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Let $(T_n)_{n\in\mathbb N}$ denote some stopping times and $(\mathcal F_t)$ a filtration continuous on the right, i.e. $$\mathcal F_t=\bigcap_{s>t}\mathcal F_s.$$ I want to show that $\inf_{n\in\mathbb N}T_n$ is a stopping time. I don't know what's wrong in my argument:

$$\{\inf_{n\in\mathbb N}T_n\leq t\}=\{\inf_{n\in \mathbb N}T_n> t\}^c=\left(\bigcap_{n\in\mathbb N}\{T_n>t\}\right)^c=\bigcup_{n\in\mathbb N}\{T_n>t\}^c\in\mathcal F_t.$$ I don't use the fact that the filtration is continuous on the right, so I think it's wrong, but I don't understand why. Any idea ?

Did
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    $$\left{\inf_{n\in \mathbb N}T_n> t\right}\ne\bigcap_{n\in\mathbb N}{T_n>t}$$ Consider for example $T_n=t+\frac1n$, then $T_n>t$ for every $n$ but $\inf T_n=t$, not $>t$. – Did Oct 13 '16 at 12:36
  • @Did: Yes sorry, but we don't have that ${\inf T_n\geq t}=\bigcap_{n\in\mathbb N}{T_n\geq t}$ ? For example, ${\sup T_n\leq t}=\bigcap_{n\in\mathbb N}{T_n\leq t}$, so why don't it works with $\inf T_n$ ? – MSE Oct 13 '16 at 12:42
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    Yes we have that -- but I am afraid you will have to much more careful and specific if you want to understand the necessity of the filtration to be continuous on the right. So, when is $T$ a stopping time for $(\mathcal F_t)$, already? – Did Oct 13 '16 at 13:09
  • @Did: $T$ is a stopping time if ${T\leq t}\in \mathcal F_t$ for all $t$. Since $T_n$ is a stopping time for all $n$, then ${T_n\geq t}\in\mathcal F_t$ for all $t$, and thus $\bigcap_{n\in\mathbb N}{T_n\geq t}\in \mathcal F_t$, therefore ${\inf_n T_n\geq t}\in \mathcal F_t$ for all $t$ and thus $\inf_n T_n$ is a stopping time. So for you this is wrong ? If yes, why ? Thank you. – MSE Oct 13 '16 at 13:15
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    First mistake: you are given a property of ${T\leqslant t}$ and you ascribe it to ${T\geqslant t}$. – Did Oct 13 '16 at 14:08
  • The identity ${\inf_{n\in\mathbb N}T_n\leq t} = \cup_{n \in \mathbb{N}} {T_n \le t}$ is obvious. – Gordon Oct 13 '16 at 14:41
  • @Did: But I also know that if $T$ is a stopping time, then ${T\geq t}\in \mathcal F_t$, that's why I conclude like this. Is it wrong ? – MSE Oct 14 '16 at 08:19
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    @Gordon "Obvious" and wrong. – Did Oct 14 '16 at 08:26
  • @MSE Yes this is wrong, because you do not know that ${T\geqslant t}$ is in $\mathcal F_t$, a priori only ${T>t}={T\leqslant t}^c$ is in $\mathcal F_t$. – Did Oct 14 '16 at 08:27
  • @Did: I see ! Thank you very much, it's clear now :-) And to have ${T\geq t}\in \mathcal F_t$ we need the continuity of $\mathcal F_t$, right ? – MSE Oct 14 '16 at 08:52
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    Indeed, without continuity this can fail. – Did Oct 14 '16 at 12:11
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    Thanks @Did. It should be ${\inf_{n \in \mathbb{N}} T_n \le t} = \cap_{k=1}^{\infty} \cup_{n\in \mathbb{N}}{T_n \le t +\frac{1}{k}}$, and then, by the continuity, we deduce that ${\inf_{n \in \mathbb{N}} T_n \le t} \in \mathscr{F}_t$. – Gordon Oct 14 '16 at 12:48
  • @Gordon maybe post the answer since it's not obvious the question is resolved (or OP can do it?) – baibo Nov 01 '19 at 09:44

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