> [!theoremb] Theorem > > Let $(X, \cm)$ be a [[Measure Space|measurable space]] and $\nu$ be a [[Signed Measure|signed measure]] on $(X, \cm)$. Then there exists a partition of $X$ into a positive set $P$ and a negative set $Q$. This decomposition is known as a **Hahn decomposition** for $\nu$. > > If $P', N'$ is another such pair, then $P \Delta P' = N \Delta N'$ is [[Null Set|null]] for $\nu$. > > *Proof*. Assume without loss of generality that $\nu < \infty$. Let > $ > m = \sup\bracs{\nu(E): E \in \cm, E \text{ positive}} > $ > and let $\seq{\nu(E_i)}$ such that $\nu(E_i) \upto m$. Then $P = \bigcup_{i \in \nat}E_i$ is also a positive set such that $\nu(P) = m$. > > Let $N = X \setminus P$. Then $N$ has no strictly positive subsets. Suppose that $E \subset N$ is a positive set, then $P \cup E$ is another positive set with $\nu(P \cup E) > \nu(P) = m$, which contradicts the fact that $m$ is an upper bound. > > Now let $E \subset N$. If $\nu(E) > 0$, then $E$ is not positive and there exists $F \subset E$ such that $\nu(F) < 0$. Let $G = E \setminus F$, then $\nu(G) > \nu(E)$. This means that *every* set of strictly positive measure contained in $N$ has a part of strictly negative measure, that can be trimmed off to yield a subset of greater measure. > > Suppose that $N$ is not negative. Let > $ > c_0 = \sup_{B \subset N}\nu(B) > 0 > $ > and take $A_1$ such that $\nu(A_1) > c_0/2$. Let > $ > c_n = \sup_{B \subset A_n}\nu(B) - \nu(A_n) > 0 > $ > and take $A_{n + 1}$ such that $\nu(A_{n + 1}) - \nu(A_n) > c_n/2$. Then > $ > \begin{align*} > \nu(A_{n + 1}) &= \sum_{k = 1}^{n}\nu(A_{k + 1}) - \nu(A_k) + \nu(A_1) - \nu(N) \\ > &> \sum_{k = 0}^{n}\frac{c_k}{2} = \frac{1}{2}\sum_{k = 0}^{n}c_k > \end{align*} > $ > and $\sum_{n \ge 0}c_n < \infty$, meaning that $c_n \to 0$. Taking $A = \bigcap_{n \ge 1}A_n$, and let > $ > c = \sup_{B \subset A}\nu(B) - \nu(A) > 0 > $ > Let $B \subset A$ such that $\nu(B) - \nu(A) > c/2$, then there exists $k \in \nat$ such that $c_k < c/2$, then > $ > \nu(B) - \nu(A_n) > \nu(B) - \nu(A) > c/2 > c_k > $ > which contradicts the definition of $c_k$ as the supremum. Therefore the assumption that $N$ is non-negative is false. > > Let $P, N$ and $P', N'$ be two such decompositions of $X$. Then $P \setminus P' \subset P$ and $P \setminus P' \subset N'$. Therefore $P \setminus P'$ is both positive and negative, and therefore null. Same with $P' \setminus P$.