Prove that first exit time is a stopping time I have seen several questions on the site similar to the one I am about the ask, however none of the previous answers was exactly considering the kind of problem I am having. Therefore, I would appreciate if someone explained the following puzzle to me. I am following the proof given in Klebaner, Fima: Introduction to Stochastic Calculus with Applications, 2nd edition, pp. 52.

Let $S(t)$ be a continuous stochastic process adapted to the filtered probability space ($\Omega$, $F$,  $\mathcal{F}$, $P$), where $\Omega$ is the sample space, $F$ is a $\sigma$-field of subsets of $\Omega$, $P$ is a probability measure defined on the elements of $F$ and $\mathcal{F}$ is a filtration of increasing $\sigma$-fields, such that $F_t$ $\subset$ $F$. 

We want to prove that for any interval $D = (a,b)$ on the real line a first exit time (defined as $\tau_d =$ inf $\{t \geq 0: S(t) \notin D \}$) is also a stopping time.

The proof: We know that the following sets are equal: $\{\tau_d &gt; t\} = \{S(u) \in D, \text{ for all } u \leq t \} = \cap_{0 \leq u \leq t} \{S(u) \in D\}$. Although we know that each of the elements in the last set belong to $F_t $, we are not able to use this knowledge, because by definition $\sigma$-algebra is closed only under countable intersections. Therefore we have to transform this set to a countable set. And now comes the argument where I am stuck. The author further argues: "Due to the continuity of $S(u)$ and $D$ being open, for any irrational $u$ with $S(u) \in D$ there is a rational $q$ with $S(q) \in D$. Therefore:

$\bigcap\limits_{0 \leq u \leq t} \{S(u) \in D\}  = \bigcap\limits_{0 \leq q \leq t} \{S(q) \in D\}.$"

I understand that the last term is a countable intersection of events in $F_t$ and therefore the event $\{\tau_d &gt; t\}$ belongs to $F_t$ and is therefore a stopping time.

Therefore my question is:

Why is the set $\cap_{0 \leq u \leq t} \{S(u) \in D\}$ substitutable by $\bigcap\limits_{0 \leq u \leq t} \{S(u) \in D\}$ encompassing only irrational numbers and further why is this set $ \bigcap\limits_{0 \leq q \leq t} \{S(q) \in D\}$ encompassing only rational numbers identical to the previous one?
Every irrational number can be approximated by a rational number. 

Source: https://www.google.co.in/url?sa=t&amp;source=web&amp;rct=j&amp;url=https://people.eecs.berkeley.edu/~wkahan/Math55/p249n17.pdf&amp;ved=0ahUKEwjc-OTgyYjSAhWBLY8KHe2jDQgQFggfMAE&amp;usg=AFQjCNHUzr1RgddnCUCEsvGaSJ5Y1XT3sA

Now because S is continuous and since D is open (as the book said), the 2 intersection sets are equal (using the above facts, prove this by showing each is a subset of the other.)
Every irrational number can be approximated by a rational number. 

Source: https://www.google.co.in/url?sa=t&amp;source=web&amp;rct=j&amp;url=https://people.eecs.berkeley.edu/~wkahan/Math55/p249n17.pdf&amp;ved=0ahUKEwjc-OTgyYjSAhWBLY8KHe2jDQgQFggfMAE&amp;usg=AFQjCNHUzr1RgddnCUCEsvGaSJ5Y1XT3sA

Now because S is continuous and since D is open (as the book said), the 2 intersection sets are equal (using the above facts, prove this by showing each is a subset of the other.)