Background: I work on a SPDE problem where in order to apply Prokhorov's theorem I need that some measure space is Polish space. And additionaly it would be good if that space is Banach space. Earlier today I was reading the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, and I have a question from the Subsection 1.2.8 titled Radon measures. The definitions given bellow are taken from the same book.

On the one hand, the space of Radon measures is defined as: $$M(\mathbb{R}^d)\equiv \{ \mu : C_0 (\mathbb{R}) \rightarrow \mathbb{R}; \mu \hspace{0.2cm} linear \hspace{0.2cm} s.t. \hspace{0.2cm} \exists c>0, |\mu (f)|\leq c ||f||_{\infty}, \forall f \in \mathcal{D}(\mathbb{R}^d)\}.$$


$C_0(\mathbb{R}^d)\equiv \{ u \in C(\mathbb{R}^d): lim_{|x|\rightarrow \infty} u(x) = 0 \}$ and $C_0(\mathbb{R}^d)=\overline{\mathcal{D}(\mathbb{R}^d)}^{||\cdot||_{\infty}}$.

As usual $\mathcal{D}(\Omega)$ stands for the space of functions from $C^{\infty}( \overline{\Omega})$ with compact support in $\Omega$.

If we further define

$||\mu||_{M(\mathbb{R}^d)}\equiv sup\{|\mu(f)|: f \in \mathcal{D}(\mathbb{R}^d),||f||_{\infty}\leq 1 \}$,

then the space $(M(\mathbb{R}^d), || \cdot ||_{M(\mathbb{R}^d)})$ is a Banach space.

On the other hand, let $\Omega$ be a bounded domain. We denote by $M(\Omega)$ the space of Radon measures defined as the dual space of $C(\overline{\Omega})$. Also in this case we know that $L^1(\Omega)\hookrightarrow M(\Omega)$ (and we know that $L^1(\Omega)$ is separable).

My questions are:

  1. Is the space of Radon measures separable - in the case $\Omega \subset \mathbb{R}^d$ and in the case $\mathbb{R}^d$? Or to be more precise is it a Polish space? I have search it in a few books and in the questions here but I didn't find any concrete answer (I maybe have missed something).
  2. Maybe some subspace of the space of Radon measure is Polish? I've read somewhere that the space of positive Radon measures is Polish but didn't find any book to confirm that.
  3. Are there some other spaces of measure-valued functions that are Polish (besides the spaces mentioned above)?

I usually avoid dealing with meaasure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.

  • 1
    $\begingroup$ You have better chances with the weak $^*$ topology. The dual unit ball of a separable Banach space is weak$^*$ compact and metrizable. $\endgroup$ – Jochen Wengenroth Sep 9 '19 at 17:28
  • 1
    $\begingroup$ @JochenWengenroth: On the other hand, the space of all Radon measures with the weak-* topology is not metrizable, though it is separable. $\endgroup$ – Nate Eldredge Sep 9 '19 at 17:32
  • 2
    $\begingroup$ It is a long established fact that the space of Radon measures on a completely regular space is naturally identifiable with the dual of the space $C^b(S)$ of bounded, continuous functions thereon (with the so-called strict topology—-the finest locally convex topology which coincides with compact convergence on the unit ball). This was shown by Buck in the locally compact case in the 50’s and extended to the general case in the 60‘s. If the underlying space is polish, then so is that of the Radon probability measures (under the weak topology induced by $C^b(S)$)—-again a celebrated result. $\endgroup$ – user131781 Sep 9 '19 at 17:55
  • 1
    $\begingroup$ @NateEldredge Do you mean as opposed to Radon probability measures? The set of Radon probability measures on a Polish space is Polish as a subspace of the weak-* topology, even though the weak-* topology isn't first-countable on the whole space $C_b(X)^*$. $\endgroup$ – Robert Furber Sep 9 '19 at 17:57
  • $\begingroup$ @RobertFurber: Right, that's what I mean. The space $M(\Omega)$ defined in the question is the space of all signed Radon measures, i.e. the dual of $C(\overline{\Omega})$ (or something like that), and this is not weak-* metrizable (nor, as you say, even first countable). $\endgroup$ – Nate Eldredge Sep 9 '19 at 18:11

With respect to the norm topology, the space of Radon measures on a domain $\Omega$ is not seperable. Indeed, for any two distinct points $x,y$ in $\Omega$, the Dirac measures $\delta_x$ and $\delta_y$ (where $\delta_x(f)=f(x)$) satisfy $\|\delta_x-\delta_y\|=2$ since you can always find a compactly supported smooth function $f$ with $f(x)=-f(y)=1=\|f\|_\infty$. Any metric space that contains uncountably many disjoint open balls cannot be seperable. Of course there are many subspaces of Radon measures that are seperable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is seperable.

