I am reading Ergodic theory with a view towards number theory and the following is from Example 2.8. in Section 2.1:
Let $X=\{0,1\}^{\mathbb N}$ with the infinite product measure $\mu=\prod_{\mathbb N}\mu_{(1/2,1/2)}$, $\mu_{(1/2,1/2)}(\{0\})=\mu_{(1/2,1/2)}(\{1\})=\frac{1}{2}$. The left shift map $\sigma: X\to X$ defined by $$ \sigma(x_0,x_1,...)=(x_1,x_2,...) $$ preserves $\mu$. The map $\phi: X\to \mathbb T$ defined by $$\phi(x_0,x_1,...)=\sum_{n=0}^\infty\frac{x_n}{2^{n+1}}$$ is measure-preserving from $(X,\mu)$ to $(\mathbb T, m_{\mathbb T})$ and $\phi(\sigma(x))=T_2(\phi(x))$. The map $\phi$ has a measurable inverse defined on all but the countable set of dyadic rationals $\mathbb Z[\frac 12]/\mathbb Z$.
My questions:
How to show that the map is measure-preserving from $(X,\mu)$ to $(\mathbb T, m_{\mathbb T})$?
Why $\phi$ does not have a measurable inverse on the dyadic rationals?
My attempt:
It suffices to show that the measure of preimage of every open set in $\mathbb T$ is preserved. To do this, first suppose that the open interval has rational endpoints, i.e., the dyadic expression terminates and we can easily show that this is true. If the endpoints are not rationals, then we can use a sequence of rationals to approximate the endpoints and take the limit. But I am willing to know whether there are other ways to show this.
I have no idea why we cannot just define the image of $\frac{m}{2^n}$ to be $(x_0,x_1,...)$ in the following way:
Write $\frac{m}{2^n}$ into $\sum_{n=0}^\infty\frac{x_n}{2^{n+1}}$ and map the $x_n$'s to $(x_0,x_1,...)$.?
Edit:
Now I understand why we need to remove the set of dyadic rationals. Since we want the map to be bijective, and every dyadic rational has two equivalent ways of expression as a binary sequence, one is of finite length and another is infinitely long.