Let $X$ be a locally compact separable & metrizable space, and $M^+(X)$ its space of positive measures (i.e. positive linear forms on the space of continuous functions on $X$, continuous on each space of continuous functions with a support included in a given compact $K$ of $X$). It is not difficult to show that if we use the vague topology on $M^+(X)$ (i.e. restriction of the weak topology defined on the algebraic dual of the space of functions with compact support defined on $X$), $M^+(X)$ is still metrizable. I found in the Treatise on Analysis of Dieudonné cap. 13.4 that it is not separable. But it seems to me that it is easy to prove that is is (see my comment below). Am I mistaken ?
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This was also posted on Math.SE. I'm reposting my answer from there. I think I'm with you; the exercise in Dieudonné seems to be in error. I certainly agree that $M^+(X)$ can be separable, for $X$ non-compact. Take for instance $X = \mathbb{N}$. Then $M^+(X)$ with the vague topology is (unless I am greatly mistaken) homeomorphic to $[0,\infty)^{\mathbb{N}}$ with the product topology. This is certainly separable; a countable dense subset is given by the set of all finitely supported rational sequences. I also think your proof sketch looks good. Another way to say it is that the compactly supported measures are vaguely dense in $M^+(X)$, since if we fix an exhaustion of $X$ by compact sets $K_n$, we have $\mu|_{K_n} \to \mu$ vaguely (since for any continuous compactly supported $f$, we have $f$ supported in $K_m$ for some $m$, and then $\mu_n(f) = \mu(f)$ for all $n \ge m$). But the set of compactly supported measures is just the union of all $M^+(K_n)$, and each of these is known to be separable. Incidentally, I found the exercise from Dieudonné on Google Books: here. I didn't actually see an assumption that $X$ be separable and metrizable, though. |
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