(Protter Stochastic Integration and Differential Equation : Theorem 3)
I do not understand the proof ...
Let $X$ be an adapted cadlag process and let $\lambda$ be an open set. Then the hitting time $T(\omega)= \inf\ \{t>0: X_t \in \lambda\}$ is a stopping time.
Proof: $\{T<t\}=\bigcup\limits_{s \in \mathbb Q \cap [0,t)} \{ X_s \in \lambda\}$ since
(1) $\lambda$ is open and
(2) $X$ has right-continuous paths
I don't understand the relevance of (2) in proving the equality ?
As I understand it we are taking the countable union of sets $\bigcup\limits_{s \in \mathbb Q \cap [0,t)} \{ \omega:X_s(\omega) \in \lambda\}$.
Now for a given $\omega$ the first few sets will be empty (since the open set will not have been hit yet), then for some $s$ we finally have $X_s(\omega) \in \lambda$. But what is the relevance of the right-continuity (2) ?
If the process $X$ is not right-continuous, then $$\{T<t\} = \bigcup_{s \in \mathbb{Q} \cap [0,t)} \{X_s \in \lambda\}$$ does, in general, not hold true. Just consider the stochastic process $(X_t)_{t \geq 0}$ defined by $$X_t(\omega) := \begin{cases}5, & t=\pi, \omega \in \Omega, \\ 0, & \text{otherwise} \end{cases}$$ and the hitting time $$T := \inf\{t > 0; X_t \in \lambda := (4,6)\},$$ then $$\{T<t\} = \begin{cases} \emptyset, & t \leq \pi, \\ \Omega, & t > \pi, \end{cases}$$ but $$\bigcup_{s \in \mathbb{Q} \cap [0,t)} \{X_s \in (4,6)\} = \emptyset \qquad \text{for all $t \geq 0$}.$$ This shows that $$\{T < t\} \neq \bigcup_{s \in \mathbb{Q} \cap [0,t)} \{X_s \in \lambda\} \qquad \text{for all} \, \, t>\pi.$$