Prove that a sequence of maps is a sequence of i.i.d. r.v. I need an help with the following exercise.
I have a probability space $([0,1], \mathcal B, dx)$, where we denote with $\mathcal B$ the borelian sets in $[0,1]$ and $dx$ is the Lebesgue measure. We can write  each $x\in [0,1]$ as $x=\sum_{i=1}^\infty \frac{a_i}{2^i}$, where $a_i\in \{0,1\}$. When $x\in [0,1]$ has two possible expansions, choose the one with infinitely many zeroes.
Define the maps $X_i : [0,1]\to \{0,1\}\quad X_i(x)=a_i$, for each $i\in \mathbb N$.
I want to prove that this maps are  i.i.d. random variables.
Now, firstly we have to prove that $X_i^{-1}((-\infty, t])\in \mathcal B\quad \forall t\in \mathbb R$, and we know that $X_i$ can assume only two values, $0$ or $1$. So:


*

*if $t<0\, X_i^{-1}((-\infty, t]) = \{\}$,

*if $t \leq 1\, X_i^{-1}((-\infty, t]) = [0,1]$,

*if $0\leq t<1\, X_i^{-1}((-\infty, t]) = \{x\in [0,1]: a_i=0\}:=A$.


Now my first problem is: who's $A$? How can I prove that $A$ is a borelian set?
I have the same problem proving that the variables are identically distributed. In fact, I have to prove that $\mathbb P (X_i=0)=dx(A)$ is the same for all $i$, but I don't know where to start, also for independence.
 A: It's best to look at the first few cases to get the general idea. If $X_1(x)=0$, then $x<\frac12$. So $[X_1=0]=[0,\frac12)$. If $X_2(x)=0$, then we have two options: either $X_1(x)=0$, so $x\in[0,\frac14)$, or $X_1(x)=1$, in which case $x\in[\frac12,\frac34)$. Hence $[X_2=0]=[0,\frac14)\cup[\frac12,\frac34)$. This suggests the following general formula which turns out to be true:
$$[X_n=0]=\bigcup_{i=0}^{2^{n-1}-1}\left[\frac{2i}{2^n},\frac{2i+1}{2^n}\right)$$
You might like to prove this general case - maybe try induction. Note that this is the disjoint union of $2^{n-1}$ intervals all of length $2^{-n}$, so the total Lebesgue measure of $[X_n=0]$ is $\frac12$. This shows that $(X_n)$ is identically distributed. To show it is independent we must show that the sets
$$[X_j=0,X_k=0],\quad[X_j=0,X_k=1],\quad[X_j=1,X_k=0],\quad[X_j=1,X_k=1]$$
all have Lebesgue measure $\frac14$ if $j\neq k$ (the number $\frac14$ follows from the fact that $P(X_j=0)=P(X_j=1)=\frac12$ for any $j$). You should be able to do this by carefully considering the intersection of any two such intervals.
A: $$\left\{ X_{i}=0\right\} =$$$$\left[0,2^{-i}\right)\cup\left[2.2^{-i},3.2^{-i}\right)\cup\left[4.2^{-i},5.2^{-i}\right)\cup\left[6.2^{-i},7.2^{-i}\right)\cup\cdots\cup\left[\left(2^{i}-2\right).2^{-i},\left(2^{i}-1\right).2^{-i}\right)$$
A disjoint union of $2^{i-1}$ Borel sets, so a Borel set itself. Secondly $X_i$ only takes values in $\{0,1\}$ so that this is enough allready to show that $X_i$ is a random variable.
Based on this we find that $P(\{X_i=0\})=\frac12$ for every $i$. So the $X_i$ are identically distributed.
In this answer a proof of independency lacks.
