Why is the joint probability measure of two random variables not defined on the product of their sample spaces? Let $X, Y$ be random variables defined on a measure space $(\Omega, \mathcal{F}, P)$. The random vector $(X, Y)$ is defined as the function $$(X, Y): \Omega \to \mathbb{R}^2, (X, Y)(\omega) \mapsto (X(\omega), Y(\omega)).$$
My question is this: why is the random vector not instead defined on the product of the measure spaces:
$$(X, Y): \Omega \times \Omega \to \mathbb{R}^2?$$
For example, if I take two iid Bernoulli random variables, then the random vector that they form is a random variable that can only take two values: $(0, 0)$ and $(1, 1)$. It seems to me that the random vector and its corresponding joint distribution should instead be defined on the product of the sample spaces.
 A: 
For example, if I take two iid Bernoulli random variables, then the random vector that they form is a random variable that can only take two values: (0,0) and (1,1). It seems to me that the random vector and its corresponding joint distribution should instead be defined on the product of the sample spaces.

The sample space is the set of all possible outcomes.  We may map this set of outcomes to various random variables, such as the random pair, the individual members, or their sum.  Also, any probability measure.
$$\begin{align}\langle X,Y\rangle &: \Omega\mapsto\{0,1\}^2\\X &: \Omega\mapsto\{0,1\}\\Y &: \Omega\mapsto\{0,1\}\\X+Y&:\Omega\mapsto \{0,1,2\}\\\Bbb P&:\Omega\mapsto[0..1]\end{align}$$
So, we are measuring the same set of outcomes to different images.  $X$ and $Y$ do not have different pre-images.

Often, for convenience, we conflate the sample space with a measured image that gathers the most information in which we're interested in identifying outcomes; and discards irrelevant data.
Indeed, this is typically a Cartesian product, such as $\{\langle 0,0\rangle, \langle 0,1\rangle, \langle 1,0\rangle, \langle 1,1\rangle\}$ .
But we do not have to do that.

Another approach.
We may take as a sample set the outcomes of a die roll (which is unbiassed and six-sided). $$\Omega=\{1,2,3,4,5,6\}$$
Let $X$ indicate "the result is even", and $Y$ indicate "the result is prime".  So $$\begin{align}X(\omega)&=\begin{cases}1&:& \omega\in\{2,4,6\}\\0&:&\omega\in\{1,3,5\}\end{cases}\\Y(\omega)&=\begin{cases}1&:& \omega\in\{2,3,5\}\\0&:&\omega\in\{1,4,6\}\end{cases}\\\langle X, Y\rangle(\omega)&=\begin{cases}\langle 1,1\rangle&:&\omega\in\{2\}\\\langle 1, 0\rangle&:&\omega \in\{4,6\}\\\langle 0,1\rangle &:&\omega\in\{3,5\}\\\langle 0,0\rangle &:& \omega\in \{1\}\end{cases}\end{align}$$
Then we have that $$\begin{align}X&:\{1,2,3,4,5,6\}\mapsto\{0,1\}\\Y&:\{1,2,3,4,5,6\}\mapsto\{0,1\}\\\langle X,Y\rangle &: \{1,2,3,4,5,6\}\mapsto\{0,1\}^2\end{align}$$
A: There is an important distinction between identical random variables (where $X = Y$) and identically distributed random variables.
In the first case, $X$ and $Y$ really are the same function on $\Omega,$
so $X(\omega) = Y(\omega)$ always.
But in the second case, $X$ and $Y$ may be quite different functions on $\Omega.$
For example, if $X$ and $Y$ are both Bernoulli variables, "identically distributed" means that the set of $\omega$ for which $X(\omega) = 1$ has exactly the same probability measure as the set of $\omega$ for which $Y(\omega) = 1.$
But it could be the case that
$$P((X,Y)=(1,0)) = P((X,Y)=(0,1)) = P((X,Y)=(0,0)) = \frac13.$$
That is, $X=1$ one-third of the time and $Y=1$ one-third of the time,
but never at the same time.
This allows us to construct identically distributed random variables
that are not independent.
It seems to me that would be very hard to do if each variable had to act on its own individual copy of $\Omega.$
How would you model correlations between identically-distributed variables?
The usual definition also allows us to define a Bernoulli variable $X$ such that
$P(X=0) = P(X=1) = \frac12$ and a second variable $Y = 1 - X,$
and then say that $X$ and $Y$ are identically distributed.
This would be impossible if identically-distributed variables had to be the same function of $\Omega.$
