I want to show that any martingale w.r.t. the filtration $\{\mathcal{F}_{t \wedge \tau}\}$ is also a martingale w.r.t. the filtration $\{\mathcal{F}_{t}\}$.
So, suppose $(X_n)_{n \geq 0}$ is a martingale w.r.t. $\{\mathcal{F}_{t \wedge \tau}\}$. That is, $X_t$ is $\mathcal{F}_{t \wedge \tau}$-measurable and integrable and $\mathbb{E}[X_t | \mathcal{F}_{t \wedge \tau}] = X_s$ for $s<t$.
Since $\mathbb{1}_{\{\tau \leq s\}} X_t$ is $\mathcal{F}_{t \wedge \tau}$-measurable, \begin{align*} \mathbb{E}[X_t \mid \mathcal{F}_{t \wedge \tau}] &= X_s \\ \mathbb{1}_{\{\tau \leq s\}} \cdot \mathbb{E}[X_t \mid \mathcal{F}_{t \wedge \tau}] &= \mathbb{1}_{\{\tau \leq s\}} X_s\\ \mathbb{E}[\mathbb{1}_{\{\tau \leq s\}}X_t \mid \mathcal{F}_{t \wedge \tau}] &= \mathbb{1}_{\{\tau \leq s\}} X_s \qquad \text{(is this step correct?)}\\ \mathbb{1}_{\{\tau \leq s\}}X_t &= \mathbb{1}_{\{\tau \leq s\}} X_s. \end{align*} Then I want to show that $(X_n)$ is indeed a martingale w.r.t. $\{\mathcal{F}_{t}\}$. With $s<t$ and $A \in \mathcal{F}_s$, \begin{align*} \mathbb{E}[\mathbb{1}_A X_t] &= \mathbb{E}[\mathbb{1}_A \mathbb{1}_{\{\tau \leq s\}}X_t] + \mathbb{E}[\mathbb{1}_A \mathbb{1}_{\{\tau > s\}}X_t] \\ &= \mathbb{E}[\mathbb{1}_A \mathbb{1}_{\{\tau \leq s\}}X_s] + \mathbb{E}[\mathbb{1}_A \mathbb{1}_{\{\tau > s\}}X_t]. \end{align*} Where I got stucked... Why should we elaborate $\mathbb{E}[\mathbb{1}_A X_t]$? How is this related to $\mathbb{E}[X_t \mid \mathcal{F}_s]$ and how to find that this is equal to $X_s$?