Suppose $X$ is an adapted process and $\tau_1, \ldots , \tau_k$ are stopping times such that $X^{\tau_1}, \ldots , X^{\tau_k}$ are all martingales. I want to show that then $X^{\tau_1 \vee \ldots \vee \tau_k}$ is a martingale.
Therefore, we have that \begin{align*} &X^{\tau_i} = \{X_{\tau_i \wedge t}: t \in \mathbb{R}_+ \} \qquad &&\text{w.r.t. the filtration $\{\mathcal{F}_t\}$.} \\ \text{and} \qquad &\mathbb{E}[X^{\tau_i}_t \mid \mathcal{F}_s] = X^{\tau_i}_s \qquad &&\text{for all $s < t$ and $1 \leq i \leq k$.} \end{align*} So, \begin{align*} X^{\tau_1 \vee \ldots \vee \tau_k} &= \{X_{(\tau_1 \vee \ldots \vee \tau_k) \wedge t}: t \in \mathbb{R}_+ \}\\ &= \{X_{(\tau_1 \vee t) \wedge \ldots \wedge (\tau_k\vee t)}: t \in \mathbb{R}_+ \}. \end{align*} In general, suppose $M$ is a right-continuous martingale and $\sigma$ is a stopping time. Then the stopped process $\{M_{\sigma \wedge t}: t \in \mathbb{R}_+ \}$ is a martingale w.r.t. the filtration $\{\mathcal{F}_t\}$. $(*)$
Hence, it suffices to elaborate the case $k=2$ and verify that $X^{\tau_1 \vee \tau_2}$ is indeed a martingale and then apply $(*)$ for all $k$.
However, I do not see how to write $X^{\tau_1 \vee \tau_2}$ in terms of $X^{\tau_1}$, $X^{\tau_2}$ and $X^{\tau_1 \wedge \tau_2}$.
