Hahn decomposition theorem

Remarks

Assume $\nu$ does not take the value $+\infty$; otherwise apply the argument to $-\nu$. Let $s$ be the supremum of $\nu(A)$ over positive sets $A$. A maximising sequence of positive sets has a positive union $P$ with $\nu(P) = s$. Suppose $N = X \setminus P$ contained a set of strictly positive measure. Then a lemma gives a positive subset of $N$ of strictly positive measure, and adding it to $P$ contradicts maximality.

Example: let $\nu(E) = \int_E f \, d\mu$ with $f \in L^1(\mu)$ real. Then $P = \set{f \ge 0}$ and $N = \set{f < 0}$ is a Hahn decomposition. The set $\set{f = 0}$ can be put into either part, which shows that the decomposition is unique only up to null sets.

Going deeper

For a student

The Hahn decomposition is the first step towards the Jordan decomposition $\nu = \nu^+ - \nu^-$, which reduces questions about signed measures to questions about measures. It is also used in the proof of the Radon–Nikodym theorem, applied to signed measures of the form $\nu - c \mu$.