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The following problem is related to work on topological dynamics, but I feel like the question is interesting on its own. I think the answer to the question below is likely to be well-known and I hope someone here will be able to help.

Here goes: a theorem of Lyapunov implies that, given atomless Borel probability measures $\mu_1,\ldots,\mu_n$ with full support on a Cantor space $X$, and $\varepsilon >0$, there exists some clopen subset $A$ of $X$ such that $|\mu_i(A)-\frac{1}{2}| < \varepsilon$ for all $i \in \{1,\ldots,n\}$. Could someone give me an example where this result fails for countably many measures $(\mu_i)$? That is, what is an example of countably many Borel probability measures $(\mu_i)$ which are all atomless and with full support, but such that for no clopen set $A$ can the sequence $(\mu_i(A))$ be arbitrarily close to the constant sequence $(\frac{1}{2})$?

I'd also be grateful for pointers to relevant litterature - I've tried to look up books on vector measures but without great success (the best that came up is Knowles' theorem but it does not seem useful here). My hope is that there is a simple, well-known example and that someone here will know it...

Thanks in advance for your help.

EDIT (March 13, 2017) Anthony Quas' example below boils down to the fact that, if there is a Dirac measure in the closure of $(\mu_i)$ then one cannot have a clopen set all of whose measures are far from both $0$ and $1$. Indeed I realize I should have added the following assumption: all measures in the closure of $(\mu_i)$ are also atomless and of full support. Is there also an easy counterexample then?

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  • $\begingroup$ For a clopen set $A$ you have that $\mu(A)=\lim \mu_n(A)$ if $\mu_n\to\mu$ in weak-$*$ sense, so no set of extra conditions that still leaves you with a dense set of measures can change anything. $\endgroup$ Commented Mar 13, 2017 at 20:56
  • $\begingroup$ I'm sorry, but I don't understand your comment: certainly, the condition that all measures in the closure of $(\mu_i)$ are nonatomic and with full support is a nontrivial addition to the fact that each $\mu_i$ satisfies those conditions. So, I am not sure what you mean? $\endgroup$ Commented Mar 14, 2017 at 9:50
  • $\begingroup$ This condition defines a (weak $*$) dense subset, so one can now just pick measures $\mu_{n,k}$ that (as $k\to\infty$) approach the measures $\mu_n$ from Anthony's original counterexample. Since $\mu_{n,k}(A)$ will be close $\mu_n(A)$ for large $k$ for any clopen $A$, this is still a counterexample to your conjecture. $\endgroup$ Commented Mar 14, 2017 at 18:23

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Sure: Let $(q_i)_{i\in\mathbb N}$ be an enumeration of the rationals in $(0,1)$. Let $\mu_i$ be the product measure on $\{0,1\}^{\mathbb N}$ such that each coordinate is 1 with probability $q_i$ and 0 with probability $1-q_i$. Now let $C$ be any clopen set. It is a union of finitely many cylinder sets: that is - that is membership in $C$ depends only on $x_1,\ldots x_n$ for some $n$. Now for any $q_i$ such that $q_i^n>\frac 34$, any $n$-cylinder set has measure in the range $[0,\frac 14)\cup (\frac 34,1]$.

EDIT: I think you can satisfy the non-atomicity and full support requirements also. Fix a measure $\mu_0$, say a (1/2,1/2) Bernoulli. Now enumerate the clopen sets $C_i$. If $\mu_0(C_i)\not\in(1/3,2/3)$, let $\mu_i=\mu_0$. Otherwise, let $\mu_i$ be the normalized version of $\lambda_i(A)=2\mu_0(C_i\cap A)+1/2\mu_0(A\setminus C_i)$. All limits are non-atomic and fully supported. Indeed, all are absolutely continuous with respect to $\mu_0$ with derivative uniformly bounded away from 0 and 1. Clearly there's no clopen set which all \mu_i$ measure as close to 1/2.

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  • $\begingroup$ Thanks! For some reason I had convinced myself that Bernoulli measures could not provide counterexamples... Unfortunately this does not seem like it will help with the original problem I was working on, but thanks to you I'll go to bed slightly less stupid tonight. $\endgroup$ Commented Mar 13, 2017 at 8:12
  • $\begingroup$ PS: I realize that when I was thinking about this I was implicitly making an additional assumption which I should have included in my question, see the edit above in case you are interested (maybe Bernoulli measures with parameters bounded away from $0$ and $1$ still provide a counterexample? I'll think about it...). $\endgroup$ Commented Mar 13, 2017 at 8:47
  • $\begingroup$ And thanks again! That helps a lot. Sorry I cannot upvote your answer twice... $\endgroup$ Commented Mar 13, 2017 at 16:27

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