# Can we show that a local martingale is closed when it is stopped at one of the localising stopping times?

Let $$M$$ be a $$(\mathcal F_{t}^{B})_{t\geq 0}-$$ local martingale starting in $$0$$ where $$(\mathcal F_{t}^{B})_{t\geq 0}$$ denotes the completed filtration of the Brownian motion. Now consider the reducing sequence $$(\tau_{n})_{n}$$ so that $$M^{\tau_{n}}$$ is a martingale. It is then stated that I may assume that $$(M_{\tau_{n} \land t})_{t\geq 0}$$ is a closed martingale (considering either $$\tau_{n}$$ or $$\tau_{n}\land n$$), $$n \in \mathbb N$$. I am not sure why closedness of the martingale $$M^{\tau_{n}}$$ or $$M^{\tau_{n}\land n}$$ holds. Could anyone help me to explain why this is the case?

Martingale $$(M_{t})_{t\geq 0}$$ is closed iff $$M_{\infty}\in \mathcal{F}_{\infty}$$ and integrable exists s.t. $$E[M_{\infty}\lvert \mathcal{F}_{t}]=M_{t}$$

• For what purpose are you making that assumption? Is there a bigger lemma of theorem that's being proved with the help of the assumption? Also, in which book is this argument being made / paragraph found? Mar 7, 2021 at 18:11
• Hint: If $N=\{N_t,t\ge 0\}$ is a (right continuous) martingale, then $N^K=\{N_{t\wedge K},t\ge 0\}$ is a right closed martingale for each $K>0$, since $N_{t\wedge K}=\mathsf{E}[N_K|\mathscr{F}_t]$. Mar 8, 2021 at 11:53

First, we have that $$\tau_n \wedge n$$ is a bounded stopping time so $$N_t := M_{(\tau_n \wedge n) \wedge t}$$ is a closed martingale because $$N_t = \mathbb{E}[N_n | \mathcal F_t]$$ for all $$t$$, i.e. $$N_\infty = N_n = M_{\tau_n \wedge n}$$.
The reason that we may assume $$(M_{\tau_n \wedge t})$$ is closed is because if it isn't, we can replace $$\tau_n$$ with $$\tilde{\tau}_n := \tau_n \wedge n$$. Then $$\tilde{\tau}_n$$ is still a localizing sequence, and we showed above that $$(M_{\tilde{\tau}_n \wedge t})$$ is a closed martingale. Since we can always replace a localizing sequence with one that makes $$(M_{\tau_n \wedge t})$$ closed, we might as well just assume that the original localizing sequence does too.