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We are given a finite measure $\mu$ and a dense set $\{\lambda_k\}$ (one can think that it is countable if this helps) on a compact. Can one approximate the integrals $\int f\,d\mu$ for all continuous $f$ by sequences (or nets) of sums of the form $\sum c_kf(\lambda_k)$, in which the coefficients do not depend on $f$? As far as I understand, this is possible if the topology on the compact is generated by a metric, but is this also true in general, or under some natural assumptions?

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Yes, this is true without further assumptions on $X$ than compactness.

There's no harm in assuming that $\mu$ is a probability measure (otherwise scale everything).

The extremal points of the compact convex set $P(X)$ of probability measures on $X$ (in the weak$^{\ast}$ topology on the space $M(X) = C(X)^{\ast}$) are precisely the point measures. Moreover, the map $\delta: X \to P(X)$ sending $x$ to the point measure $\delta_{x}$ at $x$ is continuous and injective, hence a homeomorphism onto its image because $X$ is compact.

By the Kreĭn-Mil'man theorem $P(X)$ is the closed convex hull of the point measures (the extremal points of $P(X)$). Now if $D \subset X$ is our dense subset then $\delta(D)$ is dense in $\delta(X)$, so the closed convex hull $C$ of $\delta(D)$ must be all of $P(X)$ (since $C$ is a closed set containing $\delta(D)$ we must have $C \supset \delta(X)$ hence $C \supset P(X)$ while the other inclusion is obvious).

It remains to mention that the closed convex hull is the same thing as the closure of the linear algebra convex hull (this is because the closure of a convex set is convex) and we're done by observing that the closure of a set can be described by limit points of nets.

Unpacking all this we see that $\mu$ can be approximated (in the weak$^{\ast}$ topology) by a net $\nu_i$ where each $\nu_i$ is a finite convex combination of $\delta_{d}$'s with $d \in D$. In other words $\mu(f) = \lim_{i} \nu_i (f)$ and $\nu_i(f)$ is of the desired form.


Added: As was discussed in the comments, essentially the same proof shows the analogous result for signed measures and the result can be extended to locally compact spaces as opposed to compact spaces without significant effort.

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    @Byron: that's certainly right. If I understand your notation correctly we're assuming $X$ non-Hausdorff but compact. I'd bet that $C(X/\!\!\!\sim) = C(X)$, and if that's true then $X/\!\!\!\sim$ must be compact Hausdorff because it is compact and the $\delta$-function is then continuous 1-1 from a compact to a Hausdorff space, hence a homeomorphism, so we're back to my argument. I have no intuition whatsoever about measures on $X$ upstairs.2011-08-29