Are probability measures weak-* closed? Non-duplicates
This is in a different setting, and this only deals with compact spaces which is the easy case.
Now for the question. Let $X$ be a locally compact Hausdorff space. $\mathcal{C}_0(X)$ is the set of continuous functions $f:X\to\mathbb{R}$ that go to zero at infinity, i.e. are less than any $\epsilon$ outside an appropriate compact set. One of the many Riesz representation theory gives an isometric isomorphism between $\mathcal{C}_0(X)$ and the space $M(X)$ of all complex regular Borel measures on $X$. So Banach-Alaoglu shows the ball of $M(X)$ is compact in the weak-* topology. Consider $P(X)\subseteq\{\mu:\|\mu\|\leq1\}$ the set of positive probability measures, i.e. $P(X)=\{\mu\in M(X):\mu\geq0,\mu(X)=1\}$.

Is $P(X)$ weak-* compact?

$P(X)$ is contained in the ball, so if it is weak-* closed, it is closed in a compact set, hence compact. If $X$ is compact, certainly $f\equiv1$ is in $\mathcal{C}_0$, so if $\mu_\alpha$ is a net of probabilities converging to $\mu$, $1=\mu_n(X)=\mu_n(f)\to\mu(f)=\mu(X)$, and since limits preserve inequalities for any $g\geq0$ $\mu(g)\geq0$, so $\mu\in P(X)$.
The argument for positivity works in general. The hard part with noncompact spaces is $\mu(X)=1$. It is fairly easy to prove $\mu(X)\leq1$: one merely has to work with $\epsilon$s and $\delta$s in an appropriate way. But I couldn't find any way to conclude $\mu(X)\geq1$. Proceeding by contradiction with $\mu(X)=1-\delta$ doesn't lead me anywhere. I googled for 1-2 hours and found nothing: wherever the weak-* closure was mentioned, it was just assumed. So how do I prove this last bit? Is $\mu(X)\geq1$? Or are there cases where it isn't? And if so, counterexamples?
 A: Say $X=\Bbb R$. Let $\mu_n=\delta_n$, a point mass at $n$. For $f\in C_0(\Bbb R)$ we have $$\int f\,d\mu_n=f(n)\to0\quad(n\to\infty).$$

Comment: Ian says something about "issues" like this. Of course one can get into trouble if one is not aware that this is how things are, but I submit that it's not a bug, it's a feature.
The example illustrates that $P(X)$ has a certain compactness when $X$ is compact that does not obtain when $X$ is just locally compact. This is as it "should" be. A probability measure is after all the distribution of a random variable. The family of, say, $[0,1]$-valued random variables "should" have some compactness not shared by the family of real-valued random variables! If $x_n$ is a random variable which equals $n$ with probability $1$ then we wouldn't want there to be a (real-valued!) random variable which was in any sense the limit of $x_n$.
Some day when we do want those random variables to have a limit we talk about $[-\infty,\infty]$-valued random variables instead.
The notion of a tight family of measures should be mentioned here, in regard to the question of what the compact subsets of $P(\Bbb R)$ are.
