= Solution
Write $w=p/q$ where $q>0$, assigning zero elsewhere. The envelope ensures $0\leq w/M\leq1$, and $\pi$ is nonnegative. Integrating its <joint probability density> over the auxiliary coordinate gives
$$
\int_0^1\pi(x,u)\,du=Mq(x)\frac{w(x)}M=p(x).
$$
Consequently $\int_\Omega\int_0^1\pi(x,u)\,du\,dx=\int_\Omega p(x)\,dx=1$. 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> is
$$
\pi(u\mid x)=\frac M{w(x)}\mathbf1_{[0,w(x)/M]}(u).
$$
Thus $U\mid X=x$ 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 $U\sim\operatorname{Unif}(0,1/M)$ are <independent>. An unconditional answer that they can never be <independent> would therefore be too strong.
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