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Following Rosenblatt and Yang, I say that a subset $A$ of $\mathbb Z$ has a unique mean value if for all invariant means $\lambda_1,\lambda_2$ on $\mathbb Z$, one has $\lambda_1(A)=\lambda_2(A)$.

Notice that the set of subsets having a unique mean value has the same cardinality as the set of subsets non-having a unique mean value, thanks to Andreas Thom's answer to Intrinsically measurable subsets of amenable semigroups..

Nevertheless, roughly speaking, it should be clear that most subsets of $\mathbb Z$ should not have a unique mean value. Indeed, to a have a unique mean value one needs a very particular structure, as almost-periodicity.

Question: Is there any probabilistic way to formalize the intuition that generic subsets of $\mathbb Z$ does not have a unique mean value?

Thank you in advance,

Valerio

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    $\begingroup$ "the set of subsets having a unique mean value has the same cardinality as the set of subsets having a unique mean value" - Am I reading this wrong, or is this a tautology? $\endgroup$
    – HJRW
    Commented Apr 18, 2012 at 11:34
  • $\begingroup$ sorry, I meant that the set of subsets non-having bla bla bla.. I am editing right now. $\endgroup$
    – Valerio Capraro
    Commented Apr 18, 2012 at 15:04
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    $\begingroup$ Actually, every set that doesn't have its mean value determined by Cesaro, doesn't have unique mean value (the standard construction of a mean value is just an arbitrary norm-preserving linear extension of the Cesaro mean). Now you have a lot of room to show that the sets without Cesaro mean are "generic" in any decent sense you want :) $\endgroup$
    – fedja
    Commented Nov 16, 2012 at 3:01
  • $\begingroup$ Thank you, Fedja. It's now a bit late to add this observation where I wanted to add it (the paper has already been accepted). However, I remember that my problem was that I really did not have a precise idea about how to formalise the word "generic"? What's a "generic" subset of integers? $\endgroup$
    – Valerio Capraro
    Commented Nov 16, 2012 at 9:35

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If a set $A$ has arbitrarily large gaps, meaning that there are intervals $I_k$ of lengths tending to infinity that are disjoint from $A$, then there is a mean $\lambda$ such that $\lambda(A)=0$: take $\lambda(X)$ be a (Banach) limit of $|X \cap I_k| / |I_k|$. Similarly, if $A$ contains arbitrarily long intervals, then there is a mean $\mu$ such that $\mu(A)=1$.

So, for example, the set of subsets not having a unique mean is a dense $G_\delta$ set inside $2^ \mathbb{Z}$. Also, if you randomize a set and the events that $n$ belongs to the set are independent (and don't have probability 0 or 1), then, with probability 1, the set you get will not have a unique mean.

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    $\begingroup$ How do you know that the set is $G_\delta$? $\endgroup$
    – François G. Dorais
    Commented Apr 18, 2012 at 17:24
  • $\begingroup$ I am sorry, but I do not understand your second paragraph. Could you add some details? Thank you. $\endgroup$
    – Valerio Capraro
    Commented Apr 19, 2012 at 17:48
  • $\begingroup$ The set of sequences such that <i>for all</i> $n$, <i>there exists</i> $m$ such that $A\cap [m,m+n]=\emptyset$. This is a countable intersection of open sets. $\endgroup$
    – Anthony Quas
    Commented Oct 4, 2012 at 13:22

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