Below is from Tao's lecture note and he says there exists an i.i.d. sequence of random variables $\{X_n\}_n$ such that each $X_i$ is uniformly distributed. For any given random variable $X_1$, does there exist an i.i.d. sequence$\{X_n\}_n$ ?
https://terrytao.wordpress.com/2015/10/23/275a-notes-3-the-weak-and-strong-law-of-large-numbers/#sme
Proposition 6 Let ${\varepsilon > 0}$. Then, for sufficiently large ${n}$, a proportion of at least $1-\varepsilon$ of the cube ${[-1,1]^n}$ (by ${n}$-dimensional Lebesgue measure) is contained in the annulus ${\{ x \in {\bf R}^n: (1-\varepsilon) \sqrt{n/3} \leq |x| \leq (1+\varepsilon) \sqrt{n/3} \}}$.
Proof: Let ${X_1,X_2,\dots}$ be iid random variables drawn uniformly from ${[-1,1]}.$
Best Answer
Let $P$ be the distribution of $X_1$, $\Omega=\mathbb{R}^{\mathbb{N}}$, $\mathcal{F}=\mathcal{B}(\mathbb{R})^{\otimes\mathbb{N}}$, where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$ and $\mathbb{P}=P^{\otimes\mathbb{N}}$. For each $n\in\mathbb{N}$, let $\pi_n:\Omega\to\mathbb{R}$ be defined as $\pi_n(\omega)=\omega_n$, where $\omega=\{\omega_n\}_{n\in\mathbb{N}}$. We have $\{\pi_n\}_{n\in\mathbb{N}}$ is a sequence of $\mathbb{P}$-independent random variables such that for each $n\in\mathbb{N}$, $\pi_n$ has distribution $P$, i.e., $\pi_n$ and $X_1$ have the same distribution.