Think independent random variables defined on a duplicated product space Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space on which defines two random variables
$$
X \colon \Omega \to \mathbb{R}\\
Y \colon \Omega \to \mathbb{R}
$$
Now suppose that $X$ and $Y$ are independent meaning that $\mathbb{P}(\omega \in \Omega \colon X \in A, Y \in B) = \mathbb{P}(\omega \in \Omega \colon X \in A)\, \mathbb{P}(\omega \in \Omega \colon Y \in B)$ for every $A,B \in \mathcal{B}$.
Now let us duplicate the space $(\Omega, \mathcal{F}, \mathbb{P})$ yielding $(\Omega', \mathcal{F}', \mathbb{P}')$ such that $\Omega = \Omega'$, $\mathcal{F}=\mathcal{F}'$, and $\mathbb{P} = \mathbb{P}'$. We consider the  product space $(\Omega \times \Omega', \mathcal{F}\otimes \mathcal{F}', \mathbb{P}\,\mathbb{P}')$.
We then extend those random variables to that product space by
$$
\overline{X}(\omega,\omega')\colon \Omega \times \Omega' \to \mathbb{R} = X(\omega), \\
\overline{Y}(\omega,\omega')\colon \Omega \times \Omega' \to \mathbb{R} = Y(\omega')
$$
We can verify that $\overline{X}$ and $\overline{X}$ are random variables on that product space, and they are independent.
Am I on a correct path to think about the independence of $X$ and $Y$ in this product measure sense? In short, suppose we have two indepdendent random variables on a probaiblity space, can I think them as the sliced random variables on their product measure space?
 A: I think you are confused a bit.
If you want to create two independent copies of a given radom variable with distribution $F,$ do this: start with a distribution, say $F$ on $\mathbf{R}.$ This distribution function induces a measure $\mu_F$ on the Borel set of $\mathbf{R}.$ If you want a pair of random variables with distribution $F$ each of them and that are independendent, then you will consider the probability space $(\mathbf{R}^2, \mathscr{B}_{\mathbf{R}^2}, \mu_F \otimes \mu_F)$ and to define the random variables you will then set $X(\omega) = x$ and $Y(\omega) = y$ where $\omega = (x, y).$ By definition, $X$ and $Y$ are independent and they both have distribution $F.$ Mutatis mutandis you can do the same with two probability spaces $(\mathrm{X}, \mathscr{X}, \mu)$ and $(\mathrm{Y}, \mathscr{Y}, \nu)$ and consider the product space $\mathrm{Z} = \mathrm{X} \times \mathrm{Y},$ $\mathscr{Z} = \mathscr{X} \otimes \mathscr{Y}$ and $\rho = \mu \otimes \nu$ and the random object is $T(x, y) = (x, y)$ which is a $\mathrm{Z}$-valued random object with independent coordinates and whose marginal laws are $\mu$ and $\nu,$ respectively.
If you start with two random variables $X$ and $Y$ defined on respective probability spaces $(\Omega_X, \mathscr{F}_X, \mathbf{P}_X)$ and $(\Omega_Y, \mathscr{F}_Y, \mathbf{P}_Y)$: then you can construct the product space
$$
(\Omega, \mathscr{F}, \mathbf{P}) = (\Omega_X \times \Omega_Y, \mathscr{F}_X \otimes \mathscr{F}_Y, \mathbf{P}_X \otimes \mathbf{P}_Y)
$$
and on it then define a random vector $Z(\omega) = Z(\omega_x, \omega_y) = (X(\omega_x), Y(\omega_y)) \in \mathbf{R}^2.$ Consider the projections $\mathbf{R}^2 \to \mathbf{R}$ given by $\pi_1(x,y) = x$ and $\pi_2(x,y) = y.$ In this construction, it follows that $\pi_1(Z)$ and $\pi_2(Z)$ are independent random variables defined on $\Omega$ with respective probabilities $\mathbf{P}_X$ and $\mathbf{P}_Y.$ These are your $\bar{X}$ and $\bar{Y}.$
Do every pair of independent random variables come from a product space? No. Consider uniform distribution on $[0, 1].$ Given a random number of this distribution, it can be expanded uniquely as $\sum\limits_{k = 1}^\infty b_k 2^{-k}$ where $b_k$ is either $0$ or $1$ and the expansion is unique by considering finite representations (that is, there is no infinite sequence of contiguous 1s starting some index until infinity). The random bits $(b_k)$ are independent Bernoulli with parameter $\dfrac{1}{2}$ and they do not come from a product space. However, in general is good idea to think of independent random vectors as defined on product spaces.
