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Let $\Omega = \{0,1\}^\mathbb{N}$, let $\mathcal{A}$ be the algebra generated by the open subsets of $\Omega$, where we use the product of discrete topologies, and let $\mathcal{F} = \sigma(\mathcal{A})$ be the Borel sigma-algebra.

Let $Q$ be a non-atomic, countably additive probability measure on $(\Omega, \mathcal{A})$. Let $E$ be the set of finitely additive probability measures on $(\Omega, \mathcal{F})$ that extend $Q$, i.e if $P \in E$ iff $P$ is a finitely additive probability measure on $(\Omega, \mathcal{F})$ and $P(A)=Q(A)$ for all $A \in \mathcal{A}$. $E$ is convex, and, by Krein-Milman, it has extreme points. Let $exE$ be the set of extreme points of $E$.

Is the cardinality of $exE$ infinite?

According to this paper (Theorem 1), $|exE|$ is an $\omega$-power, i.e. it is of the form $\mathfrak{n}^{\aleph_0}$, where $\mathfrak{n}$ is a cardinal. So, if I can show that $|exE|>1$, then it follows that $|exE|$ is infinite, and in fact uncountable. I know that $|exE|$ is at least $1$, because $Q$ extends to a (unique) countably additive probability measure on $(\Omega, \mathcal{F})$. The countably additive extension is an extreme point by Theorem 1 in this paper.

This is probably basic, but I'm unable to show that the countably additive extension of $Q$ is not the only member of $exE$.

(This question is cross-posted at MSE.)

Having thought about it some more, I wonder if the following argument is okay.

If $|exE|=1$, then, again by Krein-Milman, $|E|=1$ (because $E$ is the closed convex hull of its extreme points). But I happen to know that $|E|>1$ because, in addition to there being a countably additive extension of $Q$, there are also merely finitely additive extensions of $Q$. Thus, $|exE|>1$, and, by the arguments given above, it follows that $|exE|$ is infinite.

Comments and criticisms are welcomed.

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  • $\begingroup$ In order to apply the Krein-Milman theorem, you must be working in some vector space, and your set $E$ needs to be compact in some locally convex TVS. In what vector space are you working? $\endgroup$
    – user114263
    Commented Aug 23, 2018 at 0:09
  • $\begingroup$ @user114263 Right. I'm viewing sets of probabilities on $(\Omega, \mathcal{F})$ as subsets of the TVS of signed, finitely additive measures on $(\Omega, \mathcal{F})$ equipped with the weak* topology (setwise convergence). $\endgroup$
    – aduh
    Commented Aug 23, 2018 at 0:42
  • $\begingroup$ Do you mean the TVS of signed, finitely additive measures on $(\Omega, \mathcal{F})$ with bounded variation norm (that is, $L_\infty(\Omega, \mathcal{F})^*$)? Once we have that, your argument will be okay provided you can exhibit something in $E$ which isn't $Q$. Your argument says they are there. Is it obvious that there is something else in $E$ besides $Q$? Can we exhibit or cite the existence of such a thing? $\endgroup$
    – user114263
    Commented Aug 23, 2018 at 1:54
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    $\begingroup$ Your example of a $P$ which is in $E$ and not equal to $Q$ is nice. I think with this example you can cook up an elementary proof that $E$ in infinite. Take an infinite (even a continuum) $(D_\tau)_{\tau\in \mathbb{R}}$ of pairwise disjoint, countable, dense subsets of $\{0,1\}^\mathbb{N}$. For each $\tau$, construct $P_\tau$ as in your example and take a countably additive extension (let's call the extension $R_\tau$) to $\mathcal{F}$. For different $\tau$, you get different $R_\tau$, since $R_\tau(D_\tau)=1$ and the sets $D_\tau$ are pairwise disjoint. $\endgroup$
    – user114263
    Commented Aug 24, 2018 at 12:09
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    $\begingroup$ The only thing left to is to find such $(D_\tau)_{\tau \in \mathbb{R}}$. Perhaps you already know how to do this. A classical (and somewhat useful) result of Sierpinski is that there exists a collection of infinite subsets $(A_\tau)_{\tau\in \mathbb{R}}$ of $\mathbb{N}$ such that for each $\tau\neq \sigma$, $|A_\tau\cap A_\sigma|<\infty$. Now let $D_\tau$ denote the set of $(\epsilon_1, \epsilon_2, \ldots)\in \{0,1\}^\mathbb{N}$ such that $$|\{n: \ee_n=1\}\Delta A_\tau|<\infty.$$ When defined in this way, the sets $(D_\tau)_{\tau\in \mathbb{R}}$ will be pairwise disjoint, dense, and countable $\endgroup$
    – user114263
    Commented Aug 24, 2018 at 12:34

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