# Karatzas + Shreve 2.22: For optional time $S$, stopping time $T$, prove that $\mathscr{F}_{S+} \subset \mathscr{F}_T$

Prove that if $$S$$ is an optional time and $$T$$ is a positive stopping time such that with $$S \le T$$ and $$S < T$$ on $$\{S < \infty\}$$, then $$\mathscr{F}_{S+} \subset \mathscr{F}_T$$.

The following is the given solution:

Consider any $$A \in \mathscr{F}_{S+}$$, which means that $$\forall t \ge 0, A \cap \{S \le t\} \in \mathscr{F}_{t+}$$. Then:

\begin{align*} A &= \left( \bigcup_{r \in \mathbb{Q}} [A \cap \{S < r < T \}] \right) \cup [A \cap \{S = \infty\}] \\ \end{align*}

For the left clause, $$A \cap \{S < r < T\} = A \cap \{S < r \} \cap \{T > r\}$$ is in $$\mathscr{F}_T$$ because $$A \cap \{S < r\} \in \mathscr{F}_r$$.

For the right clause, $$A \cap \{S = \infty\} = [A \cap \{S = \infty\} ] \cap \{T = \infty\}$$ is in $$\mathscr{F}_T$$. Then the conclusion follows.

All of that makes sense except I don't see how we can conclude that $$A \cap \{S < r\} \in \mathscr{F}_r$$

• I get that part. I don't get the earlier part how we say that $A \cap \{S < r\} \in \mathscr{F}_t$
– clay
Commented Mar 22, 2022 at 5:59
• I think there is a typo. In fact no $t$ has been specified. So it should read $A\cap \{S<r\} \in \mathcal F_r$. Commented Mar 22, 2022 at 6:06
• yikes, you are right, that should have been $\mathscr{F}_r$. I fixed it. However, I still don't understand how you get that. Of course $\{ S < r \} \in \mathscr{F}_r$, that is the optional time property. But why is $A \cap \{S < r\} \in \mathscr{F}_r$?
– clay
Commented Mar 22, 2022 at 13:17
• Since $A \in \mathscr{F}_{S+}$, that means for any $t \ge 0$, we have $A \cap \{S \le t\} \in \mathscr{F}_{t+}$. We can change notation to get for any $r \ge 0$, we have $A \cap \{S \le r\} \in \mathscr{F}_{r+}$. How can you infer from that that $A \cap \{S < r\} \in \mathscr{F}_r$? Can't we have $A \in \mathscr{F}_{r+}$ but $A \not\in\ \mathscr{F}_r$?
– clay
Commented Mar 22, 2022 at 13:22

## 1 Answer

This follows from the definition of $$\mathcal{F_{S+}}$$. Recall that $$A \cap \{S \leq t\} \in \mathcal{F_{t+}},\forall t\geq 0$$. Now, $$A \cap \{S < r\} = A \cap ( \cup_{n=1}^{\infty} \{S\leq r-\frac{1}{n}\}) = \cup_{n=1}^{\infty} (A \cap \{ S \leq r- \frac{1}{n} \})$$. But $$A \cap \{ S \leq r- \frac{1}{n} \} \in \mathcal{F_{(r-\frac{1}{n})+}} \subset \mathcal{F_{r}} \forall n$$. Therefore, the union is in $$\mathcal{F_{r}}$$.

• Thank you! That's rather simple in hindsight after you've explained it :)
– clay
Commented Mar 26, 2022 at 18:50