For two measures $\nu, \mu$ that are defined on the same measure space with $\nu = \int f d\mu$, it is a well known result, mostly used in the context of the radon nikodym theorem, that for an integrable $g$: $$\int g d\nu = \int g d(\int f d\mu) = \int g f d\mu.$$
My question is if there is some equivalent or related statement if two measures are not defined on the same space. In particular, I am interested in the following setting: Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space, $X:(\Omega, \mathcal{A})\rightarrow (\mathcal{X}, \mathcal{F})$ and $Y:(\Omega, \mathcal{A})\rightarrow (\mathcal{Y}, \mathcal{G})$ be two random variables that map into different spaces. Now assume that a regular conditional distribution exists (let $F\in \mathcal{F})$: $$\mathbb{P}_Y(F) = \int_{\mathcal{X}} \mathbb{P}_{Y|X=x}(F) d\mathbb{P}_X(x).$$
Now my question is, given I have an integral like this: $$\int_{\mathcal{Y}}h(y) d\mathbb{P}_Y(y)= \int_{\mathcal{Y}}h(y)d(\int\mathbb{P}_{Y|X=x} d\mathbb{P}_X)= \int ... d\mathbb{P}_X,$$
can I somehow write it as an integral where I integrate against the measure $d\mathbb{P}_X$? How would that look like if it was possible? Do I need the Radon Nikodym theorem somewhere?
$$\int_{\mathcal{Y}}h(y) d\mathbb{P}_Y=\int_{\mathcal{X}}d\mathbb{P}_X\cdot\int_{\mathcal{Y}}h(y)d\mathbb{P}_{Y/X}$$ In general: $$\int_{\mathcal{X\times Y}}f(x,y) d\mathbb{P}_{Y\circ X}=\int_{\mathcal{X}}d\mathbb{P}_X\cdot\int_{\mathcal{Y}}f(x,y)d\mathbb{P}_{Y/X}$$