# Separability of sets of positive measures

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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Perhaps I should give more explanation to why I see it is separable. The reasoning is quite elementary : cover $X$ with an increasing sequence of relatively compact open sets $(U_n)$. In each $M^+(cls(U_n))$ take a dense sequence of positive measures for the vague topology. Extend canonically each measure in this sequence to $X$. Then all of these extensions for all $n$ is a dense sequence in $M^+(X)$. Right ? – brunoh 0 secs ago –  brunoh Oct 12 '11 at 17:05
Perhaps Dieudonné said it is not separable in the norm topology. –  Gerald Edgar Oct 12 '11 at 20:48
Thank you for your comment but nope. He specifically mentioned separable for the weak (vague) one (12.4 ex.4) –  brunoh Oct 12 '11 at 21:26
I'm missing something. A positive measure is supposed to be a positive linear form on which space of continuous functions on $X$? All continuous functions? All bounded continuous functions, compact support, etc? In particular, are these measures supposed to be finite, $\sigma$-finite, ? –  Nate Eldredge Oct 13 '11 at 5:11
Why should it be obvious that $M^{+}(X)$ is metrizable? I think this fails for the weak$^{*}$ topology. –  Benjamin Hayes Oct 13 '11 at 5:59

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.