The space of Radon measures on a domain $\Omega$ is seperable in the weak$^*$ topology (This is probably the remark you allude to have read somewhere.) Indeed, consider the countable set $M_Q$ of measures of the form $\sum_{x \in S} a_x \delta_x$ where the coefficients $a_x$ are rational and $S$ runs over finite sets of points with rational coordinates. This $M_Q$ is countable and weak$^*$ dense. Also the embedding of $L^1$ as space of measures with absolutely continuous to Lebesgue measure is dense, and this gives another proof of weak$^*$ seperability.

| cite | improve this answer | |

No. Let $\Omega=[0,1]$. If $x\in[0,1]$, let $\delta_x$ be the point mass at $x$. They are all Radon measures. It is not that hard to show that $\|\delta_x-\delta_{x'}\|=2$. So you can construct an uncountable family of disjoint open balls. Since a separable metrizable space has a countable basis, this shows that the space in question is not separable.

If there exists a subspace $S$ with a countable dense subset $\{\mu_n|n\in\mathbb{N}\}$, then every element of $S$ will be absolutely continuous with respect to $\sum_n 2^{-n}\bar{\mu}_n$ where $\bar{\mu}_n$ is the normalization of $\mu_n$ to having total mass $1$. So separable subspaces are exactly those included in a $L_1$-space for some measure.

Unbounded domains may bring additional complications.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. Nice simple example in the first paragraph that explains the problem of separability. Could you elaborate more your second paragraph? $S$ is subspace with some countable dense subset of measures... Knowing that sounds useful. Also I should be using a bounded domain, but just for curiosity, what would be additional complications in the unbounded $\mathbb{R}^d$ case? $\endgroup$ – Mark Sep 10 '19 at 8:21
  • $\begingroup$ I'm just not sure that this will give you the variation norm in general. If it does, then it still holds that a subspace is separable if and only if it is a subspace of $L_1(\tau)$ for an appropriate measure $\tau$. All that is needed that the underlying measurable space is countably generated. $\endgroup$ – Michael Greinecker Sep 10 '19 at 9:25
  • $\begingroup$ Thanks for the follow up. I am not sure if my measurable space is countably generated but I guess I could add it as a request. It is interesting that you mentioned variational norm. Why do I need a variation norm? This is connected to the fact that in my SPDE problem I have a solution that is in BV space - and that space is not separable so I can't use it for the Prokhorov's theorem. But I know that BV functions are functions whose distributional derivative is finite Radon measure. I just don't know how to use that. $\endgroup$ – Mark Sep 10 '19 at 9:43
  • 1
    $\begingroup$ Yes, the Borel $\sigma$-algebra in Euclidean spaces (or any separable metric space) is countably generated. In a bounded domain, one can do some nice approximations that show you can basically use all bounded measurable functions in the definition f your norm and that gives you the variation norm; basically one can use Lusin's theorem to show it is enough to use continuous functions and then Stone-Weierstrass to go to smooth functions with compact support. This might fail for general domains. I simply don't know, those parts are outside my knowledge. $\endgroup$ – Michael Greinecker Sep 10 '19 at 9:50
  • $\begingroup$ Reminder of Lusin's theorem was nice. Especially that the classical statement has property that looks similar to tightness which I am trying to prove. By the end of the week I will know more about my problem and I'll maybe sent you additional question. Thanks again. $\endgroup$ – Mark Sep 10 '19 at 10:36

The other answers very adequately explain why the norm topology is not Polish except for trivial cases, so this answer is about the weak-* topology. Also, most results in the literature are about the space of Radon probability measures, not signed measures or positive measures, so I'll concentrate on this case.

If $\Omega \subseteq \mathbb{R}$ is bounded, then $\overline{\Omega}$ is compact and metrizable. It is generally the case that if $X$ is compact and metrizable, $C(X)$ is separable, and from this it follows that the unit ball of $C(X)^*$ is compact and metrizable in the weak-* topology (a.k.a. $\sigma(C(X)^*,C(X))$), and therefore Polish. Since the set of Radon probability measures is a closed subset of this unit ball, it is itself Polish. This is the easiest case to prove.

