Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 108 (Topics in Ergodic Theory)

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.

The statement $D-lim{a}_n=a$ means convergence in density of a sequence: for every $\epsilon >0$, the proportion of $n\le N$ with $|a_n-a|\ge \epsilon$ tends to zero. The statement $C-lim{a}_n=a$ means Cesaro convergence of a sequence, $N^{-1}{\sum }_{n=1}^N{a}_n\to a$.

Suppose $|a_n-a|\le M$. Splitting into the indices where $|a_n-a|<\epsilon$ and its complement givesThus density convergence implies Cesaro convergence of the absolute deviations to zero. Conversely, the Markov inequality givesproving the reverse implication.

The product characterization of weak mixing says that if $T$ is weakly mixing, then $T\times S$ is ergodic exactly when $S$ is ergodic; in particular, ergodicity of $T\times T$ characterizes weak mixing.

Suppose $U_{T}f=\lambda f$. Unitarity of the Koopman operator gives $|\lambda |=1$. Weak mixing implies ergodicity, so $|f|$ is almost everywhere constant. On the product,is invariant under $T\times T$. Since the square is ergodic, $F$ is constant, which forces $f$ to be constant almost everywhere. Thus there are no nonconstant Koopman eigenfunctions.

Finally use the density-one correlation characterization of a weakly mixing measure-preserving transformation. For each positive-measure pair $(A,B)$, the integers $n$ for which $\mu (T^{-n}A\cap B)>0$ form a density-one set after discarding finitely many terms; the corresponding sets for $(A,B)$ and $(A,C)$ therefore intersect, proving simultaneous hitting. Conversely, the simultaneous-hitting property is precisely the simultaneous hitting characterization of weak mixing, so it implies weak mixing.

For a finite measurable partition $\xi$,Its conditional entropy of finite measurable partitions relative to $\eta$ isThe concavity of $-t\log t$, equivalently conditioning reduces entropy, gives

The atoms of $T^{-1}\xi$ are $T^{-1}A$ with the same measures as the atoms $A$ of $\xi$. Hence $\boxed{H_{\mu }(T^{-1}\xi )=H_{\mu }(\xi )}$.

SetThe chain rule and invariance give $a_{N+M}\le a_N+a_M$, so $(a_N)$ is a subadditive sequence. ThereforeThe Kolmogorov-Sinai entropy is $h_{\mu }(T)={\sup }_{\xi }{h}_{\mu }(T,\xi )$ over finite partitions.

Taking $F=\{0,\dots ,N-1\}$ immediately shows that the infimum over arbitrary finite $F$ is at most $h_{\mu }(T,\xi )$. For the reverse inequality, apply Shearer's inequality to translates of a fixed finite $F$ inside a long interval. Every interior coordinate is covered $|F|$ times, while only $O(\max F)$ boundary coordinates are lost. Subadditivity bounds the boundary contribution; division by the interval length and passage to the limit giveTaking the infimum proves

For a finite partition $\xi$, putThe system is K-mixing when every measurable $A$ becomes uniformly asymptotically independent of this remote future:Taking $\xi =\{B,X\setminus B\}$ makes $T^{-n}B\in {\mathcal{F}}_n(\xi )$, soThus K-mixing implies mixing.

The tail sigma-algebra of a measurable partition isBy the reverse martingale convergence theorem,in $L^1$. The uniform independence in the definition of K-mixing is equivalent to the limit being the constant $\mu (A)$ for every $A$. This holds exactly when every $\mathcal{T}(\xi )$-measurable set has measure zero or one. Hence the system is K-mixing if and only if every finite partition has trivial tail sigma-algebra.

Let $\mathcal{P}=\{A:h_{\mu }(T,{\xi }_A)=0\}$. Complements preserve the binary partition. If $A,B\in \mathcal{P}$, then ${\xi }_{A\cup B}$ is coarser than ${\xi }_A\vee {\xi }_B$, so subadditivity gives zero entropy rate. Thus $\mathcal{P}$ is an algebra. For $A={\bigcup }_{j\ge 1}{A}_j$, let $A^{(m)}={\bigcup }_{j\le m}{A}_j$. The entropy metric continuity boundtends to zero because $\mu (A△A^{(m)})\to 0$. Hence $A\in \mathcal{P}$, proving that $\mathcal{P}$ is a sigma-algebra: the Pinsker sigma-algebra.

If $A$ belongs modulo null sets to $\mathcal{T}(\xi )$ for a finite $\xi$, remote-future approximations make the entropy rate of ${\xi }_A$ zero. Conversely, if $h_{\mu }(T,{\xi }_A)=0$, the conditional-entropy formula for entropy rate givesThus $A$ is measurable modulo null sets from its strict future. Iterating this fact makes it measurable from every remote future, so $A\in \mathcal{T}({\xi }_A)$ modulo null sets. ThereforeThis is the Tail characterization of the Pinsker sigma-algebra.