Describe boundary points of set of finitely additive probability measures

convex-analysisfunctional-analysismeasure-theoryprobability theory

Let $X$ be a set, and let $\mathcal X$ be its powerset. Consider the space $[0,1]^\mathcal X$ equipped with product topology. It is compact and contains the set $\mathscr P$ of all finitely additive probability measures as a closed (hence, compact) and convex subset.

By the Krein-Milman theorem, $\mathscr P$ is the closed, convex hull of its extreme points. It's easy to describe these extreme points explicitly.

Claim. A probability measure $p$ is an extreme point of $\mathscr P$ iff it is 0-1 valued (iff $p = \delta_x$ for some $x \in X$, where $\delta_x$ is point mass at $x$).

I am wondering:

Is there an explicit description of the boundary points of $\mathscr P$ as well?


Proof of Claim. The assertion in parentheses follows immediately from the fact that the domain of $p$ is the powerset of $X$. If $p$ is 0-1 valued and $\delta_x = p = \lambda p_1 + (1-\lambda)p_2$, with $\lambda \in (0,1)$, then $p_1\{x\} = p_2\{x\}=1$, which implies $p_1 = p_2 = p$, so $p$ is an extreme point. If $p$ is not 0-1 valued, so that $p(A), p(A^c) \in (0,1)$ for some $A \in \mathcal X$, then
$$p = p(A)p(\cdot \mid A) + p(A^c)p(\cdot \mid A^c),$$
so $p$ is not an extreme point.

Best Answer

Is there an explicit description of the boundary points of $\mathscr P$ as well?

There is, and it’s very simple: every point in $\mathscr P$ is a boundary point because $\mathscr P$ has an empty interior.

To see this, suppose for the sake of contradiction that there exist an open subset $U$ of $[0,1]^{\mathcal X}$ and some $p\in\mathscr P$ such that $p\in U\subseteq\mathscr P$. By the definition of the product topology, there is some finite collection $E_1,\ldots,E_n$ (where $n\in\mathbb N$) of subsets of $X$ such that if $q\in[0,1]^{\mathcal X}$ and $|q(E_i)-p(E_i)|$ is sufficiently small for every $i\in\{1,\ldots,n\}$, then $q\in U\subseteq\mathscr P$. But this is impossible, since this would imply that $(1-\varepsilon)p$, viewed as an element of $[0,1]^{\mathcal X}$, was contained in $U$, and hence in $\mathscr P$, for $\varepsilon>0$ small enough, despite the fact that $(1-\varepsilon)p$ is not a finitely additive probability measure.

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