Question: Show that $\mathbb{E}[\xi|\mathscr{F}_{\sigma}] = \mathbb{E}[\xi|\mathscr{F}_{\sigma \wedge \tau}] $ on $\{ \sigma \le \tau\}$ for every $\xi \in L^1$. Note that $\mathscr{F}_{\tau} =\{A \in \mathscr{F} : A \cap \{ \tau \le n \} \in \mathscr{F}_n \}$ and $\tau, \sigma$ are stopping times wrt to the same filtration.
Attempt: I was able to prove that both $\{ \sigma < \tau \}$ and $\{\sigma = \tau \}$ are $\in \mathscr{F}_{\sigma \wedge \tau}$ which implies that $\{ \sigma \le \tau\} = \{ \sigma < \tau \} \cup \{\sigma = \tau \}\in \mathscr{F}_{\sigma \wedge \tau}$. However, I'm unsure about how to show that two conditional expectations are equivalent to each other with respect to a certain set. Am I suppose to prove that $\mathscr{F}_{\sigma} = \mathscr{F}_{\sigma \wedge \tau}$ on $\{\sigma \le \tau\}$ which implies that the conditional expectations are equal? If so, how could I make this proof more rigorous?
Let $E=\{\sigma \leq \tau\}$. What you are asked to prove is equivalent to $\int_{A\cap E} \mathbb E (\xi |\mathcal F_{\sigma})=\int_{A\cap E} \mathbb E (\xi |\mathcal F_{\sigma \wedge \tau})$ for every $A \in \mathcal F_{\sigma \wedge \tau}$.
[If$X$ and $Y$ are measurable w.r.t. $\mathcal G$ then $X=Y$ a.s on a set $E$ in $\mathcal G$ iff $\int_{A\cap E} XdP=\int_{A\cap E} YdP$ for all $A \in \mathcal G$].
But $A \cap E$ belongs to $\mathcal F_{\sigma \wedge \tau}$ so the conclusion follow from definition of conditional expectation.