Adapted Cadlag processes : hitting times are stopping times?

Question

(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 (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 $\bigcup\limits_{x \in \mathbb Q \cap [0,t)} \{ \omega:X_s(\omega) \in \lambda\}$

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) ?

Answer 1

If the process $X$ is not right-continuous, then $$\{T 0; X_t \in \lambda := (4,6)\},$$ then $$\{T \pi, \end{cases}$$ 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 $$\{T < t\} \neq \bigcup_{s \in \mathbb{Q} \cap [0,t)} \{X_s \in \lambda\} \qquad \text{for all} \, \, t>\pi.$$