Suppose we have a Markov chain $(X_n)_{n \geq 0}$ with state space $S$. Suppose that $(\pi_i)_{i \in S}$ is a limit distribution. Then is $(\pi_i)_{i \in S}$ an invariant distribution ?
I know the answer is yes if $S$ is finite, but what about the infinite case ?
On
If we define the limit distribution $\nu$ by $d(\mu P^n, \nu) \to 0$ with some distance $d$ in the distribution space and the initial distribution $\mu$, then what we want to prove is $d(\nu, \nu P) = 0$.
This is true if $P$ reduces the distance, i.e. if $d(\mu_1 P , \mu_2 P) \le d(\mu_1, \mu_2)$, because
$$d(\nu,\nu P) \le d(\nu, \mu P^n) + d(\mu P^n, \nu P) \le d(\nu, \mu P^n) + d(\mu P^{n-1}, \nu) \to 0$$
For the total variation distance $d(\mu_1 P , \mu_2 P) \le d(\mu_1, \mu_2)$ is true quite generally, see lemma 3 in this. I am not sure if this is generally true for other kinds of distance
Consider a Markov chain on the integers where it moves right with probability 1. The limit distribution is all zeros, but this doesn't qualify as an invariant distribution because it needs at least one positive entry.