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Let $(\Omega,\mathscr F,(\mathscr F_t)_{t\geq 0},\mathsf P)$ be a complete filtered probability space and $X = (X_t)_{t\geq 0}$ be a cadlag stochastic process with value in a Polish space $E$. Is it true that for any closed subset $A$ of $E$ the first hitting time $$ \tau = \inf\{t\geq 0:X_t\in A\} $$ is a stopping time, i.e. $\{\tau\leq t\}\in \mathscr F_t$ for all $t\geq 0$; and what if $A$ is an open set.

As an example, we can consider $X$ with values in $\mathbb R$. It holds that the first hitting time of a closed half-line $$ \tau = \inf\{t\geq 0:X_t\in[K,\infty)\} $$ is a stopping time. Can we apply the following argument:

Let $\tau = \inf\{t\geq 0:X_t\in (K,\infty)\}$ then $$ \{\tau\leq t\} = \bigcup\limits_{n=1}^\infty\{\tau_n\leq t\}\in \mathscr F_t $$ where $\tau_n = \inf\{t\geq 0:X_t\in[K+1/n,\infty)\}$ and hence $\{\tau_n\leq t\}\in \mathscr F_t$.

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    @ Ilya :In your greyed box, if true, shouldn't it be an intersection (instead of an union), as if $\omega$ is in $\{\tau\le t\}$ then it is in $\{\tau_n\le t\}$ ? Best regards2012-02-20
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    @ Ilya : Anyway I don't think it is true as $\tau_n$ doesn't converge a.s. to $\tau$. But to $\tau'=inf\{t\ge 0, X_t\in [K,+\infty)\}$, best regards.2012-02-20
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    @TheBridge: thanks for mentioning it, there was a typo. I use this sequence because $$ (K,\infty) = \bigcup\limits_{n=1}^\infty[K+1/n,\infty). $$ The previous version of course was not correct, fixed it now.2012-02-20

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The measurability of hitting times is a subtle and complex problem, i.e., a pain in the neck. Of course, the raw $\sigma$-fields are pretty much hopeless, even if the process has continuous paths and $A$ is open. (Take two sample paths that start on the boundary of $A$; one stays put while the other immediately heads into $A$. If $\tau_A$ were a stopping time, then $1[\tau_A=0]$ should be a function of the initial position only.)

It is often assumed that the filtration is complete and right continuous, then things work out better. In this case, for open $A$ and $t>0$ we have $$\{\tau_A where $Q_t$ is the set of rational numbers in $[0,t)$. This, and the right-continuity of the fields shows that $\tau_A$ is a stopping time.

For general Borel sets, I suggest reading The measurability of hitting times by Richard Bass, Electron. Comm. Probab. 15 (2010) 99-105; 16 (2011) 189-191 for an elementary (but not easy) exposition of quite general results about hitting times. Note that even Prof. Bass made a subtle error, later corrected. The paper and correction are number 130 on his home page:

http://bass.math.uconn.edu/biblio.html

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    Thank you very much for these references, it seems to be something that I needed. Let me also put a link here to your perfect answer to the [related question of mine on MO a year ago.](http://mathoverflow.net/questions/50154/reachability-for-markov-process/50241#50241)2012-02-21
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    Glad to be of help.2012-02-21
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    Thanks. But the link seems down (fixed while I'm writing this). Also, if the filtration $\{\mathcal F_t\}$ is such that $\mathcal F_t$ is generated by random variables of the form $B_1(s_1,\cdot),\cdots, B_m(s_m,\cdot),s_k\le t$ in which $(B_1,\cdots,B_m)$ is a multidimensional Brownian motion, then are we guaranteed to say that $\{\mathcal F_t\}$ is right continuous?2017-03-27
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    By "raw" $\sigma$-fields I mean exactly those ${\cal F}_t$ in your definition. So, no, they aren't right continuous.2017-03-27