Let $X_1 , X_2,..$ be independent r.v's, defined on some probability space $(\Omega, \mathcal{F},P)$, s.t.
$$P(X_n = 0) = P(X_n = 1) =1/2 \text{ } \forall n$$
Let $U = \sum_{n=1}^{\infty}X_n 2^{-n}$.
(i) Show that the distribution of $U$ is the Lebesgue measure. (ii) Construct an independent sequence $U_1,U_2,..$ on $((0,1),\mathcal{B},\lambda)$ s.t. the distribution of each $U_i$ is $\lambda$ where $\lambda$ is Lebesgue.
My attempt: Let $E_n = \{ X_i = T_i | i=1,..,n \text{ } \& X_{n+1}<T_{n+1}\}$. To find the distribution of $U$, I need $P(X_1 X_2... < T_1 T_2...)$ where $X_i , T_i \in \{0,1\}$. Then $$\{X_1 X_2... < T_1 T_2...\} = \coprod_{n=1}^{n=\infty}\{E_n \cap \{T_{n+1}=1\}\}$$ where $\coprod$ is the disjoint union.
$$P\{E_n \cap \{T_{n+1}=1\}\} = 1_{\{T_{n+1}=1\}}.(\frac{1}{2})^{n+1}$$ and thus:
$$P\{X_1 X_2... < T_1 T_2...\} = \sum_{n=1}^{\infty}1_{\{T_{n+1}=1\}}.(\frac{1}{2})^{n+1}$$
Am I thinking in the right direction? and if yes, I'm not sure which set on $\mathbb{R}$ I should be looking at to take it's Lebesgue measure to be equal to the distribution obtained above? Thanks and appreciate a hint.