In a local martingale, does the sequence of stopping times is important? Let $(M_t)$ a local martingale. Does : 
1) There exist an increasing sequence of stopping time $(\tau_k)_k$ s.t. $\tau_k\to \infty $ a.s when $k\to \infty $ s.t. the process $(M_{\tau_k\wedge t})_t$ is a martingale 
or 
2) For any increasing sequence of stopping time $(\tau_k)_k$ s.t. $\tau_k\to \infty $ a.s. when $k\to \infty $, the process $(M_{\tau_k\wedge t})_t$ is a martingale ?
 A: Consider the local martingale $X_t$ constructed in these notes, which consists of standard Brownian motion stopped upon hitting 1 and sped up to fit all of time into $(0,1)$.  If you take $\tau_k = k$ for $k = 1, 2, 3, \dots$, then $\tau_k \uparrow \infty$, but $X_{\tau_k \wedge t}$ is the same as $X_t$ for all $k$, and is not a martingale.
A: There is some flexibility available. Suppose $(\tau_n)$ is a localizing sequence of stopping times for the lcoal martingale $M$. If $(\tau'_n)$ is a second sequence of stopping times with $\tau_n'\le\tau_n$ for all $n$, and $\lim_n\tau'_n=\infty$ a.s., then $(\tau'_n)$ localizes $M$ as well. Also, if $M$ is a local martingale such that $t\mapsto M_t$ is continuous a.s., then $\sigma_n:=\inf\{t: |M_t|>n\}$, $n=1,2,\ldots$, is a localizing sequence for $M$. The proof of this last statement is a nice exercise in one's understanding of the notions involved.
A: (2) does not make sense because it will automatically imply that $M$ is a martingale. For example, let $\tau_k=\infty$ for all $k$, then $(\tau_k)$ is an increasing sequence of stopping times such that $\tau_k\rightarrow\infty$. In this case, the stopped process $M^{\tau_k}$ is simply $M$. Hence, requiring that $M^{\tau_k}$ is a martingale is equivalent to requiring that $M$ is a martingale.
