Assume you have a probability space with a filtration, $(\Omega,\mathcal{F},P,\{\mathcal{F}_t\})$. Assume that the stochastic process $X_t$ is adapted to this filtration, and is jointly measurable with respect to $\mathcal{B}(\mathbb{R})\otimes \mathcal{F}$. Also assume that $P(\int_0^t|X_s|ds<\infty, \forall t)=1$. Then the integral $\int_0^tX_sds$ makes sense for almost all $\omega$. But I am wondering if this is an adapted process?
In order to find the answer I assume we must look at simple functions approximating $\int_0^tX_s ds$. I am able to do this pointwise for each $\omega$, but then I don't know how to use it for measurability with respect to $\mathcal{F}_t$. Maybe it is not the case that it is measurable?
The reason I am asking about this, is that when an Itô-process is defined, this is one part that comes up in the process. The Itô-process consist of two parts an integral with respect to a Brownian motion and this part, and integral with respect to the lebesgue measure. I assume that an ito-process is adapted, and hence this part which is an integral with respect to the lebesgue measure should be adapted?
This integral is described on page 43 and 44 here: http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf
Update:
I have gotten closer to a solution: If I can show that $X(t,\omega)$ is measurable with respect to $\mathcal{B}([0,t])\otimes \mathcal{F}_t$, then I will be done, because then I can look at the positive and negative part, and by Tonelli's theorem $\int_0^t X_s ds$ will be $\mathcal{F}_t$-measurable.
What I have is that $X(t,\omega)$ is measurable with respect to $\mathcal{B}(\mathbb{R})\otimes \mathcal{F}$. If I restict the t-values it is easy to see that $X(s,\omega)$ is measurable with respect to $\mathcal{B}([0,t])\otimes \mathcal{F}$ when only considering the relevant values in the integral $s \le t$. So I end up with that for $s\le t$, $X_s$ is $\mathcal{F}_t$ measurable when holding s fixed, and as a joint function $X(s,\omega)$ is $\mathcal{B}([0,t])\otimes \mathcal{F}$-measurable. Is there then any way to conclude that it is also $\mathcal{B}([0,t])\otimes \mathcal{F}_t$ measurable?
The property you're looking for, namely that $X_t$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_t$-measurable for all $t$, is called progressive measurability. Now your $X$ may not be progressively measurable, but as an adapted (and measurable) process, it has a progressively measurable modification $Y$, and it is easy to see that modification does not change the time integral in the sense that $P(\int_0^tX_sds=\int_0^tY_sds\,\forall t\geq0)=1$.
Edit: for the last claim: For fixed $t$ and $A\in\mathcal{F}$, $E\int_0^tX_sdsI_A=E\int_0^tY_sdsI_A$, since $X$ and $Y$ differ only on a $\lambda\otimes P$-measure $0$ set. Therefore, the random variables $\int_0^tX_sds$ and $\int_0^tY_sds$ are equal almost surely. It then also follows that $P(\int_0^tX_sds=\int_0^tY_sds\,\forall t\in\mathbb{Q})=1$, and since the integral processes are continuous, this is enough.