This question was asked on Math.SE here, but received no replies after several months. So I have posted it here, though with somewhat revised structuring of the question.
Let $V$ be a real vector space, and for any $S \subset V$ write $\mathrm{hull}(S)$ for the convex hull of $S$.
Given $S \subset V$, it is easy to see that for any $\mathbf{x} \in S$ the following statements are equivalent:
- $\mathbf{x}$ is not an extreme point of $\mathrm{hull}(S)$;
- there exists $n \geq 2$, points $\mathbf{x}_1,\ldots,\mathbf{x}_n \in S \setminus \{\mathbf{x}\}$ and values $\lambda_1,\ldots,\lambda_n \in [0,1]$ such that $\sum_{i=1}^n \lambda_i=1\,$ and $\,\sum_{i=1}^n \lambda_i\mathbf{x}_i=\mathbf{x}$.
Is there a term for a set $S \subset V$ with the property that for all $\mathbf{x} \in S$, if $\mathbf{x}$ is not an extreme point of $\mathrm{hull}(S)$ then there exist $\mathbf{x}_1,\mathbf{x}_2 \in S \setminus \{\mathbf{x}\}$ and $\lambda \in [0,1]$ such that $\lambda\mathbf{x}_1 + (1-\lambda)\mathbf{x}_2=\mathbf{x}$?
In other words, is there a term for sets $S \subset V$ with the property that in the further-above characterisation of non-extremity, one can simply fix $n=2$?
Obviously, by definition, any convex set has this property; but there are also non-convex sets with this property.
[Some trivial examples that are nonetheless interesting to observe: An arbitrary union of open line segments has this property. A set consisting of the three corners of a triangle has this property; but a set consisting of the three corners of a triangle plus one interior point of the triangle does not have this property. Nonetheless, a set consisting of the boundary of a triangle has this property, and so does a set consisting of the boundary of a triangle plus any subset of the interior of the triangle.]
Motivation from ergodic theory. Let $(X,\mathcal{X})$ be a measurable space and let $f \colon X \to X$ be a measurable map. An $f$-invariant measure is a probability measure $\mu$ on $X$ such that $\mu(A)=\mu(f^{-1}(A))$ for all $A \in \mathcal{X}$. An $f$-ergodic measure is an $f$-invariant measure $\mu$ with the additional property that the following equivalent statements hold:
- for any $A \in \mathcal{X}$ with $f^{-1}(A)=A$, we have $\mu(A) \in \{0,1\}$;
- for any $A \in \mathcal{X}$ with $\mu(f^{-1}(A) \triangle A)=0$, we have $\mu(A) \in \{0,1\}$;
- the only $f$-invariant measure that is absolutely continuous with respect to $\mu$ is $\mu$ itself.
Given the equivalence of these three statements, the following well-known result is not difficult to show:
Proposition 1. The $f$-ergodic measures are precisely the extreme points of the convex set of $f$-invariant measures.
But sometimes, when we have a map with an invariant measure, we like to consider the map as only defined modulo null sets under this measure. So it might be useful to have some kind of convex-geometric characterisation of ergodicity that doesn't require measure-preserving maps to be defined more precisely than in this "mod null sets" sense.
With this goal in mind, I have obtained (by essentially the same proof as Proposition 1) the following generalisation of Proposition 1:
Proposition 2. Let $S$ be a set of probability measures on $(X,\mathcal{X})$, and suppose we have an $S$-indexed family $(f_\mu)_{\mu \in S}$ of measurable maps $f_\mu \colon X \to X$ such that for each $\mu \in S$, $\mu$ is $f_\mu$-invariant. Suppose furthermore that for every $\mu \in S$ and every probability measure $\nu$ on $(X,\mathcal{X})$ that is absolutely continuous with respect to $\mu$, we have $$ \nu \in S \ \ \Leftrightarrow \ \ \text{$\nu$ is $f_\mu$-invariant.} $$ Then $S$ has precisely the property I am asking about in this MO question, and for each $\mu \in S$, $\mu$ is $f_\mu$-ergodic if and only if $\mu$ is an extreme point of $\mathrm{hull}(S)$.
(In my Math.SE question I gave a version of this proposition for general Markov operators.)
