If $T_1, T_2$ are stopping times such that $T_1 \le T_2$ almost surely, then $\mathcal{F}_{T_1} \subset \mathcal{F}_{T_2}$

Question

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and $\left(\mathcal{F}_n, n \in \mathbb N\right)$ a filtration. If $T$ is a stopping time w.r.t. $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$, one defines the information one possesses at time $T$ as the following $\sigma$-field: $$ \mathcal{F}_T := \left\{A \in \mathcal{F}: A \cap\{T=n\} \in \mathcal{F}_n, \forall n \in \mathbb{N}\right\}. $$

I'm trying to prove below theorem from a Wikipedia page, i.e.,

Theorem Let $T_1, T_2$ be stopping times w.r.t. $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$. If $T_1 \le T_2$ almost surely then $\mathcal{F}_{T_1} \subset \mathcal{F}_{T_2}$.

The strategy in the first case does not work in the second case due to the appearance of $N$. Could you elaborate on how to finish the proof?

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My attempt Let $A \in \mathcal{F}_{T_1}$ and $n \in \mathbb N$. We want to prove $A \cap \{T_2=n\} \in \mathcal F_n$.

• First, we assume $T_1 \le T_2$ surely, i.e., $T_1(\omega) \le T_2 (\omega)$ for all $\omega \in \Omega$. Then $$ \begin{align} A \cap \{T_2 =n\} &= A \cap \{T_2 =n\} \cap \bigcup_{k \in \mathbb N \cup\{+\infty\}} \{T_1 =k\} \\ &= \left( \bigcup_{k \le n} (A \cap \{T_1 =k\} ) \right ) \cap \{T_2 =n\}. \end{align} $$ We have $A \cap \{T_1 =k\} \in \mathcal F_k \subset \mathcal F_n$ for all $k \le n$. It follows that $A \cap \{T_2 =n\} \in \mathcal F_n$.

• Second, we assume $T_1 \le T_2$ almost surely, i.e., there is $N \in \mathcal F$ such that $\mathbb P(N)=0$ and $T_1 (\omega) \le T_2 (\omega)$ for all $\omega \in N^c :=\Omega \setminus N$. Then $$ \begin{align} A \cap \{T_2 =n\} &= \left [ \left( \bigcup_{k \le n} (A \cap \{T_1 =k\} ) \right ) \cap \{T_2 =n\} \cap N^c \right ] \bigcup [A \cap \{T_2 =n\} \cap N ]. \end{align} $$

Answer 1

It appears that there is a hypothesis missing from the Wikipedia article. Specifically, it is standard to assume that the filtration is augmented with the sub-null sets of $\mathcal F$. More precisely, we assume that if $B \in \mathcal F$ with $\mathbb{P}(B) = 0$ and $A \subseteq B$, then $A \in \mathcal F_n$ for all $n$.

To finish your proof of the almost sure case, we can use the same argument as in the case of the sure case to show that the set enclosed in brackets is $\mathcal F_n$-measurable. Then $A \cap \{T_2 = n\} \cap N \subseteq N$ and $\mathbb{P}(N) = 0$, so $A \cap \{T_2 = n\} \cap N \in \mathcal F_n$ as well.