Different defintion of local martingales

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?

• Does Definition (2) require the filtration to be right continuous? Commented Jun 18, 2020 at 15:33
• @user6247850 thank you for the comment. In the lecture notes it is not written explicitly but in general I think we assume our filtrations to satisfy the usual conditions, so yes, the filtration is right continuous. Do you think that at some point in the proof we need right continuity of the filtration? Commented Jun 18, 2020 at 16:57
• I had a proof that used it, but the answer already posted is better. I was going to use $\tilde \tau_n := \inf_{k \ge n} \tau_k$, but that's not necessarily a stopping time if the filtration isn't right continuous. Commented Jun 18, 2020 at 17:35

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 $$M^{\sigma\vee\tau}$$.