Measurability of an integral Let $\{X_t\}_{t\ge 0}$ be an adapted $\mathbb{R}$-valued stochastic process on some filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},\mathbb{P}\}$ such that for each $\omega\in\Omega$ and $t\ge 0$ the integral $\int_0^t X_s(\omega)ds$ exists (Lebesgue integration). Why is this integral $\mathcal{F}$-$\mathfrak{B}(\mathbb{R})$-measurable (as a function in $\omega$) if the process $X$ is progressively measurable? Does someone know how to prove this? 
Thank you!
 A: example
Without assuming some "joint" measurability of $X_t(\omega)$ in $t$ and $\omega$ you are out of luck.  
We use the Continuum Hypothesis.  There is a set
$A \subseteq [0,1] \times [0,1]$
such that:
$\qquad$For each $t \in [0,1]$,$\qquad \{\omega \in [0,1] : (t,\omega) \in A\}$ is countable,  
$\qquad$For each $\omega \in [0,1]$,$\qquad \{t \in [0,1] : (t,\omega) \in A\}$ is co-countable.

 On the space $[0,\omega_1)$ of countable ordinals, the set $\{(s,t) : s > t\}$ has these properties; transfer it to $\mathbb R$ by CH.

Now let $\Omega = [0,1]$, $\mathcal F = \mathfrak B$, the Borel sets in $[0,1]$.  Also let $\mathcal F_t = \mathcal F$ for all $t \in [0,1]$.
Let $\mathbb P$ be Lebesgue measure.  
Choose a non-measurable function $h : [0,1] \to [0,1]$.
Define:
$$
X_t(\omega) := \begin{cases}
h(\omega),\qquad (t,\omega) \in A
\\
0,\qquad (t,\omega) \not\in A
\end{cases}
$$
Now for each fixed $t$, we have $X_t(\omega)=0$ except for countably many $\omega$, so $X_t$ is $\mathcal F_t$-measurable.  [I hope this is your definiton of "adapted $\mathbb R$-valued stochastic process".]  
On the other hand, for each fixed $\omega$, we have $X_t(\omega)=h(\omega)$ except for countably many $t$.  Integrating a function that is constant a.e., we get
$$
\int_0^1 X_t(\omega)\;dt = h(\omega)
$$
which is a non-measurable function of $\omega$.
