Let $X$ be a set.
  
  
  
  

Let $\mathcal{R}$ be a set of subsets of $X$ such that
  
  
  

$\{\} \in \mathcal{R}$
  
  
and  
  

For all members $A$ and $B$ of $\mathcal{R}$, $\;\; (A\cup B)-(A\cap B) \; \in \; \mathcal{R} \;\;$.
  
  
and  
  

For all sequences $\; \langle A\_0,A\_1,A\_2,A\_3,...\rangle \;$ of members of $\mathcal{R}$, $\;\;\;\; \displaystyle\bigcap\_{n=0}^{\infty} \; A\_n \;\; \in \;\; \mathcal{R}$
  
  
  

.
  
  
  
  

Let $\;\; \phi : \mathcal{R} \to \mathbb{R} \;\;$ be such that for all pairwise disjoint
  

sequences $\; \langle A\_0,A\_1,A\_2,A\_3,...\rangle \;$ of members 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 member $P$ of $\mathcal{R}$ such that
>    
> 
>  for all members $A$ of $\mathcal{R}$, $\;\; A\cap P \; \in \; \mathcal{R} \;\;$ and $\; \phi(A-P\hspace{.01 in}) \leq 0 \leq \phi(A\cap P\hspace{.01 in}) \;\;$?
> 
> 
> 


  




Do signed measures on sigma-rings always have a Hahn decomposition?
Your axioms are:


1. $\emptyset\in\mathcal R$
2. $A,B\in\mathcal R \implies (A\cup B)\setminus (A\cap B)\in\mathcal R$
3. $A\_n\in\mathcal R \implies \bigcap\_n A\_n\in\mathcal R$


Then 2 and 3 show that $A,B\in\mathcal R \implies A\setminus B\in\mathcal R$. Then 2 shows that $\mathcal R$ is closed under disjoint unions, so combined with 3, we see that $A,B\in\mathcal R \implies A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B) \in \mathcal R$. So actually your axioms describe a "$\delta$-ring" (see <http://en.wikipedia.org/wiki/Delta-ring> ) and not a $\sigma$-ring.


Here then is a counter-example to your question. Let $X=\mathbb N$ with $\mathcal R$ being the collection of all finite subsets of $X$. Define $\phi$ by
`\[ \phi(A) = |A\cap\{\text{evens}\}| - |A\cap\{\text{odds}\}|. \]`
This is a "measure" in your sense. But then the only choice for $P$ would be the set of even numbers, and that's not in $\mathcal R$.


For a positive result, note that for any $A\in\mathcal R$, the collection $\mathcal R\_A = \{ A\cap B : B\in\mathcal R \}$ is a $\sigma$-algebra on $A$, and $\phi$ restricted to $A$ is a signed measure in the usual sense. So there is a Hahn-Decomposition for $A$. You might think one could glue the Hahn decompositions together. Perhaps getting $P\subseteq X$ such that for all $B\in\mathcal R$, we did have $B\cap P, B\setminus P \in\mathcal R$ and $\phi(B\setminus P) \leq 0 \leq \phi(B\cap P)$. (This is true in my example). But my gut reaction suggests that this probably can't be done in general (I await a good counter-example from someone else!)


