Given a filtered space $\left(S, \mathcal{\Sigma}, \{\mathcal{\Sigma_n: n\ge 0}\}, \mathbb{P}\right)$, and a stopping time $\tau: S \to \{0, 1,2, \ldots\}$ with respect to $\{\mathcal{\Sigma_n: n\ge 0}\}$, suppose that there exist $m\in\mathbb{N}$ and $\epsilon$ such that, for every $n\ge 0$: $$\mathbb{E}\left[\mathbb{I}_{\tau \le n+m} | \mathcal{\Sigma_n}\right] > \epsilon \quad \text{a.s.}.$$ Prove that $\mathbb{E}[\tau]<\infty$.
I think one should use $E[\tau] < \infty \Longleftrightarrow \sum\limits_{n=0}^\infty \mathbb{P}(\tau \ge n)< \infty.$ However, I cannot find a tight enough bound for $\mathbb{P}(\tau \ge n)$. The best I can do is $$\mathbb{P}(\tau > n+m) = 1- \mathbb{P}(\tau \le n+m) = 1 - \mathbb{E}\left[\mathbb{I}_{\tau \le n+m}\right] = 1-\mathbb{E}\mathbb{E}\left[\mathbb{I}_{\tau \le n+m} | \mathcal{\Sigma_n}\right] \le 1- \mathbb{E}[\epsilon] =1 - \epsilon.$$ A hint is provided:
For every $k = 1, 2, 3, \ldots,$ $$\mathbb{P}(\tau > (k+1)m) = \mathbb{P}(\tau > (k+1)m \quad \text{and} \quad \tau > km).$$ Use induction to show that $\mathbb{P}(\tau > km) \le (1-\epsilon)^k.$
Since $\{\tau>km\} \in \mathcal{\Sigma_{km}}:$
$$\mathbb{P}(\tau \le (k+1)m, \tau >km) =\mathbb{E}[I_{\tau \le (k+1)m, \tau >km}]= \mathbb{E}\mathbb{E}[I_{\tau \le (k+1)m, \tau >km}|\mathcal{\Sigma_{km}}] = \mathbb{E}\mathbb{E}[I_{\tau \le (k+1)m} I_{\tau >km}|\mathcal{\Sigma_{km}}] = \mathbb{E}\left[I_{\tau >km}\mathbb{E}[I_{\tau \le (k+1)m} |\mathcal{\Sigma_{km}}]\right] > \mathbb{E}\left[I_{\tau >km}\epsilon\right] = \mathbb{P}(\tau>km)\epsilon.$$ On the other hand $$ \mathbb{P}(\tau \le (k+1)m, \tau >km) = \mathbb{P}(km < \tau \le (k+1)m ) = \mathbb{P}(\tau>km) - \mathbb{P}(\tau>(k+1)m), $$ hence $$ \mathbb{P}(\tau>km) - \mathbb{P}(\tau>(k+1)m) > \mathbb{P}(\tau>km)\epsilon, $$ or $$ \mathbb{P}(\tau>(k+1)m) \le (1-\epsilon)\mathbb{P}(\tau>km). $$ Induction gives the desired result.