Limit of decreasing sequences of markov time (stopping time) is markov time?

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Let $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t \geqslant 0}, \mathbb{P})$ be a filtered probability space and let $\tau_n \geqslant \tau_{n+1}$ be a markov time (stopping time) with respect to filtering $\{\mathcal{F}_t\}_{t\geqslant 0}$. Also for every $\omega \in \Omega$ next limit is exist: $$lim_{n\to\infty} \tau_n = \tau_0$$ Will $\tau_0$ stopping time or not with respect to same filtering?

I proved following: $\bigcap_{m \geqslant 1} \bigcup_{n \geqslant 1} \bigcap_{k \geqslant n} \{ \omega : \tau_k \leqslant t + \frac{1}{m} \} = \{ \tau_0 \leqslant t \}$

If left side of equal is measurable with respect to $ \mathcal{F}_t $ then $\tau_0$ - stopping time. A set from left side $ \{ \omega : \tau_k \leqslant t + \frac{1}{m} \} $ for every $ m \geqslant 1$ and every $ k \geqslant 1$ is measurable with respect to $ \mathcal{F}_{t+\frac{1}{m}} $. Therefore union of intersection (with index $k$) of these sets is measurable with respect to $ \mathcal{F}_{t+\frac{1}{m}} $ by definition of sigma-algebra.

And there is a first question: will the second intersection (with index $m$) of these sets, measurable with respect to $ \mathcal{F}_{t+\frac{1}{m}} $ ,be measurable with respect to $ \mathcal{F}_t $?

It seems non-measurable, so a second question following: Is there example of $\tau_n$ and $\{\mathcal{F}_t\}_{t \geqslant 0}$ in any book that can demonstrate it?