I'm reading about stopping times in this note, i.e.,
Let $(\Omega, \mathcal{F}, P)$ be a probability space, and $\mathcal{F}_{\bullet}=\left(\mathcal{F}_t \subseteq \mathcal{F}: t \in T\right)$ be a filtration on this probability space for an ordered index set $T \subseteq \mathbb{R}$ considered as time.
Lemma 1.11. Let $\tau, \tau_1, \tau_2$ be stopping times adapted to a filtration $\mathcal{F}_{\bullet}$. If $\tau_1 \leqslant \tau_2$ almost surely, then $\mathcal{F}_{\tau_1} \subseteq \mathcal{F}_{\tau_2}$.
Proof. Recall, that for any $t \geqslant 0$, we have $\{\tau \leqslant t\} \in \mathcal{F}_t$. From the hypothesis $\tau_1 \leqslant \tau_2$ a.s., we get $\left\{\tau_2 \leqslant t\right\} \subseteq\left\{\tau_1 \leqslant t\right\}$ a.s., where both events belong to $\mathcal{F}_t$ since they are stopping times. The result follows since for any $A \in \mathcal{F}_{\tau_1}$ and $t \in T$, we can write $A \cap\left\{\tau_2 \leqslant t\right\}=A \cap\left\{\tau_1 \leqslant t\right\} \cap\left\{\tau_2 \leqslant t\right\} \in \mathcal{F}_t$.
My understanding The inclusion $\left\{\tau_2 \leqslant t\right\} \subseteq\left\{\tau_1 \leqslant t\right\}$ holds almost surely, i.e., it may not hold on a $P$-null set.
Could you explain how the author obtains the surely inclusion $A \cap\left\{\tau_2 \leqslant t\right\}=A \cap\left\{\tau_1 \leqslant t\right\} \cap\left\{\tau_2 \leqslant t\right\}$?
If $\tau _1\leqslant \tau _2$ then $$ \tau _2\leqslant t\implies \tau _1\leqslant t\\[1em] \therefore\quad \{\tau _2\leqslant t\}=\{\tau _1\leqslant \tau _2\leqslant t\}=\{\tau _1\leqslant t\,\land\, \tau _2\leqslant t\}=\{\tau _1\leqslant t\}\cap \{\tau _2\leqslant t\} $$
If $\tau _1\leqslant \tau _2$ a.s. then a similar reasoning show that $$ \{\tau _2\leqslant t\}=\{\tau _1\leqslant t\}\cap \{\tau _2\leqslant t\}\;\text{ a.s.} $$ That is, the above equality becomes an almost sure equality.