Suppose that $T$ is a stopping time such that for some positive integer $\alpha$ and some $\epsilon>0$ we have for every $n$: $$ P(T\leq n+\alpha|\mathcal{F}_n)>\epsilon \text{ a.s.} $$
I have shown that for some $\epsilon\in(0,1)$ and some $\alpha \in \mathbb{N}$ that $P(T>k\alpha)\leq (1-\epsilon)^k$. I would now like to show that this implies $E(T)<\infty$.
If I let $A_k=\{ T>k\alpha \}$, then $\sum\limits_{k=1}^\infty P(A_k) < \infty.$ So Borel-Cantelli tells us that $P(A_k \text{ i.o.})=0$. I was hoping to use this to show that $T$ must be bounded above a.s., but I am not quite sure how to do that.
Thank you for any help.
Hint: Note that $$E[T] = E\bigg[ \sum_{n=0}^{\infty} 1_{\{T>n\}} \bigg] = \sum_{n=0}^{\infty} P(T>n) \leq \sum_{k=0}^{\infty} \alpha P(T>k\alpha)$$ This is a standard trick, and doesn't depend on the fact that $T$ is a stopping time.