Irreducible aperiodic Markov chain $P$ with invariant distribution implies $p_{ij}^{(n)} \rightarrow \pi_j$

Question

I am reading Norris' book on Markov chains, and there is a theorem that sais :

Let $P$ be irreducible and aperiodic, and suppose that $P$ has an invariant distribution $\pi$. Let $\lambda$ be any distribution. Suppose that $(X_n)_{n\ge 0}$ is $Markov(\lambda,P)$. Then
$P(X_n=j)\rightarrow \pi_j$ as $n \rightarrow \infty$ for all $j$.
In particular, $p_{ij}^{(n)} \rightarrow \pi_j$ as $n \rightarrow\infty$ for all $i,j$.

I more or less understand the first statement, but I don't see why it implies the second one (the one after the "In particular..."). It should be pretty straight forward since he just claims that this follows, but I don't get why exactly...

Answer 1

When he says $P(X_n=j) \to \pi_j$ he actually means $P(X_n=j|X_0=i) \to \pi_j$ for every $i$. (The limit is independent of $i$). Since $p_{ij}^{(n)}=P(X_n=j|X_0=i)$ we get $p_{ij}^{(n)} \to \pi_j$.