# Karatzas Shreve exercise 1.8

Let $$X$$ be a stochastic process (parametrised by positive reals) whose sample paths are RCLL (finite left hand limits everywhere, right continuous everywhere) almost surely. Let $$\{\mathcal{F}_t:t\geq 0\}$$ be a filtration satisfying $$\mathcal{F}_t^X\subset \mathcal{F}_t$$, $$t\geq 0$$, and also $$\mathcal{F}_{t_0}$$ is complete under $$P$$. Show that the event that $$X$$ is continuous on $$[0,t_0)$$ measurable with respect to $$\mathcal{F}_{t_0}$$.

I understand that an RCLL function is continuous if and only if its restriction to a countable dense set is uniformly continuous, and if we fix the countable dense set to be the rationals in $$[0,t_0)$$, then we can show that the event of the restriction being uniformly continuous is measurable with respect to $$\mathcal{F}_{t_0}$$. If $$N\in \mathcal{F}$$ denotes the event that the function is not RCLL, then $$N$$ may not be measurable with respect to $$\mathcal{F}_{t_0}$$, even though $$\mathcal{F}_{t_0}$$ is complete because completeness only says that every subset of a $$P$$-null measurable set in $$\mathcal{F}_{t_0}$$ is also measurable with respect to the same $$\sigma$$-algebra. The solution given in the book just writes that $$\{\text{function is continuous}\}=\{\text{Restriction to rationals is uniformly continuous}\}\cap N^c$$, which somehow seems to assume that $$N$$ is measurable with respect to $$\mathcal{F}_{t_0}$$.

• That is not what the exercise in Karatzas and Shreve asks us to show. The exercise specifically asks us prove that the event is $\mathcal{F}_{t_0}$-measurable under the condition (among other conditions) that "$\mathcal{F}_{t_0}$ contains all $P$-null sets of $\mathcal{F}$." May 7, 2021 at 0:04

"...even though $$\mathcal{F}_{t_0}$$ is complete because completeness only says that every subset of a $$P$$-null measurable set in $$\mathcal{F}_{t_0}$$ is also measurable with respect to the same $$\sigma$$-algebra.."

Notice the difference between considering the completion of $$(\Omega,\mathcal{F},P)$$ and the completion of $$(\Omega,\mathcal{F}_{t_0},P)$$. If we were considering the latter one, your objection would be valid. However, usually completeness in the context of filtrations means that $$\mathcal{F}_0$$ should contain all the $$(\Omega,\mathcal{F},P)$$-null sets, cf this blog page.

Given that the solution in the book relies on the second interpretation, I think it's safe to assume that that's what Karatzas and Shreve mean to assume here, even if the way they formulate it is a bit misleading.

• $N$ is a null set with respect to $\mathcal{F}$; why can't it happen that there is no null set in $\mathcal{F}_{t_0}$ containing $N$? Nov 14, 2018 at 19:26
• Wikipedia says that a complete space is a measure space in which every subset of every null set is measurable (having measure zero). Here we are given that $\mathcal{F}_{t_0}$ is complete, so to get that $N$ is measurable, we need to embed $N$ inside a set which is $P$-null and is already known to be measurable with respect to $\mathcal{F}_{t_0}$. So doesn't your argument begin by assuming what we need to show? Nov 14, 2018 at 19:34
• Okay, for filtrations to make sense, we need to start with some probability space $(\Omega,\mathcal{F},P)$ such that $\mathcal{F}_{t} \subseteq \mathcal{F}$, right? Now saying that some $\mathcal{F}_{t}$ is complete means precisely that it contains all the null sets, where a null set is defined to be a subset of some set in $\mathcal{F}$ with measure zero. Nov 14, 2018 at 19:41
• Okay, I think I see now what you problem is. Going to rewrite my answer. Nov 14, 2018 at 19:47
• So does the expression '$\mathcal{F}_{t_0}$ is complete under $P$' mean something different than '$(\Omega,\mathcal{F}_{t_0}, P)$ is complete? Nov 14, 2018 at 19:51