An equality in stopping time.

Question

In a proof,I need the following equality:

Suppose $\tau,\sigma$ are two stopping time and $A$ is a event.Then: $$(A\cap\{\sigma\le\tau\})\cap\{\tau\le t\}=(A\cap\{\color{red}{\sigma}\le t\})\cap\{\color{red}{\sigma}\wedge t\le\color{red}{\tau}\wedge t\}\cap \{\tau\le t\}$$

I could deal with the "$\subset$",but I have no idea for the opposite direction.

Answer 1

Obviously, the equality holds true if we show that

$$\omega \in \{\sigma \leq \tau\} \Leftrightarrow \omega \in \{\sigma \leq t\} \cap \{\sigma \wedge t \leq \tau \wedge t\}$$

for any $\omega$ such that $\tau(\omega) \leq t$.

"$\subseteq$": First of all, $\sigma(\omega) \leq \tau(\omega) \leq t$ implies $\omega \in \{\sigma \leq t\}$. Moreover, this shows that $$ \sigma(\omega) \wedge t = \sigma(\omega) \leq \tau(\omega) = \tau(\omega) \wedge t$$, i.e. $\omega \in \{\sigma \wedge t \leq \tau \wedge t\}$.

"$\supseteq$": As $\sigma(\omega) \leq t$ and $\tau(\omega) \leq t$, we find that $$\underbrace{\sigma(\omega) \wedge t}_{\sigma(\omega)} \leq \underbrace{\tau(\omega) \wedge t}_{\tau(\omega)}.$$ Hence, $\omega \in \{\sigma \leq \tau\}$.