How to prove $\mathcal F_T \subseteq \sigma(\bigcup_n \mathcal F_{T_n})$? I'm doing the following exercise about stopping time:

Let $I \subseteq \mathbb N$ and $(\mathcal F_n)$ be a filtration. A random variable $T$ taking values in $I \cup \{\infty\}$ is a stopping time if and only if $\{T = n\} \in \mathcal F_n$ for all $n \in I$.

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*Let $(T_n)$ be a decreasing sequence of stopping times that converges to $T$. Show that $\mathcal{F}_{T} = \bigcap_n \mathcal F_{T_n}$. Hint: if $T_{n} \downarrow T$ are all stopping times, they are integer-valued (or infinite), so that $T_{n}(\omega)=T(\omega)$ for $n$ large enough.


*Let $(T_n)$ be an increasing sequence of stopping times that converges to a finite $T$. Show that $\mathcal F_T = \sigma(\bigcup_n \mathcal F_{T_n})$.

I'm stuck at proving $\mathcal F_T \subseteq \sigma(\bigcup_n \mathcal F_{T_n})$ in the second claim. How can I proceed to finish the proof? Many thanks!

My attempt:

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*Because $T \le T_n$, $\mathcal F_T \subseteq \mathcal F_{T_n}$. Hence $\mathcal F_T \subseteq \bigcap_n \mathcal F_{T_n}$. It remains to show that $\bigcap_n \mathcal F_{T_n} \subseteq \mathcal F_T$. First, we have $\{T=n\} = \bigcup_{m \in \mathbb N} \bigcap_{k \ge m} \{T_k=n\}$. Let $\Lambda \in \bigcap_n \mathcal F_{T_n}$. Then $$\Lambda \cap \{T=n\} = \bigcup_{m \in \mathbb N} \bigcap_{k \ge m} \left [\{T_k=n\} \cap \Lambda \right ]$$ It follows from $\Lambda \in \mathcal F_{T_k}$ for all $k$ that $\{T_k=n\} \cap \Lambda \in \mathcal F_n$ for all $k$. Then $\Lambda \cap \{T=n\} \in \mathcal F_n$ for all $n \in I$ and thus $\Lambda \in \mathcal F_T$. The claim then follows.


*Because $T_n \le T$, $\mathcal F_{T_n} \subseteq \mathcal F_T$. Then $\bigcup_n \mathcal F_{T_n} \subseteq \mathcal F_T$ and thus $\sigma(\bigcup_n \mathcal F_{T_n}) \subseteq \sigma(\mathcal F_T)$. This means $\sigma(\bigcup_n \mathcal F_{T_n}) \subseteq \mathcal F_T$. It remains to show that $\mathcal F_T \subseteq \sigma(\bigcup_n \mathcal F_{T_n})$.
 A: I adapted @saz's answer here to my question. All credits go to @saz.

Because $T_n \le T$, $\mathcal F_{T_n} \subseteq \mathcal F_T$. Then $\bigcup_n \mathcal F_{T_n} \subseteq \mathcal F_T$ and thus $\sigma(\bigcup_n \mathcal F_{T_n}) \subseteq \sigma(\mathcal F_T)$. This means $\sigma(\bigcup_n \mathcal F_{T_n}) \subseteq \mathcal F_T$. It remains to show that $\mathcal F_T \subseteq \sigma(\bigcup_n \mathcal F_{T_n})$. First, we need a lemma:

$\textbf{Lemma:}$ If $S \le T$ are stopping times, then $\Lambda \cap  \{S=T\} \in \mathcal F_S$ for all $\Lambda \in \mathcal F_T$.
$\textbf{Proof:}$ We have $[\Lambda \cap  \{S=T\}] \cap \{T=n\} = [\Lambda \cap  \{T=n\}] \cap \{S=n\}$. By definition, $\Lambda \cap  \{T=n\} \in \mathcal F_n$ and  $\{S=n\}\in \mathcal F_n$. Hence $[\Lambda \cap  \{S=T\}] \cap \{T=n\} \in \mathcal F_n$ for all $n \in I$. As a result, $\Lambda \cap  \{S=T\} \in \mathcal F_T$ by definition.

Let $\Lambda \in \mathcal F_T$. Then $\Lambda = \Lambda \cap \Omega = \Lambda \cap \bigcup_{n} \{T_n = T\} =  \bigcup_{n} [\Lambda \cap \{T_n = T\}]$. By lemma, $\Lambda \cap \{T_n = T\} \in \mathcal F_{T_n}$. It follows that $\Lambda \in \sigma(\bigcup_n \mathcal F_{T_n})$. The claim then follows.
