# Ergodic transformation on a atomless measure space

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.

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 $$1$$.