I am trying to understand a proof of the following claim:
If $P$ is irreducible and $\pi$ any stationary distribution, then $\pi_i = \frac{1}{m_i}$, where $m_i$ is the mean return time to state $i$.
Here's the proof:
Suppose $\pi$ is stationary for $P$, and let $X$ be a Markov chain with initial distribution $\pi$ and transition matrix $P$. Let $V_i(n)$ be the number of visits to state $i$ before time $n$. Then, by stationarity, $\mathbb{P}(X_n=i) = \pi_i$ for all $n$, and $$ \frac{\mathbb{E}[V_i(n)]}{n} = \frac{1}{n} \sum_{r=0}^{n-1}\mathbb{E}[1\{X_r=i\}] = \frac{1}{n} \sum_{r=0}^{n-1}\mathbb{P}(X_r=i) = \pi_i \tag{$1$} \label{ref1} $$ From the ergodic theorem, for any $\epsilon$, $$ \mathbb{P}\left(\left|\frac{V_i(n)}{n} - \frac{1}{m_i} \right| > \epsilon\right) < \epsilon \tag{$2$} \label{ref2} $$ for large enough $n$ (since almost sure convergence implies convergence in probability).
But since $\frac{V_i(n)}{n}$ is bounded between $0$ and $1$, it follows from \eqref{ref2} that $$ \boxed{\frac{\mathbb{E}[V_i(n)]}{n} \to \frac{1}{m_i} \quad \text{as} \quad n \to \infty} $$ Comparing to \eqref{ref1}, we obtain $\pi_i = \frac{1}{m_i}$.
Why does the boxed statement follow from \eqref{ref2}?