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It's known that, given $\Gamma \in \mathcal B (\mathbb R ^d)$ and $X = > (X_t)_{t\geq 0}$ with right-continuous path, the random time

$$T_{\Gamma} = \inf \{ t\geq 0 : X_t (\omega) \in \Gamma\} $$

  • is an optional time if $\Gamma$ is open,
  • it's a stopping time if $\Gamma$ is closed.

this result comes from I. Karatzas, S. E. Shreve, Brownian Motion and Stochastic Calculus.

My first question is whether this result holds also for left-continuous paths processes.

More generally, if the filtration $(\mathcal F_t)_{t\geq 0}$ is right-continuous, then for any adapted, optional process $X = (X_t)_{t\geq 0}$ and any borelien $\Gamma$, $T_{\Gamma}$ is a stopping time.

A result from P. A. Meyer, Dellacherie, Probabilit├ęs et Potentiel. (I wish hardly to see this proof so if anyone has a numerical version of this book I'd appreciate it very much).

An exercice which serves as an exemple is the following.

Let be $\alpha, \beta \in \mathbb R$ such that $\alpha < \beta $ and $x \in [\alpha, \beta ]$. Consider the random time

$$T_x = \inf \{ t\geq 0 : x+ B_t \notin [\alpha, \beta]\},$$ where $B=(B_t)_{t\geq 0}$ is a standard brownian motion in $\mathbb R$ starting from zero. Show that $T_x$ is a stopping time and $\mathbb P -\text{a.e.}$ finite.

The solution I've seen say so.

Consider $\tilde{ T_x} = \inf \{ t\geq 0 : x+ B_t \in (-\infty,\alpha[\cup ]\beta, +\infty) \},$ then $\{T_x \leq t\}=\{\tilde{T_x} \leq t\}$ without justifying. HERE the first question about this example.

Then we start to consider $\{T_x \leq t\}^c$

\begin{align} \{T_x > t\}&= \{\forall s \in [0,t], x+B_s \in [\alpha, \beta] \}\cup \{x+B_t =\alpha\}^c \cup \{x+B_t =\alpha\}^c \\ &= \underset{\{q\in [0,t],\\ q \in \mathbb Q\}}{\cap}\{x+B_q \in [\alpha, \beta]\}\end{align}

since $\{x+B_q \in [\alpha, \beta]\}\in \mathcal F_q \subset \mathcal F_t, \ \forall q \in \mathbb Q $, we conclude that $T_x$ is a stopping time.

In addition, in the solution one remarks that $\{x+B_t =\alpha\}^c ,\ \{x+B_t =\alpha\}^c \in \mathcal F_0$.

The point is I did not understand why the first line is right even less why the last remark is needed, obviously supposing that there is no mistake and no redundancy. I don't see why the current version wold be correct. More precisely, why would not be correct to write $\{T_x > t\}= \{\forall s \in [0,t], x+B_s \in [\alpha, \beta] \}$ simply.

I didn't get neither why must we index the intersection by rational numbers. Is it really necessary? The only answer I find to myself is that is just a technical issue related countable sets and indexation. But we could simply "index ate" by $s \in \mathbb R$, don't we ? Or should it be technically wrong by some reason I am unaware?

Edit: I made showing that $T_x < + \infty \ \mathbb P -\text{a.e.}$ subject of another question

I would appretiate if someone could help me whith all those questions.

Thanks in advance.

share|improve this question
If I might offer a suggestion, you may not really want to see the proof in Probabilités et Potentiel. It is capacity theoretic and hard. There is a recent, elementary proof of the Debut Theorem by Bass here: arxiv.org/pdf/1001.3619v2.pdf –  Chris Janjigian Oct 23 '13 at 1:47
@ChrisJanjigian : Thanks a lot Chris. Do you have a pdf version of Probabilités et Potentiel? I would like to see it anyway and other staff too. –  Paul Oct 23 '13 at 2:28
The part after "Yet" should be omitted since you asked it in a separate question. –  Did Oct 23 '13 at 6:49
@Did: Sure, I forgot it. TY –  Paul Oct 23 '13 at 7:47

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