First is the definition of stopping time in our class of martingale:
Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n})_{n \in \mathbb{N}}, \mathbb P\right)$ be a filtered probability space. A random variable $T$ taking values in $I \cup\{\infty\}$ is a stopping time (or optional time) if $\{\omega: T(\omega)=n\} \in \mathcal{F}_{n}$ for all $n \in I$.
Let $T$ be a stopping time. The sigma field $\mathcal{F}_{T}$, sometimes called "the past before $T$" is $$\mathcal{F}_{T}=\left\{\Lambda \in \mathcal{F} \mid \forall n \in I:\Lambda \cap\{T=n\} \in \mathcal{F}_{n} \right\}$$
Then we have a proposition:
If $T_{1} \le T_{2}$ are stopping times, then $\mathcal{F}_{T_{1}} \subseteq \mathcal{F}_{T_{2}}$.
Could you please verify if my proof is correct? Many thanks!
My attempt:
Let $\Lambda \in \mathcal F_{T_1}$. Then $\forall n \in I: \Lambda \cap \{T_1 = n\} \in \mathcal F_n$. On the other hand, we have $$\begin{aligned} \Lambda \cap \{T_2 = n\} &\overset{(1)}{=} \big ( \Lambda \cap \{T_2 = n\} \big ) \cap \bigcup_{k \in I \cup \{\infty\}} \{T_1 = k\} \\ &= \bigcup_{k \in I \cup \{\infty\}} \big ( \Lambda \cap \{T_2 = n\} \cap \{T_1 = k\} \big ) \\ &\overset{(2)}{=} \bigcup_{k \in I \cap \{0,\ldots,n\}} \big ( \Lambda \cap \{T_1 = k\} \cap \{T_2 = n\} \big )\\\end{aligned}$$
$(1)$ follows from $(\{T_1 = k\})_{k \in I \cup \{\infty\}}$ is a partition of $\Omega$.
$(2)$ follows from $T_2 (\omega) \le n \implies T_1 (\omega) \le n$.
Because $\Lambda \cap \{T_1 = k\} \in \mathcal F_k \subseteq \mathcal F_n$ for all $k \le n$ and $ \{T_2 = n\} \in \mathcal F_n$, we have $ \Lambda \cap \{T_1 = k\} \cap \{T_2 = n\} \in \mathcal F_n$. Hence $\Lambda \cap \{T_2 = n\} \in \mathcal F_n$ and thus $\Lambda \in \mathcal F_{T_2}$. As a result, $\mathcal F_{T_1} \subseteq \mathcal F_{T_2}$.