Preparation: Assume that \mu does not take the value - \infty (otherwise decompose according to - \mu ). As mentioned above, a negative set is a set A \in \Sigma such that \mu(B) \leq 0 for every \Sigma -measurable subset B \subseteq A .
Claim: Suppose that D \in \Sigma satisfies \mu(D) \leq 0 . Then there is a negative set A \subseteq D such that \mu(A) \leq \mu(D) .
Proof of the claim: Define A_{0} := D .
Inductively assume for n \in \mathbb{N}_{0} that A_{n} \subseteq D has been constructed. Let : t_{n} := \sup(\{ \mu(B) \mid B \in \Sigma ~ \text{and} ~ B \subseteq A_{n} \}) denote the
supremum of \mu(B) over all the \Sigma -measurable subsets B of A_{n} . This supremum might
a priori be infinite. As the empty set \varnothing is a possible candidate for B in the definition of t_{n} , and as \mu(\varnothing) = 0 , we have t_{n} \geq 0 . By the definition of t_{n} , there then exists a \Sigma -measurable subset B_{n} \subseteq A_{n} satisfying : \mu(B_{n}) \geq \min \! \left( 1,\frac{t_{n}}{2} \right). Set A_{n + 1} := A_{n} \setminus B_{n} to finish the induction step. Finally, define : A := D \Bigg\backslash \bigcup_{n = 0}^{\infty} B_{n}. As the sets (B_{n})_{n = 0}^{\infty} are disjoint subsets of D , it follows from the
sigma additivity of the signed measure \mu that : \mu(D) = \mu(A) + \sum_{n = 0}^{\infty} \mu(B_{n}) \geq \mu(A) + \sum_{n = 0}^{\infty} \min \! \left( 1,\frac{t_{n}}{2} \right)\geq \mu(A). This shows that \mu(A) \leq \mu(D) . Assume A were not a negative set. This means that there would exist a \Sigma -measurable subset B \subseteq A that satisfies \mu(B) > 0 . Then t_{n} \geq \mu(B) for every n \in \mathbb{N}_{0} , so the
series on the right would have to diverge to + \infty , implying that \mu(D) = + \infty , which is a contradiction, since \mu(D) \leq 0 . Therefore, A must be a negative set.
Construction of the decomposition: Set N_{0} = \varnothing . Inductively, given N_{n} , define : s_{n} := \inf(\{ \mu(D) \mid D \in \Sigma ~ \text{and} ~ D \subseteq X \setminus N_{n} \}). as the
infimum of \mu(D) over all the \Sigma -measurable subsets D of X \setminus N_{n} . This infimum might
a priori be - \infty . As \varnothing is a possible candidate for D in the definition of s_{n} , and as \mu(\varnothing) = 0 , we have s_{n} \leq 0 . Hence, there exists a \Sigma -measurable subset D_{n} \subseteq X \setminus N_{n} such that : \mu(D_{n}) \leq \max \! \left( \frac{s_{n}}{2},- 1 \right) \leq 0. By the claim above, there is a negative set A_{n} \subseteq D_{n} such that \mu(A_{n}) \leq \mu(D_{n}) . Set N_{n + 1} := N_{n} \cup A_{n} to finish the induction step. Finally, define : N := \bigcup_{n = 0}^{\infty} A_{n}. As the sets (A_{n})_{n = 0}^{\infty} are disjoint, we have for every \Sigma -measurable subset B \subseteq N that : \mu(B) = \sum_{n = 0}^{\infty} \mu(B \cap A_{n}) by the sigma additivity of \mu . In particular, this shows that N is a negative set. Next, define P := X \setminus N . If P were not a positive set, there would exist a \Sigma -measurable subset D \subseteq P with \mu(D) . Then s_{n} \leq \mu(D) for all n \in \mathbb{N}_{0} and : \mu(N) = \sum_{n = 0}^{\infty} \mu(A_{n}) \leq \sum_{n = 0}^{\infty} \max \! \left( \frac{s_{n}}{2},- 1 \right) = - \infty, which is not allowed for \mu . Therefore, P is a positive set.
Proof of the uniqueness statement: Suppose that (N',P') is another Hahn decomposition of X . Then P \cap N' is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to N \cap P' . As : P \triangle P' = N \triangle N' = (P \cap N') \cup (N \cap P'), this completes the proof.
Q.E.D. ==References==