# Showing that the event: $X$ is continuous on $[0,t_0)$ belongs to the given filtration. [duplicate]

Let $( \Omega, F, P)$ be a probability space.

Let $X$ be a stochastic process defined on such probability space whose sample paths are RCLL (right continuous on $[0, \infty)$ with finite left hand limits on $(0, \infty)$. Let $A$ be the event that $X$ is continuous on $[0,t_0)$.

I am tasked in showing that $A \in F^X_{t_0}$, where $F^X_{t_0}:= \sigma(X_s ; 0 \le s \le t) = \{ \{ X_s \in B \}\subset \Omega : B \in \mathcal{B}(\Bbb{R}), 0 \le s \le t \}$

Some may recognize this exercise as one of the first from Brownian Motion and Stochastic Calculus by Karatzas and Shreve, I am in fact trying to self study this book through august.

My problem here is that I can't figure out exactly what $A$ represents, I know that for a fixed $w \in \Omega$ I have that $X_s(w)$ is RCLL but I can't figure out how to start.

Your second equality in the definition of $F^X_{t_0}$ is incorrect - $F^X_{t_0}$ is the $\sigma$-field generated by events of the form $\{X_s\in B\}$ for $0\le s\le t_0$ and $B\in\mathcal B(\mathbb R)$, but it contains many more events, too. For example, $\{X_{s_1}\in B_1,X_{s_2}\in B_2,\ldots,X_{s_n}\in B_n\}\in F^X_{t_0}$ for all $s_1,\ldots,s_n\in[0,t_0]$ and $B_1,\ldots,B_n\in\mathcal B(\mathbb R)$. Or we could take unions instead, such as $\bigcup_{i=1}^\infty\{X_{s_i}\in B_i\}$. Or we could take functions of multiple $X_s$'s, i.e. something like $\{X_s+X_t\in B\}$.

Informally, an event is $F^X_{t_0}$-measurable if we know whether or not it occurs after observing the process up to time $t_0$. There is a small caveat: we can only make countable many choices. An event of the form $\{X_s\in B_s\text{ for }s\in[0,t_0]\}$, where $B_s\in\mathcal B(\mathbb R)$ for all $s$, is not measurable. However, as long as we stick to countable combinations, we can do whatever we like.

As for what $A$ represents: we already know that the sample paths of $X$ are RCLL, and $A$ is the event that up to time $t_0$, all left and right limits coincide. Specifically,

\begin{align*} A &=\{\omega\in\Omega:t\mapsto X_t(\omega)\text{ is continuous on }[0,t_0)\}\\ &=\{\omega\in\Omega:\lim_{s\uparrow t}X_s(\omega)=\lim_{s\downarrow t}X_s(\omega)\text{ for all }t\in(0,t_0)\}. \end{align*}

Using our heuristic description of $F^X_{t_0}$, it seems clear that $A\in F^X_{t_0}$: if you observe the process up to time $t_0$, you know whether or not it was continuous, right? Well, we have to be careful. If $X$ was a process with arbitrary sample paths rather than RCLL, then in fact we would have a negative answer. The reason is we need to know that $X$ is continuous at every point, and as there are uncountably many points, this is too much information for this $\sigma$-field to be able to track.

However, in the RCLL case, things become a little easier. RCLL functions cannot have removable discontinuities, so if the restriction of $X$ to a dense subset of $[0,t]$ was (uniformly) continuous with $t<t_0$, then the function on all of $[0,t]$ must be as well, and then we can let $t\uparrow t_0$ in an appropriate way. In particular, using the fact that $\mathbb Q$ is a countable dense subset of $\mathbb R$, we find that

\begin{align*} A&=\bigcap_{t\in(0,t_0)\cap\mathbb Q}\{\omega\in\Omega:\forall \varepsilon\in(0,\infty)\cap\mathbb Q,\,\exists\delta\in(0,\infty)\cap\mathbb Q\text{ such that }\\ &\qquad\qquad\qquad\forall u\in[0,t]\cap\mathbb Q,\forall s\in(u-\delta,u+\delta)\cap\mathbb Q\cap[0,t],\,|X_s(\omega)-X_u(\omega)|<\varepsilon\}\\ &=\bigcap_{t\in(0,t_0)\cap\mathbb Q}\bigcap_{\varepsilon\in(0,\infty)\cap\mathbb Q}\bigcup_{\delta\in(0,\infty)\cap\mathbb Q}\bigcap_{u\in[0,t]\cap\mathbb Q}\bigcap_{s\in(u-\delta,u+\delta)\cap\mathbb Q\cap[0,t]}\{|X_s-X_u|<\varepsilon\} \end{align*} which is $F^X_{t_0}$-measurable.

(Note that this has been edited to reflect comments by John Dawkins.)

• But taking $t_0=4$ (for example), the path $\omega$ with $X_t(\omega)=0$ for $t\in[0,\pi)$ but $X_t(\omega)=1$ for $t\in[\pi,4)$ is cadlag and an element of $A$, but is not continuous. Jul 26, 2017 at 16:48
• "Let $A$ be the event that $X$ is continuous on $[0,t_0).$" How is $\omega\in A$ for your example? Jul 26, 2017 at 17:41
• More carefully, the path I suggested is in the event $$\bigcap_{t\in(0,t_0)\cap\mathbb Q}\bigcap_{\varepsilon\in(0,\infty)\cap\mathbb Q}\bigcup_{\delta\in(0,t)\cap\mathbb Q}\bigcap_{s\in(t-\delta,t+\delta)\cap\mathbb Q}\{|X_t-X_s|<\varepsilon\}$$ but it is not continuous on $[0,4)$. The point being that you have not quite captured $A$. Jul 26, 2017 at 17:55
• No. Let's fix $t=7/2$, $\varepsilon=1/2$ and suppose there exists such $\delta$. Then choose $u\in[0,t]\cap\mathbb Q$ such that $\pi-\delta/3<u<\pi$. Choose $s\in[0,t]\cap\mathbb Q$ such that $\pi<s<\pi+\delta/3$. Then $s\in(u-\delta,u+\delta)$ but $|X_s(\omega)-X_u(\omega)|=|1-0|=1\ge\varepsilon$. Jul 26, 2017 at 21:59
• No, remember by definition that $A$ is the set of continuous paths. The trick is to write $A$ in a clever way which relies only on countable unions and intersections of events we know to be measurable. John Dawkins' point was that my initial attempt did not actually capture $A$, so I refined it slightly so that it did. In words, I have written that $A$ is the set of RCLL paths which are uniformly continuous on the restriction to $[0,t]\cap\mathbb Q$ for all $t<t_0$, which is true. Jul 26, 2017 at 22:02