Let $X,Y : \Omega \rightarrow \mathbb{R}$ be independent r.v.'s and $f$ continuous. Then $A \subset \Omega$
$\int_{A} f(X,Y) dP = \int_{(X,Y)(A)} f(z) dP_{X,Y}(z) = \int_{(X,Y)(A)} f(x,y) dP_X(x) dP_Y(y) $ by independence.
Now, what I would have expected is that the result is
$\int_{X(A)} \int_{Y(A)} f(x,y) dP_X(x) dP_Y(y) = \int_{X(A) \times Y(A)} f(z) dP_{(X,Y)}(z)$ by independence, so it boils down to the question whether $(X,Y)(A)= X(A) \times Y(A)$ which is probably wrong. Thus, I was wondering whether it is possible to simplify the first expression further in order to get two integrals instead of just one or whether this is impossible?
With an arbitrary measurable $A\subseteq\Omega$ you won't get far: $$\int_{A}f\left(X\left(\omega\right),Y\left(\omega\right)\right)dP\left(\omega\right)=\int f\left(X\left(\omega\right),Y\left(\omega\right)\right)1_{A}\left(\omega\right)dP\left(\omega\right)$$ and there is no way to write this as: $$\int_{B}f\left(x,y\right)dP_{X,Y}\left(x,y\right)$$ unless $A=\left\{ \left(X,Y\right)\in B\right\} $ for some Borelset $B\subseteq\mathbb{R}^{2}$.
In general if $Z:\Omega\to\mathbb R^n$ then measure $P_Z$ induced by $Z$ "comes in" if functions of the form $h\circ Z$ are integrated:$$\int h\circ ZdP=\int hdP_Z$$