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How big is the weak-* closure of the set of all (finite) convex combinations of Bernoulli measures among all invariant probability measures?

I mean, we are in the symbolic space $\{1,2,\ldots,d\}^{\mathbb{N}}$ and I would like to know, how big the (weak-*) closure of the set $C$ is, where $$C = \left\{\sum_{i=1}^n \alpha_i \eta_i: n \in \mathbb{N}; \sum_{i=1}^n \alpha_i=1; \eta_i = \mathrm{Ber}(p^i_1,\ldots,p^i_d) \right\}.$$

Of course, any reference towards this topic may be useful.

Thanks a lot for your attention

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    $\begingroup$ What is your measure of "bigness"? de Finetti's theorem states that the convex mixtures of Bernoulli measures are precisely the measures that are invariant under finite permutations of the indices. $\endgroup$ – Algernon Feb 24 '16 at 17:58
  • $\begingroup$ @Algernon "Bigness" in the sense this set can be almost everything or, on the opposite, has so many properties, is so specific that only a few invariant measures can be there... $\endgroup$ – Bruno Brogni Uggioni Feb 24 '16 at 18:01
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    $\begingroup$ This is precisely the set of all measures of the form $\int m\, d\mathbb{P}(m)$ where $\mathbb{P}$ is a Borel probability measure on the set of Bernoulli measures (a closed set). So, it's a closed face of the set of invariant measures. $\endgroup$ – Ian Morris Feb 24 '16 at 18:02
  • $\begingroup$ @IanMorris, sorry, but $\int m d\mathbb{P}(m)$ is kind of a ergodic decomposition? I mean, a Borel probability on the space of Bernoulli measures... I don't understand how this integral you wrote works... $\endgroup$ – Bruno Brogni Uggioni Feb 24 '16 at 18:09
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    $\begingroup$ I don't see anything complicated about this limit. Write each term in the sequence as $\int m\,d\mathbb{P}_n$ where $\mathbb{P}_n=\sum_{i=1}^{M_n} \alpha_i^n \delta_{Ber_i^n}$, then $\eta = \int m\,d\mathbb{P}$ where $\mathbb{P}$ is the weak-* limit of the measures $m_n$. If $\mathbb{P}$ is a point mass then this is a Bernoulli measure, if not then $\eta$ is not ergodic. $\endgroup$ – Ian Morris Feb 25 '16 at 15:39
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The set of Bernoulli measures $B$ is closed, and the closure of $C$ is precisely the set of measures of the form $\int m\,d\mathbb{P}$ where $\mathbb{P}$ is a Borel probability measure on the set of Bernoulli measures.

Let me unpack that statement a little. Let $\mathcal{M}_\sigma$ denote the set of shift-invariant Borel probability measures on $\Sigma_d:=\{1,\ldots,d\}^{\mathbb{N}}$. This we equip with the weak-* topology, which is the smallest topology on $\mathcal{M}_\sigma$ such that for every continuous function $f \colon \Sigma_d \to \mathbb{R}$, the map $m \mapsto \int f\,dm$ is a continuous function from $\mathcal{M}_\sigma$ to $\mathbb{R}$.

This topology (which is metrisable) defines a Borel $\sigma$-algebra on $\mathcal{M}_\sigma$. A linear combination of measures $\sum_{i=1}^n\alpha_im_i \in \mathcal{M}_\sigma$ satisfies $$\int f\,d\left(\sum_{i=1}^n \alpha_i m_i\right) = \sum_{i=1}^n \alpha_i \int f\,dm_i$$ for every $f \in C(\Sigma_d)$. More generally, if $\mathbb{P}$ is a Borel probability measure on the compact metrisable space $\mathcal{M}_\sigma$, then defining $$\int f\,d\mu:=\iint f \,dm\,d\mathbb{P}(m)$$ defines a measure $\mu$ on $\Sigma_d$, which we abbreviate to $\mu=\int m\,d\mathbb{P}(m)$.

Now, it is easy to check that the set of Bernoulli measures is closed in the weak-* topology: one uses the fact that a measure is $(p_1,\ldots,p_d)$-Bernoulli if and only if every cylinder set satisfies an equation in the weights, $$\mu([x_1\cdots x_n])=p_{x_1}\cdots p_{x_n},$$ and $\mu([x_1\cdots x_n])$ is just $\int \chi \,d\mu$ where $\chi$ is the (continuous) indicator function of the (clopen) cylinder set $[x_1\cdots x_n]$. The set $C$ is thus the set of all measures of the form $\int m\,d\mathbb{P}$ where $\mathbb{P}$ is an atomic probability measure supported on the closed set $B$. Since $B$ is closed, the closure of $C$ is the set of all measures of the form $\int m\,d\mathbb{P}$ where $\mathbb{P}$ is an arbitrary Borel probability measure on $B$.

I am not quite sure how to answer the question "How big is the closure of $C$?" but I can at least say that it is an infinite-dimensional sub-simplex of $\mathcal{M}_\sigma$, and its only ergodic elements are Bernoulli measures.

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  • $\begingroup$ thanks a lot. I really appreciate this part: its only ergodic elements are Bernoulli measures. Good answer. I will try to understand it properly, thanks – $\endgroup$ – Bruno Brogni Uggioni Feb 25 '16 at 11:15
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Maybe it is good to note that a similar question: What is the weak-$^*$ closure of the set $$ D=\{\mu\in\mathcal{M}_\sigma: \mu\text{ is isomorphic to some }\nu\in C\}? $$ has a dramatically different answer: $D$ is dense in $\mathcal{M}_\sigma$. This is a corollary to a result of Jean-Paul Thouvenot and Benjy Weiss (still unpublished as far as I know). Weiss announced it during a talk in Prague last year and mentioned that Dan Rudolph also knew that.

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