$\phi: (X, \mu) \to (\Bbb T, m_{\Bbb T})$ given by $\phi(\mathbf{x}) = \sum_{n=0}^\infty \frac{x_n}{2^{n+1}}$ is measure-preserving 
Let $X = \{0,1\}^\Bbb{N}$ with the product measure $\mu = \prod_{\Bbb N} \nu$, where $\nu(\{0\}) = \nu(\{1\}) = \frac12$. The map $\sigma:X\to X$ by $\sigma(x_0, x_1, \ldots) = (x_1, x_2, \ldots)$ is the left-shift operator on $X$. It is easy to see that $\sigma$ preserves $\mu$, i.e., for every measurable subset $B \in \mathscr B$ of the measure space $(X,\mathscr B, \mu)$, we have $\mu(B) = \mu(\sigma^{-1}(B))$. Define a map $\phi:X\to \Bbb T$ by $$\phi(x_0, x_1, x_2, \ldots) := \sum_{n=0}^\infty \frac{x_n}{2^{n+1}}$$

I would like to show that $\phi$ is measure-preserving from $(X, \mu)$ to $(\Bbb T, m_{\Bbb T})$.
Remark. $\mu$ and $m_{\Bbb T}$ are probability measures, and $\Bbb T = \Bbb R/\Bbb Z \cong S^1$.

We need to show that $$\mu(\phi^{-1}(B)) = m_{\Bbb T} (B)$$ for all Borel-measurable subsets of $\Bbb T$. It is enough to prove the result for the generators of the Borel $\sigma$-algebra of $\Bbb T$, i.e., the open sets in $\Bbb T$. What do open sets in $\Bbb T$ look like? I need to understand better $m_{\Bbb T}$ and open sets in $\Bbb T$ to tackle the problem. Thanks for any help! I'd appreciate a detailed explanation if possible.

Reference: Ergodic Theory with a view towards Number Theory by Manfred Einsiedler and Thomas Ward.
 A: Call $p$ the canonical projection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Z}$. By definition of the quotient topology, the open sets in $\mathbb{R}/\mathbb{Z}$ are the subsets of $\mathbb{R}/\mathbb{Z}$ whose premiages by $p$ are open sets on $\mathbb{R}$.
One can check that the open sets in $\mathbb{R}/\mathbb{Z}$ are the union of « open intervals » of $\mathbb{R}/\mathbb{Z}$ (namely images by $p$ of open intervals of $\mathbb{R}$). Any such union can be written as a countable union of « open intervals » simply by looking at the connected components.
Hence it suffices to look at « open intervals » of $\mathbb{R}/\mathbb{Z}$. Since any open interval can be written as a countable union of disjoint dyadic intervals, it suffices to look at dyadic intervals.
A: The projections $X_n(\omega)=\omega_n$ on $\Omega=\{0,1\}^{\mathbb{N}}$, with the measure structure $(\Omega,(2^{\{0,1\}})^{\otimes\mathbb{N}},\mu)$, form an i.i.d sequence of standard Bernoulli random variables. The map $\phi:\omega\mapsto \sum_{n\in\mathbb{N}}2^{-n}\omega_n$ induced the Lebesgue measure on $([0,1],\mathscr{B}([0,1]))$ (see for example the Lemma in this posting), that is $$m=\mu\circ\phi^{-1}$$
which is basically what the OP is asking about (seem comments related to the dynamics on $\mathbb{R}\mod 1$ and the unit circle $\mathbb{S}^1$).
The shift operator $S$ on $\Omega$ translate into the map $T:x\mapsto\{2x\}$ on $[0,1]$ where $\{x\}:=x-\lfloor x\rfloor=x\mod1$, that is
$$\phi\circ S = T\circ \phi$$
Since $\mu$ and $m$ are $S$ and $T$ invariant respectively, the dynamical systems $(\Omega,\mathscr{F},\mu,T)$ and $([0,1],\mathscr{B}([0,1]), m,T)$ are conjugate (some minor details about bijectively need to be addressed, see comments).

Comments:

*

*The map is 1-1 $\mu$-almust surely. For example, consider the set $R$ that admit a unique binary expansion. $Q=[0,1]\setminus R$ is countable and hence of Lebesgue measure $0$ and so,  and $S'=\phi^{-1}(Q)$ is also countable and hence also of $\mu$-measure $0$.
The map $\phi:S'\rightarrow R$ is bijective $\sum_nr_n=\infty$ and let $S'=\phi^{-1}([0,1]\setminus R)$. As $Q=[0,1]\setminus R$ is countable, $S'$

*The unit circle $\mathbb{S}^1$ and the unit interval $[0,1)$ are measurable isometric: $f:\theta\mapsto e^{2\pi i\theta}$ is one such measurable isomorphism. If $\lambda$ is the arch length on $\mathbb{S}^1$, and $m$ is Lebesgue measure on $[0,1)$ then $m\circ f^{-1}=\frac{1}{2\phi}\lambda$ and $\lambda\circ f=2\pi m$.

*arch length is invariant under the map $D:z\mapsto z^2$ on $\mathbb{S}^1$; moreover
$$ D\circ f = f \circ T$$
Thus, $([0,1],\mathscr{B}([0,1]),m,T)$ and $(\mathbb{S}^1,\mathscr{B}(\mathbb{S}^1),\lambda/2\pi,D)$ are conjugate.

