Hahn Decomposition Theorem

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\}$

1.   $\exists N\in \tilde{N}\Big(\exists P \in \tilde{P}\Big( N\cap P=\emptyset \; \mbox{ and }\; N\cup P = X\Big)\Big)$
2.   $\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. $\square$