Does $\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n}) = \mathcal{F}_S$ hold for every stopping time $S$?

Suppose $S$ is a stopping time. How do I show that $\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n}) = \mathcal{F}_S$?

This is a VERY basic question but I'm very confused. The inclusion $\sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n}) \subseteq \mathcal{F}_S$ is trivial, but the reverse direction is impossible for me.

Trying to answer myself:

Suppose $A \in \mathcal{F}_S \implies A \cap \{S = i\} \in \mathcal{F}_i \quad \forall i\in \mathbb{N}$

$A = \cup_{i = 0}^\infty (A \cap \{S = i\}) \cup (A \cap\{S=\infty\})$

Fix $i$ and choose any $n > i$.

Now $A \cap\{S = i\} \cap\{S\wedge n \leq k\} = A \cap \{S = i\}$ for $k \ge i$ and is the empty set otherwise.

As $\mathcal{F}_i \subseteq \mathcal{F}_k$ for $k \ge i$, we have that $A \cap \{S=i\} \in \mathcal{F}_{S\wedge n} \subset \sigma(\cup_{n \in \mathbb{N}}\mathcal{F}_{S\wedge n})$

I can't figure out why the the event $A \cap \{S = \infty \} \in \sigma(\cup_{n=0}^\infty \mathcal{F}_{S \wedge n})$, which is all I would need to conclude.

Please help me. This is astronomically demoralizing, because I think this should be really simple.

Solution 1:

Let $(\Omega,\mathcal{F})$ be a measurable space and $(\mathcal{F}_n)_{n \in \mathbb{N}}$ a filtration. If $\mathcal{F}$ is strictly bigger than $\mathcal{F}_{\infty} := \sigma(\bigcup_{n \in \mathbb{N}} \mathcal{F}_n)$, then the assertion fails to hold; see the comments for a counterexample. Therefore I will assume in the following that $$\mathcal{F} = \mathcal{F}_{\infty}.$$ Under this assumption, we are going to show that $$\mathcal{F}_{\tau} = \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_{\tau \wedge n} \right)$$ for any $(\mathcal{F}_n)_n$-stopping time $\tau:\Omega \to \mathbb{N} \cup \{\infty\}$. Recall that, by definition, $$\mathcal{F}_{\tau} = \{A \in \mathcal{F}; \forall n \in \mathbb{N}: A \cap \{\tau \leq n\} \in \mathcal{F}_n\}.$$

We start with some preparations.

Lemma 1: If $\mathcal{A}$ is a $\sigma$-algebra and $\Omega_0 \in \mathcal{A}$, then $$\mathcal{A} = (\mathcal{A} \cap \Omega_0) \cup (\mathcal{A} \cap \Omega_0^c).$$

Proof: Set $\tilde{\mathcal{A}} := (\mathcal{A} \cap \Omega_0) \cup (\mathcal{A} \cap \Omega_0^c)$. If $A \in \mathcal{A}$, then $$A = (A \cap \Omega_0) \cup (A \cap \Omega_0^c) \in \tilde{\mathcal{A}}$$ On the other hand, if $A \in \tilde{\mathcal{A}}$, i.e. $$A = (B \cap \Omega_0) \cup (C \cap \Omega_0^c)$$ for $B,C \in \mathcal{A}$, then $A \in \mathcal{A}$ because $\Omega_0 \in \mathcal{A}$.

Lemma 2: If $\tau$ is a stopping time, then $\{\tau=\infty\} \in \mathcal{F}_{\tau}$ and $\{\tau=\infty\} \in \sigma(\bigcup_n \mathcal{F}_{n \wedge \tau})$.

Proof: The first statement is obvious from the definition of $\mathcal{F}_{\tau}$. For the second one, we note that $$\{\tau=\infty\} = \bigcap_{n \in \mathbb{N}} \{\tau>n\}.$$ Since $\{\tau>n\} \in \mathcal{F}_n$ and $\{\tau>n\} \in \mathcal{F}_{\tau}$, we have $\{\tau>n\} \in \mathcal{F}_n \cap \mathcal{F}_{\tau} = \mathcal{F}_{n \wedge \tau}$. Hence, $\{\tau=\infty\} \in \sigma(\bigcup_n \mathcal{F}_{n \wedge \tau})$.

Lemma 3: Let $\mathcal{G}$ be any family of sets and $\tau$ any mapping. Then $$\sigma(\mathcal{G}) \cap \{\tau=\infty\} = \sigma(\mathcal{G} \cap \{\tau=\infty\}).$$

Proof: Define $T:\{\tau=\infty\} \to \Omega, \omega \mapsto \omega$, then the assertion reads $$T^{-1}(\sigma(\mathcal{G})) = \sigma(T^{-1}(\mathcal{G})),$$ this identity holds for any mapping $T$ (see this question).

