I am currently reading Kakutani–Rokhlin lemma and faced a problem which is given below :---

Let $(X,\mathscr B,\mu,T)$ be an invertible measure preserving system such that $\mu(\{x\})=0,\forall x\in X$. Suppose $T$ is ergodic we have to show, $$\mu\bigg(\bigcup_{k\in \Bbb Z,k\not=0}\{x\in X:T^k(x)=x\}\bigg)=0.$$

I don't how do I start with this problem. Any help will be appreciated.


1 Answer 1


I believe the correct condition instead of "$\forall x\in X: \mu(\{x\})=0$" is that $\mu$ is atom-free.

You want to show that for every invertible measurable map $T$ on a measurable space $(X,\mathscr{B})$, the set of $T$-periodic points has measure zero with respect to every atom-free $T$-ergodic measure $\mu$.

If this is not the case, there must be an integer $n>0$ such that the set $P_n$ of points having period $n$ has positive measure. By ergodicity, the measure of $P_n$ must in fact be $1$. Since $\mu$ is atom-free, there must be a measurable set $A\subsetneq P_n$ such that $0<\mu(A)<1/n$. The set $E:=A\cup T^{-1}(A)\cup\cdots\cup T^{-(n-1)}(A)$ will then be an invariant set with $0<\mu(E)<1$, contradicting ergodicity. Q.E.D.

If you only require that there are no singleton atoms, then the conclusion does not hold. For example, take $X:=[0,1]$ with $\mathscr{B}$ being the $\sigma$-algebra generated by singletons, and for each $A\in\mathscr{B}$, let \begin{align} \mu(A) &:= \begin{cases} 0 & \text{if $A$ is countable,} \\ 1 & \text{if $X\setminus A$ is countable.} \end{cases} \end{align} The identity map $T:x\mapsto x$ is measurable, measure-preserving and ergodic. It has no singleton atoms, but it is not non-atomic because the complement of any countable set is an atom. On the other hand, every point is periodic under $T$ with period $1$.


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