Post Undeleted by Matthew Daws
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Your axioms are: I am really not sure this actually answers the question-- I await clarification as to whether

• $\emptyset\in\mathcal R$
• $A,B\in\mathcal R \implies (A\cup B)\setminus (A\cap B)\in\mathcal R$
• $A_n\in\mathcal R \implies \bigcap_n A_n\in\mathcal R$
• 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 really do have 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 $\sigma$-algebra, or something \sigma$-ring. Here then is a little weaker..counter-example to your question. I think this is asking about Let$X=\mathbb N$with$\mathcal R$being the usual concept collection of all finite subsets of$X$. Define$\phi$by$\phi(A) = |A\cap\{\text{evens}\}| - |A\cap\{\text{odds}\}|.$This is a "measure" in your sense. But then the only choice for$\sigma$-algebra P$ would be the set of even numbers, and a signed measurethat's not in $\mathcal R$.Actually

For a positive result, as your measure must take always finite valuesnote that for any $A\in\mathcal R$, this is even slightly more restrictive than the notion of "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 " which I've seenin the usual sense.

A textbook I like So there is "Foundations of Modern Analysis" by Friedman-- a Hahn-Decomposition for $A$. You might think one could glue the proof in section 1.10 makes no assumptions on Hahn decompositions together. Perhaps getting $P\subseteq X$ or such that for all $\phi$.
As your measure takes finite valuesB\in\mathcal R$, also you could follow Rudinwe did have$B\cap P, "Real B\setminus P \in\mathcal R$and complex analysis" Chapter 6$\phi(B\setminus P) \leq 0 \leq \phi(B\cap P)\$.