Stochastic process with right-continuous paths, intersection of borel preimages Let $X=(X_t)_{t\in \mathbb{R^+}}$ be a stochastic process with right continuous paths adapted to the filtration $\mathbb F=(\mathcal F_t)_{t\in \mathbb{R^+}}$ and assume $B$ is a Borel set. 
I have stumbled upon these ideas and I am not sure to what extent they are valid and whether one implies the other:


*

*Is it correct to assume, based on right-continuity, that $${\{X_s \in B\}}= \bigcap_{\varepsilon>0}{\{X_{s+\varepsilon} \in B\}} $$ 
and, if so, how can it be shown rigorously?

*This second one I have seen many a times in slightly different forms:
$${\bigcap_{s  \in I} \{X_s \in B\}}= \bigcap_{s \in\mathbb{Q} \cap I}{\{X_{s} \in B\}}$$ where $I$ is, say, an interval. I understand that it has to do with the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ but could someone show it in detail? Does right-continuity matter at all? What about 
$${\bigcup_{s  \in I} \{X_s \in B\}}= \bigcup_{s \in\mathbb{Q} \cap I}{\{X_{s} \in B\}}$$ ?
 A: 
$\{X_s \in B\} = \bigcap_{\epsilon>0} \{X_{s+\epsilon} \in B\}$

This identity does, in general, not hold true; even for very nice sets $B$. Note that this would mean that $X_s(\omega) \in B$ for some $s \geq 0$ implies $X_{s+\epsilon}(\omega) \in B$ for all $\epsilon >0$, i.e. after hitting the set $B$ the path will stay in $B$.
Counterexample: Consider the deterministic process $X_t := t$ and $B:=\{0\}$. Then $\{X_0 \in B\} = \Omega$, but $\{X_{\epsilon} \in B\} = \emptyset$ for all $\epsilon>0$.

$\bigcup_{s \in I} \{X_s \in B\} = \bigcup_{s \in \mathbb{Q} \cap I} \{X_s \in B\}$

Whether such an identity holds depends where much on the set $B$. For general Borel sets $B$ this does, in general, not hold true.
Example: Consider again $X_t := t$. If we define $B := \mathbb{R} \backslash \mathbb{Q}$, then $$\bigcup_{s \in \mathbb{Q} \cap I} \{X_s \in B\} = \emptyset$$ for any interval $I$. However, $\{X_s \in B\} = \Omega$ for $s \geq 0$, $s \in \mathbb{R} \backslash \mathbb{Q}$, and therefore $$\bigcup_{s \in I} \{X_s \in B\} = \Omega.$$
From now on, I'll assume for simplicity that $I=[0,\infty)$; for the more general setting see the remark at the end of my answer.
The identity does hold true for open sets $B$. The inclusion "$\supseteq$" is trivial; to prove "$\subseteq$" we have to use the right-continuity. If $X_s(\omega) \in B$ for some $s \in I$, then we can choose $\epsilon>0$ such that $B(X_s(\omega),\epsilon) \subseteq B$ for all $\epsilon>0$. Now the right-continuity implies that there exists $\delta>0$ such that $$|X_s(\omega)-X_t(\omega)| < \epsilon \qquad \text{for all $t \in [s,s+\delta]$}$$ and therefore $X_t(\omega) \in B(X_s(\omega),\epsilon) \subseteq B$ for $t \in [s,s+\delta]$.

$\bigcap_{s \in I} \{X_s \in B\} = \bigcap_{s \in \mathbb{Q} \cap I} \{X_s \in B\}$.

This holds, for instance, if $B$ is closed and $(X_t)_{t \geq 0}$ is right-continuous. The inclusion "$\subseteq$" is obvious. To prove "$\supseteq$" suppose that $X_s(\omega) \in B$ for all $s \in \mathbb{Q} \cap I$. Now for $t \in I$ choose a sequence $s_n \in \mathbb{Q} \cap I$ such that $s_n \downarrow t$. Then, by the right-continuity,
$$X_t(\omega) = \lim_{n \to \infty} X_{s_n}(\omega).$$
Since $X_{s_n}(\omega) \in B$ it follows from the closedness of $B$ that $X_t(\omega) \in B$.
If $B$ is not closed, then we can, in general, not expect that such an identity holds.
Remark: In the proofs we have used that for any irrational $s \in I$ there exists a sequence $s_n \downarrow s$, $s_n \in \mathbb{Q} \cap I$. Therefore, the statements remain valid if we replace $I=[0,\infty)$ by one of the following sets:


*

*$[a,b)$ for $a \in \mathbb{R}$, $b \in \mathbb{R} \cup \{\infty\}$,

*$(a,b)$ for $a \in \mathbb{R} \cup \{-\infty\}$, $b \in \mathbb{R} \cup \{\infty\}$,

*$[a,q]$ for $a \in \mathbb{R}$, $q \in \mathbb{Q}$

*$(a,q]$ for $a \in \mathbb{R} \cup \{-\infty\}$, $q \in \mathbb{Q}$

