Random variable and probability space Can two random variables be considered independent even if their original domains are different probability space?
For example, let $X$ and $Y$ be two random variables with $$X: (\Omega, \mathcal{A})\to (E,\mathcal{E})$$ and $$Y: (\Delta, \mathcal{K})\to (F,\mathcal{F}).$$ How can I say and prove that $X$ and $Y$ are independent random variables?
If not possible, can I compare them someway?
 A: Random variables map a sample space to a real number. So I think you mean 
\begin{align}
&X:\Omega \rightarrow \mathbb{R}\\
&Y:\Delta \rightarrow \mathbb{R}
\end{align}
The functions that define random variables must be measurable, so that $\{\omega \in \Omega : X(\omega) \leq x\}$ is measurable for all $x \in \mathbb{R}$, and $\{v \in \Delta : Y(v) \leq y\}$ is measurable for all $y \in \mathbb{R}$. 
Independence of random variables $X$ and $Y$ means that $P[X\leq x, Y\leq y]=P[X\leq x]P[Y\leq y]$ for all $x,y\in \mathbb{R}$.  For the left-hand-side of this definition to make sense, the random variables must be on the same probability space. Also, it is assumed that the same probability measure over this space is used to measure the events $\{X\leq x, Y\leq y\}$, $\{X\leq x\}$, and $\{Y\leq y\}$.  So we have $X:\Omega\rightarrow\mathbb{R}$, $Y:\Omega\rightarrow\mathbb{R}$ and
$$ P[X\leq x, Y\leq y] = P[\{\omega \in \Omega : X(\omega) \leq x, Y(\omega) \leq y\}]$$
Formally, we have
$$\{\omega \in \Omega : X(\omega)\leq x, Y(\omega)\leq y\} = \{\omega \in \Omega : X(\omega) \leq x\} \cap \{\omega \in \Omega : Y(\omega) \leq y\} $$
This subset of $\Omega$ is the intersection of two measurable subsets $\{\omega \in \Omega: X(\omega) \leq x\}$ and $\{\omega \in \Omega: Y(\omega)\leq y\}$, and so it is itself measurable. 
If $X$ and $Y$ are on different probability spaces, you can still compare their cumulative distribution functions  $F_X:\mathbb{R}\rightarrow\mathbb{R}$ and $F_Y:\mathbb{R}\rightarrow\mathbb{R}$: 
\begin{align}
F_X(x) &= P[X\leq x] \quad, \forall x \in \mathbb{R}\\
F_Y(y) &= P[Y  \leq y] \quad, \forall y \in \mathbb{R}
\end{align}
For example, if $F_X(x)=F_Y(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$ for all $x \in \mathbb{R}$, then the random variables have the same distribution (and in this case it is a Gaussian distribution with zero mean and unit variance). 
