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 :)



Does the Radon-Nikodym derivative commute with integration?
Provided $x\mapsto f(x,B)$ is Borel measurable for each $B\in\Sigma$, you can define
$$
\mu(B):=\int\_{\Bbb R} f(x,B) dx,
$$ and check (using Tonelli's theorem) that $\mu$ is a measure.
If $H\ge 0$ is a measurable function on $\Omega$,
then
$$
\int\_\Omega H d\mu=\int\_{\Bbb R}\left[\int\_\Omega H(\omega) d\_\omega f(x,\omega)\right] dx,
$$
by the usual simple function approximation.


