Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and $X:\Omega \rightarrow \mathcal{X}$ and $Y:\Omega \rightarrow \mathcal{Y}$ random variables. Let $f:\mathcal{Y}\rightarrow \mathbb{R}$ be a measurable function. Moreover there exists a regular conditional distribution such that: $$\mathbb{P}_Y(E)=\int_{\mathcal{X}} \mathbb{P}_{Y|X=x}(E)\mathbb{P}_X(dx)$$
Now let us consider the following integral: $$ \int_{\mathcal{Y}}f(y)\mathbb{P}_Y(dy).$$
My question is, if I can "plug in" the first equation into the second to get something like: $$\int_{\mathcal{Y}}f(y) \int_{\mathcal{X}} \mathbb{P}_{Y|X=x}(dy)\mathbb{P}_X(dx)$$
If this is allowed which integration has to be done first and how could I proof that statement? Or are there any books where this statement is proven?
The correct is: $$\mathbb{E}[f(Y)]=\int_{\mathcal{Y}}f(y)d\mathbb{P}_Y=\int_{\mathcal{X}}d\mathbb{P}_X\int_{\mathcal{Y}}f(y)d\mathbb{P}_{Y/X}$$ The demonstration is easy: $$\begin{array}{ll} \int_{\mathcal{Y}}f(y)d\mathbb{P}_{Y/X}=\mathbb{E}[f(Y)\mid X] & \hspace{20pt}\text{(Doob 1953, Theorem 9.1)}\\ \mathbb{E}\left[\mathbb{E}[f(Y)\mid X]\right]=\mathbb{E}[f(Y)] & \hspace{20pt}\text{(Definition of Conditional Expectation)} \end{array}$$ More general: $$\mathbb{E}[f(X,Y)]=\int_{\mathcal{X}\times\mathcal{Y}}f(x,y)d\mathbb{P}_{X\circ Y}=\int_{\mathcal{X}}d\mathbb{P}_X\int_{\mathcal{Y}}f(x,y)d\mathbb{P}_{Y/X}$$