Suppose $f:\mathbb{R}\to[0,\infty)$ is Borel measurable and the integral of $f$ is strictly positive and finite. Moreover we have a r.v.'s $X$ and $Y$, where $Y$ is uniformly(0,1) distributed and independent of $X$. Furthermore we assume that the distribution of $X$ is given by
$P_X(A)=\frac{\int_Af(x)dx}{\int_\mathbb{R}f(x)dx}$
Then I want to calculate the distribution Function of $Z:=(X,f(X)Y)$, i.e.
$P(Z\in A)$
for a Borel set $A=A_1\times A_2\in \mathbb{R}\times\mathbb{R}$.
$P(Z\in A)=P(X\in A_1,f(X)Y\in A_2)$
I know that $X$ and $Y$ are independent, but I think $X,f(X)Y$ are not in general. So how can I compute this? It's an exercise in my probability book. The solution should be, that $Z$ is uniformly subgraph$(f)$ distributed, which means.
$\mathcal{U}_{subgraph(f)}(B)=\frac{\lambda(B)}{\lambda(subgraph(f))}$
where $\lambda$ is the lebesgue measure and $subgraph(f):=\{(x,y)\in \mathbb{R}^2;0\le y\le f(x)\}$. I know that $\lambda(subgraph(f))=\int_\mathbb{R}f(x)dx$.