Theorem: Let $(\mathcal{F}_n)_{n \in \mathbb{N}}$ be a filtration and $\tau:\Omega \to \mathbb{N} \cup \{\infty\}$ a stopping time. Then $$\mathcal{F}_{\tau} = \sigma \left( \bigcup_{n \in \mathbb{N}} \mathcal{F}_{n \wedge \tau}\right).$$

Proof: Because of Lemma 1 and Lemma 2, it suffices to show that \begin{align} \mathcal{F}_{\tau} \cap \{\tau<\infty\} &= \sigma \left( \bigcup_{n \in \mathbb{N}_0} \mathcal{F}_{n \wedge \tau}\right) \cap \{\tau<\infty\} \tag{1} \\ \mathcal{F}_{\tau} \cap \{\tau=\infty\} &= \sigma \left( \bigcup_{n \in \mathbb{N}_0} \mathcal{F}_{n \wedge \tau}\right) \cap \{\tau=\infty\} \tag{2}. \end{align} Proof of $(1)$: Since $\mathcal{F}_{n \wedge \tau} \subseteq \mathcal{F}_{\tau}$, it suffices to prove '$\subseteq$'. Take $F' = F \cap \{\tau<\infty\}$ for $F \in \mathcal{F}_{\tau}$. Then $$F' = \bigcup_{n \in \mathbb{N}} \underbrace{F \cap \{\tau \leq n\}}_{\in \mathcal{F}_{n \wedge \tau}} \cap \{\tau<\infty\} \in \sigma \left( \bigcup_{n \in \mathbb{N}_0} \mathcal{F}_{n \wedge \tau}\right) \cap \{\tau<\infty\}.$$ Proof of $(2)$: By the definition of $\mathcal{F}_{\tau}$, we have \begin{equation} \mathcal{F}_{\tau} \cap \{\tau=\infty\} = \mathcal{F}_{\infty} \cap \{\tau=\infty\}; \tag{3} \end{equation} '$\subseteq$' is obvious; for the other inclusion take $F' = F \cap \{\tau=\infty\}$ for $F \in \mathcal{F}_{\infty}$, then $F \cap \{\tau=\infty\} \in \mathcal{F}_{\tau}$ (any subset of $\{\tau=\infty\}$ is in $\mathcal{F}_{\tau}$) and so $$F' = (F \cap \{\tau=\infty\}) \cap \{\tau=\infty\} \in \mathcal{F}_{\tau} \cap \{\tau=\infty\}.$$ Moreover, \begin{equation} \mathcal{F}_{n \wedge \tau} \cap \{\tau=\infty\} = \mathcal{F}_{n} \cap \{\tau=\infty\}\tag{4} \end{equation} Indeed: Again '$\subseteq$' is obvious. If $F' = F \cap \{\tau=\infty\}$ for $F \in \mathcal{F}_n$, then $$(F \cap \{\tau>n\}) \cap \{\tau \wedge n \leq k\} \in \mathcal{F}_k$$ and so $F \cap \{\tau>n\} \in \mathcal{F}_{\tau \wedge n}$; thus, $$F' = (F \cap \{\tau>n\}) \cap \{\tau=\infty\} \in \mathcal{F}_{n \wedge \tau} \cap \{\tau=\infty\}.$$ Finally, \begin{align*} \mathcal{F}_{\tau} \cap \{\tau=\infty\} \stackrel{(3)}{=} \mathcal{F}_{\infty} \cap \{\tau=\infty\} &= \sigma(\mathcal{F}_n; n \in \mathbb{N}) \cap \{\tau=\infty\} \\ &\stackrel{\text{Lem 3}}{=} \sigma(\mathcal{F}_n \cap \{\tau=\infty\}; n \in \mathbb{N}) \\ &\stackrel{\text{(4)}}{=} \sigma(\mathcal{F}_{n \wedge \tau} \cap \{\tau=\infty\}; n \in \mathbb{N}) \\ &\stackrel{\text{Lem 3}}{=} \sigma(\mathcal{F}_{n \wedge \tau}; n \in \mathbb{N}) \cap \{\tau=\infty\}, \end{align*} which proves $(2)$.

Solution 2:

Note that $A \cap \{S = \infty \}$ = $\cap_{i \in \mathbb{N}}A \cap \{S > i \}$

Fixing $i$, we see that $A \cap \{S > i \} \in \mathcal{F}_i$ [I'M NOT SURE IF THIS IS RIGHT]

Fix $n = i$. We have that $ A \cap \{S > i \} \cap \{S \wedge n \leq k \} = A \cap \{S > i \}$ if $k \ge i$ and is the empty set otherwise.

As $\mathcal{F}_i \subseteq \mathcal{F}_k$ for all $k \ge i$, we FINALLY conclude that $A \cap \{S > i \} \in \mathcal{F}_{S\wedge n}$.

Therefore we have that $A$ is acountable union of sets inside the sigma algebra genererated by the $\mathcal{F}_n$.

Can someone please tell me if this is correct?