Is the space of Radon measures a Polish space or at least separable?

Question

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 [1] by Malek, Necas, Rokyta, Ruzicka, 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: $$M(\mathbb{R}^d)\equiv \{ \mu : C_0 (\mathbb{R}) \to \mathbb{R}; \mu \text{ is linear}\wedge \exists c>0, |\mu (f)|\le c \|f\|_{\infty}, \forall f \in \mathcal{D}(\mathbb{R}^d)\}.$$ Here

• $C_0(\mathbb{R}^d)\equiv \{ u \in C(\mathbb{R}^d): \lim_{|x|\to \infty} u(x) = 0 \}$ 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 $\big(M(\mathbb{R}^d), \| \cdot \|_{M(\mathbb{R}^d)}\big)$ 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 measure-valued spaces so I don't know much about them. Help with this would be great (and I definitely need it). Thanks in advance.

Reference

[1] Jindřich Nečas, Josef Málek, Mirko Rokyta, Michael Růžička, Weak and measure-valued solutions to evolutionary PDEs, Applied Mathematics and Mathematical Computation. 13. London: Chapman & Hall, pp. vii+317 (1996), ISBN:0-412-57750-X, MR1409366, Zbl 0851.35002.

Answer 1

With respect to the norm topology, the space of Radon measures on a domain $\Omega$ is not separable. 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 separable. Of course there are many subspaces of Radon measures that are separable in the norm topology, e.g., as you noted $L^1$ naturally embeds as a subspace and is separable.

The space of Radon measures on a domain $\Omega$ is separable 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$^*$ separability.

Answer 2

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.

Answer 3

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.

Answer 4

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.