measurability question with regard to a stochastic process Here are two related exercise from Karatzas and Shreve
Let $X$ be a process, every sample path of which is right continuous with left limits.
Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A\in\mathcal{F}_{t_0}^X$ (that is the $\sigma$-algebra generated by $X$)
Embarassingly, I am not sure how to do this. If I write down 
$\bigcap_{\epsilon\in \{1,1/2,1/3,...\}}\bigcup_{q\in \mathbb{Q}^+}\bigcap_{p\in\mathbb{Q},0<t-p\leq q} \{|X_t-X_p|<\epsilon\} $
Is this equivalent to $X$ is continuous at $t$? (If so, this is measurable w.r.t $\mathcal{F}^X_t$) but this wouldn't be helpful, because, we want left continuity on an interval, and still it wouldn't be a countable union of events which are measurable.

If $X$ is only almost surely RCLL, then $A$ can fail to be in $\mathcal{F}_{t_0}^X$ but if $\mathcal{F}_t$ is a filtration such that $\mathcal{F}_{t_0}^X\subset\mathcal{F}_t$, $t\geq 0$, and $\mathcal{F}_{t_0}$ contains all null sets, then $A\in \mathcal{F}_{t_0}$
I assume this depend on the previous part. Will skip for the time being.
 A: The set you wrote is the event that $X_t$ is left-continuous at $t > 0$.  However, as you commented, it wouldn't work because  you need one more intersection over $t \in [0, t_0)$ to say $X_t$ is left-continuous for all $t \in [0, t_0)$ but it may not be expressed as a countable intersection.  This is overcome by employing uniform continuity instead, where RCLL property plays a key role.
So let's first show the following lemma:

An RCLL function $f \colon [0, +\infty) \to \mathbb{R}$ is continuous on $[0, t_0)$ if and only if the following holds:
  $$
\forall \varepsilon > 0, \ \exists \delta > 0, \ 
\forall q_1, q_2 \in \mathbb{Q} \cap [0, t_0), \ 
\lvert q_1 - q_2 \rvert < \delta \implies \lvert f(q_1) - f(q_2) \rvert < \varepsilon
\qquad \text{(1)}
$$

Once this lemma is proved, it's easy to see that
$$
A = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}}
    \bigcap_{\substack{q_1, q_2 \in \mathbb{Q} \cap [0, t_0)\\
                       \lvert q_1 - q_2 \rvert < 1/m}}
    \biggl\{ \lvert X_{q_1} - X_{q_2} \rvert < \frac{1}{n} \biggr\}
$$
and, hence, $A$ is $\mathcal{F}_{t_0}^X$-measurable.  To prove the lemma, let's define $\widetilde{f} \colon [0, t_0] \to \mathbb{R}$ by
$$
\widetilde{f}(t) = \begin{cases}
  f(t) & \text{if}\ t \in [0, t_0), \\
  \displaystyle\lim_{s \uparrow t_0} f(s) & \text{if}\ t = t_0.
\end{cases}
$$
This $\widetilde{f}$ is well-defined because $f$ has finite left-hand limits.
First, suppose that $f$ is continuous on $[0, t_0)$.  Then, obviously $\widetilde{f}$ is continuous on $[0, t_0]$ but, since $[0, t_0]$ is compact, it's also uniformly continuous, i.e.:
$$
\forall \varepsilon > 0, \ \exists \delta > 0, \
\forall s, t \in [0, t_0], \ \lvert s - t \rvert < \delta \implies
\lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert < \varepsilon.
$$
Condition $(1)$ immediately follows from this.
Conversely, suppose that condition $(1)$ holds.  We shall show that $\widetilde{f}$ is uniformly continuous on $[0, t_0]$.  For any fixed $\varepsilon > 0$, by condition $(1)$ and the definition of $\widetilde{f}$, we can pick $\delta_1, \delta_2 > 0$ so that
$$
\forall q_1, q_2 \in \mathbb{Q} \cap [0, t_0), \ 
\lvert q_1 - q_2 \rvert < \delta_1 \implies
\lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert < \frac{\varepsilon}{3}
$$
and
$$
\forall q \in \mathbb{Q} \cap [0, t_0], \
\lvert q - t_0 \rvert < \delta_2 \implies
\lvert \widetilde{f}(q) - \widetilde{f}(t_0) \rvert < \frac{2\varepsilon}{3},
$$
respectively.  Then, by choosing $\delta = \min\{\delta_1, \delta_2\}$, we shall show that, for all $s, t \in [0, t_0]$, $\lvert s - t \rvert < \delta$ implies $\lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert < \varepsilon$.  Since it obviously holds when $s = t$, without loss of generality, we may assume that $s < t$.  Let's consider two cases: $t < t_0$ and $t = t_0$.  If $t < t_0$, the right continuity of $f$ guarantees the existence of $q_1 \in \mathbb{Q} \cap (s, t)$ and $q_2 \in \mathbb{Q} \cap (t, t_0)$ such that
$$
s < q_1 < s + \delta, \qquad
\lvert \widetilde{f}(s) - \widetilde{f}(q_1) \rvert < \frac{\varepsilon}{3}, \\
t < q_2 < t + q_1 - s, \qquad
\lvert \widetilde{f}(t) - \widetilde{f}(q_2) \rvert < \frac{\varepsilon}{3}.
$$
Then, noting that $\lvert q_1 - q_2 \rvert < \lvert s - t \rvert < \delta \le \delta_1$, we have $\lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert < \varepsilon / 3$ by the selection of $\delta_1$.  Hence,
$$
\lvert \widetilde{f}(s) - \widetilde{f}(t) \rvert \le
\lvert \widetilde{f}(s) - \widetilde{f}(q_1) \rvert +
\lvert \widetilde{f}(q_1) - \widetilde{f}(q_2) \rvert +
\lvert \widetilde{f}(q_2) - \widetilde{f}(t) \rvert <
\varepsilon.
$$
For the case where $t = t_0$, pick $q \in \mathbb{Q} \cap (s, t_0)$ in the same way as $q_1$ above.  Then, by noting that $\lvert q - t_0 \rvert < \lvert s - t_0 \rvert < \delta \le \delta_2$ and by the selection of $\delta_2$, we have
$$
\lvert \widetilde{f}(s) - \widetilde{f}(t_0) \rvert \le
\lvert \widetilde{f}(s) - \widetilde{f}(q) \rvert +
\lvert \widetilde{f}(q) - \widetilde{f}(t_0) \rvert < \varepsilon.
$$
We have shown that $\widetilde{f}$ is uniformly continuous on $[0, t_0]$, which implies the continuity of $f$ on $[0, t_0)$.     Q.E.D.
Please see p.38 for the answer to the latter question.
