Edit

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

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.

The answer is "yes".

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$\phi(B\setminus P) \leq 0 \leq \phi(B\cap P)$.

Neither proof (This is short-- I'm not going to type out the details here! Why did you ask?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!)

Edit: I am really not sure this actually answers the question-- I await clarification as to whether we really do have a $\sigma$-algebra, or something a little weaker...

I think this is asking about the usual concept of a $\sigma$-algebra and a signed measure. Actually, as your measure must take always finite values, this is even slightly more restrictive than the notion of "signed measure" which I've seen.

The answer is "yes".

A textbook I like is "Foundations of Modern Analysis" by Friedman-- the proof in section 1.10 makes no assumptions on $X$ or $\phi$.

As your measure takes finite values, also you could follow Rudin, "Real and complex analysis" Chapter 6.

Neither proof is short-- I'm not going to type out the details here! Why did you ask?