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 ?

  • $\begingroup$ In the Treatise on analysis of Dieudonne volume 2 second edition, chapter 12.4 exercise 4 it is asked to show it is not separable. But I think I find a simple proof that it is : using a cover of $X$ by an increasing chain of relatively compact open set $(U_n)$, on each of the closure of each $U_n$, the set of positive measures with the vague topology is metrizable and separable. Given a positive measure $\mu$ on $X$ and a function $f$ with compact support $K$ in $X$, there is a $U_n$ containing $K$, (to be continued in the next comment ...) $\endgroup$ – brunoh Oct 10 '11 at 13:51
  • $\begingroup$ and a positive measure $\mu_n$ on the closure of $U_n$ that can be canonically extended to $X$ with $\mu_n(f)$ as close of $\mu(f)$ that we want. Therefore, canonical extensions of a dense numerable set of positive measures in each set of positive measures on the closure of each $U_n$ give a dense numerable set of positive measures in $M^+(X)$. Am I mistaken and where ? $\endgroup$ – brunoh Oct 10 '11 at 13:58
  • $\begingroup$ Why is the set of measures on a compact metric space separable? $\endgroup$ – N. S. Oct 15 '11 at 18:04
  • $\begingroup$ Can you precisely state the exercise from Dieudonne? Obviously it is not true that $M^+(X)$ is always non-separable; for instance if $X$ is a finite set. $\endgroup$ – Nate Eldredge Oct 15 '11 at 18:32
  • $\begingroup$ @ Nate obviously you are right ! Dieudonné asked to show that it is not separable if X is not compact. $\endgroup$ – brunoh Oct 15 '11 at 21:13

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.

  • $\begingroup$ thank you Nate for taking the time to answer me very clearly. By the way, I have posted the same question with a bounty on math overflow ... Dieudonné said at the beginning of chapter 13 that he will say locally compact for spaces that are locally compact metrizable and separable for the sake of simplicity. All his teachings about measures and integration happen using this kind of spaces only. $\endgroup$ – brunoh Oct 16 '11 at 7:25
  • $\begingroup$ @user14860: You certainly do need separability and metrizability in order to have $M^+(K)$ separable and metrizable. This is because $K$ embeds into $M^+(K)$ by sending $x$ to the point measure at $x$. $\endgroup$ – t.b. Oct 16 '11 at 8:22
  • $\begingroup$ @t.b. Dieudonné clearly states that $X$ is metrizable and separable as an hypothesis, you are right. But your comment makes me think that to truly get an embedding of $X$ into $M^+(X)$ you need "enough" continuous functions (like the space $X$ to be a normal space). I do not know what happen for a topological space in general, so I am not sure about the "need" part of your comment in very general situations. $\endgroup$ – brunoh Oct 16 '11 at 9:26
  • $\begingroup$ @user14860: I was only speaking of locally compact (Hausdorff) spaces since that was given in the statement of the problem. I tried to point out that Nate's argument only works with this second countability assumption on $X$ in addition to local compactness since for non-metrizable compact spaces $M^+(X)$ is not separable. I agree that you need some hypotheses for general topological spaces to get an injective map from $X$ to $M^+(X)$ (complete regularity seems necessary and sufficient at a quick glance), but in order for it to be a homeomorphism onto its image, you probably need more. $\endgroup$ – t.b. Oct 16 '11 at 9:53
  • $\begingroup$ @t.b. I think I understand better your comment. Clearly the argument of Nate (which is the same as mine) works only with second countability assumption on $X$ and local compactness in order to get the covering of $X$ by an increasing sequence of relatively compact open sets. And this hypothesis was stated by Dieudonné at the beginning of chapter 13. I am still not sure that the fact that $M^+(X)$ is separable necessarily implies everything for $X$ (second countability, and so metrizability ...) $\endgroup$ – brunoh Oct 16 '11 at 10:17

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