Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space and $\mathcal G = (\mathcal G_t, t \ge 0)$ a filtration. Let $M$ be a real-valued continuous local martingale w.r.t. $\mathcal G$. Let $(\tau_n, n \in \mathbb N)$ be a localizing sequence of $M$, i.e.,
- $\tau_n : \Omega \to \mathbb R_{\ge 0}$ is a stopping time for all $n$.
- $\tau_n \uparrow \infty$ a.s.
- $M^{\tau_n} := (M_{t \wedge \tau_n}, t\ge 0)$ is a martingale for all $n$.
We define a process $X$ by $X_t := M_t^2 - \langle M \rangle_t$ for all $t \ge 0$. Clearly, $X$ is a continuous local martingale w.r.t. $\mathcal G$. We have $$ \begin{align} X^{\tau_n}_t &= M_{t \wedge \tau_n}^2 - \langle M \rangle_{t \wedge \tau_n} \\ &= (M^{\tau_n}_t)^2 - \langle M^{\tau_n} \rangle_{t} \\ &= ((M^{\tau_n})^2 - \langle M^{\tau_n} \rangle)_t. \end{align} $$
By assumption, $M^{\tau_n}$ is a martingale, so is $(M^{\tau_n})^2 - \langle M^{\tau_n} \rangle$. Hence $X^{\tau_n}$ is a martingale and thus $(\tau_n, n \in \mathbb N)$ is also a localizing sequence of $X$.
Assume $M^{\tau_n}$ is uniformly integrable for all $n$. Is $X^{\tau_n}$ uniformly integrable for all $n$?