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Hi maybe someone on here can help me. I have been stuck on showing this fact for several months. I asked this question in the stack exchange and it has floated around for a while but to no avail.

The question I would like to answer is: If my filtration $\{\mathcal{F}_t\}_{t \geq 0}$ satisfies the usual conditions, and a cadlag process is adapted to that filtration, then that process is progressively measurable. The reason I would like to know this is because I am trying to figure out the proof of Theorem 6 in Protter's Stochastic Integration and Differential Equations. I have a hunch that what may have been used is the statement in the title. I've seen it used a couple of places already, but I cannot figure out why it is true. For example in this question https://math.stackexchange.com/questions/1682370/measurability-of-a-stopped-random-variable, both answers use the statement.

Here is Protter's theorem 6: Let $\{\mathcal{F}_t\}_{t \geq 0}$ be a filtration satisfying the usual conditions. Let $T$ be a finite stopping time for the filtration $\{\mathcal{F}_t\}_{t \geq 0}$. The $\{A \in \mathcal{F}: \text{ for each }t \geq 0, A \cap T^{-1}([0,t]) \in \mathcal{F}_t\} = \sigma( \{X_T: X \text{ is a cadlag process adapted to } \{\mathcal{F}_t\}_{t \geq 0} \})$

In Protter's theorem, he claims that

As far as I'm aware, cadlag means that ALMOST every path is cadlag. I showed that if a process $X$ has each sample path cadlag and is adapted, then it is progressively measurable by using the the functions $X^n(s,\omega) = X(0,\omega)\mathbf{1}_{\{0\}}(s)+\sum\limits_{k = 1}^{2^n}X(tk/2^n,\omega)\mathbf{1}_{(t(k-1)/2^n, tk/2^n]}(s)$ on $[0,t]\times \Omega$.

$X^n$ is $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$ measurable and by right continuity, $X^n$ converges everywhere to $X$ on $[0,t]\times \Omega$, so X is progressively measurable.

However, if we drop the assumption that $X$ has every path cadlag and instead just almost every path is cadlag (ie $X$ is a cadlag process) we can let $N = \{\omega \in \Omega: X(\cdot,\omega) \text{ is not cadlag}\}$. Then N is a null set. Even if $\mathcal{F}_t$ is complete, $N$ is a measurable null set in $\mathcal{F}_t$ and therefore $[0,t]\times N$ is null in $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$. Since $X^n$ converges to $X$ pointwise on $([0,t]\times N)^c$ we have that $X^n$ converges to $X$ almost everywhere. Now, if we don't know that the product measure space ($[0,t] \times \Omega, \mathcal{B}([0,t]) \bigotimes \mathcal{F}_t) $ is complete, then we can't say that the limit $X$ is $\mathcal{B}([0,t]) \bigotimes \mathcal{F}_t$ measurable.

So if anyone can clear up my confusion I would be very appreciative. I've been stumped for weeks now. I think that there must be another way to show the result than how I did with the simpler case above using sequences.

In theorem 6 of Protter, I was able to show that regardless of whether $\mathcal{F}_t$ is complete or not, $\mathcal{G}^* = \sigma(\{X_T: X \text{ is everywhere cadlag and adapted to } \{\mathcal{F}_t\}\}) = \{A \in \mathcal{F}: \text{ for each } t \geq 0, A \cap T^{-1}([0,t]) \in \mathcal{F_t}\} = \mathcal{F}_T$

However, I still cannot show the analogue for $\mathcal{F_t}$ satisfying the usual conditions and $X_t$ cadlag almost everywhere.

**Edit

I could show Protter's theorem 6 by following along in Bass' book using simpler stopping times that approximate $T$, but still have not solved the question I posted here. So now, I do not need a solution to Protter's theorem, but to the statement in the title.

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  • $\begingroup$ Could you please include the exact statement of Theorem 6? And if a proof is given, could you include the specific part of the proof that you don't understand or think may have a gap? $\endgroup$ – Nate Eldredge Aug 17 '18 at 15:18

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