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Sep 10, 2019 at 14:27 comment added Mark Thanks again. $G_{\delta}$ sets got me interested. And I will definitely check out the book you have recommended.
Sep 10, 2019 at 13:25 comment added Robert Furber 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.
Sep 10, 2019 at 13:23 comment added Robert Furber 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.
Sep 10, 2019 at 12:00 comment added Mark 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.
Sep 10, 2019 at 11:45 comment added Robert Furber @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 $C_b(\mathbb{R})$.
Sep 10, 2019 at 8:42 comment added Mark 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.
Sep 9, 2019 at 18:35 history answered Robert Furber CC BY-SA 4.0