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According to Definition 2.7 in Ergodic Theory: with a view towards Number Theory, the systems $(X, \mathcal{B}_X, \mu, T)$ and $(Y, \mathcal{B}_Y, \nu, S)$ are isomorphic when there is a $X' \in \mathcal{B}_X$ and a $Y' \in \mathcal{B}_Y$ with $\mu(X') = \nu(Y') = 1$, $TX' \subset X'$, $SY' \subset Y'$, and an invertible measure-preserving $\phi: X' \to Y'$ $$ \phi \circ T(x) = S \circ \phi(x) $$ for all $x \in X'$. The book does not define isomorphism of measure spaces. I assume that it is the same as Definition 2.7, with the identity in place of $S$ and $T$.

Now, Exercise 2.1.1 asks me to show that $(\mathbb{T}, \mathcal{B}_{\mathbb{T}}, m_{\mathbb{T}})$ is isomorphic to $(\mathbb{T}^2, \mathcal{B}_{\mathbb{T}^2}, m_{\mathbb{T}^2})$, where $\mathbb{T} = \mathbb{R}/\mathbb{Z}$, and the measures $m_{\mathbb{T}}$ and $m_{\mathbb{T}^2}$ are the Haar measures in $\mathbb{T}$ and $\mathbb{T}^2$ with their usual group structure.

How do I do that?


This is not "homework", but it could be...

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This is totally overkill, but something you should know: Let $(X,\Sigma,\mu)$ be a probability space, where $X$ is an uncountable Polish space (completely metrizable and second countable) and $\Sigma$ the Borel $\sigma$-algebra on $X$. If $\mu$ has no atom then $(X,\Sigma,\mu)$ is isomorphic (in your sense) to $[0,1]$ with Lebesgue measure. You can find this in Kechris, Classical Descriptive Set Theory, Theorem 17.41, p.116. –  t.b. Oct 18 '11 at 23:51
    
@t.b.: I am getting acquainted to Lebesgue spaces... It is a big beast for me right now. I wondered if there was a direct proof for this case, since it was exercise 2.1.1... :-) –  André Caldas Oct 19 '11 at 2:25
    
I see :) I don't have the time to write up the details right now, but I would do the following lines: Identify $\mathbb{T}$ with $[0,1]$ and Lebesgue measure. Equip the Cantor set $C = \{0,1\}^{\mathbb{N}}$ with the product measure $\mu = \left[\frac{1}{2}(\delta_{0} + \delta_{1})\right]^{\mathbb{N}}$. Then map a binary sequence $x = (x_n)$ to $\varphi(x) = \sum_{n=1}^{\infty} 2^{-n} x_n$ and observe that this map is measure-preserving (the binary sequences starting with $(y_1,y_2,\cdots,y_n)$ are mapped to an interval $[y,y+2^{-n}]$ in $[0,1]$, while in the product measure these sequences... –  t.b. Oct 19 '11 at 8:22
    
...have measure $2^{-n}$). There is a measurable and measure-preserving right inverse $[0,1] \to C$ defined everywhere except at the dyadic rationals, so $[0,1]$ and $C$ are isomorphic. Now observe that $C \cong C \times C$ and map $C \times C \to [0,1]^2$ and check that you get a measure-preserving isomorphism. This roughly amounts to cutting up the interval $[0,1]$ successively into $2^n$ intervals with dyadic endpoints and $[0,1]^2$ into squares with dyadic corners and matching them up appropriately. –  t.b. Oct 19 '11 at 8:27
    
@t.b.: Very nice! Thank you. –  André Caldas Oct 19 '11 at 13:52
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