Does the composition of a non-degenerate random variable-valued function with itself induce dependence? (this is a revision of Does the composition of a random variable-valued function with itself induce dependence?)
Say I have a probability space $(\mathbb{R}, \Sigma, \mu)$ and a function $f$ of the form $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ such that for any distinct $x_1,x_2 \in \mathbb{R}$ the functions $f(x_1, -)$ and $f(x_2, -)$ are non-degenerate independent random variables over $(\mathbb{R}, \Sigma, \mu)$. 
Now let's define the random variables $G$ and $H$ as:
\begin{align}
G(y) = f(f(x_1, y), y) \\
H(y) = f(f(x_2, y), y) \\
\end{align}
Are $G$ and $H$ independent?
 A: Let $\Omega$ be a general sample space (possibly different from the set $[0,1]$). Let $f:\mathbb{R}\times \Omega \rightarrow \mathbb{R}$ be a function that satisfies your assumptions: For outcomes $y$ in the sample space $\Omega$, the random variables $f(x,y)$ (indexed by $x \in \mathbb{R}$) are non-generate and pairwise independent. In particular $f(x,y)$ is a measurable function of the outcome $y$ for each $x \in \mathbb{R}$. For simplicity of notation define $V_x(y) = f(x,y)$.  Non-degenerate implies that for each $x \in \mathbb{R}$ there is a threshold $h_x$ such that 
$$P[V_x(y)>h_x] \in (0,1)$$
Define the function $g:\mathbb{R}\times \Omega \rightarrow \mathbb{R}$ by 
$$ g(x,y) = \left\{ \begin{array}{ll}
3 &\mbox{ if $x=0$ and $V_0(y)>h_0$} \\
2 & \mbox{ if $x=0$ and $V_0(y)\leq h_0$}\\ 
0 & \mbox{if $x\neq 0$ and $V_x(y)>h_x$}\\ 
1 & \mbox{if $x \neq 0$ and $V_x(y)\leq h_x$} 
\end{array}\right.$$
Then for each distinct $x_1,x_2 \in \mathbb{R}$ the functions $g(x_1,y)$ and $g(x_2,y)$ are measurable functions of $y$, non-degenerate, and are independent random variables. However for $x_1=1$ and $x_2=2$: 
$$G(y)=g(g(1,y),y) =\left\{\begin{array}{ll}
g(0,y) & \mbox{if  $V_1(y)>h_1$} \\
g(1,y) & \mbox{ if $V_1(y)\leq h_1$}
\end{array}\right.$$
$$H(y)=g(g(2,y),y) =\left\{\begin{array}{ll}
g(0,y) & \mbox{if  $V_2(y)>h_2$} \\
g(1,y) & \mbox{ if $V_2(y)\leq h_2$}
\end{array}\right.$$
Then 
$$P[H =3]=P[V_2(y)>h_2]P[V_0(y)>h_0] > 0 $$
but 
$$P[H=3|G=2]=0$$
So $G$ and $H$ are not independent. 
