Let $A \times \Theta \subseteq \mathbf{R}^2$ be an open subset, $(\Omega, \mu)$ a measure space and $f: A \times \Theta \times \Omega \rightarrow \mathbf{R}$ a map.
Under which conditions can we state that
$$\frac{d}{d \theta} \int_\Omega \int_A f(x, a, \theta) d\mu(x) da = \int_\Omega \int_A \frac{\partial}{\partial \theta} f(x, a, \theta) d\mu(x) da$$
Based on Interchange the order of integral and derivative, I would expect the conditions that
$\theta \rightarrow f(x, a, \theta)$ is differentiable for a.e. $x \in \Omega, a \in A$
There exists an integrable function $g$ such that
$$\mid \frac{\partial}{\partial \theta}f(x,a, \theta) \mid < g(x, a)$$
Am I correct? And how can I prove it?
This follows immediately from the result for one integral and Fubini. You just need to take the product measure on $\Omega \times A$.