Let $$\operatorname{frac}(x):=x-\lfloor x\rfloor\;\;\;\text{for }x\ge0$$ and $$\theta(x):=\operatorname{frac}(2x)\;\;\;\text{for }x\in[0,1)$$ denote the Bernoulli shift. Now define $$X(x):=\lfloor 2x\rfloor=\left.\begin{cases}0&\displaystyle\text{, if }x\in\left[0,\frac12\right)\\1&\displaystyle\text{, if }x\in\left[\frac12,1\right)\end{cases}\right\}\;\;\;\text{for }x\in[0,1)$$ and $$X_n:=X\circ\theta^{n-1}\;\;\;\text{for }n\in\mathbb N.$$
How can we show that $(X_n)_{n\in\mathbb N}$, considered as a process on the probability space $([0,1),\mathcal (B[0,1)),\operatorname P)$, where $\operatorname P$ is the Lebesgue measure on $\mathcal B([0,1))$, is an independent stationary process and each $X_n$ is Bernoulli distributed with parameter $1/2$?
I guess this is really simple, but I even fail to show the claim about the distributions. Clearly, $$\operatorname P\left[X_1=0\right]=\operatorname P\left[\left[0,\frac12\right)\right]=\frac12\tag1.$$ So, at least $X_1$ has the claimed distribution.
But how can we easily prove the other claims? What's the "trick"?
EDIT: The stationarity is clearly trivial, if we can show the independence and distribution claim ...
Hint: $\ X_n(x)\ $ is the $\ n^\text{th}\ $ bit in the binary expansion of $\ x\ $ (i.e. the coefficient of $\ \frac{1}{2^n}\ $ in that expansion). What interval must $\ x\ $ belong to if $\ X_i=a_i\ $ for $\ i=0,1,\dots,n-1\ $, and $\ a_i\ $ is some arbitrary sequence of zeroes and ones?
Response to query in comments: $$\ P\left[X_1=a_1, X_2=a_2, \dots, X_n=a_n\right]=\text{Lebesgue measure of }\left[\sum_{i=1}^n\frac{a_i}{2^i}, \sum_{i=1}^n\frac{a_i}{2^i}+\frac{1}{2^n}\right)=\ ?\ $$