Let $I\subseteq\overline{\mathbb R}$, $(\mathcal F_t)_{t\in I}$ be a filtration on a probability space $(\Omega,\mathcal A,\operatorname P)$ and $\sigma,\tau:\Omega\to I\cup\{\sup I\}$ be $(\mathcal F_t)_{t\in I}$-stopping times. Moreover, let $\mathcal F_\tau$ denote the $\sigma$-algebra of the $\tau$-past, i.e. $$\mathcal F_\tau=\{A\in\mathcal A:A\cap\{\tau\le t\}\in\mathcal F_t\},$$ and $\left.\mathcal F\right|_A:=\{A\cap B:B\in\mathcal F\}$ denote the trace of a $\sigma$-algebra $\mathcal F$ on a set $A$.
Now, if $X$ is an integrable random variable on $(\Omega,\mathcal A,\operatorname P)$, are we able to show that $$\operatorname E\left[X\mid\mathcal F_\sigma\right]=\operatorname E\left[X\mid\mathcal F_{\sigma\:\wedge\:\tau}\right]\tag1$$ on $\{\sigma\le\tau\}$?
Since $$\left.\mathcal F_{\sigma}\right|_{\{\:\sigma\:\le\:\tau\:\}}=\left.\mathcal F_{\sigma\:\wedge\:\tau}\right|_{\{\:\sigma\:\le\:\tau\:\}}\tag1,$$ the desired claim would immediately follow from the the local property of conditional expectation if we could show that $\{\sigma\le\tau\}\in\mathcal F_\sigma$ and $\{\sigma\le\tau\}\in\mathcal F_{\sigma\:\wedge\:\tau}$.
Now $\mathcal F_{\sigma\:\wedge\:\tau}\subseteq\mathcal F_\sigma$. So, it is sufficient to show $\{\sigma\le\tau\}\in\mathcal F_{\sigma\:\wedge\:\tau}$. But is this really true?
Best Answer
Proof of $\{\sigma \leq \tau\} \in \mathcal F_{\sigma \wedge \tau}$:
We have to show that $\{\sigma \leq \tau\} \cap \{\sigma \wedge \tau \leq t\} \in \mathcal F_t$ for all $t$.
Write this as $$\{\tau >t, \sigma \leq t\} \cup \{\sigma \leq t, \tau \leq t, \sigma \leq \tau\} $$
(by considering the cases $\tau \leq t$ and $\tau >t$).
Now, $\{\sigma \leq t, \tau \leq t, \sigma \leq \tau\}=\{\sigma \leq t, \tau \leq t\} \setminus \{\sigma \leq t, \tau \leq t, \sigma > \tau\}$.
Finally, write $\{\sigma \leq t, \tau \leq t, \sigma > \tau\}$ as $$\bigcup_{r \in \mathbb Q, r <t} \{\sigma \leq t, \tau \leq t, \sigma > r>\tau\} \,.$$
Since $\{\sigma \leq t, \tau \leq t, \sigma > r>\tau\}$ is the intersection of $\{\sigma \leq t\}, \{\tau \leq t\}, \{\sigma > r\}$ and $\{\tau <r\}$ we are done. [The last event is the union over $n$ of $\tau \leq r-\frac 1 n$].