Let $X_t$ be a local martingale. Let $\tau_n$ be a localizing sequence.
We then know that $$E[X_{t\land \tau_n}|\mathcal F_s]=X_{s\land\tau_n}$$ $1)$ But $X_{s\land\tau_n}$ is measurable w.r.t wich sigma algebra? is it $\mathcal F_s$ measurable?
$2)$ Moreover, if i have $E[e^{X_{t\land \tau_n} + t\land \tau_n}|\mathcal F_s]=1$ can I bring on the other side $t\land \tau_n$ obtaining: $E[e^{X_{t\land \tau_n}}|\mathcal F_s]=e^{- t\land \tau_n}$? I think that it is not possible because $\tau_n$ is not $\mathcal F_s$ measurable, but then is measurable w.r.t. to what?
Yes. It is measurable w.r.t $\mathcal F_{s \wedge \tau_n}$ and $F_{s \wedge \tau_n} \subset \mathcal F_s$.
$e^{t\wedge \tau_n}$ is measurable w.r.t. $\mathcal F_t$. So it is legitimate to bring it to RHS only when you know that $t \leq s$.