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Let $X$ be a standard Borel space, so that the space of Borel probability measures on $X$ is also a standard Borel space. We denote it by $\mathcal P(X)$.

In this paper for any family of probability measures $P\subset \mathcal P(X)$ its strong convex hull is defined as $$ \operatorname{sco}P:=\left\{\left.\int_{\mathcal P(X)}q\;\nu(\mathrm dq)\;\right|\;\nu\in \mathcal P(\mathcal P(X)): \nu^*(P) = 1\right\} \subset \mathcal P(X). $$ which is exactly the set of all convex combinations of elements of $P$: here $\nu^*(P)$ is the outer $\nu$-measure of the set $P$. Since $\operatorname{sco}$ is a monotone map, I guess it admits at least one fixpoint which can be perhaps referred to as a convex closure of $P$. I'm not sure though whether the fixpoint is unique.

My question concerns the literature on convex hulls and convex closures of families of measures e.g. on Borel spaces. The aforementioned paper only contains a single reference to "Probability and Potential" by Dellacherie and Meyer, however I have only volume C available in my library, and there I did not find by any means a comprehensive study of these concepts. Any hint is greatly appreciated.

In particular, besides the uniqueness of the fixpoint I am interested whether the limit $$ \operatorname{clo} P := \bigcup_{n\in \Bbb N}\operatorname{sco}^n P $$ is a fixpoint of $\operatorname{sco}$.

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  • $\begingroup$ Why is this "exactly the set of all convex combinations"? And wouldn't that answer the question since taking convex hulls is an idempotent operation? $\endgroup$ – Michael Greinecker Dec 27 '13 at 17:40
  • $\begingroup$ @MichaelGreinecker: by a convex combination I mean the integral over a "weighting" measure $\nu$ which gives a full outer measure to $P$ (at least I'm not aware of any other definition to be used here). Can you say that $\operatorname{sco}$ is idempotent? $\endgroup$ – Ilya Dec 27 '13 at 18:51
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    $\begingroup$ The study of this sort of convex hull is the domain of Choquet theory. In fact assuming $\operatorname{sco}$ is idempotent, the Choquet theorem implies that convex sets are exactly the fixed points of $\operatorname{sco}$. The wiki page has some good references en.wikipedia.org/wiki/Choquet_theory $\endgroup$ – D. Kelleher Dec 27 '13 at 22:09
  • $\begingroup$ @D.Kelleher: can you suggest any sufficient conditions for the idempotence of $\operatorname{sco}$? Also, the Choquet's theorem you are talking about can be found in "Lecture notes on" I guess, isn't it? $\endgroup$ – Ilya Dec 28 '13 at 13:03
  • $\begingroup$ @Ilya The Choquet theorem is on the wiki page (though it should be in "Lecture notes on" too), and essentially says that a convex set is $\operatorname{sco} P$ for some $P$. I would think that it should be idempotent pretty generally, but i don't know off hand, and generalizing the proof from convex combinations runs into measure issues. $\endgroup$ – D. Kelleher Dec 28 '13 at 15:06
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See the notion of "measure-convex set" ...

MR2269765 Dostál, Petr; Lukeš, Jaroslav; Spurný, Jiří Measure convex and measure extremal sets. Canad. Math. Bull. 49 (2006), no. 4, 536–548.

MR1009196 Rosenthal, Haskell Martingale proofs of a general integral representation theorem. Analysis at Urbana, Vol. II (Urbana, IL, 1986–1987), 294–356, London Math. Soc. Lecture Note Ser., 138, Cambridge Univ. Press, Cambridge, 1989.

also see MO post Integral in a σ−convex set.

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  • $\begingroup$ Thanks, Gerald - I'll look into this. Do you know how can I find the second paper? $\endgroup$ – Ilya Dec 28 '13 at 12:57
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This is more a comment but I don't have that privilege. If I understand your question correctly, then two very comprehensive papers on the topologies of spaces of probability measures by T. Banakh should be of interest. These originally appeared in russian but are now easily available in english translation (arxiv: 1112.6161 and 1206.1727).

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  • $\begingroup$ Thanks, the second paper seems to provide interesting results regarding the connections between barycentric convexity and other types of convexity. $\endgroup$ – Ilya Jan 2 '14 at 11:14
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Many thanks to Gerald Edgar and D. Kelleher for their answers. I am (and was) a little familiar with the Choquet's theory and tried to find the answer to OP and similar questions there (mostly in "Lecture notes on Choquet's theorem"), but my search was not very successful. By no means it implies that the answer can't be found there - just perhaps my familiarity with the Choquet's theory is insufficient.

Inspired by the fact that the paper by Dostal et al. (Gerald's first reference) mentions a concept of a barycentre which also appears in the book "Probabilities and Potential, C" by Dellacherie and Meyer (English version, 1988), I've decided to look one more time into this book with focus on this keyword. There the Theorem in Chapter XI, par. 33 (p. 196) shows that for all strongly convex sets $P$ it holds that $\operatorname{sco}P = P$ and for all analytic sets $P$ the set $\operatorname{sco}P$ is analytic and strongly convex (in fact, it is the strongly convex envelope of $P$ being and intersection of all strongly convex supersets of $P$). The authors put a remark that the result is "most probably due to Fremlin", and Dostal et al. also contain a reference to Fremlin's work so perhaps for the origins I'll have to look into that direction.

In particular, we have now that $\operatorname{sco}$ maps the class of analytic subsets of $\mathcal P(X)$ into itself, and it is idempotent on this class. Together with the comments by D. Kelleher it may give a nice characterizaion of fixed points of $\operatorname{sco}$.

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