Uniqueness of the transformation turning random variables into IID uniform

We have two random variable $X:\Omega \to \mathbb R$ and $Y: \Omega \to \mathbb R^d, d \in \mathbb N$, $F_Y$ is the density function of $Y$ and $F_{X|Y=y}$ is a regular density function of $X$ conditional on a value of $Y$. $F_Y$ is continuous and $F_{X|Y=y}$ is continuous w.r.t. $x, P_Y$-a.s.

It can be shown that random variables $F_Y(Y), F_{X|Y = y}(X,Y)$ are IID uniform on $[0,1]$.

Also I can show that among continuous distribution functions only $F_Y$ can transform $Y$ in $[0,1]$ uniform.

I need to show now that if we have a $G_2(x,y)$ continuous w.r.t. $x, P_Y$-a.s., such that $F_Y(Y), G_2(X,Y)$ are IID uniform on $[0,1]$, then $G_2=F_{X|Y = y}, P_Y$-a.s.

• "Also I can show that among continuous functions only $F_Y$ can transform $Y$ in $[0,1]$ uniform." I doubt that (try $1-F_Y$). – Did Aug 28 '15 at 10:13
• @Did, you got me, yes I needed to say "among continuous DISTRIBUTION functions", $1-F_Y$ being decreasing is not a distribution function, I corrected the question. – Sergey Zykov Aug 28 '15 at 10:18
• Do you know the basic proof approach to showing uniqueness? If so, it might help to add in what you've done so far. – user237392 Sep 9 '15 at 10:35