# 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

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For your first question: "The limit of a measurable function is measurable": This statement doesn't make sense; the limit of a function is not a function. You may be thinking of the fact that the limit of a sequence of measurable functions is a measurable function. But you also must keep in mind what $\sigma$-algebra the functions are measurable with respect to. In particular, for $s > t_0$, $X_s$ is not necessarily a measurable function with respect to $\mathcal{F}_{t_0}^X$. –  Nate Eldredge Feb 23 '12 at 16:55
@ Nate Eldredge: I edited my question. Obviously I meant that, sorry for being inexact. –  math Feb 24 '12 at 9:17

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 $$\{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0)\} =\bigcup_{n=\lceil 1/t_0\rceil}^\infty \{\omega: t\mapsto X_t(\omega)\mbox{ continuous on }[0,t_0-1/n]\}.$$

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 $r,s,k$.

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@ Byron Schmuland: Thanks for your answer. Perhaps the question was not that clear. What I really don't get is the difference between the two questions. Why does $A$ fail to be measurable in the second question. Actually I think this is more a linguistic problem. I wrote down a much more precise proof of claim 1. Perhaps I should delete the "right?". –  math Feb 24 '12 at 9:21
@Math : IMO this is not "language problem" but there is a material difference, as in the second claim paths are only almost surely RCCL, and not RCLL for every $\omega$ as in the first claim. –  TheBridge Feb 24 '12 at 11:59
@ TheBridge: This was also my understanding. I was just not quit sure if this is right. Anyway thanks for clear the uncertainness. –  math Feb 24 '12 at 13:12
Where did you use the assumption that paths have finite left hand limits? Is it extraneous? –  Mark Aug 19 '13 at 13:47
@Mark The left limits are not needed here. –  Byron Schmuland Aug 19 '13 at 17:54