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Let $(X,\mu,\sigma)$ be a subshift on a finite alphabet, which we assume to be mixing. That is, for any cylinders $C, C'$ we have $\mu(\sigma^{-n}C\cap C')\to\mu(C)\mu(C')$ as $n\to+\infty$. We also assume $h_\mu(\sigma)$, the metric entropy of our subshift, to be positive.

Let $F_n$ denote the set of all cylinders of the form $[x_1=i_1,\dots,x_n=i_n]\subset X$.

Question. Is there a $\gamma\in(0,1)$ and $c>0$ such that for any $n\ge1$ and any union $U\subset F_n$ such that $|U|\ge c\cdot |F_n|/n$ we can partition $U$ into $U'$ and $U''$ in such a way that $$ \gamma \le \mu(U')/\mu(U'')\le 1/\gamma? $$ If that's not necessarily true in this generality, what extra condition do we need?

For instance, if $X$ is a transitive subshift of finite type (or, more generally, a transitive sofic subshift) and $\mu$ is the unique measure of maximal entropy for $X$, then it is known that $\mu[x_1=i_1,\dots,x_n=i_n]\asymp \alpha^n$, where $\alpha=\exp(-h_\mu(\sigma))\in(0,1)$, so we can just split any $U$ into approximately equal parts (i.e., $||U'|-|U''||\le1$), and this'll do.

Of course, the Shannon-McMillan-Breiman theorem ensures that $\mu[x_1=i_1,\dots,x_n=i_n]\approx \alpha^n$ for most cylinders, but this is just too crude for my purposes, since the kind of subsets $U$ I'm dealing with have very small measure $\mu$.

EDIT added: a lower bound of the size of $U$.

EDIT #2 added: in view of Anthony's counterexample, let's assume $\mu$ to be a measure of maximal entropy for $X$. (Unique if it helps.)

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  • $\begingroup$ Let me see if I understand the question. I think you're asking the following: let $F_n$ be the finite $\sigma$-algebra of unions of $n$-cylinder sets. You're asking is it true that for each $n$, there is an $m>n$ such that each $U$ in $F_n$ can be roughly equally partitioned into two sets in $F_m$? $\endgroup$
    – Anthony Quas
    Commented Sep 10, 2017 at 6:58
  • $\begingroup$ No, there is no partitioning of individual cylinders! All I want is to split $U$ into two roughly equal sets (wrt $\mu$). Hypothetically, one can imagine a silly situation when $U$ is comprised of one massive cylinder and a bunch of tiny ones; then such a partition would be impossible. I don't think we can actually get this, but don't know how to prove it. $\endgroup$
    – Nikita Sidorov
    Commented Sep 10, 2017 at 11:56
  • $\begingroup$ So I'm still not understanding the question. Is the idea that you want to take an element of $F_n$ and you want to partition it into two other elements of $F_n$? $\endgroup$
    – Anthony Quas
    Commented Sep 10, 2017 at 12:17
  • $\begingroup$ More specifically, can you be specific about what kinds of sets $U'$ and $U''$ are? Does $|U_n|$ mean the number of $n$-cylinder sets in $U_n$? $\endgroup$
    – Anthony Quas
    Commented Sep 10, 2017 at 12:22
  • $\begingroup$ Is the idea that you want to take an element of $F_n$ and you want to partition it into two other elements of $F_n$? - yes $\endgroup$
    – Nikita Sidorov
    Commented Sep 10, 2017 at 12:47

1 Answer 1

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Let $\mu$ be a Bernoulli measure on $\{0,1\}^{\mathbb Z}$ where 1 has probability 9/10. Let $U$ be the union of cylinder sets of length $n$ consisting of the cylinder set of all 1's together with all cylinder sets with at most $n/2$ 1's (so that $|U|\ge \frac 12|F_n|$).

Let $C=[1111111111]$, the cylinder set of $n$ 1's. Then $\mu(C)=(9/10)^n$. On the other hand, $\mu(U\setminus C)\le 2^n (9/10)^{n/2}(1/10)^{n/2}\le (2/3)^n\mu(C)$.

Hence there is no way to split $U$ into two parts of approximately equal mass as required.

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  • $\begingroup$ Thanks, Anthony. I wonder if an extra assumption will make a difference... OK, I have modified my question again. Let's assume $\mu$ to be a measure of maximal entropy.(Unique if it helps; such are all the examples I have in mind at the moment anyway.) $\endgroup$
    – Nikita Sidorov
    Commented Sep 10, 2017 at 17:15
  • $\begingroup$ I'm pretty sure the answer will still be negative, but I'll try and come up with something precise. $\endgroup$
    – Anthony Quas
    Commented Sep 10, 2017 at 17:19
  • $\begingroup$ Am I right that for a transitive sofic shift we do have $\mu(C)\asymp \alpha^n$ for all $C\in F_n$ if mu is the MME? If I am, this means your counterexample should be non-sofic. Also, for a $\beta$-shift with the Parry measure $\mu$, we do have $\mu(C)\asymp \beta^{-n}$, even when $\beta$ is transcendental. . $\endgroup$
    – Nikita Sidorov
    Commented Sep 10, 2017 at 17:28
  • $\begingroup$ I agree that a counter-example would be non-sofic. But if $\beta$ is very Louisville, I think $\mu(C)$ can be much smaller than $\beta^{-n}$, can't it? $\endgroup$
    – Anthony Quas
    Commented Sep 10, 2017 at 20:17
  • $\begingroup$ You mean for a $\beta$-shift? No, it doesn't matter if $\beta$ is Liouville. We have $|F_n|\asymp \beta^n$, and the length of each interval on $[0,1]$ is $\asymp \beta^{-n}$ as shown by R\'enyi. Then the density of the Parry measure is bounded (which is Parry's result), whence $\mu(C)\asymp \beta^{-n}$ for any $C\in F_n$ with constants depending on $\beta$ only. $\endgroup$
    – Nikita Sidorov
    Commented Sep 10, 2017 at 22:43

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