Jordan decomposition of a signed measure

Remarks

Existence comes from a Hahn decomposition $X = P \cup N$: put $\nu^+(E) = \nu(E \cap P)$ and $\nu^-(E) = -\nu(E \cap N)$. These are measures, they are concentrated on the disjoint sets $P$ and $N$, and their difference is $\nu$. Since $\nu$ takes at most one infinite value, one of them is finite. Uniqueness: any such pair gives a Hahn decomposition, and two Hahn decompositions differ by a $\nu$-null set.

Example: for $\nu(E) = \int_E f \, d\mu$ with $f \in L^1(\mu)$ real, $\nu^+(E) = \int_E f^+ \, d\mu$ and $\nu^-(E) = \int_E f^- \, d\mu$, where $f^+ = \max(f, 0)$ and $f^- = \max(-f, 0)$.

Mutual singularity is what makes the decomposition unique. Without it, $\nu = (\nu^+ + \mu) - (\nu^- + \mu)$ for any finite measure $\mu$.

Going deeper

For a student

The Jordan decomposition reduces signed measures to ordinary measures: one integrates against $\nu$ by integrating against $\nu^+$ and $\nu^-$. The sum $\nu^+ + \nu^-$ is the total variation $\abs{\nu}$, the natural size of $\nu$. The name comes from the Jordan decomposition of a function of bounded variation as a difference of increasing functions. For the measure $dF$ of such a function $F$ the two decompositions correspond.