Show that a certain set is measurable I just started to working through Karatzas and Shreve "Brownian Motion and Stochastic Calculus". Solving an exercise there's some questions around. I want to prove.

Let $X$ be a stochastic process, every sample path of which is RCLL. Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A\in \mathcal{F}_{t_0}^X$, where the latter is defined as, $\mathcal{F}_{t_0}^X:=\sigma{(\{X_s;0\le s\le t_0\})}$

What I did (sketch):
Write $A$ as, 
$$ A=\{\omega \in \Omega; \lim_{s\uparrow t}X_s(\omega)=\lim_{s\downarrow t}X_s(\omega)\}$$
for $t\in [0,t_0)$. Since the Limit of a sequence of measurable functions is measurable, this is in $\mathcal{F}_{t_0}^X$. For more details see the answer of Byron Schmuland below. 
Now here's the part which is unclear. There's a second claim, very similar to the first one:  

Let $X$ be a stochastic process whose sample paths are RCLL a.s. Let $A$ be the event that $X$ is continuous on $[0,t_0)$. Show that $A$ can fail to be in $\mathcal{F}_{t_0}^X$. But if $\{\mathcal{F_t;t\ge 0}\}$ is a filtration satisfying $\mathcal{F}^X_t\subset\mathcal{F}_t,t\ge 0$ and $\mathcal{F}_{t_0}$
  is complete under P, then $A \in \mathcal{F}_{t_0}$.

There's a sample solution to this second question. But what I don't get is the difference between the two questions. What exactly is the difference? As mentioned in my comment after Byron Schmulands answer I think this is more a linguistic problem.  
cheers
math
 A: These measurability results have to be checked carefully, as 
intuition is a pretty poor guide. Ultimately, the set we are 
interested in must be described in terms of countable unions and intersections
of sets known to lie in  ${\cal F}^X_{t_0}$.
Let's start by noting that $f$ is continuous on $[0,t_0)$ 
if and only if $f$ is continuous on $[0,t_0-\varepsilon]$ for all small $\varepsilon>0$.
More formally, we can write
$$\{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0)\}
=\bigcap_{n=\lceil 1/t_0\rceil}^\infty 
\{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0-1/n]\}.$$
Hence it suffices to replace $[0,t_0)$ with a closed interval $[0,t^*]$
 with $t^*<t_0$. 
The advantage is that $f$ is continuous on $[0,t^*]$ if and only if 
$f$ is uniformly continuous. Also, a right continuous function $f$ is 
 uniformly continuous on $[0,t^*]$ if and only if $f$ is uniformly continuous on $\mathbb{Q}\cap[0,t^*]$.
That's because a uniformly continuous $f$ on $\mathbb{Q}\cap[0,t^*]$ has a unique
continuous extension to $[0,t^*]$, and the right continuity guarantees that we recover the 
original function.
This means we only need to check countably many values of the function. More formally, we can write the set
$$\{\omega: t\mapsto X_t(\omega)\text{ is  continuous on }[0,t^*]\}$$
as $$\bigcap_{k\geq 1}\bigcup_{\ell\geq 1}\bigcap_{|s-r|<1/\ell} 
\{\omega: |X_r(\omega)-X_s(\omega)|<1/k\},$$
where the inner intersection is over the countable set $s,r\in\mathbb{Q}\cap[0,t^*]$. 
This gives the desired result, since $\{\omega: |X_r(\omega)-X_s(\omega)|<1/k\}\in{\cal F}^X_{t_0}$ for all choices of $r,s,k$. 
