Suppose I have a measurable space $(\Omega, \Sigma)$ and a function $f: \mathbb{R} \times \mathbb{\Sigma} \rightarrow [0,1]$ such that for any $x \in \mathbb{R}$ the tuple $(\Omega, \Sigma, f(x, \_))$ is a measure space.

Is it the case that $\int_{x \in \mathbb{R}} \int_{\Omega} df(\_, x) dx = \int_{\Omega} d\int_{x \in \mathbb{R}} f(\_, x) dx$?

I suspect that this comes down to the question of whether we can interchange the Radon-Nikodym derivative and integration operations, but I am not sure. If this question doesn't make sense because of a missing assumption please let me know :)

Suppose I have a measurable space $(\Omega, \Sigma)$ and a function $f: \mathbb{R} \times \mathbb{\Sigma} \rightarrow [0,1]$ such that for any $x \in \mathbb{R}$ the tuple $(\Omega, \Sigma, f(x, \_))$ is a measure space.

Is it the case that $$\int_{\mathbb{R}} \left[ \int_{\Omega} df(x, \_) \right] dx = \int_{\Omega} d \left[ \int_{\mathbb{R}} f(x, \_) dx \right] \ ?$$

I suspect that this comes down to the question of whether we can interchange the Radon-Nikodym derivative and integration operations, but I am not sure. If this question doesn't make sense because of a missing assumption please let me know :)