I am struggling with 1.7 exercise from the Karatzas, Shreve "Brownian motion and stoch. calulus".




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Denote by $\mathcal{F}^X\_{t\_0}$ the natural filtration corresponding to a process $X:[0,\infty)\times \Omega \to \mathbb{R}^k$. Suppose that every sample path of $X$ is RCLL (**R**ight **C**ontinuous on $[0,\infty)$ and with finite **L**eft-hand **L**imits on $(0,\infty)$)


Introduce an event $A\subset\Omega$ that $X$ is continuous on $[0,t\_0)$ (we have an underlying probability space $(\Omega, \mathcal{F}, P)$). **The task is to show that $A\in \mathcal{F}^X\_{t\_0}.$** 




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I see that $$A=\left\{\omega\in\Omega\,|\, \forall t\in [0,t\_0)\,:\, \lim\_{s\to t^{-}} X(s,\omega) = X(t,\omega)\right\}.$$
However I have no clue how to prove that this is somehow in the minimal $\sigma-$algerbra $\mathcal{F}^X\_{t\_0}$ what I interpret as the minimal $\sigma-$algerbra spaned by the sets of the form $$\left\{\omega\in\Omega\,|\,X(t,\omega)\in B\right\}$$ where $0\leq t \leq t\_0$ and $B$ is an arbitrary Borel subset of $\mathbb{R}^k.$


I thought to somehow interpret the limiting property $\lim\_{s\to t^{-}} X(s,\omega) = X(t,\omega)$ with a set-theoretical approach as follows:


\begin{gather\*}
\{\omega\in\Omega\, | \, \lim\_{s\to t^{-}} X(s,\omega) = X(t,\omega)\} =\\ \bigcap\_{n\in\mathbb{N}} \bigcap\_{m\in\mathbb{N}, m\geq m^\*\_n}\{\omega\in\Omega\,|\,\forall s\in(t-1/m,t)\,:\, ||X(t,\omega)-X(s,\omega)||<1/n\} 
\end{gather\*}


Above the index $m^\*\_n$ comes from a usual sequential understanding of left-hand limits.


However, I am not able to make a step forward as I do not know how to deal with the continuous character of $t,s$ variables above (i.e. it is not easy to show measurability in $\Omega$ unless I can somehow make discrete unions/intersections...)




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Hopefully, someone can help. Thx in advance!



Filtration exercise
Continuity at zero is included in the RCLL property. 
If $X(\cdot,\omega)$ is discontinuous at some $t \in (0,t\_0)$, then the left and right limits at $t$ (which exist) must differ. Thus there must exist an integer $n>0$ such that 
$$
\|\lim\_{s\to t^{-}} X(s,\omega)-\lim\_{s\to t^{+}} X(s,\omega)\|>1/n \,.
$$
Since the limits exists, this implies that for every integer $k \ge 1$ there exist rational numbers $q\_k,r\_k$ such that
 $$
(\*) \; \, \; q\_k<r\_k<q\_k+1/k \quad \mbox{ yet } 
\quad \|X(q\_k,\omega)-X(r\_k,\omega)|>1/n \; .
$$


Conversely, If $X(\cdot,\omega)$ is continuous on $[0,t\_0)$ then it is uniformly continuous there (since replacing the value at $t\_0$ by the left limit yields a continuous function on a closed interval) so (\*) must fail for some $k \ge 1$.


Therefore, $X(\cdot,\omega)$ is continuous on $[0,t\_0)$ if and only if for every
 $n \ge 1$
there exists $k \ge 1$ such that for all pairs of rationals $q,r \in (0,t\_0)$ with $q<r<q+1/k$, we have $\|X(q,\omega)-X(r,\omega)| \le 1/n$.


Writing this in set theory notation shows that $A\in \mathcal{F}^X\_{t\_0}.$


