_{2}\subsetneq A_{1}$ such that
$\nu(a_{2})-\nu(a_{1})>\frac{1}{n_{1}}$ . We may once again use property (2) to get
at least one measurable set $B\subsetneq A_{1}$ such that
$0<\nu(A_{1})<\nu(B)$ (notice that $\nu(a_{2})-\nu(a_{1})>\frac{1}{n_{1}}$ implies that
$\nu(A_{2})>\nu(A_{1})>0$) and then we can once again define
\[ n_{2}=\min\left\{ n\in \mathbb{N}\setminus 0 \; : \; \exists B\in
2^{A_{2}}\cap \mathcal{M}\Big(\nu(B)-\nu(A_2)>\frac{1}{n}\right\}. \]
We can proceed in this way to define (recursively) a sequence
$A_{j}$ such that $\infty\ne \nu(A_{1})<\nu(A_{2})<\ldots$ and
$A_{1}\supsetneq A_{2}\supsetneq A_{3}\supsetneq \ldots$ using this
and
proposition we get
$\lim\limits_{j\to\infty} \nu(A_{j})=\nu\left( \bigcap_{j=1}^{\infty}A_{j}
\right)$. Now denote $A=\bigcup_{j=1}^{\infty} A_{j}$, then $A$ is
measurable and
\[ \infty
>\nu(A)=\lim_{j\to\infty}\nu(A_{j})=\nu(A_{1})+
\sum\limits_{j=2}^{\infty}\nu(A_{j})-\nu(A_{j-1})>\nu(A_{1})+\sum\limits_{j=2}^{\infty}
\frac{1}{n_{j}} >0. \]
Hence $\sum\limits_{j=1}^{\infty}\frac{1}{n_{j}}$ converge to a
finite value and, of course, $\lim\limits_{j\to\infty}
\frac{1}{n_{j}}=0$. In addition $\nu(A)>0$ and $A\subset N$,
therefore, by property (2), we can find $E\in 2^{A}\cap
\mathcal{M}$ such that $\nu(E)-\nu(A)>0$ and from this we
conclude the existence of a positive integer $t$ (sufficiently
large) such that $\nu(E)-\nu(A)>\frac{1}{t}$. Since
$\lim\limits_{j\to\infty}\nu(A_{j})=\nu(A)$, we get
$\lim\limits_{j\to\infty}\nu(E)-\nu(A_{j})>\frac{1}{t}$, so
there exists sufficiently large positive integer $k$ such that
\[ m>k \Longrightarrow \nu(E)-\nu(A_{m})>\frac{1}{t}.\]
Since $\lim\limits_{j\to\infty} n_{j}=\infty$ we can choose
positive integer $j_{0}$ such that $n_{j_{0}}>\max\left\{ t,k
\right\}$ and then get $\frac{1}{t}>\frac{1}{n_{j_{0}}}$ and
$\nu(E)-\nu(A_{j_0})>\frac{1}{n_{j_{0}}}$, therefore we get
$\nu(E)-\nu(A_{j_{0}})>\frac{1}{t}>\frac{1}{n_{j_{0}}}$. From
this we get
\[ n_{j_{0}}>t\in \left\{ n\in \mathbb{N}\setminus \left\{
0 \right\} \; :\; \exists B\in 2^{A_{j_{0}}}\cap
\mathcal{M}\Big(\nu(B)-\nu(A_{j_{0}})
>\frac{1}{n} \right\} \]
which contradict our definition of $n_{j_{0}}$, therefore
$N=X\setminus P $ must be $\nu$-negative.
Part 2
If $X=P\sqcup N$ and $X=P^{'}\sqcup N^{'}$, then $P\setminus
P^{'}\in N^{'}$ and $P^{'}\setminus P\in N$. Hence the sets
$P\setminus P^{'}$ and $P^{'}\setminus P$ are both $\nu$-negatives and
$\nu$-positives, this implies that $\nu(P^{'}\setminus
P)=\nu(P\setminus P^{'})=0$ and
\[ \nu(P\triangle P^{'})=\nu\left( (p\setminus P^{'})\cup
(P^{'}\setminus P) \right) = \nu(P\setminus
P^{'})+\nu(P^{'}\setminus P)=0. \]
The prove that $0=\nu(N\triangle N^{'})$ use a similar argument.
$\square$