Set of probability measures compact in weak$^{*}$-topology? Let $\mathcal{P}$ be the set of all Borel probability measures on $\mathbb{R}_{+}$ with compact support and equipped with the weak$^{*}$-topology which is deduced from the norm of total variation. I think that $\mathcal{P}$ is convex and compact in the weak$^{*}$-topology, where the latter holds because any sequence of probability measures on $\mathbb{R}_{+}$ has a subsequence which converges in weak* topolpogy. Is that correct and is it used and if yes where, that the measures have compact support?
EDIT: Let $K$ be a compact set in $\mathbb{R}_{+}$ and let $\mathcal{P}$ be the set of all Borel probability measures on $\mathbb{R}_{+}$ with compact support in $K$and equipped with the weak$^{*}$-topology. Is it now true that $\mathcal{P}$ is compact in the weak$^{*}$-topology?
Secondly, is it correct that Dirac measures are the extreme points of $\mathcal{P}$? In Example 8.16 in http://www.math.caltech.edu/simon_chp8.pdf it is assumed that the measures are on a compact set, so does the same hold when we assume that measures have only compact support instead?
 A: $\mathcal{P}$ is not compact.  (Hint: If $\mu$ is any probability measure on $\mathbb{R}_{+}$, but not necessarily with finite support, then the measures $(\mu_{n})_{n \in \mathbb{N}} \subseteq \mathcal{P}$ given by $\mu_{n}(A) = \mu([0,n])^{-1} \mu(A \cap [0,n])$ converge weakly-$*$ to $\mu$ as $n \to \infty$.  That proves it's not closed.  It isn't pre-compact either --- the Dirac masses $(\delta_{n})_{n \in \mathbb{N}}$ do not have a weakly-$*$ convergent subsequence.)
If we restrict attention to a fixed compact set $K \subseteq \mathbb{R}_{+}$, then the previous counter-examples don't work and, indeed, the situation changes.  Let $\mathcal{P}_{K}$ denote the family of all probability measures in $\mathbb{R}_{+}$ with support contained in $K$.  Then $\mathcal{P}_{K}$ is compact.  There are a number of ways to see this (e.g. Banach-Alaoglu Theorem or Portmanteau Theorem).  The key point is that we can find continuous functions on $\mathbb{R}_{+}$ that are identically one on $K$.  Therefore, there is no opportunity to lose mass (i.e. weak-$*$ limit points of sequences in $\mathcal{P}_{K}$ are still probability measures, unlike the $(\delta_{n})_{n \in \mathbb{N}}$ example).
The only extreme points of $\mathcal{P}$ (or $\mathcal{P}_{K}$ for that matter) are Dirac masses.  (Hint: If $\mu \in \mathcal{P}$ and there are disjoint sets $A_{1}$ and $A_{2}$ with $\min\{\mu(A_{1}),\mu(A_{2})\} > 0$, then we can write $\mu = (1 - \lambda) \mu_{1} + \lambda \mu_{2}$, where $\mu_{i}(A) = \mu(A_{i})^{-1} \mu(A \cap A_{i})$.)
A useful way to think about the extreme points question is that we can write $\mu = \mu * \delta_{0}$ (trivially), which writes as
\begin{equation*}
\mu(A) = \int_{\mathbb{R}_{+}} \delta_{0}(A - x) \mu(dx) = \int_{\mathbb{R}_{+}} \delta_{x}(A) \mu(dx)
\end{equation*}
which is a "generalized convex combination."  When $\mu = \sum_{i = 1}^{N} \lambda_{i} \delta_{x_{i}}$ for some $\{x_{1},\dots,x_{N}\}$ and $(\lambda_{1},\dots,\lambda_{N})$, then this literally becomes a convex combination of Dirac masses.  But probability measures, in general, can be thought of as a tool for forming "generalized convex combinations."  The decomposition above suggests that the only way $\mu$ could be an extreme point is if $\mu = \delta_{x_{0}}$ for some $x_{0} \in \mathbb{R}_{+}$.  (The previous paragraph is a rigorous, if less enlightening, way to prove this.)
