In most literature I read (Karatzas, Shreve, or Le Gall) and also in wikipedia a local martingale is defined in a filtered probability space $(\Omega,\mathscr{F}, (\mathscr{F}_t)_t, \mathbb{P})$:
Definition. (1) A local martingale $M$ is an adapted process so that there exists an increasing sequence of stopping times $(\tau_n)_n$ with $\tau_n\rightarrow\infty$ and for every $n$: $M^{\tau_n}$ is a martingale.
Of course the requirements for the sequence of stopping times only must hold almost surely. However in my lecture notes the requirement that the sequence is increasing was not written, so I began to wonder, if this is a mistake or if we can really drop this.
Definition. (2) A local martingale $M$ is an adapted process so that there exists a sequence of stopping times $(\tau_n)_n$ with $\tau_n\rightarrow\infty$ and for every $n$: $M^{\tau_n}$ is a martingale.
It is clear that "$(1)\implies(2)$". How do I show the converse, respectively is it possible to show the converse? My initial guess was to define $\tilde{\tau}_n=\vee_{k=1}^n\tau_k$, but I failed to show that $M^{\tilde{\tau}_n}$ is a martingale. Defining $\tilde{\tau}_n=\wedge_{k=1}^n\tau_k$ would solve this problem, but then the sequence is not increasing anymore. So I start to believe that "$(2)\implies(1)$" does not hold, is there an example that demonstrates this?
Observe that if $\sigma $ and $\tau$ are stopping times then $M^{\sigma\vee\tau}+M^{\sigma\wedge\tau} = M^\sigma+M^\tau$. Thus, if $M^\sigma$, $M^\tau$, and $M^{\sigma\wedge\tau}$ are all martingales then so is $M^{\sigma\vee\tau}$.