If $I$ is countable, $\tau$ is a stopping time iff $\forall t\in I, (\tau=t)\in \mathcal F_t$ 
Let $(\Omega, \mathcal F,P)$ be a probability space, $I$ a countable subset of $\mathbb R$ and $(\mathcal F_i)_{i\in I}$ a filtration.
Let $\tau$ be a random variable with values in $I\cup \{\infty\}$. Prove that $\tau$ is a stopping time if and only if $\forall t\in I, (\tau=t)\in \mathcal F_t$

I've found the implication $\tau$ stopping time $\implies \forall t\in I, (\tau=t)\in \mathcal F_t$ surprisingly tedious to prove. Give some $t\in I$, I considered two cases, whether $t$ can be approached from below by elements of $I$ or not. 


*

*If it can, $$\displaystyle (\tau = t)= (\tau \leq t) \cap \bigcap_{\substack{a\in I\\a<t}} (\tau \leq a)^c \in \mathcal F_t$$
(note that countability of $I$ comes into play).

*It it cannot, there is some $\epsilon >0$ such that $(t-\epsilon,t)\cap I =\emptyset$. If $(-\infty,t)\cap I =\emptyset$, we're done. Otherwise, I consider $u=\sup \{u<t, u\in I\}$. Then I checked two cases again:


*

*If $u\in I$, $(\tau = t) = (\tau \leq t) \cap (\tau > u) \in \mathcal F_t$

*If $u\notin I$, $$\displaystyle (\tau = t)= (\tau \leq t) \cap \bigcap_{\substack{a\in I\\a<u}} (\tau \leq a)^c \in \mathcal F_t$$

This proof is quite dull and lengthy, do you know a more concise one ?


In the second case $\{\tau=t\}=\{\tau\le t\}\setminus\{\tau\le t-\epsilon\}$.
