Mean value theorem inside the Expectation Consider a stochastic process $X_t$ with continuous paths. I'd like to apply the mean value theorem inside the expectation, i.e. write something like
$$ \operatorname{E} \left[ \int_0^t X_s \, \mathrm{d}s\right] = t \operatorname{E} [X_{\xi_{\omega}}] \leq  t \operatorname{E} \left[ \sup_{\xi_{\omega} \in (0,t)} X_{\xi_{\omega}}\right].$$
Is it enough to require something like $\operatorname{E} [ \sup_{s \in (0,t)} |X_s|] < \infty$ for this reasoning to be valid or does one need additional assumptions? I'm concerned about the measurability of $\omega \mapsto X_{\xi_{\omega}}(\omega)$.
 A: The mapping $\omega \mapsto X_{\xi(\omega)}(\omega)$ is measurable if the corresponding process is jointly measurable, i.e. if
$$(t,\omega) \mapsto X(t,\omega) \tag{1}$$
is $\mathcal{B}([0,T]) \otimes \mathcal{A} /\mathcal{B}(\mathbb{R})$-measurable for $T>0$. This can be proved in the following way:
Let $f$ be a continuously differentiable function, then by Taylor's formula
$$f(t)-f(s) = f'(\xi) \cdot (t-s)$$
for some intermediate value $\xi \in (s,t)$. This is the Lagrange form of the remainder term. On the other hand, we can also write
$$f(t)-f(s) = \int_s^t f'(r) \, dr.$$
This shows that
$$f'(\xi) = \frac{1}{t-s} \int_s^t f'(r) \, dr. \tag{2}$$
In your case, we have $f(t) = \int_0^t X_r(\omega) \, dr$; hence, by $(2)$,
$$X_{\xi(\omega)}(\omega)= f'(\xi(\omega)) = \frac{1}{t} \int_0^t X_r(\omega) \, dr.$$
Now it follows from $(1)$ and Fubini's theorem that
$$\omega \mapsto \frac{1}{t} \int_0^t X_r(\omega) \, dr$$
is measurable for each $t>0$. Finally, note that condition $(1)$ is automatically satisfied since the process has continuous sample paths. (Hint: Approximate the process with simple functions.)
