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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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    In fact, I think this should even be true for general locally compact spaces $X$ but the argument becomes a bit more subtle (it is still true that the point measures are the extremal points of $P(X)$ but the latter is no longer compact).2011-08-28
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    That's very nice. One thing I'm not comfortable with: each $\nu_i$ is should a finite convex combination of $\alpha_i \delta_{d_i}$ for some elements $d_i \in D$ and constants $\alpha_i \in \mathbb{S}^1$, no? The constants seem necessary to me because multiplying point measures by them still yields extreme points of $P(X)$.2011-08-28
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    Thanks. For me, a probability measure is a *positive* measure (so $\nu(1_{X}) = 1$) hence the $\alpha_i$'s must be $1$ (note that the positive measures form a weak$^{\ast}$-closed cone in $M(X)$ and evaluation at $1_{X}$ is continuous, hence what I denoted by $P(X)$ still is compact because it is a closed subset of the unit ball). However, it is also true that the extremal points of the unit ball are precisely the $\alpha \delta_x$ with $\alpha \in \mathbb{S}^1$ and you'd get the same result for *signed* measures by the same argument. Does this address what you're saying?2011-08-28
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    Yup. I mistakenly thought that $P(X)$ denoted complex (or signed) measures of total variation 1. It seems also that the argument generalizes readily for $X$ locally compact and $f$ which vanishes at infinity (simply by passing to the one-point compactification).2011-08-28
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    Yes, that's right, thanks for the confirmation of the locally compact case. I didn't mean to say the bit (as in "a bit more subtle") was significant, I just hadn't checked (again) at that point. :)2011-08-28
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    @Theo Just to clarify, you are working with the Baire $\sigma$-field (not Borel) and you assume that $X$ is Hausdorff, right?2011-08-29
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    @Byron: Yes, I assume that $X$ is Hausdorff (always when dealing with measures). I'm not sure what (uniqueness?) subtlety you have in mind but I was thinking of the Borel $\sigma$-field (not Baire) (using the *inner regular* finite signed Borel measures), as e.g. in theorem 6.3.4 on page 237 of Pedersen's Analysis Now. (I'm sweeping the assumption that the total variation of a signed measure is should be an inner regular Borel measure under the rug in the added bit), see also the remarks on the last page of [Arveson's notes](http://math.berkeley.edu/~arveson/Dvi/rieszMarkov.pdf).2011-08-29
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    @Theo Oh, I'm just trying to follow the argument and I'm referring to Bogachev's book where $C(X)^∗$ is the space of Baire measures. And it seems to me that the result is true even if $X$ is not Hausdorff, because $C(X)$ will be. We can still define delta measures and they are the only extreme measures, even though singletons may not be measurable. All we lose, as far as I can see, is that your map $\delta$ will not be injective. But I haven't checked all the details...2011-08-29
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    @Byron: I really don't know what happens in the non-Hausdorff case. My reflex would be to quotient out the equivalence relation $x \sim y$ if and only if $\delta_x = \delta_y$ and see if I can say anything about the quotient space. In my argument I'm appealing to Bogachev's Theorem 7.11.3 on page 116.2011-08-29
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    @Theo My vague feeling is that passing from $X$ to $C(X)$ and $M(X)$ automatically "quotients out" or "lumps together" such $x$ and $y$.2011-08-29
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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