Let $(X, B, \mu, T)$ be a measure preserving system. For any set $B$ of positive measure, $E = \{n \in \Bbb N |\; \mu(B \; \cap \;T^{-n}B) > 0\}$ is syndetic.
This exercise comes from Einseidler and Ward. The exercise before is the "uniform" mean ergodic theorem which is proved basically the same way as the mean ergodic theorem, and they say it should be used in the proof. Can someone help me get started? Thanks in advance!
On
There is also an elementary proof.
Assume for contradiction that there is a dynamical system (X,B,μ,Τ) and A ∈ B with μ(Α)>0, such that the set E isn't syndetic.
Then its complement, say R={n∈N:μ(Α∩Τ^(-n)A)=0}, contains arbitrarily large intervals, i.e. sets of the form J={m,m+1,...,m+M}, m,M ∈ N, 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. (This is true because |(n_j-n_i)-n_j|<l(J_j)/2)
However, this implies that μ(A∩Τ^(-(n_j-n_i))A)=0, for all i<j, i,j ∈ N, which is a contradiction. Can you see why?
Hint: If the set was not syndetic, then the sequence $$ \frac1n\sum_{k=0}^{n-1}\mu(B\cap T^{-k}B) $$ would have zero as an accumulation point (take larger and larger gaps). But $$ \lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\mu(B\cap T^{-k}B)=\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\int_B(\chi_B\circ T^k)\,d\mu=\int_B\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}(\chi_B\circ T^k)\,d\mu, $$ using the dominated convergence theorem.