Measurability From countable to Uncountable sequences family of functions I am studying Protter Stochastic Integration and Differential Equations. In a theorem on hitting times Protter has the following theorem.
Let $X$ be an adapted cadlag stochastic process, and $\Lambda$ be an open set. Then the hitting time of $\Lambda$ is a stopping time. 
The proof of the theorem makes use of the fact that the hitting time defined by,
$T(\omega) = \inf \{t > 0 : X_t \in \Lambda\}$.
The proof claims $\{T(\omega) < t\} = \bigcup_{s \in \mathbb{Q} \cap [0,t)} \{X_s \in \Lambda \}$
I understand that $\{x : \inf f_n(x) < \alpha \} = \bigcup_n \{x: f_n(x) < \alpha\}$ for a countable family of measurable functions $f_n$. I am not sure how we extend this idea to an uncountable family of right continuous measurable functions.
 A: The inclusion
$$\{T(\omega)<t\} \supseteq \bigcup_{s\in\mathbb Q \cap [0,t)} \{X_s(\omega)\in\Lambda\}$$
holds clearly even without $\Lambda$ being open and $X$ being cadlag. For the other inclusion we need right-continuity and openess are indeed essential. If $\{T(\omega)<t\}$ for some $t$, then there is some time $s\in\mathbb R$ with $s<t$ and $X_s\in\Lambda$. As $\Lambda$ is open and $X$ right-continuous, increasing $s$ a bit does not result in $X$ leaving $\Lambda$. Hence, we find some $r\in\mathbb Q\cap [0,t)$ with $X_r\in\Lambda$. This proves the second inclusion.
A: It has been a while that I posted this question. I unfortunately did not quite understand the proof on a second visit to the same question. Below is my proof. Is this right?
The proof uses the fact that the indexing set is $\mathbb{R}$. We will use also the fact that the infimum of a set is the same as the infimum of its closure and the fact that $X$ is cadlag. 
    \begin{align}
 &\inf\{ t \in \mathbb{R}^+ : X_t(\omega) \in \Lambda\} = \inf \{q \in \mathbb{Q}^+ : X_q(\omega) \in \Lambda\} \\
 &\{\omega : T(\omega) < \alpha \} = \{\omega: \inf\{t > 0, X_t(\omega) \in \Lambda \} < \alpha \}  = \{\omega: \inf\{q \in \mathbb{Q}^+, X_q(\omega) \in \Lambda \} < \alpha \}
 \end{align}
The next step is to show the following equality $$S_l = \{\omega: \inf\{q \in \mathbb{Q}^+, X_q(\omega) \in \Lambda \} < \alpha \} = \bigcup_{q \in \mathbb{Q}^+ \bigcap [0,\alpha)} \{\omega: X_q(\omega) \in \Lambda \} = S_r$$
    The argument is pretty straight forward. If $\omega \in S_l$ then $q \in \mathbb{Q}^+, q < \alpha$ and $X_q(\omega) \in \Lambda \Rightarrow \omega \in S_r \Rightarrow S_l \subseteq S_r$. Similarly if $\omega \in S_r$ the same argument applies, i.e. $q \in Q^+, q < \alpha, X_q(\omega) \in \Lambda \Rightarrow S_r \subseteq S_l$. 
The rest is to prove the measurability, here it is anyway.
Now that we have reduced the uncountable system of sets to a countable union, we can now apply the usual measure theoretic stuff to prove the rest. Since $X_t$ are all cadlag and adapted $\{\omega: X_t(\omega) \in \Lambda\} = \{\omega: \omega \in X_t^{-1}(\Lambda) \} \in \mathcal{F}_t$. 
