Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 108 / 1

Assume first that $\mu (A\cap T^{-n}A)=0$ for every $n\ge 1$. For $n>m$ the measure-preserving transformation property givesThus the sets $T^{-n}A$, $n\ge 0$, are pairwise disjoint and all have measure $\mu (A)>0$, impossible in a probability space. Hence

The Poincare recurrence theorem states that almost every $x\in A$ returns to $A$ infinitely often. LetThe sets $T^{-n}E$ are pairwise disjoint: if $x$ belonged to the $m$th and $n$th inverse images with $m<n$, then $T^{m}x\in E$ would return to $E\subseteq A$ after $n-m$ steps. Invariance and finiteness therefore give $\mu (E)=0$. A point of $A$ with only finitely many returns belongs, after its last return, to some $T^{-k}E$. The countable union of these null sets is null, proving the theorem.

Now let $S\subseteq \mathbb{Z}$ have $d^{\ast }(S)>0$. Put $x=1_S\in \{0,1{\}}^{\mathbb{Z}}$ and let $\sigma$ be the bilateral shift. Choose intervals $I_j$ with $|I_j|\to \infty$ and $|S\cap I_j|/|I_j|\to d^{\ast }(S)$. A weak-star limit ofexists on the compact shift space. Boundary terms show that the limit $\mu$ is $\sigma$-invariant. For the cylinder $A=\{y:y_0=1\}$,The stated polynomial recurrence theorem supplies $n>0$ with $\mu (A\cap {\sigma }^{-|P(n)|}A)>0$. Hence that cylinder intersection meets the orbit closure of $x$; because it is open, some shift ${\sigma }^{m}x$ lies in it. Therefore $m,m+|P(n)|\in S$. Orient these two integers according to the sign of $P(n)$ to obtainThis is the Furstenberg correspondence principle.