# Is the space of Radon measures a Prohorov space?

Consider the spaces $C_c(\mathbb{R})$ of compactly supported continuous functions equipped with the inductive limit topology and the Banach space $C_0(\mathbb{R}) = \overline{C_c(\mathbb{R})}^{\, _{||.||_\infty}}$ of continuous functions vanishing at infinity equipped with the sup norm. The dual $M = C_c'$ is the space of Radon measures on $\mathbb{R}$ and $M_f = C_0' \subseteq M$ the subspace of finite Radon measures. Equip $M$ and $M_f$ with their weak-* topology. Now consider the space of probability measures on $M$ resp. $M_f$ equipped with the weak topology.

I want to know whether $M$ and $M_f$ are (sequential) Prohorov spaces, i.e. if every compact set (or sequence) of probability measures on $M$ resp. $M_f$ is tight.

Bogachev, Measure Theory II, Remark 8.10.15 mentions that $\mathcal{D}'(\mathbb{R})$ (the space of distributions) is Prohorov. So, since $M \subseteq \mathcal{D}'$ is closed, it follows that $M$ is also Prohorov. On the other hand, Proposition 8.10.19 says that the dual of any infinite-dimensional Banach space equipped with its weak-* topology is never Prohorov (by applying the Baire category theorem). So, it follows that $M_f$ is not Prohorov, which is somehow confusing.

I don't have any intuition for that fact. I always thought that Banach spaces are not such beasts. Is it somehow related to the distinction between Banach spaces and nuclear spaces?

Correction: $M$ is dense in $\mathcal{D'}$! (Thanks to @weather for the correction). [I have misleadingly thought of the continuous injection $M \to \mathcal{D}'$ as an isomorphism.] So there is no confusion anymore.

The theorem of Prohorov states that polish spaces, i.e. complete separable metric spaces, satisfy your condition. Another result states that the space of probability measures on a polish space when provided with the weak topology generated by the bounded, continuous functions. Hence your first question has a positive answer if you are prepared to use a stronger weak topology.

There seems to be some confusion in your other question. The weak topologies in the two spaces involved---distributions and measures---are distinct and, far from being closed, $M$ is dense in your space of distributions. This leaves the question of whether it has the Prohorov property open. Given the result on dual spaces you mention, I suspect that the answer will be negative---the (presumably crucial) difference to the case of distributions is that the latter space is nuclear.

• Thank you for the clarification. I will correct the statement above. Do you know how to construct such a stronger "weak topology" on $M$ which makes $M$ Polish? I need also sequences like $\delta_0-\delta_{1/n}$ to converge to $0$. I had also a look at the construction of a weaker metrizable topology here p.3/4, under which the Borel sets on $M$ remain the same. $M$ is Souslin via the continuous surjection $M_+\times M_+\to M$, $(\mu,\nu)\mapsto\mu-\nu$, where $M_+$ is the Polish space of positive measures. But it doesn't help me so far.
Feb 13, 2015 at 12:58
• As I said in my answer, you take the weak topology generated by the bounded, continuous functions. The restriction of this to the probability measures makes this a polish space and the natural injection of $I$ is a homeomorphism. This works for any completely regular spaces, although you won't, of course, get a polish space in the general situation. A good reference is Kechris, Classical descriptive set theory. Feb 13, 2015 at 15:46
• Sorry, but I don't understand. I need to consider probability measures $P(M)$ on the space of (possibly unbounded) Radon measures $M$ and not the subset of probability measures $P(\mathbb{R}) \subseteq M$. In order to speak of continuous bounded functions on $M$, I need a topology on $M$ first. I can't equip $M$ with the topology generated by $C_b(\mathbb{R})$ since in particular for the Lebesgue measure $\lambda \in M$ and $f \equiv 1$ the value $\int f \, d\lambda$ is infinite. What do you mean by $I$?
• It is probably better to work in the abastract situation. If $S$ is a completely regular space, the the natural topology on the space of probability measures on $Sis the weak topology generated by the continuous bounded functions. One can then ask whether on this space tightness and relative weak compactness coincide. As I understand it, you are interested in the cases where$S\$ is the real line, the measures thereon, or the bounded measures. I tried to shed some light on this but may have misunderstood you, in which case I apologise. Feb 13, 2015 at 21:48