More generally, for every Borel probability measure on a Polish space $X$, we can map it into the dual space of the C$^*$-algebra of bounded continuous functions $C_b(X)$ by $$ \mu \mapsto \left( f \mapsto \int_X f \mathrm{d}\mu \right) $$ If we topologize the image of this with the weak-* topology coming from $C_b(X)$, we get a Polish topology. This is true even though the unit ball of $C_b(X)^*$ need not even be weak-* first-countable (this already happens for $X = \mathbb{R}$, although as $\mathbb{R}$ is locally compact you can use the weak-* topology coming from $C_0(\mathbb{R})$ instead). One place that this is proved is in Kechris's Classical Descriptive Set Theory Theorem 17.23 on page 112.

One last thing - a measure is sometimes defined to be Radon if it is locally finite and inner regular with respect to compact sets. This is true for all $\sigma$-finite measures on Polish spaces, so does not define a distinct type of measure in this case.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. Because I have SPDE problem, somehow I have a feeling that I will work with the Radon probability measures. Based on your answer, it looks that I will probably use the weak* topology also (coming from the $C_0$ or $C_0^{\infty}$). By the end of the week I will know far more details and send here additional questions. $\endgroup$ – Mark Sep 10 '19 at 8:42
  • $\begingroup$ @Mark For that special case, you can adapt the proof for a compact metric space. The norm topology on $C_0(\mathbb{R})$ is separable, so the unit ball of $C_0(\mathbb{R})^*$ is compact and metrizable (so Polish). The set of "subprobability measures" is closed, and therefore also Polish. The difficulty in this case is that the set of probability measures is not closed. But luckily it is a $G_\delta$ set (an intersection of open sets) - take a sequence of functions $(f_i)_{i \in \mathbb{N}}$ in $C_0(\mathbb{R})$ converging pointwise to the constant $1$ function in $C_b(\mathbb{R})$. $\endgroup$ – Robert Furber Sep 10 '19 at 11:45
  • $\begingroup$ Thanks for the follow up. I wasn't thinking of a subprobability measures and $G_{\delta}$ sets. Those may be useful. I need to try a few things that I've read in all of the answers. I will know more than and post additional things here. $\endgroup$ – Mark Sep 10 '19 at 12:00
  • $\begingroup$ Sorry, I got distracted while typing the rest of my comment. Here it is: The sequence of functions $(f_i)_{i \in \mathbb{N}}$ should also satisfy the property that for all $x \in \mathbb{R}$ and $i \in \mathbb{N}$, $f_i(x) \leq f_{i+1}(x)$. Write $\mathrm{ev} : C_0(\mathbb{R}) \rightarrow C_0(\mathbb{R})^{**}$ for the double dual embedding. Then the pointwise limit $\Phi$ of $\mathrm{ev}(f_i)$ is lower semicontinuous, so the set $\Phi^{-1}([1,\infty))$ is $G_\delta$, and the intersection of $\Phi^{-1}([1,\infty))$ with the subprobability is the probability measures. $\endgroup$ – Robert Furber Sep 10 '19 at 13:23
  • $\begingroup$ The point of proving the set of probability measures is $G_\delta$ is that a $G_\delta$ subset of a Polish space is Polish, so it proves that the set of probability measures is Polish. For this, and many other useful facts about Polish spaces, I recommend Kechris's book that I mentioned in the answer. $\endgroup$ – Robert Furber Sep 10 '19 at 13:25

To get separability you may be interested in considering transportation cost or Wasserstein distance instead of other metrics. Then you will get separability in many interesting cases (see the book of Villani, Optimal transport), but the distance will be defined only between measures of the same total mass and with finite moments.

| cite | improve this answer | |
  • $\begingroup$ Mikhail Ostrovskii, last summer one professor of mine gave me notes that on the different problem use space of probability measures with finite first moments and the Wasserstein distance $W_1$. At that moment I was using other technique so I wasn't much interested in it. And I completely forget about those notes unfortunately. But a few days ago I found them again and after reading them I figure it out that their technique could also work on my problem. Next week I will work on that. If it is not too much of a hustle for you, I would send you a question or two about that? Thanks in advance. $\endgroup$ – Mark Feb 21 at 10:43
  • $\begingroup$ I can look at the questions, but I am not an expert on $W_1$. What I am working on, you can see from my papers: front.math.ucdavis.edu/1902.10334 and front.math.ucdavis.edu/1907.01155 $\endgroup$ – Mikhail Ostrovskii Feb 22 at 18:23
  • $\begingroup$ As far as I can see, we don't have much in common in the area of interest. But maybe I do not see the whole picture. However, I am sure that I could learn something useful from you. I work on PDEs and SPDEs. I will send a question. If you can help, that's good, if not, that's ok too. Anyway, thanks for taking the time for this. $\endgroup$ – Mark Feb 23 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.