Uniformly integrable martingale problem: typo (wrong filtration)? I am working on the following problem: Let $(X_n,{\cal F}_n)_{n\in\mathbb N}$ be a uniformly integrable martingale, let $M\in{\mathbb R}_{>0}$ and let $\tau=\inf\{n\in\mathbb N:|X_n|\ge M\}.$ (Here, $(X_n,{\cal F}_n)$ being a martingale means $({\cal F}_n)_{n\in\mathbb N}$ is a filtration and $(X_n)_{n\in\mathbb N}$ is a martingale with respect to $({\cal F}_n).$) Prove that $(X_{n\land\tau},{\cal F}_{n\land\tau})$ is a uniformly integrable martingale.
Currently, I am more concerned with the filtration $({\cal F}_{n\land\tau})$ than with showing uniform integrability. So far, I have shown $(X_{n\land\tau},{\cal F}_n)$ is a martingale. However, I am concerned that $({\cal F}_{n\land\tau})$ might not be a filtration: For all natural numbers $n,k,$ if $k<n$ then $\{n\land\tau\le k\}=\{\tau\le k\},$ and if $k\ge n$ then $\{n\land\tau\le k\}=\Omega,$ so
$${\cal F}_{n\land\tau}=\{A\in{\cal F}:(\forall k)~(A\cap\{n\land\tau\le k\}\in{\cal F}_k)\}=\{A\in{\cal F}:(\forall k<n)~(A\cap\{\tau\le k\}\in{\cal F}_k)\};$$
thus $\forall n\quad{\cal F}_{(n+1)\land\tau}\subseteq{\cal F}_{n\land\tau},$ which is the opposite inclusion. Thus, if $({\cal F}_{n\land\tau})$ is a filtration, then $\forall n\quad{\cal F}_{n\land\tau}={\cal F}_{(n+1)\land\tau},$ which seems unlikely to be always true.
So, is this a typo? I tried finding a counterexample where $\exists n\quad{\cal F}_{n\land\tau}\ne{\cal F}_{(n+1)\land\tau},$ but I had trouble. Did the authors mean to use e.g. $({\cal F_n})_{n\in\mathbb N}$ or ${\cal F}^{X_{\bullet\land\tau}}$ (natural filtration for $X_{\bullet\land\tau}$)?
 A: No, it's not a typo; $(\mathcal{F}_{n \wedge \tau})_{n \in \mathbb{N}}$ is a filtration.  It holds for any two stopping times $S \leq T$ that $\mathcal{F}_S \subseteq \mathcal{F}_T$ (see e.g. this question). Applying this for $S= \tau \wedge n$ and $T=\tau \wedge (n+1)$ gives immediately that $(\mathcal{F}_{n \wedge \tau})_{n \in \mathbb{N}}$ is a filtration.
The second "=" in your display is wrong. Since $\{n \wedge \tau \leq k\} = \Omega$ for $k \geq n$ and $\{n \wedge \tau \leq k\} = \{\tau \leq k\}$ for $k<n$, we have
\begin{align*} \mathcal{F}_{n \wedge \tau} &= \{A \in \mathcal{F}; \forall k < n:A \cap \{\tau \leq k\} \in \mathcal{F}_k, \forall k \geq n: A \cap \Omega \in \mathcal{F}_k\} \\ &= \{A \in \mathcal{F}_{\color{red}{n}}; \forall k<n :A \cap \{\tau \leq k\} \in \mathcal{F}_k\}. \end{align*}
From this we see that $\mathcal{F}_{n \wedge \tau} \subseteq \mathcal{F}_n$ whereas $\mathcal{F}_{(n+1) \wedge \tau}$ may contain sets from $\mathcal{F}_{n+1} \backslash \mathcal{F}_n$.
