$T=\sup_{n \in \mathbb{N}}T_n$ implies $\mathcal{F}_T=\sigma(\bigcup_{n \in \mathbb{N}}{\mathcal{F}_{T_n}})$

Asked Jul 29 '19 at 14:31

Active Mar 30 '21 at 19:21

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Let $(T_n)_n$ be a sequence of stopping times for the filtration $(\mathcal{F}_k)_k$. Let $T=\sup_{n \in \mathbb{N}}T_n$. Prove that $\mathcal{F}_T=\sigma(\bigcup_{n \in \mathbb{N}}{\mathcal{F}_{T_n}})$ where $\forall n \in \mathbb{N},$ $\mathcal{F}_{T_n}=\left\{K;\forall p \in \mathbb{N},K \cap \left\{T_n\leq p \right\} \in \mathcal{F}_p\right\}.$

If $K \in \bigcup_{n \in \mathbb{N}}{\mathcal{F}_{T_n}}$ then there exists $n_0 \in \mathbb{N}$ such that $K \in \mathcal{F}_{T_{n_0}}.$ $\forall p \in \mathbb{N},K \cap\left\{T \leq p\right\}=K \cap\left\{T\leq p \right\} \cap\left\{T_{n_0}\leq p \right\} \in \mathcal{F}_p$ then $\sigma(\bigcup_{n \in \mathbb{N}}{\mathcal{F}_{T_n}})\subset \mathcal{F}_T$.

for the other one if $K \in \mathcal{F}_T,$ then $\forall p \in \mathbb{N},K \cap \bigcap_{n \in \mathbb{N}}\left\{T_n\leq p \right\} \in \mathcal{F}_p$ .

How to deduce that $K \in \sigma(\bigcup_{n \in \mathbb{N}} \mathcal{F}_{T_n})$?

edited Jul 29 '19 at 14:37

Mars Plastic

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asked Jul 29 '19 at 14:31

Do you assume $T<\infty$? – saz Jul 29 '19 at 17:35
@saz No not necessary – Jul 29 '19 at 19:52
I see, thanks for the clarification. Out of curiosity: Where did you find this assertion? – saz Jul 31 '19 at 19:26
See here for the particular case that $T_n = T \wedge n$ for a stopping time $T$. – saz Feb 06 '20 at 16:18

$T=\sup_{n \in \mathbb{N}}T_n$ implies $\mathcal{F}_T=\sigma(\bigcup_{n \in \mathbb{N}}{\mathcal{F}_{T_n}})$

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