For a stopping time does $E[(\tau \wedge t)1_A]=E[(\tau \wedge s)1_A] $ for $s \le t $ and any $A \in \mathcal F_{\tau \wedge s }$? I have the following question as stated in the title:

For a stopping time $\tau $ does $E[(\tau \wedge t)1_A]=E[(\tau \wedge s)1_A] $ for $s \le t $ and any $A \in \mathcal F_{\tau \wedge s }$?

Here $\mathcal F_{\tau \wedge s } $ is the $\sigma $-algebra of $\tau \wedge s $-past.

The context is that I'm trying to show that $(B^2_{\tau \wedge t } - \tau \wedge t,\mathcal F_{\tau \wedge t })$ is a martingale and the question is here is a second step in showing that $E[\tau \wedge t|\mathcal F_{\tau \wedge s} ]=\tau \wedge s$.
Thanks in advance!

We have
\begin{alignat*}{2}
\mathbb{E}[(\tau \wedge t)\mathbf{1}_A] & = \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge t)\mathbf{1}_A\;|\;\mathcal F_{\tau \wedge s }\right]\Big]\quad\text{by the tower property}\\
& = \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge t)\;|\;\mathcal F_{\tau \wedge s }\right]\mathbf{1}_A\Big]\quad\text{by measurability}\\
& \neq \mathbb{E}\Big[\mathbb{E}\left[(\tau \wedge s)\;|\;\mathcal F_{\tau \wedge s }\right]\mathbf{1}_A\Big]\\
& = \mathbb{E}\left[(\tau \wedge s)\mathbf{1}_A\right],
\end{alignat*}
so answer seems no. Counterexample: $\tau=\infty$ almost surely.

If $B$ is a standard Brownian motion, then $\mathbb E\left[ B^2_{t}\right] = {t}$ and $\mathbb E\left[ B^2_{t}\;\;|\;\; \mathcal{F}_s\right] = B_s^2 + {t-s}$.
What you need to show seems
$$
\mathbb E\left[ B_{\tau\wedge t}^2 - {\tau\wedge t} \;\;|\;\; \mathcal{F}_{\tau\wedge s}\right] = B_{\tau\wedge s}^2 - {\tau\wedge s}
$$
for which you need $(B_t-B_s)\amalg\mathcal F_s$ and $(B_t-B_s)\sim B_{t-s}$.
