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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 $\mathcal{F}_{t_0}$.

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    $\begingroup$ 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 $\mathcal{F}$." $\endgroup$
    – Rushi
    Commented May 7, 2021 at 0:04

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"...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.

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  • $\begingroup$ $N$ 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 $N$? $\endgroup$
    – Manan
    Commented Nov 14, 2018 at 19:26
  • $\begingroup$ 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? $\endgroup$
    – Manan
    Commented Nov 14, 2018 at 19:34
  • $\begingroup$ 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. $\endgroup$
    – user159517
    Commented Nov 14, 2018 at 19:41
  • $\begingroup$ Okay, I think I see now what you problem is. Going to rewrite my answer. $\endgroup$
    – user159517
    Commented Nov 14, 2018 at 19:47
  • $\begingroup$ So does the expression '$\mathcal{F}_{t_0}$ is complete under $P$' mean something different than '$(\Omega,\mathcal{F}_{t_0}, P)$ is complete? $\endgroup$
    – Manan
    Commented Nov 14, 2018 at 19:51

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