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

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 (Right Continuous on $[0,\infty)$ and with finite Left-hand Limits 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 $A\in \mathcal{F}^X_{t_0}.$

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...)

Hopefully, someone can help. Thx in advance!


1 Answer 1


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 $A\in \mathcal{F}^X_{t_0}.$

  • $\begingroup$ Thx for the compact and straightforward solution! $\endgroup$
    – Alex
    Jun 10, 2019 at 7:02

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