# How to show that a particular distribution is the Lebesgue measure

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 $$E_n = \{ X_i = T_i | i=1,..,n \text{ } \& X_{n+1}. To find the distribution of $$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.

• Show $P(U \in (a, b]) = b - a$ for $a$ and $b$ dyadic rationals and then use continuity of the probability $P(A_n) \to P(A)$ when $A_n$ monotonically converges to $A.$ – Will M. Mar 13 '19 at 15:40

As stated in the hint but using the fact that $$(k/2^m , (k+1)/2^m]$$ form a $$\pi$$ system, we can show that $$P\{\sum_{n=1}^{\infty}\frac{X_n}{2^n}\in (k/2^m , (k+1)/2^m]\} = \frac{1}{2^m} = \lambda((k/2^m , (k+1)/2^m])$$ and thus the two measures agree on $$\mathcal{B}$$. To see the above equality, there is only one way with positive probability s.t. $$\sum_{n=1}^{\infty}\frac{X_n}{2^n}\in (k/2^m , (k+1)/2^m]$$ where in $$0's$$ and $$1's$$ are assigned to $$n=1,..,m$$ and that happens with probability $$1/2^m$$.
$$U$$ comes from an infinite coin toss, so I think to construct independent $$U_i$$ whose distribution is Lebesgue, we can take independent infinite coin tosses.