For stopping times $\tau_k,\mathcal{F}_{\sup_{k \in \mathbb{N}^*}\tau_k}=\sigma(\bigcup_{k \in \mathbb{N}^*}\mathcal{F}_{\tau_k})$?

Question

$(\tau_k)_{k \in \mathbb{N}^*}$ is a sequence of stopping times (taking values in $\overline{\mathbb{N}}$) for the filtration $(\mathcal{F}_n)_{n \in \mathbb{N}^*}.$ Let $\tau=\sup_{k \in \mathbb{N}^*} \tau_k.$

One can prove that $\mathcal{F}_{\max(\tau_1,\tau_2)}=\sigma(\mathcal{F}_{\tau_1} \cup \mathcal{F}_{\tau_2})$ and more generally that for $k \in \mathbb{N}^*,\mathcal{F}_{\max_{1\leq q \leq k}\tau_q}=\sigma(\bigcup_{1 \leq q \leq k}\mathcal{F}_{\tau_q}).$

Does this mean that $\mathcal{F}_{\tau}=\sigma(\bigcup_{q \in \mathbb{N^*}}\mathcal{F}_{\tau_q})$ ? (Obviously $\sigma(\bigcup_{q \in \mathbb{N}^*}\mathcal{F}_{\tau_q}) \subset \mathcal{F}_{\tau}$).

This question was asked several times : here and here without a clear proof.

We can also find particular cases of $\tau_k$: here and here.

Answer 1

Concering your question, the following facts may be useful.

(S. N. Cohen and R. J. Elliott, Stochastic Calculus and Applications, 2nd Ed., Springer, 2015.) Lemma 6.2.14 If $ \{T_n\}_{n\in\mathbb{N}} $ is a nondecreasing sequence of stoping times and $ T=\lim_n T_n $, then $ \mathcal{F}_{T-}=\bigvee_n \mathcal{F}_{T_n-} $.

Corollary 6.2.15 Suppose $ \{T_n\}_{n\in\mathbb{N}} $ is a nondecreasing sequence of stoping times and $ T=\lim_n T_n $. If $ T_n<T $ on $ \{0<T<\infty\} $ then $ \mathcal{F}_{T-}=\bigvee_n \mathcal{F}_{T_n} $.

(S. W. He et al., Semimartingale Theory and Stochastic Calculus, Sci. Press and CRC(1992).) Corollary 3.5. 6) If $ (T_n) $ is a stationary sequence of stopping times, then \begin{equation*} \mathcal{F}_{\bigvee\limits_nT_n}=\bigvee_n \mathcal{F}_{T_n} =\Big\{ \bigcup_{n=1}^\infty A_n:A_n\in \mathcal{F}_{T_n},n\ge 1, A_kA_j=\emptyset, k\ne j. \Big\} \end{equation*} Remark: $ (T_n) $ is stationary, i.e. for each $ \omega\in\Omega$ there exists a natural number $ n(\omega) $, such that $ T_n(\omega)=T_{n(\omega)}(\omega) $ when $ n\ge n(\omega) $.