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Given a real valued stochastic process $X_t,t\in [0,\infty)$ adapted to a filtration $F_t$, it will be quite tempting to claim that $$\tau(\omega):=\inf\{t\ge 0\mid X_t(\omega)\in B\}$$ in which $B$ is a Borel set is a stopping time with regard to $F_t$, because at first glance it seems to just tell you the moment at which the particle hits the set $B$ "for the first time". This is what I initially anticipated too. However, much to my surprise, in this post @Did gave a counterexample.

Nevertheless, in this Wikipedia article, it is claimed that when $X_t$ is the Brownian motion, $F_t$ is the filtration it generates and $B=(a,\infty)$, then $\tau$ is indeed a stopping time which corresponds to the stopping rule: "stop as soon as the Brownian motion exceeds the value $a$." It thus leads me to believe that there may exist some sufficient conditions on $X_t$, $F_t$, or $B$ under which $\tau$ is indeed a stopping time. Could anyone enlighten me further? Thanks.

EDIT
It might be worth mentioning that I am particularly interested in the following conditions:
1). $X_t$ is a (generalised) Itô process, meaning it comes from a multidimensional Brownian motion $(B_1,\cdots,B_m)$.
2). $F_t$ is the sigma algebra generated by that multidimensional Brownian motion, in particular, $F_t$ is generated by r.v.s of the form $B_1(s_1,\cdot),\cdots,B_n(s_n,\cdot);s_k\le t$. (By the way, Is this sufficient to make $F_t=F_t^+$?)
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Vim
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  • @saz's comment on my answer over there provides one such condition: the right continuity of the filtration. It also suggests natural sufficient conditions such as $B$ being open and the paths of $X$ being continuous. – Did Mar 26 '17 at 10:23
  • @Did I saw that comment, thanks. In particular, I'm only interested in the case where $X$ is continuous a.s. since I've been dealing Itô processes. So is it true in general for Itô processes, without reducing the generality of the other components (i.e. still assuming the original filtration and arbitrary Borel set)? I'm still struggling with the case where $B=[a,\infty)$ though. – Vim Mar 26 '17 at 10:34
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    If $B=[a,\infty)$, then $[\tau\leqslant t]$ is the event that for every positive integer $n$, there exists $t_n\leqslant t+\frac1n$ such that $X(t_n)\geqslant a$. Can you then show that the event $[\forall s\leqslant t,X(s)<a]$ cannot happen, assuming that $X$ is path continuous? This answers your question. – Did Mar 26 '17 at 11:00
  • @Did oh yes of course, it's just a matter of taking intersection and using the density of $\Bbb Q$ and the continuity of $X$. What has me stuck is taking the complement... thanks, I understand this question now. A follow up question: since we have effectively proved the result for $B$ being general open intervals, i think it's not hard to furthering it for $B$ being général Borel sets, does the result really hold then? – Vim Mar 26 '17 at 11:06
  • @Did wait a minute, it seems the counterexample you constructed in that answer is also path-continuous isn't it? And there $B$ is just two half intervals united, everything is fine but $\alpha$ isn't a stopping time. – Vim Mar 26 '17 at 11:13
  • @Did take a look at my argument s'il vous plait: assuming path continuity of $X_t$ and $B$ open, then note that $$[\tau\le t]=[\forall m\in\Bbb N, \exists s<t+1/m,\text{s.t.}, X_s\in B] = \cap_{m\ge 0}\cup_{s<t+1/m}X_s^{-1}(B)$$ But by continuity of $X_t$ we have $$\cup_{s<t+1/m}X_s^{-1}(B) = \cup_{s<t+1/m,,s\in\Bbb Q}X_s^{-1}(B)$$ hence proving the result. Now returning to your counterexample, $X_t$ is clearly path continuous by your construction, and $B=(-\infty,-1)\cup(1,\infty)$ there is clearly open, however $\tau$ ($\alpha$ there) isn't a stopping time. – Vim Mar 26 '17 at 11:37
  • @Did ...and the scariest of all is I cannot identify any flaw either in your answer or in my proof. So one of us must have been missing something (considering you are much more sophisticated in this area, the hidden flaw is most likely on my part, though). – Vim Mar 26 '17 at 11:39
