Applying Tonelli's theorem with a composition of two functions. Consider the following.
We have the two maps $\Psi: \ \mathbb{R} ^2\times \Omega \to \mathbb R_+$, and $h(\omega)=(U(\omega),V(\omega)): \Omega \to \mathbb R^2$ and suppose further that $h$ has distribution $\nu(du,dv)$ on $\mathbb R^2$. [To be clear both $P $ and $\nu $ are probability measures.]
Tonelli's theorem would apply if we tried to calculate the integral 
$\int \Psi((u,v),\omega)\nu(du,dv) \otimes P(d \omega)$ as the iterated integral $\int \nu(du,dv)\int \Psi((u,v),\omega)P(d \omega)$. 

But does this mean 
  $$\int \nu(du,dv)\int \Psi((u,v),\omega)P(d \omega)=\int \Psi((U(\omega),V(\omega)),\omega)P(d \omega)$$
  ?

If we look at $$\int \Psi((u,v),\omega)\nu(du,dv) \otimes P(d \omega)$$
This would seem to say that 
$$\int \Psi((U(\omega),V(\omega)),\omega)P(d \omega)=\int \Psi((U(\omega),V(\omega)),\omega_0)P(d \omega)\otimes P(d \omega_0)$$
Is this true? I have some trepidation since it feels like we integrate some function $f(\omega) $ of one variable by renaming some instances of $\omega $ to $\omega_0 $ and then integrate $f $ with respect to the product measure $P \otimes P$.
Most grateful for any help provided!
 A: No, the two sides are not identical.
On the LHS, the construction of the product measure $\nu \otimes P$ assumes that the $\sigma$-algebras on $\mathbb R^2$ and $\Omega$ are independent even though the distribution $\nu = P \circ h^{-1}$ is constructed from $P$, analogous to how identically distributed random variables can still be independent. The integration is done over all points $\bigl((u,v), \omega\bigr) \in \mathbb R^2 \times \Omega$, where sets consisting of points that fail to satisfy $(u,v) = h(\omega)$ may still have positive probability.
This is in contrast to the RHS, where your location in $\mathbb R^2$ is completely determined by your location in $\Omega$. If we also consider it as an integration over the set product $\mathbb R^2 \times \Omega$, we are essentially only integrating over points satisfying $(u,v) = h(\omega)$. Thus the divergence.
As an example, consider $\Omega = \mathbb R^2$ both equipped with the uniform probability measure on $[0,1]^2$, and let $h$ be the identity function. For the function
$$
\delta(\omega_1, \omega_2)
= \begin{cases} 1 & \text{if $\omega_1 = \omega_2$} \\
0 & \text{if $\omega_1 \neq \omega_2$,} \end{cases}$$
we have
$$ \int \delta \, d(\nu \otimes P) = 0 \neq 1 = \int \delta(h(\omega), \omega) \, P(d\omega).$$
