Theorem about coupling and independence of random variables

Question

I am reading a book of E. Rio and I found there a theorem (without a proof) about coupling. Please see below.

Theorem: Let $(\xi_i)_{i \in \mathbb{Z}}$ be a sequence of random variables with values in some Polish space $X$. Assume that $(\Omega, \mathcal{T}, \mathbb{P})$ is rich enough to contain a random variable $U$ with uniform distribution over $[0, 1]$, independent of $(\xi_i)_{i \in \mathbb{Z}}$. Let $\mathcal{F_0} = \sigma(\xi_i: i \le 0)$ and $\mathcal{G}_n = \sigma(\xi_i: i \ge n)$. Then one can construct a sequence $(\xi^*_i)_{i \in \mathbb{Z}}$ with the same joint distribution as the initial sequence $(\xi_i )_{i \in \mathbb{Z}}$, independent of $\mathcal{F}_0$ and measurable with respect to the $\sigma$-field generated by $U$ and $(\xi^*_i)_{i \in \mathbb{Z}}$, in such a way that, for any positive integer $n$, $$\mathbb{P}(\xi_k \neq \xi^*_k \text{for some } k \ge n \mid \mathcal{F_0}) = \text{esssup}(|\mathbb{P}(B \mid \mathcal{F_0})−\mathbb{P}(B)| \colon B \in \mathcal{G}_n).$$

Question 1: I am wondering what are the conditions these theorem to hold? Is my understanding correct that the only condition that $\mathcal{X} = (\Omega, \mathcal{T}, \mathbb{P})$ should be reach enough to contain a uniform r.v. independent of $(\xi_i)_{i \in \mathbb{Z}}$?

Question 2: Could anyone provide some intuition about this condition (i.e., what does it mean, why is it crucial here, why uniform)? I would be grateful for any example where this condition doesn't hold.

Answer 1

It is not meant to be a real answer. Probably only a proof of the theorem would clarify what's going on.

Question 1: I see another assumption of the theorem. The random variables $\xi_i$ are taking values on a Polish space $\mathcal{X}$.

Question 2: Why do we ask the random variable $U$ to be an $\mathrm{Unif}([0,1])$ independent of $(\xi_i)_{i\in\mathbb{Z}}$?

Fact:

If $U$ is $\mathrm{Unif}([0,1])$ and $F$ is some cumulative distribution function, then the random variable $F^{-1}(U)$ follows the distribution $F$, i.e. $$\mathbb{P}\bigl(F^{-1}(U)\le x\bigr) = F(x),\quad\forall x\in\mathbb{R},$$ where $F^{-1}$ is the generalized inverse function of $F$.

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In other words, with such an $U$ one can define (on the same probability space) real-valued random variables with desired distributions. So, the existence of a random variable $U\sim\mathrm{Unif}([0,1])$ in a probability space guarantees the existence of real-valued random variables with any distribution. Furthermore, note that if $U$ is independent of another random variable $\xi$ with cdf $F_\xi$, then the random variable $\xi':=F_\xi^{-1}(U)$ follows the same distribution of $\xi$ and is also independent of $\xi$. Note the analogy with our theorem.

The theorem claims that there exists another sequence $(\xi'_i)_{i\in\mathbb{Z}}$ with the same joint distribution of $(\xi_i)_{i\in\mathbb{Z}}\dots$ It is clear that a constructive proof of our theorem needs to construct random variables taking values on the Polish space $\mathcal{X}$ with desired distributions.

Anyway, the statement of the theorem is general as it doesn't require the Polish space $\mathcal{X}$ to be $\mathbb{R}$, but I bet that its proof uses the existence of $U$ to define such a $(\xi'_i)_{i\in\mathbb{Z}}$ in a constructive way.

And the second part of the question? What happens if our probability space is not rich enough? Consider the probability space $\bigl(\Omega, \mathcal{F}, \mathbb{P}\bigr)$ $$\Omega = \{T,H\},\\ \mathcal{F} = \bigl\{\emptyset, \{T\}, \{H\}, \Omega\bigr\}, \\ \mathbb{P}(\{T\}) = \frac{1}{3},\quad\mathbb{P}(\{H\}) = \frac{2}{3}.$$ Consider the random variable $\xi:\Omega\to\mathbb{R}$ $$\xi(T)=0,\quad\xi(H)=1$$ This probability space is too poor, we cannot even find a $\xi'$ with the same distribution and independent of $\xi$, as the only choice is $\xi'=\xi$, but clearly they are not independent.