Separability of the set 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 believe it is separable but Dieudonné thought the contrary. I am mistaken ? 
 A: I think I'm with you.
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.
