Hahn decomposition for signed measures, unique up to total-variation-null sets

A signed measure omits at least one infinite sign. Replacing $\nu$ by $-\nu$ swaps positive and negative sets, so first work in the case $\nu (E)<+\mathrm{\infty }$ for every measurable $E$. Put $m=\sup \{\nu (A):A\in \mathcal{A},A\text{ positive}\}\in [0,+\mathrm{\infty }]$; the empty set is admissible. If $m<+\mathrm{\infty }$, for every $n\ge 1$ the family of positive sets with $\nu (A)>m-1\mathrm{/}n$ is nonempty. If $m=+\mathrm{\infty }$, use instead the nonempty family with $\nu (A)>n$. AC selects $A_n$ from these families simultaneously. Set $P={\bigcup }_{n\ge 1}{A}_n$. This is the optimizing-sequence use of AC; the positive-subset supplier [L2] has its separate stated AC use.

For measurable $E\subseteq P$, set $B_n=E\cap (A_n\setminus {\bigcup }_{1\le j<n}{A}_j)$. These are disjoint measurable subsets of positive sets, with union $E$. Thus countable additivity gives $\nu (E)={\sum }_n\nu (B_n)\ge 0$. Hence $P$ is positive and $\nu (P)$ is finite in the chosen sign convention. For each $n$, positivity of $P\setminus A_n$ gives $\nu (P)\ge \nu (A_n)$. The branch $m=+\mathrm{\infty }$ would imply $\nu (P)>n$ for every $n$, impossible. Therefore $m$ is finite and $m-1\mathrm{/}n<\nu (A_n)\le \nu (P)\le m$, whence $\nu (P)=m$.

Put $N=X\setminus P$. If a measurable $E\subseteq N$ had $\nu (E)>0$, its value would be finite by the chosen sign convention. By [L2] it would contain a positive subset $Q$ with $\nu (Q)\ge \nu (E)>0$. Disjoint positive sets have positive union, by splitting each measurable subset between them; hence $P\cup Q$ would be positive and $\nu (P\cup Q)=m+\nu (Q)>m$, a contradiction. Thus every measurable subset of $N$ has nonpositive value, so $N$ is negative.

If the original sign was reversed, interchange the two sets just constructed; they then form a positive/negative partition for the original measure. For any two decompositions of that measure, $P\setminus P^{\mathrm{\prime }}$ lies in both a positive set $P$ and a negative set $N^{\mathrm{\prime }}$, and similarly $P^{\mathrm{\prime }}\setminus P$ lies in $P^{\mathrm{\prime }}\cap N$. Every measurable subset of either difference therefore has value zero. Splitting a measurable subset of their disjoint union into its two parts proves that $P\mathrm{△}{P}^{\mathrm{\prime }}$ is null. By [L3] its total variation is zero. This uniqueness argument itself makes no sequence choices.