# 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 measure.

If $U\subseteq X$ 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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Why all these inverses and rings? The universal properties imply that $H(X)$ is the set of all $\mathbb{R}_{\geq 0}$-linear maps $\{\text{measures on }X\} \to \mathbb{R}$. – Martin Brandenburg Mar 7 '12 at 14:45
Of course not every such map is an evaluation, just take some $\mathbb{R}_{\geq 0}$-linear combination of evaluations: If $U,V$ are measurable subsets, then $\mu \mapsto \mu(U) + \mu(V)$ is an example. – Martin Brandenburg Mar 7 '12 at 14:51
It's a good point; my question wasn't even clear to myself. First of all, I want $H(X)$ for my own purposes, but there's a better answer to your question, though maybe the answer is obvious. Suppose I want to quotient $\mathcal{O}(X)$ by the ideal $I=${$\mu | \mu(U)=0$ for all $U\in\mathcal{M}_X$}? Is $I=0$? – David Spivak Mar 7 '12 at 14:52
But then I can ask whether every element of $H(X)$ is a linear combination of evaluations.. – David Spivak Mar 7 '12 at 14:54
Crossposted to math.SE: math.stackexchange.com/questions/119638/… – Theo Buehler Mar 13 '12 at 11:25