> [!theorem] Theorem > > Let $(X, \cm)$ be a [[Measure Space|measurable space]] and $\nu$ be a [[Signed Measure|signed measure]] on $(X, \cm)$. Then there exists unique positive measures $\nu^+$ and $\mu^-$ such that they are [[Mutual Singularity|mutually singular]] and $\nu = \nu^+ - \nu^-$. > > The measures $\nu^+$ and $\nu^-$ are the **positive** and **negative variations** of $\nu$, and $\nu = \nu^+ - \nu^-$ is the **Jordan decomposition** of $\nu$. The sum $|\nu| = \nu^+ - \nu^-$ is the **total variation** of $\nu$. > > Moreover, $\nu(E) = \int_E (\chi_{P} - \chi_{N}) d|\nu|$ can be represented as an [[Integral|integral]] with respect to its total variation. > > *Proof*. Let $X = P \cup N$ be the [[Hahn Decomposition Theorem|Hahn decomposition]] for $\nu$ and define $\nu^+(E) = \nu(E \cap P)$ and $\nu^-(E) = -\nu(E \cap N)$, then $\nu^+ \perp \nu^-$ and $\nu = \nu^+ - \nu^-$. > > Let $\nu = \mu^+ - \mu^-$ be another decomposition of $\nu$ into positive measures such that $\mu^+ \perp \mu^-$. Let $E, F$ be a partition of $X$ such that $F$ is $\mu^+$-null and $E$ is $\mu^-$-null. Then $X = E \cup F$ is another Hahn decomposition for $\nu$, and $P \Delta E = N \Delta F$ are $\nu$-null. Therefore for any $A \in \cm$, > $ > \nu(A \cap P) = \nu^+(A \cap P) = \mu^+(A \cap E) > $ > and > $ > \nu(A \cap N) = \nu^-(A \cap N) = \mu^-(A \cap E) > $ > So $\nu^+ = \mu^+$ and $\nu^- = \mu^-$. Therefore the decomposition is unique. > >