Fix a probability space $(\Omega,\mathcal F,\mathbb P)$, and let $X,Y:\Omega\to\mathbb R$ be random variables with laws $\mu_X,\mu_Y$ respectively. Let $\pi$ be a probability measure on $\mathbb R^2$ with marginal distributions $\mu_X,\mu_Y$. Does there exist a random variable $Z:\Omega\to\mathbb R^2$ such that $\mu_Z=\pi$?
Intuitively this seems true, and if it is true, it seems like something that would be a classical well-known fact in probability. However, I have not read any such statement anywhere.
Not necessarily. Let $\Omega = \{0,1\}$ have two points and have the probability measure assign equal weight to both of them. Let $X$ and $Y$ both be the inclusion function into $\mathbb R.$ Then these are two random variables with the Beroulli distribution $p=1/2.$ But it's pretty clear we can't, for instance, make a RV $Z:\Omega\to \mathbb R^2$ that has the distribution of $(Z_1,Z_2)$ where $Z_1, Z_2$ are independent Bernoulli's with $p=1/2$ since $(Z_1,Z_2)$ can take at most two values.