Say, we have some $\sigma$-algebra on $\mathbb N$ and let $\mathbb P$ be the set of all probability measures on it. We know that $\mathbb P$ is convex, so I wonder how do the extreme points look like.
I think we can do it with atoms. Let $\mathcal{A}=\{A_1, A_2, \ldots \}$ be the set of all atoms of $\mathcal{F}$ (we know $\mathcal{A}$ is countable). Now we define
$$T_n(E) :=
\begin{cases}
1 & \text{if $A_n \subset E$} \\
0 & \text{else} \\
\end{cases}$$
Then $\mathbb{T}=\{T_1, T_2, \ldots \} \subset \mathbb{P}$ is the set of all extreme points. We can prove it by showing that if $T=\lambda P_1+(1-\lambda)P_2$ with $0<\lambda<1$ then $T=P_1=P_2$ which follows from the fact what each $E\in\mathcal{F}$ is a disjunct union of atoms.
Does it make sense? Did I miss anything?
EDIT: I have shown that each $T\in \mathbb{T}$ is an extreme point yet got stuck in proving that each extreme point is in $\mathbb{T}$.
So let $P\in\mathbb{P}$ be an extreme point. That is $$\forall P_1, P_2 \in\mathbb{P} \ \ \forall \lambda \in (0,1) \ \ (P=\lambda P_1+(1-\lambda)P_2 \implies T=P_1=P_2)$$
And we need to show that $$\exists A\in\mathcal{A}:P(A)=1$$
This would imply that $P\in \mathbb{T}$.
I've tried proving both this and the contrapositive statement yet with no success. Any help is hugely appreciated.
Best Answer
You're close. Here is a hint. To prove that $\mathbb{T}$ contains all the extreme measures, you might look for a contradiction: Assume $P$ is an extreme measure not in $\mathbb{T}$. Use the fact that $\mathbb{N}$ is the union of the set of atoms $\mathcal{A}$, and that those atoms are pair-wise disjoint. Then, apply the properties of probability measures to construct two distinct measures generating $P$.
Hope this helps.