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.
 A: 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 $1$. 
