I've been told the following result:
Prop. Let $X$ be a markov chain on $S$ (not necessarily finite) and transistion matrix $P=(p_{i,j})_{i,j\in S}$. Then if for some $i\in S$ we have that for all $j\in S$, $p^{(n)}_{i,j}\rightarrow \pi_j$ as $n\rightarrow \infty$, then $(\pi_j)_{j\in S}$ in an invariant measure (i.e. $\pi P=P$). Moreover if $S$ is finite then it's not only a measure but a probability distribution.
Well, if $S$ is finite then everything is easy to me, but i cannot understand the case in which $S$ is not finite. Indeed the argument that i came up with is the following:
$$\pi_j=\displaystyle \lim_{n\rightarrow \infty} p_{i,j}^{(n)}=\lim_{n\rightarrow \infty} \sum_{k\in S}p_{i,k}^{(n-1)}p_{k,j}=\sum_{k\in S} \lim_{n\rightarrow \infty} p_{i,k}^{(n-1)}p_{k,j}=\sum_{k\in S} \pi_{k}p_{k,j} $$
The problem here is the third equality; if $S$ in not finite then i'm exchanging the limit with a series and i don't know how to justify it. I also think that in this setting monotone/dominated convergence theorems cannot help beacause i'm not able to see that series as integral w.r.t some measure on which i can apply those theorems.
So is this result correct but require a different proof? Or is the result not correct for $S$ countable? Or is the limit exchange i pointed out correct and it's just me that i can't understand it?
The result is correct. Let $S = \{1, 2, 3, ...\}$ be countably infinite.
1) First prove that $$ \sum_{j \in S} \pi_j \leq 1$$
2) Next prove that $\pi \geq \pi P$ (via the hint in my comment above), equivalently:
$$ \pi_j \geq \sum_{k \in S} \pi_k P_{kj} \quad \forall j \in S \quad (Eq. 1)$$
3) Suppose (Eq. 1) holds with strict inequality for at least one $j\in S$. Summing over all $j \in S$ gives $$ \sum_{j \in S} \pi_j > \sum_{j \in S} \sum_{k \in S} \pi_k P_{kj}$$ where we have used the fact that the sums are finite to eliminate the $\infty \geq \infty$ case. Switch the sums of the nonnegative terms (Fubini-Tonelli) to reach a contradiction.