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Let $x(t)$ be some random path $t\in[a,b]\subset\mathbb{R}$. I.e. $x:\Omega\rightarrow\mathbb{R}^{[a,b]}$ etc.

When is $\int_a^b x(t)dt$ defined?

If $x(t)$ is Brownian motion, I know it's ok.

A much simpler example is $x(0)=1$, waits an exponential length of time $T$ and then jumps down to zero. Thus $\int_0^\infty x(t)dt$ has the same distribution as $T$.

I think the question this: Under what conditions is a stochastic process Lebesgue integrable on some interval?

Is it just that the pre-image of measurable sets in $\mathbb{R}$ are measurable w.r.t. the product measure on $\Omega\times[a,b]$? How does this translate to properties of the stochastic process, e.g. bounded variation, etc.? Any hints or references are appreciated!

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    $\begingroup$ Well... I guess it depends on what you want. When you ask when $\int_a^b x(t)dt$ is defined, do you want to have it defined for a.a $\omega$ or for every $\omega$? If $\omega\in \Omega$ is fixed then $x_{\omega}(t):=x(t,\omega)\in \mathbb{R}$ is a function in the usual sense. Hence, you need to ensure $x_{\omega}$ is Lebesgue integrable, if you want it for a.a. or every $\omega$ is a matter of what you need it for. $\endgroup$ – Martingalo Apr 24 '16 at 12:24
  • $\begingroup$ Good question. I was assuming that almost always defined would be ok. Is that what is standard? I'm sure there is a well-developed theory out there for integrals of this kind. I'd be happy to have any insight. This is not for any purpose other than trying to understand under what conditions stochastic processes give integrable paths. $\endgroup$ – jdods Apr 24 '16 at 13:04
  • $\begingroup$ Well, this depends on the properties of its sample paths and what is interesting to see is if the set $A\subset\Omega$ for which $x(t,\omega)$, $\omega\in A$ is integrable, $P(A)=1$ or not. As another example... If $X$ is a stochastic process with integrable trajectories and $\mathcal{F}^X$ is the filtration generated by $X$. Then any other stochastic process $Y$ which is adapted to $\mathcal{F}^X$ has locally integrable trajectories as well. Therefore, if $W$ is a Brownian motion, any process $X$ which can be written as a Borel-measurable function of $W$ will have integrable trajectories. $\endgroup$ – Martingalo Apr 24 '16 at 13:17
  • $\begingroup$ That makes sense and goes along with what I've read about stochastic calculus thus far. Is it possible for a process that can't be expressed as a measurable function of Brownian motion to have measurable paths? $\endgroup$ – jdods Apr 24 '16 at 13:44
  • $\begingroup$ Of course, both cases. But this is something which you either impose "Let $X$ be a process with a.s. integrable trajectories" or you prove it ad hoc, as for instance, the example used above, "since this proces can be written as.... then...." $\endgroup$ – Martingalo Apr 24 '16 at 13:57

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