The ring generated by measures

Suppose $X$ is a space equipped with a $\sigma$-algebra $\mathcal{M}_X$. Then the set of measures on $X$ is closed under addition and scalar multiplication by elements of ${\mathbb R}$. Formally adjoining an additive inverse $-\mu$ for each measure $\mu$, we get an ${\mathbb R}$-module, which I'll call $V_X$. The ring of polynomials (with coefficients in ${\mathbb R}$) in these measures will be denoted $\mathcal{O}(X)=\text{Sym}(V_X)$, and called the ring of quasi-measures (for lack of a better term). We may denote an element of $\mathcal{O}(X)$ by $\mu$, even if it is a quasi-measure and not a genuine measureIf $U\subseteq X$ is a measurable subset then evaluation at $U$ gives a ring homomorphism $$e^~_U\colon\mathcal{O}(X)\to{\mathbb R},\hspace{.3in}e_U^~(\mu):=\mu(U).$$ (This homomorphism is formally clear, because $\mathcal{O}(X)$ is the free commutative ring on the measures; the idea is that a polynomial in measures can be evaluated by adding and multiplying values in the obvious way.)

Let $H(X)=\text{Hom}^~_{\bf Rings}(\mathcal{O}(X),{\mathbb R})$. Then we have a function $$e\colon\mathcal{M}_X\to H(X),$$ which takes every measurable set $U\subseteq R$ and returns $e_U\in H(X)$.

My question is: what can we say about the function $e$? For example, is it surjective? Is $H(X)$ in some way generated by the image of $e$? Is there any restriction we can put on $\mathcal{M}$ such that we can say something nice about $e$? If $e_U^~(\mu)=e_U^~(\nu)$ for every $U\in\mathcal{M}_X$, can we say that $\mu=\nu$ in $\mathcal{O}_X$?

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It would have been nice to mention that you posted the exact same question on MO –  t.b. Mar 13 '12 at 11:24
maiasm, if you are not David Spivak then you have plagiarized his question. Otherwise, see t.b.'s comment. –  anon Mar 13 '12 at 11:34
What is a polynomial on measures? –  Martin Argerami Mar 13 '12 at 18:59