Show that $\alpha = \inf\{t\,;\,|X_t|>1\}$ is a stopping time Let $\{X_t, t \ge 0\}$ be a continuous stochastic process and adapted to the filtration $\{\mathcal{F}_t,t\ge 0 \}$ and consider
$$
\alpha = \inf\{t, |X_t|>1\},
$$ 
the first time the the process $X_t$ leaves the interval $[-1,1]$. Then can you help me to show that $\alpha$ is in fact stopping time?
 A: The proof below is not correct. See an example by Did. The flaw in the proof: in fact we have
$$
  \{\alpha>t\} = \bigcup_n\{\alpha \geq t+\frac1n\} = \bigcup_{n}\bigcap_{s\in [0,t+\frac1n]}\{|X_s|\leq 1\}
$$
and there appear events $\{|X_s|\leq 1\}$ for $s>t$.

By the definition, a random variable $\tau:\Omega\to [0,\infty]$ is a stopping time if and only if
$$
  \{\tau \leq t\}\in \mathscr F_t
$$
for any $t\in [0,\infty)$. We have in your case
$$
  \{\alpha \leq t\} = \Omega\setminus \{\alpha > t\} = \Omega\setminus \bigcap_{s\in [0,t]}\{|X_s|\leq 1\} = \Omega\setminus\bigcap_{s\in \Bbb Q\cap [0,t]}\left\{|X_s|\leq 1\right\} \in \mathscr F_t
$$
where we pass to the intersection over rational numbers only since $X$ has continuous trajectories.
A: 
Then can you help me to show that $\alpha$ is in fact stopping time?

Well, this would be difficult, since $\alpha$ is not always a stopping time for the natural filtration $(\mathcal F^X_t)_{t\geqslant0}$ of the process $(X_t)_{t\geqslant0}$, defined by $\mathcal F^X_t=\sigma(X_s\,;\,0\leqslant s\leqslant t)$.
For a counterexample, consider some event $A$ such that $0<P(A)<1$ and define the process $(X_t)_{t\geqslant0}$ as follows: 


*

*On $A$, $X_t=t$ for every $t\geqslant0$.

*On $\Omega\setminus A$, $X_t=t$ for every $0\leqslant t\leqslant1$ and $X_t=2-t$ for every $t\geqslant1$.


Then $\alpha=\inf\{t\,;\,|X_t|\gt1\}$ equals $\alpha=1$ on $A$ and $\alpha=3$ on $\Omega\setminus A$.
Thus, $\{\alpha\leqslant1\}=A$. For every $t\leqslant1$, the random variable $X_t$ is deterministic, hence $\mathcal F^X_1=\{\varnothing,\Omega\}$ is the trivial sigma-algebra, but the event $A$ is not in $\mathcal F^X_1$ hence $\{\alpha\leqslant1\}\notin\mathcal F^X_1$, which proves that $\alpha$ is not a stopping time for the filtration $(\mathcal F^X_t)_{t\geqslant0}$.