  • You are missing that a continuity argument that works for B closed may not work for B open. – Did Mar 26 '17 at 11:40
  • @Did Sorry I am not getting it and I realised I actually misinterpreted your second comment: I thought you were asking me to convert the case $B=[a,\infty)$ to the case $B=(-\infty,a)$ by taking complements, however after some serious thought I now found it hard to deal with closed $B$, because the continuity argument won't work anymore if I take complements of each $X^{-1}s(B)$ inside the union $\cup{s<t+1/m}X_s^{-1}(B)$. Moreover, my argument (with which I can't find any problem) in fact works exactly for open $B$. Now I can deal with open $B$ but am lost with $B=[a,\infty)$... – Vim Mar 26 '17 at 12:11
  • @Did ... per your comment I understand why $[\forall s\leqslant t,X(s)<a]\cap[\tau\leqslant t]$ is empty since $X(t)<a$ isn't possible under this assumption, but how does it lead to the $F_t$-measurability of $[\tau\le t]$? – Vim Mar 26 '17 at 12:13
  • Sorry but I never suggested to consider complements, actually there are several arguments in your comments quite unrelated to what I actually wrote hence I would suggest to read very carefully what I wrote and to stick to the arguments I actually used rather than to others, even seemingly similar ones. Re the measurability you ask about at the end of your last comment, it follows from the fact that, if $\tau=\inf{t\mid X_t\geqslant a}$ and if $X$ is path continuous then $$[\tau\leqslant t]=\bigcup_{q\in\mathbb Q,q>0}[X_{t-q}\geqslant a]$$ Let me recommend, to see if you really understood ... – Did Mar 26 '17 at 13:09
  • ... the argument, to try to understand why, if $\sigma=\inf{t\mid X_t>a}$, even if $X$ is path continuous, then, in general, $$[\sigma\leqslant t]\ne\bigcup_{s\leqslant t}[X_s>a]$$ – Did Mar 26 '17 at 13:09
  • @Did 1). I don't think $$[\tau\leqslant t]=\bigcup_{q\in\mathbb Q,q>0}[X_{t-q}\geqslant a]$$ is true. In particular, when $\tau = t$, there need not exist any $s<t$ such that $X_s \ge a$, e.g. what if the path $\omega$ has a unique global maximum $a$ at $t$? 2). I understand why $$[\sigma\leqslant t]\ne\bigcup_{s\leqslant t}[X_s>a]$$, for instance if $X(\cdot,\omega)$ is a strictly increasing path that reaches $a$ at $t$ then $\tau(\omega)=t$ yet for all $s\le t$, $X_s\le a$. But this is different from my argument, which is $$[\tau\le t]= \bigcap_{m\in\Bbb N}\bigcup_{s<t+1/m}X_s^{-1}(B).$$ – Vim Mar 26 '17 at 21:48
  • *Typo: replace $q>0$ by $q\geqslant0$. // If your argument involves an uncountable union, it is worthless to show measurability. – Did Mar 26 '17 at 22:19
  • @Did I'm afraid this doesn't make much difference. For example, if $\omega$ is a path such that $X_s$ reaches a unique global maximum $a$ at the point $t-\sqrt 2$, then there still doesn't exist any $q\ge 0$ such that $X_{t-q}\ge a$? – Vim Mar 26 '17 at 22:26
  • @Did per my argument: sorry I didn't show it complete: actually due to path continuity and openness of $B$ I can have $$\bigcup_{s<t+1/m}X_s^{-1}(B)=\bigcup_{s<t+1/m,s\in\Bbb Q}X_s^{-1}(B)$$ which is a countable union and is measurable with regard to $F_{t}^+$. But since $F_t$ is generated by path-continuity processes (more specifically, you can think of it as generated by independent Brownian motions) I think $F_t=F_t^+$ should be okay. – Vim Mar 26 '17 at 22:30
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    Maybe helpful: http://math.stackexchange.com/questions/111273/hitting-open-sets/111450#111450 –  Mar 27 '17 at 00:23
  • @ByronSchmuland thanks again for your reference to the début theorem. However, now I become even more confused: is it true that any adapted path-continuous process $X_t$ is predictable wrt the filtration it generates? I guess so because we can know $X_t$ from $X_s, s\to t-$. Now, predictability implies progressive measurability, and by the début theorem every hitting time of a progressively measurable process is a stopping time. But this clearly contradicts the counterexample by Did. – Vim Mar 27 '17 at 04:13

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