# Definition of predictable stopping time

I am looking for a counter-example to $$\tau \text{ is a predictable stopping time} \iff \forall t ~~~ \lbrace \tau \leq t \rbrace \in \bigcap_{l

I have already shown that the only-if is true. I found the (almost) same question here: Definition of predictable process
But I can't understand the answer, I don't see why if we have $$X_t = X_{t-}$$ a.e. then $$X_t \in \mathcal{F_{t-}}$$
As I see it, what we're trying to do here is describe the elements of a sigma-algebra/filtration, and that doesn't depend on any measurement property placed on this measurable space.

Because the filtration is complete, any almost impossible event is in $$\mathcal{F}_{t-}$$ including any event contained in the event $$X_t \neq X_{t-}$$ since $$X_t = X_{t-}$$ a.s. It then follows that $$X_t \in \mathcal{F}_{t-}$$ from gluing together the part where it is equal to $$X_{t-}$$ and the part where it isn't but has to be measurable anyway because it has probability 0.
Regarding why the equivalence does not hold: note that your definition only means that one can predict the events about time $$t$$ at time $$t-$$. Predictability states essentially that random variables that are measurable at time $$t+$$ are measurable at time $$t$$ where random variables are considered measurable at time $$t+$$ if they are measurable at every time $$t'$$ strictly after $$t$$. The key here is that $$t-$$ and $$t+$$ are not real times themselves like $$t$$ but only limits so that the relation between $$t$$ and $$t-$$ can be different from the relation between $$t+$$ and $$t$$.
Do you mean $$\sigma(\cup_{l rather than $$\cap_{l? The latter is $$\mathcal F_{0+}=\mathcal F_0$$ (under the usual conditions).
Here's a counter-example of the type (I think) you're seeking. Let $$(\mathcal F_t)_{t\ge 0}$$ be the natural filtration (completed) of a unit-rate Poisson process $$N=(N_t)_{t\ge 0}$$. Let $$\tau$$ be the first jump time of $$N$$. Then $$\tau$$ is a totally inaccessible stopping time, so not predictable, But $$P(\tau=t)=0$$ for each $$t\ge 0$$, so $$\{\tau\le t\}$$ differs from $$\{\tau by a null set. And $$\{\tau so $$\{\tau\le t\}\in\mathcal F_{t-}$$ as well.