Suppose $X,Y,U$ are random variables on some probability space such that $U$ is independent of $(X,Y)$. Prove there exists a measurable function $f: \mathbb{R} \times [0,1] \rightarrow \mathbb{R}$ such that $(X,f(X,U))$ is equal in distribution to $(X,Y)$.
So far I can only come up with a few loose ideas. I suspect you have to use the information captured by the C.D.F. of $Y$ and extract a function from it. The issue I have is how using $X$ is even necessary when there appears to be no notable relation between them. Any help would be appreciated.
$Y|X=x$ has the same distribution of $f(x,U)$ for some measurable $f$ (take $f(x,.) $ the inverse of the CDF of $Y|X=x$ for an 'explicit' choice).
Then $(X,Y), (X,f(X,Y))$ have the same distributions.