 Solution

Because $X=\mathbb{N}$ is a countable set and $\mu (X)=1$, some $x\in X$ satisfies $\mu (\{x\})>0$. Invariance and the inclusion $\{T^{n}x\}\subseteq T^{-1}\{T^{n+1}x\}$ giveIf the forward orbit contained infinitely many distinct points, their disjoint singletons would each have measure at least $\mu (\{x\})>0$, contradicting that $\mu$ is a probability measure. The orbit therefore eventually enters a finite periodic orbit $Y$.

Now $T(Y)=Y$, so $Y\subseteq T^{-1}Y$. Since $\mu$ is invariant,Thus $Y$ belongs to the invariant sigma-algebra modulo null sets. It has positive measure, so the fact that $T$ is an ergodic transformation forcesThis is the ergodic invariant probability on a countable state space has finite cyclic support.