Let
- $(\Omega,\mathcal A)$ be a measurable space
- $I\subseteq[0,\infty)$
- $\mathbb F=(\mathcal F_t)_{t\in I}$ be a filtration on $(\Omega,\mathcal A)$
- $\tau$ be a $\mathbb F$-stopping time
- $\mathcal F_\tau:=\left\{A\in\mathcal A:A\cap\left\{\tau\le t\right\}\in\mathcal F_t\;\text{for all }t\in I\right\}$
- $Y:(\Omega,\mathcal F_\tau)\to\left(\mathbb R,\mathcal B\left(\mathbb R\right)\right)$ be a random variable
- $s\in I$ and $Z:=1_{\left\{\tau=s\right\}}Y$
Can we show, that $Z$ is $\mathcal F_s$-measurable? Clearly, we've got $$\left\{Z\in B\right\}=\left\{\tau=s\right\}\cap\underbrace{\left\{Y\in B\right\}}_{\in\mathcal F_\tau}\cup\underbrace{\left\{\tau=s\right\}^c\cap\left\{0\in B\right\}}_{\in\mathcal F_s}$$ for all $B\in\mathcal B(\mathbb R)$. Moreover, $$\left\{\tau=s\right\}\cap\left\{\tau\le t\right\}=\begin{cases}\emptyset&\text{, if }t<s\\\left\{\tau=s\right\}&\text{, if }t\ge s\end{cases}\;\;\;\text{for all }t\in I\;.$$ Since $\emptyset\in\mathcal F_t$, for all $t\in I$, and $\left\{\tau=s\right\}\in\mathcal F_s\subseteq\mathcal F_t$, for all $s\le t$, we've got $$\left\{\tau=s\right\}\in\mathcal F_\tau\;.$$ So, $$A:=\left\{\tau=s\right\}\cap\left\{Y\in B\right\}\in\mathcal F_\tau\;.$$ But that immediately yields $$\mathcal F_s\ni A\cap\left\{\tau\le s\right\}=A\;.$$
Did I made any mistake?