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. $E[M_{\infty}\lvert \mathcal{F}_{t}]=M_{t}$
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.