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Consider a probability distribution on $\mathbb{R}^k$, say $\mu$. Then there is a sequence of probability measures $\mu_n$ that converge weakly to $\mu$ so that each of them is discrete (takes finitely values).

Question: Assume we replace $\mathbb{R}^k$ by any any locally compact Hausdorf group $G$. Is the previous still valid? If so, could anyone please provide a useful reference?

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    $\begingroup$ The answer to the question as posed is trivially yes ($\widetilde{\mu}=\mu(K)\delta_a+(1-\mu(K))\delta_b$ for some $a\in K$, $b\notin K$), but you probably meant something else. $\endgroup$
    – Christian Remling
    Commented Jun 28, 2015 at 5:04
  • $\begingroup$ You are absolutely correct, mending it. Thanks! $\endgroup$
    – TOM
    Commented Jun 29, 2015 at 6:14

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Since the answer above addresses the problem for signed measures rather than probabilities let me add to it by noting that both versions are true and are direct consequences of versions of the Hahn-Banach theorem (bipolar theorem). The appropriate duality is that between the space of bounded, continuous functions on a completely regular space and that of the bounded, Radon measures thereon. Thus neither local compactness nor a group structure is directly relevant.

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I guess the prime reference for questions like this is Bogachev's "Measure Theory I & II". Theorem 8.9.4 ii) states that the space of Baire measures is separable if $X$ is separable and the proof uses discrete measures for a dense set in $X$.

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