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Edited: Thanks to etienne.

I start with a compact metric space $(X,d)$. Then I consider the collection of finite measure $\mathcal{M}$ on $X$ and I equip $\mathcal{M}$ with the topology of weak convergence. This means that $\mu_n$ converges to $\mu$ if and only if for all $f \in C(X, \mathbb{R})$ (continuous functions from $X$ to $\mathbb{R}$), $$ \int f d\mu_n \to \int f d\mu. $$ Now my question is what does a general continuous function from $\mathcal{M}$ to $\mathbb{R}$ look like? By definition for $f \in C(X, \mathbb{R})$, $$ F(\mu) := \int f d\mu $$ is continuous but is also linear so this cannot be all continuous functions. Can we describe the entire set in some way?

Ben
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  • The functions of the form $F(\mu)=\int f, d\mu$ are linear, so any limit of such functions will again be linear. For example, you will never reach the function $G(\mu)=\mu (X)^2$. – Etienne Mar 10 '14 at 19:40
  • Ok. That's a really good point. What about functions of integrals then? $$ F(\mu) \sim G \left( \int f d\mu \right)$$ where G is some continuous function on $\mathbb{R}$. – Ben Mar 10 '14 at 19:55
  • You need at least all sums of functions of this type, and all uniform limits of these sum. Then ... I don't know. – Etienne Mar 10 '14 at 20:30

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If you restrict yourself to $\mathcal M_1=\{ \mu\in\mathcal M;\; \Vert\mu\Vert\leq 1\}$, then any continuous function on $\mathcal M_1$ is a uniform limit of a sum of products of functions of the form $F(\mu)=\int f\, d\mu$. This follows from the Stone-Weierstrass theorem.

Etienne
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  • That's a nice result. What I'm most interested in are the bounded continuous functions on all of $\mathcal{M}$. Then it's clear to me that this class can not be the one you stated above since on the sequence of measures $\mu_n = n \mu$ such functions grow exponentially with $n$. But I still wonder if there is a nice characterization of them. – Ben Mar 11 '14 at 22:51