Need help in proving a set has bounded gaps.

Question

Let $(X,B,\mu)$ be a probability space and $T\colon X\rightarrow X$ is measure preserving. Let $A\in B$ such that $\mu(A)>0$. Then I am asked to prove the following claims:

1. The set of positive integers $n$ such that $\mu(T^{-n}(A)\cap A)>0$ is infinite.
2. The above set of positive integers $n$ has bounded gaps (i.e. if the set consists of $n_1<n_2<\dots$ then there is a constant $K$ such that $n_{j+1}-n_j\leqslant K$ for each $j$).

I have done the first part, need help in the second part. Thank you so much.

Answer 1

For sake of completeness, I also give the answer for the first claim.

1. Let $E:=A\cap \bigcap_{j\geqslant 1}T^{-j}(A^c)$. The sets $T^{-p}E, p\geqslant 1$ are pairwise disjoint hence $\mu(E)=0$. This gives that $\mu\left(A\cap\bigcup_{j\geqslant 1}T^{-j}(A)\right)=\mu(A)>0$. In particular, this proves that there exists $n$ such that $\mu(A\cap T^{-n}(A))>0$. Let $n:=\sup\{k, \mu(A\cap T^{—k}(A))>0\}$ and assume it is bounded, say by the integer $M$. The same reasoning with $T$ replaced by $T^{M+1}$ yields a contradiction.

2. Let $\{n_k\}$ be the sequence given by the first point, and let $I:=\{0\}\cup\{n_k,k\in\Bbb N\}$. We define $$S:=A\cap \bigcap_{j\notin I}T^{-j}(A^c)\cap \bigcup_{k=1}^{+\infty}T^{-n_k}(A).$$

   • Claim: $\mu(S)=\mu(A)$.
     Indeed, $\mu(S)=\mu\left(A\cap \bigcup_{k=1}^{+\infty}T^{-n_k}(A)\right)-\mu\left(A\cap \bigcup_{k=1}^{+\infty}T^{-n_k}(A)\cap\bigcup_{i\notin I}T^{-j}(A)\right)$; the first term is $\mu\left(A\cap \bigcup_{j=1}^{+\infty}T^{-j}(A)\right)=\mu(A)$ and the second $0$.

   • Fix an integer $k$. Then the family $T^{-j}S,j\in\{0,\dots,n_{k+1}-n_k-1\}$ consists of pairwise disjoint sets. Hence $\mu(S)(n_{k+1}-n_k)\leqslant 1$.

   • Combining the last two points, we get that for each $k$, $n_{k+1}-n_k\leqslant \frac 1{\mu(A)}$.

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Remarks:

1. The bound of $n_{k+1}-n_k$ is explicit in term of the measure of $A$.
2. It's sharp, which can be seen taking $X:=\{a,b\}$, $\mu(a)=\mu(b)=1/2$ and $T(a)=b$, $T(b)=a$.
3. An it is quite intuitive: if $A$ is small in measure, we have to wait more time once $A\cap T^{-n}A$ has a positive measure.

Answer 2

The answer that Davide suggests for the second part is not correct since it implies that the gaps are bounded by a constant which depends on the measure μ(Α) alone. However, this is not the case, as one can see by considering the rotation Rx = x + 1/k for any k ∈ N, defined on the unit circle (or [0,1]/~ where we identify 0 with 1) with the Lebesgue measure and choosing A=[0,1/4).

The result is an immediate consequence of Von Neumann's mean ergodic theorem.

Alternatively, assume there is a dynamical system (X,B,μ,Τ) and A ∈ B with μ(Α)>0, such that the set you mention doesn't have bounded gaps. Then its complement, R={n∈N:μ(Α∩Τ^(-n)A)=0}, contains arbitrarily large intervals, J={m,m+1,...,m+M], with l(i):=M (its length) and c(J):=m+[M/2] (its center). Let n_0 ∈ N. By assumption there exists a J_1 ⊂ R, with l(J_1)>=n_0 and c(J_1)=n_1, for some n_1 ∈ N and then there exists a J_2 ⊂ R, with l(J_2)>=2(n_0+n_1) and c(J_2)=n_2, some n_2 ∈ N. In the same manner there is a J_m ⊂ R, with l(J_m+1)>=2(n_0+n_1+...+n_m) and c(J_m+1)=n_m+1, some n_m+1 ∈ N, for any m ∈ N. Now notice that for i<j, i,j ∈ N you have that n_j-n_i ∈ J_j.

You should be able to conclude.