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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – yada
    Feb 13, 2015 at 12:58
  • $\begingroup$ 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. $\endgroup$
    – weather
    Feb 13, 2015 at 15:46
  • $\begingroup$ 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 $I$? $\endgroup$
    – yada
    Feb 13, 2015 at 17:27
  • $\begingroup$ 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. $\endgroup$
    – weather
    Feb 13, 2015 at 21:48
  • $\begingroup$ Sorry, can't edit out the TeX mistakes in the above commebt. $\endgroup$
    – weather
    Feb 13, 2015 at 21:54

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