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.