Let $X$ be a Banach space, and for any compactly supported Borel probability measure $\mathbb{P}$ on $X$, define the mean $\mu_\mathbb{P}$ by $\mu_\mathbb{P}=\int_X x \, \mathbb{P}(dx)$.

I want to say that a Borel set $S \subset X$ is continuum-convex if every compactly supported Borel probability measure $\mathbb{P}$ on $X$ with $\mathbb{P}(S)=1$ has $\mu_\mathbb{P} \in S$.

Is there a recognised term for what I have called "continuum-convex"?

The point behind this definition:

The standard "set-theoretic" definition of a "convex subset of a vector space" is a subset that is closed under finite convex combinations, i.e. under weighted averages where the weighting is discrete. I want a stronger notion where the weighting is allowed to be completely general, i.e. not necessarily discrete (hence the term "continuum").

Note that

• if $S$ is continuum-convex then $S$ is convex;
• if $S$ is closed and convex then $S$ is continuum-convex.

However, neither of these implications is reversible. For example:

Suppose that $X$ is the completion of the vector space $\mathcal{M}(M)$ of finite signed measures on a compact metric space $M$ where $\mathcal{M}(M)$ is equipped with a norm $\|\cdot\|_{\mathcal{M}(M)}$ whose topology is the topology of weak convergence. (E.g. one can find a sequence of continuous functions $g_n \colon M \to [0,2^{-n}]$ such that $\|\mu\|_{\mathcal{M}(M)}:=\sum_n |\int_M g_n \, d\mu|$ is such a norm.) Then,

• the set of all finitely supported probability measures on $M$ is convex, but not continuum-convex;
• given a non-closed Borel set $A \subset M$ and a non-empty interval $I \subset (0,1]$, the set of all probability measures assigning to $A$ a value in $I$ is not closed in $X$, but nonetheless is continuum-convex.