I am following Klenke's book on Probability Theory, and stumbles upon this exercice (Exercise 1.4.4).
Let $\Omega=\{0,1\}^\mathbb{N}$ and let $\mathcal{A}=(2^{\{0,1\}})^{\otimes\mathbb{N}}$ be the $\sigma$-algebra generated by the cylinder sets $$ \Bigl\{[\omega_1,\dots,\omega_n]:n\geqslant 1,\omega_1,\dots\omega_n\in\{0,1\}\Bigr\} $$ Further, let $\mu=(\frac{1}{2}\delta_0+\frac{1}{2}\delta_1)^{\otimes\mathbb{N}}$ be the Bernoulli measure on $\Omega$ with equal weights on $0$ and $1$. For all $n\in\Omega$, let $X_n:\Omega\to\{0,1\},\omega\mapsto\omega_n$ be the $n$th coordinate map. Finally, let $$ U(\omega)=\sum_{n=1}^\infty X_n(\omega)2^{-n} $$
1) Show that $\mathcal{A}=\sigma(X_n:n\geqslant 1)$.
2) Show that $U$ is $\mathcal{A}$-$\mathcal{B}([0,1])$-measurable.
3) Determine the image measure $\mu\circ U^{-1}$ on $([0,1],\mathcal{B}([0,1])$.
4) Determine an $\Omega_0$ in $\mathcal{A}$ such that $\tilde{U}=U\mid_{\Omega_0}$ is bijective.
5) Show that $\tilde{U}^{-1}$ is $\mathcal{B}([0,1])$-$\mathcal{A}$-measurable.
6) Give an interpretation of the map $X_n\circ \tilde{U}^{-1}$.
My thoughts so far:
1) Let $n\geqslant 1$ an integer. We have $X_n^{-1}(\emptyset)=\emptyset$, $X_n^{-1}(\{0\})=[\omega_1,\dots,\omega_{n-1},0]:=A_n$, $X_n^{-1}(\{1\})=[\omega_1,\dots,\omega_{n-1},1]=A_n^c$, $X_n^{-1}(\{0,1\})=\Omega$. Thus, we have $\sigma(X_n)=\{\emptyset, A_n, A_n^c,\Omega\}$.
$$ \begin{align} \sigma(X_n:n\geqslant 1) &=\sigma\Biggl(\bigcup_{n\geqslant 1}\sigma\big(X_n\big)\Biggr)\\ &=\sigma\Biggl(\bigcup_{n\geqslant 1}\{\emptyset,A_n,A_n^c,\Omega\}\Biggr)\\ &=\sigma\Bigl(\{\emptyset,A_1,\dots,A_n,\dots,A_1^c,\dots,A_n^c,\dots,\Omega\}\Bigr)\\ &=\sigma\Bigl(\{A_1,\dots,A_n,\dots\}\Bigr) \end{align} $$ The cylinder set $[\omega_1,\dots,\omega_n]=\{\omega'\in\Omega,\omega'_i=\omega_i~\forall i=1,\dots,n\}$ is equal to $$ \Biggl(\bigcap_{i\in I} A_{\omega_i}\Biggr)\bigcap\Biggl(\bigcap_{i\in J} A_{\omega_i}^c\Biggr) $$ where $I=\{i\in\{1,\dots,n\}:\omega_i=0\}$ and $J=\{1,\dots,n\}\setminus I$ which is $\sigma(X_n:n\geqslant1)$-measurable, thus $\mathcal{A}\subset\sigma(X_n:n\geqslant1)$.
On the other hand, $A_n=[\omega_1,\dots,\omega_{n-1},0]$ for $\omega_1,\dots,\omega_{n-1}\in\{0,1\}$ is $\mathcal{A}$-measurable. Thus $\mathcal{A}=\sigma(X_n:n\geqslant 1)$.
2) Let $f_N:\omega\mapsto\sum_{n=1}^N X_n(\omega)2^{-n}$ for $N\geqslant 1$. $f_N$ is $\mathcal{B}([0,1])$-measurable as it is a simple function. The sequence converges (even increases) pointwise to $U$, thus $U$ is $\mathcal{B}([0,1])$-measurable.
3) I do not know how to properly answer this question. I thought about this: let $a\in[0,1]$. Let $N=\min\limits_{n\geqslant 1}\Bigl(\sum_{m=1}^n 2^{-m}\geqslant a\Bigr)$. Then for $U(\omega)$ to be in $[a,1]$ for a given $\omega\in\Omega$, the $N$ first terms have to be equal to $1$. It would mean that $\mu\circ U^{-1}([a,1])=2^{-N}(\frac{1}{2}\delta_0+\frac{1}{2}\delta_1)^{\otimes\mathbb{N}}$. We also have $N=\lceil-\log_2(1-a)\rceil$.
But I do not know if this is all there is to say, and I do not know if this completely answers the question. If I had to go further, I would guess that the image measure is actually the uniform distribution on [0,1].
4) We note that $\omega_1=\{1,0,1,1,\dots,1,\dots\}$ and $\omega_2=\{1,1,0,0,\dots,0,\dots\}$ are such that $U(\omega_1)=U(\omega_2)$. So my guess is that $\Omega_0=\{\omega\in\Omega:\{n:\omega_n=0\}\text{ is infinite}\}$. Let $x\in[0,1]$. To obtain $x$ from $\tilde{U}(\omega)$, we can "deconstruct" the binary expansion of $x$ with the functions $(f_n)_{n\geqslant 1}$ with $$ \begin{align} &f_1=\mathbb{1}_{[\frac{1}{2},1)}\\ &f_2=\mathbb{1}_{[\frac{1}{4},\frac{1}{2})\cup[\frac{3}{4},1)}\\ &\dots \end{align} $$ and so on. Then $\tilde{U}^{-1}(x)=(f_n)_{n\in\mathbb{N}}$. and thus $\tilde{U}$ is a bijection.
5) $f_n$ is measurable for all $n\geqslant 1$ so $\tilde{U}$ is measurable (or does this works only for finite number of coordinates ?)
6) It gives the $n$th digit in the binary expansion of $x\in[0,1]$.
Can you give some feedback on the way I tried to answer those questions as well as some details for the questions I have concerns about ?
Thank you !