Enlarging a probability space to get independence Am reading Lemma 5.2 from this paper by Dvoretzky:
Let $X$ be a real random variable on a probability space $(\Omega,\mathcal A,P)$ satisfying $|X|\leq 1$,  let $\mathcal F=\sigma(X)$, and let $\mathcal G$ be a sub-$\sigma$-algebra of $\mathcal A$. Then
$$E\Big[\big|E[X|\mathcal G]-E[X]\big|\Big]\leq 4\alpha(\mathcal F,\mathcal G) \quad\quad\quad (1)$$
where $\alpha(\mathcal F,\mathcal G)=\sup_{F\in\mathcal F, G\in\mathcal G} \big|P(F\cap G)-P(F)P(G)\big|$.
At some point in the proof the author writes: "Let $\tilde{X}$ be a random variable with the same distribution as $X$ and independent of $\mathcal G$ (if need be the probability space can be enlarged in order to carry such a random variable)."
I don't see how to achieve this result, and how to do it without changing the quantities that appear in $(1)$. Any ideas on how to proceed?
Thanks a lot for your help.
EDIT: I tried to write down the details below. Any comment is very appreciated.

Let $Y$ be a random variable on a probaility space $(\Omega',\mathcal A',P')$ having the same distribution as $X$. On $\Omega \times \Omega'$ with the product $\sigma-$ field and the measure $P\times P'$ define $\tilde{X} (\omega,\omega')=Y(\omega')$. Then $\tilde{X}$ has the same distribution as $X$ and it is independent of $\mathcal G$.
