Proof [Theorem 3.3, Fol99]. By flipping the sign of $\mu$, assume without loss of generality that $\mu < \infty$.

(1): Let $M = \sup\bracs{\mu(P)|P \in \cm \text{ positive}}$, then there exists positive sets $\seq{P_n}\subset \cm$ such that $\mu(P_{n}) \upto M$. Let $P = \bigcup_{n \in \natp}P_{n}$, then $P$ is positive by Lemma 16.1.5, and $\sup_{n \in \natp}\mu(P_{n}) \le \mu(P) \le M$.

(3): Let $N = X \setminus P$, then $P$ is positive and $X = P \sqcup N$.

(2): Suppose for contradiction that $N$ is not negative, then

Let $A_{1} \in \cm$ such that $A_{1} \subset N$ and $\mu(A_{1}) > 0$. Since $A_{1}\cap P = \emptyset$ and $\mu(P) = M$, $A_{1}$ cannot be positive, so $M_{1} = m(A_{1}) - \mu(A_{1}) > 0$. Let $n \in \natp$ and suppose inductively that $\bracs{A_n}_{1}^{n}$ has been constructed such that

1. For each $1 \le k \le n$, $M_{k} = m(A_{k}) - \mu(A_{k}) > 0$.

2. For each $1 \le k \le n - 1$, $A_{k+1}\subset A_{k}$ with $\mu(A_{k+1}) - \mu(A_{k}) > M_{k}/2$.

Let $A_{n} \in \cm$ with $A_{n+1}\subset A_{n}$ such that $\mu(A_{n+1}) - \mu(A_{n}) > M_{n}/2$. Since $A_{n+1}\cap P = \emptyset$ and $\mu(P) = M$, $A_{n+1}$ cannot be positive, so

Let $A = \bigcap_{n \in \natp}A_{n}$, then since $0 < \mu(A_{2}) < \infty$, $\mu(A) = \limv{n}\mu(A_{n})$ by Proposition 16.1.4, so

Since $A \cap P = \emptyset$ and $\mu(P) = M$, $A$ cannot be positive. Thus there exists $B \in \cm$ with $B \subset A$ with $\mu(B) - \mu(A) > 0$. As $M_{n} \to 0$ as $n \to \infty$, there exists $n \in \natp$ such that

As $\mu(B) - \mu(A_{n}) > \mu(B) - \mu(A)$, this contradicts the definition of $m(A_{n})$.

Therefore $N$ must be negative, and $X = P \sqcup N$ is the desired decomposition.

(U): Since $X = P \sqcup N = P' \sqcup N'$,

is a union of two null sets, which is null. Likewise,

is also a null set.$\square$