Suppose $a(t)$ is an irreducible aperiodic discrete time Markov chain with $n$ states, transition matrix $A = (p_{ij})_{0 \leq i,j < n}$ and initial state $0$. Suppose, $\alpha \in (0;1)$. How can I find $\lim_{t \to \infty}P(|\{0 \leq \tau \leq t| a(\tau) = 0\}| \leq \alpha t)^{\frac{1}{t}}$?
Suppose $\pi(x)$ is the stationary distribution of this Markov chain. Then $\frac{|\{0 \leq \tau \leq t| a(\tau) = 0\}|}{t}$ converges almost surely to $\pi(0)$, because of the Theorem C.1 from "Markov Chains and Mixing Times" by David A. Levin, which states:
Suppose $X$ is the set of all states of an irreducible aperiodic finite-state Markov chain $a(t)$, $\pi$ is the stationary distribution of $a(t)$ and $f: X \to \mathbb{R}$ is some arbitrary function. Then $\frac{\sum_{s=0}^t f(a(t))}{t}$ converges almost surely to $\sum_{x \in X}f(x)\pi(x)$.
From that we can conclude, that if $\pi(0) \leq \alpha$, then $\lim_{t \to \infty}P(|\{0 \leq \tau \leq t| a(\tau) = 0\}| \leq \alpha t)^{\frac{1}{t}} = 1$. However, this argument does not work for $\alpha < \pi(0)$. How can it be done here?