Let $\{ M (t) \mid t \in [0,T] \}$ be a martingale and $\{ \tau_n \mid n = 1, 2, \ldots\}$ be an increasing sequence of stopping times such that $\tau_n \rightarrow \infty$ as $n \rightarrow \infty$. Without additional assumptions, is the family of stopped processes $\{ M (T \wedge \tau_n ) \mid n = 1, 2, \ldots \}$ always uniformly integrable?
It seems to me that by Doob's stopping theorem $M (T \wedge \tau_n) = \mathrm{E} [ M (T) \mid {\cal F}_{T \wedge \tau_n} ]$. So the uniform integrability holds. Is this argument true or not? Any hints are really appreciated.
If $T$ is a general stopping time, then no, as you can select e.g. $T=\infty$ and the family you claim to be uniformly integrable is $(M_{\tau_n})$ for some increasing sequence of stopping times $(\tau_n)$. Choosing $\tau_n = n$ would yield that $(M_n)_{n\ge1}$ is uniformly integrable for all martingales $M$, which is absurd.
If $T$ is a bounded stopping time (in particular, if $T$ is a deterministic constant), then your argument is valid: As you note, $M_{T\land \tau_n} = E(M_{T} | \mathcal{F}_{T\land \tau_n})$ by the optional sampling theorem, and as $(E(X|\mathcal{G}))$, with $\mathcal{G}$ ranging in the set of all sub-$\sigma$-algebras of some $\sigma$-algebra is uniformly integrable, the result holds.