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If we have a stochastic process $\{X_t\}$ on a filtered probability space $(\Omega, \mathcal{F},P,\mathbb{F})$. The integral process $\int_0^t|X_s|ds$ often comes up in stochastic analysis. But in order for this process to be well-defined the sample-paths have to be lebesgue-meausrable? What are sufficient conditions for this to happen?

I guess it is easy to see adaptiveness is not enough, by choosing a non-measurable function, deterministic. But is joint-measurability enough? That is, if the process is: $\mathcal{B}(\mathbb{R}_+)\times \mathcal{F}$-measurable? Is progressive measurability enough? I know that cádlag is enough, but this seems too strong, because it seems that one often needs the integral of processes which are not cádlág.

The problem I have with joint measurability is that it is ok if each singleton $\{\omega\}$ is $\mathcal{F}$-measurable. But does this need to be the case?, even if $\mathcal{F}$ is complete? Is it possible to show that if the process is jointly measurable, then a.s. it has measurable sample-paths?

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Yes, jointly measurable is sufficient. This is one of the assertions of Fubini's theorem: if $f(x,y)$ is jointly measurable on the product of two $\sigma$-finite measure spaces, then the section $f(\cdot, y)$ is measurable for every $y$.

You don't need completeness of $\mathcal{F}$ nor measurability of singletons, and if you're assuming joint measurability with respect to the product $\sigma$-field $\mathcal{B}(\mathbb{R}^+) \times \mathcal{F}$, then in fact you get that the sample paths are Borel.

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  • $\begingroup$ Thank you very much! Just one quick question, when I looked up this in my book it seem it would hold for all y, not just almost every y, do you agree? $\endgroup$
    – user119615
    Commented Aug 15, 2016 at 4:54
  • $\begingroup$ @user119615: Yes, you're right. I was thinking of something else. $\endgroup$ Commented Aug 15, 2016 at 5:03

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