$X_n$ is an irreducible and aperiodic Markov chain with transition matrix $P$. There is a function $f$ s.t. $Pf(x)\leq\alpha f(x)$ for $x\notin B$, $B$ a finite set and $\alpha<1$. $f(x)>M$ on $x\notin B$. Show $$E_x(\tau_B)<\infty$$
I can show that $Y_n:= \alpha^{-n}f(X_n)$ is a super-martingale off of $B$ and let $\tau = \tau_B$. So by optional stopping:
$$\begin{align} f(x)\geq E_x[Y_{n\wedge \tau}]&= E_x[f(X_{\tau\wedge n}\frac{1}{\alpha^{\tau\wedge n}})] \\ \implies f(x) &\geq \liminf_{n\to \infty} E_x[f(X_{\tau\wedge n}\frac{1}{\alpha^{\tau\wedge n}})] \\ &\geq E_x[\liminf_{n\to \infty} f(X_{\tau\wedge n}\frac{1}{\alpha^{\tau\wedge n}})] \text{ By Fatou's lemma} \\ &\geq E_x M(1/\alpha)^\infty.1_{\tau=\infty} + E_x (1/\alpha)^\tau f(X_\tau) .1_{\tau<\infty} \end{align} $$
Since $\alpha<1$, I can see that $P(\tau = \infty)= 0$ but not sure how to go from $\infty>f(x)>E_x [(1/\alpha)^\tau f(X_\tau) .1_{\tau<\infty}]$ to saying $E_x(\tau)<\infty$?