= 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 $\{T^nx\}\subseteq T^{-1}\{T^{n+1}x\}$ give
$$
\mu(\{T^{n+1}x\})
=\mu(T^{-1}\{T^{n+1}x\})
\geq\mu(\{T^nx\}).
$$
If 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,
$$
\mu(T^{-1}Y\setminus Y)=\mu(T^{-1}Y)-\mu(Y)=0.
$$
Thus $Y$ belongs to the <invariant sigma-algebra> modulo null sets. It has positive measure, so the fact that $T$ is an <ergodic transformation> forces
$$
\boxed{\mu(Y)=1}.
$$
This is the <ergodic invariant probability on a countable state space has finite cyclic support>.
Back to article page
