 Solution

Write $w=p/q$ where $q>0$, assigning zero elsewhere. The envelope ensures $0\le w/M\le 1$, and $\pi$ is nonnegative. Integrating its joint probability density over the auxiliary coordinate givesConsequently ${\int }_{\Omega }{\int }_0^1\pi (x,u)dudx={\int }_{\Omega }p(x)dx=1$. This proves normalization of a probability distribution; the change of integration order is justified by Tonelli's theorem.

The coordinates are generally dependent. For $p(x)>0$, their conditional probability density isThus $U\mid X=x$ has a uniform distribution on an interval whose endpoint depends on $x$. They are independent exactly when $w(x)$ is constant for $p$-almost every $x$, since only then do these conditional laws agree. In particular, $p=q$ is an exception: $X\sim p$ and $U\sim \mathrm{Unif}(0,1/M)$ are independent. An unconditional answer that they can never be independent would therefore be too strong.