# About stopping times and progressively measurable process

Given a filtered probability space $\left ( \mathbb P, \mathcal F, \left(\mathcal F_t \right)_{t \geq 0}, \Omega\right)$, consider a process $\phi = \left( \phi_t \right)_{t\geq 0}$ $\mathcal F_t$- progressively measurable such that $\mathbb E \left\{ \int _0 ^t \left |\phi_s \right |^2 ds\right\}< \infty$ for all $t \geq 0$. I would like to show that the stopping time sequence $\left(\tau_p \right)_{p\geq0}$ defined by

$$\tau_p := \inf \{ t\geq 0: \left |\phi_s \right | \geq p \}$$

is such that $\tau_p \nearrow \infty$ $\mathbb P$- a.e as $p \rightarrow \infty$

Edit: As "pgassiat" notice, the definition of $\tau_p$ is mistaked. But, I believe it can be changed in order to work. Could you help me on it please ?

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Something's missing. As written, take e.g. $\phi_s=s^{-1/4}$ deterministic, then $\tau_p=0$ for all $p$. – pgassiat Jan 15 '13 at 20:10

I belive I've found the solution for my own question. Changing the definition of $\tau_p$ for
$$\tau_p := \inf \{ t\geq 0: \int_0 ^t \left |\phi_s \right |^2 ds\geq p \}$$
and supposing by contratiction there is a constant $M >0$ such that $$\mathbb P \left\{ \tau_p \leq M, \forall p \geq 0 \right\} >0 \ ,$$ we have for all $p \geq 0$
My mistake, did. I've put $p$ where it was $M$ almost everywhere (not in the mathematical sense! =) ). Please, see the edited version and say me if you agree now. – Paul Jan 15 '13 at 22:20