Do signed measures on delta-rings always have a Hahn decomposition?

2012-03-01T23:33:18Z

(aka, what I'd meant to ask here)

Let $X$ be a set. $\:$ Let $\mathcal{R}$ be a non-empty subset of $2^X$. $\:$ Suppose $\mathcal{R}$ is a delta-ring.

Let $\: \phi : \mathcal{R} \to \mathbb{R} \:$ be such that for all sequences $\: \langle A_0,A_1,A_2,A_3,...\rangle \:$ of pairwise disjoint elements of $\mathcal{R}$,

if $\;\;\;\; \displaystyle\bigcup_{n=0}^{\infty} \; A_n \;\; \in \;\; \mathcal{R} \;\;\;\;$ then $\;\;\;\; \displaystyle\sum_{n=0}^{\infty} \; \phi(A_n) \;\; = \;\; \phi\left(\displaystyle\bigcup_{n=0}^{\infty} \; A_n\right) \;\;\;\;$.

Does it follow that there exists a subset $P\hspace{.02 in}$ of $X$ (which is not necessarily a member of $\mathcal{R}$)
such that for all members $A$ of $\mathcal{R}$, $\;\;\; A\cap P \: \in \: \mathcal{R} \;\;$ and $\; \phi(A-P\hspace{.02 in}) \leq 0 \leq \phi(A\cap P\hspace{.02 in}) \;\;\;$?

Do signed measures on delta-rings always have a Hahn decomposition? - MathOverflow

Do signed measures on delta-rings always have a Hahn decomposition?