I'm reading through Ergodic Theory: With a View Towards Number Theory with a friend, and I am requesting a proof check (the friend is asleep). Also, I'm a noob to measure theory/ergodic theory, so this might be a pretty trivial terminology check 😛
Problem: Show that the measure preserving system $(\mathbb{T}, \mathfrak{B}_{\mathbb{T}}, m_{\mathbb{T}}, T_4)$, where $T_4(x) = 4x \pmod{1}$, is measurably isomorphic to the product system $(\mathbb{T}^2, \mathfrak{B}_{\mathbb{T}^2}, m_{\mathbb{T}^2}, T_2 \times T_2)$, where $T_2(x) = 2x \pmod{1}$.
Solution:
Define $\mathfrak{B}_{\mathbb{T}} \ni X' = \mathbb{T} \setminus \mathbb{Z}[\frac{1}{2}]/\mathbb{Z}$ and $\mathfrak{B}_{\mathbb{T}^2} \ni Y' = \left(\mathbb{T} \setminus \mathbb{Z}[\frac{1}{2}]/\mathbb{Z}\right)^2$. Clearly $T_4X' \subseteq X'$ and $(T_2\times T_2) Y' \subseteq Y'$, and as the dyadic rationals are countable (this deals with repeated $1$'s), $m_{\mathbb{T}}(X') = m_{\mathbb{T}^2}(Y') = 1$.
We use the bijective map $\phi : X' \to Y'$ where $.\overline{b1b2b3b4}\ldots \mapsto (.\overline{b1b3}\ldots, .\overline{b2b4}\ldots)$
Now, for any $x \in X'$, we may write $x$ as $\sum_{n=1}^{\infty}\frac{b_n}{2^n}$ where $b_n \in \{0, 1\}$, so
$\phi \circ T(x) = \phi\left({4{\sum_{n=1}^{\infty}\frac{b_n}{2^n}}}\right) = \phi\left({\sum_{n=3}^{\infty}\frac{b_{n}}{2^{n-2}}}\right) = \left({\sum_{n=2}^\infty\frac{b_{2n-1}}{2^{2n-1}}, \sum_{n=2}^{\infty}\frac{b_{2n}}{2^{2n}}}\right) = (T_2\times T_2)\left(\sum_{n=1}^\infty\frac{b_{2n-1}}{2^{2n-1}}, \sum_{n=1}^{\infty}\frac{b_{2n}}{2^{2n}}\right) = (T_2\times T_2)\circ\phi(x) \pmod{1}$
$\blacksquare$
Best Answer
I don't have much experience of measure theory but I think I could understand what it means.
I assumed $\mathbb{T}:=\mathbb{R}/\mathbb{Z}$. I noticed two points:
・$\phi$ is not bijective because $0.10101010..$ and $0.01010101..$ are both sent to $(0,0)$. $(\mathbb{R}-\mathbb{Q})/\mathbb{Z}$ may work better (but I haven't checked enough).
・I think you need to verify that the map $\phi$ preserve the measure because it is required to show "measurably isomorphic":
I think it means for any $U \subset X', \mu_{\mathbb{T}}(T_4(U)) = \mu_{\mathbb{T}^2}((T_2 \times T_2 )(\phi(U)))$
(but I am not so familiar with the topic so I am not sure.)