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Feb 26, 2016 at 17:17 answer added Dominik Kwietniak timeline score: 4
Feb 25, 2016 at 16:01 comment added Ian Morris Sorry, "where $\mathbb{P}$ is the weak-* limit of the measures $\mathbb{P}_n$".
Feb 25, 2016 at 15:39 comment added Ian Morris 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.
Feb 25, 2016 at 15:28 comment added Bruno Brogni Uggioni Dear @IanMorris, every measure in $C$ is a proper linear combination of invariant measures, but, in the closure of $C$, for instance the $\eta$ I wrote before: $$\eta = \lim_{n}\sum_{i=1}^{M_{n}}\alpha_{i}^{n}Ber_{i}^{n},$$ you may have $M_{n}\rightarrow \infty$, so, it looks like you can have complicated measures on the closure...
Feb 25, 2016 at 13:56 comment added Ian Morris If an invariant measure is a proper linear combination of two non-identical invariant measures, it is not ergodic. This is Theorem 6.10.iii in Walters' book. Every measure $\eta$ in the closure of $C$ is a proper linear combination of invariant measures unless the associated measure $\mathbb{P}$ is a Dirac measure, in which case $\eta$ is a Bernoulli measure.
Feb 25, 2016 at 13:00 comment added Bruno Brogni Uggioni Dear @NoahStein, yes, professor Ian Morris is right, each $i$ refers to a Bernoulli measure $\eta_{i}$
Feb 25, 2016 at 12:54 comment added Bruno Brogni Uggioni Dear @IanMorris, are you sure that the only ergodic measures on that space are Bernoulli measures? It means that if a ergodic measure $\eta$ satisfies: $$\eta = \lim_{n}\sum_{i=1}^{M_{n}}\alpha_{i}^{n}Ber_{i}^{n}$$ so it must be Bernoulli... well, I will try more, but I can't see this even when I suppose $\eta$ mixing... thanks for your attention
Feb 24, 2016 at 23:25 vote accept Bruno Brogni Uggioni
Feb 24, 2016 at 19:00 history edited Michael Hardy CC BY-SA 3.0
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Feb 24, 2016 at 18:47 comment added Ian Morris Noah, I'm not sure I understand: doesn't $\eta_i$ satisfy $\eta_i([x_1\cdots x_n])=p^i_{x_1}\cdots p^i_{x_n}$ for each $x_1,\ldots,x_n$?
Feb 24, 2016 at 18:34 comment added Noah Stein Could you clarify whether you actually care about invariantness, or where you meant to put it? You mention it, but then use distinct $p^i_j$ for different $j$ which is not invariant.
Feb 24, 2016 at 18:28 answer added Ian Morris timeline score: 3
Feb 24, 2016 at 18:14 comment added Ian Morris I'll expand on this as an answer.
Feb 24, 2016 at 18:09 comment added Bruno Brogni Uggioni @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...
Feb 24, 2016 at 18:02 comment added Ian Morris 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.
Feb 24, 2016 at 18:01 comment added Bruno Brogni Uggioni @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...
Feb 24, 2016 at 17:58 comment added Algernon 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.
Feb 24, 2016 at 17:48 history asked Bruno Brogni Uggioni CC BY-SA 3.0