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...

  • $\begingroup$ 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. $\endgroup$ – t.b. Oct 18 '11 at 23:51
  • $\begingroup$ @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... :-) $\endgroup$ – André Caldas Oct 19 '11 at 2:25
  • $\begingroup$ 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... $\endgroup$ – t.b. Oct 19 '11 at 8:22
  • $\begingroup$ ...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. $\endgroup$ – t.b. Oct 19 '11 at 8:27
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    $\begingroup$ @Davide: I haven't forgotten but I lost a (rather) long answer in a browser crash and I wasn't too motivated to write it again. I'll do it in the next few days and I'll ping you. Thanks for reminding me. $\endgroup$ – t.b. Dec 6 '11 at 10:12

I am not sure with my answer and it seems that it's too late to add it, but for the future reference for anyone searching for the answer, I would like to add this. Since I cannot understand the idea of non-atomic spaces, I would make a different argument.

The idea is to make use of a symbolic representation introduced in example 2. 8. Here one sided Bernoulli shift, which is also known as full one-sided 2-shift, is introduced as an isomorphic measure-preserving system to the circle with the doubling map. We may predict that this idea can be carried over to the case of a torus.

Also, if we define an isomorphism as a measure space by the existence of an invertible measure preserving map, which is in fact equivalent to the definition you've suggested, we expect that an isomorphism as a measure-preserving system is a isomorphism as a measure space because we can just forget the commutativity of diagram condition.

So now we construct 3 maps for the 4 spaces, namely $(\mathbb{T}, \mathscr{B}_\mathbb{T}, m_\mathbb{T}),\\ (X:=\{0, 1\}^\mathbb{N}, \mathscr{B}_X, \mu:=\Pi_\mathbb{N} \mu_\left(1/2,1/2\right)),\\ (Y:=\{(0, 0),(1, 0),(0, 1)(1, 1)\}^\mathbb{N}, \mathscr{B}_Y, \nu:=\Pi_\mathbb{N} \nu_\left(1/2,1/2\right)\times \nu_\left(1/2,1/2\right)),\\ (\mathbb{T}^2, \mathscr{B}_{\mathbb{T}^2}, m_{\mathbb{T}^2}) $

The map from X to the circle is in the page 23. Next the map from Y to the torus can be defined similarly as follows:

$\psi:Y\rightarrow \mathbb{T}^2$ by $\psi\left((x_0,y_0),(x_1,y_1), ...\right)=(\sum^{\infty}_{n=0}\frac{x_n}{2^{n+1}}, \sum^{\infty}_{n=0}\frac{y_n}{2^{n+1}})$

Remark that this is also an isomorphism as a measure-preserving system. To check that this map is measure-preserving is checked in the same way as the case of the circle.

Now we ignore the iterated map structure, and construct the isomorphism between X and Y.

We define $\sigma :X\rightarrow Y$ by $\sigma\left(x_0,x_1,x_2,x_3, ...\right)=((x_0,x_1),(x_2,x_3),...)$. This map is invertible measure-preserving map, although it does not commute with the iterated shifting maps.

In fact, using the similar construction would solve the exercise 2.1.2, and in turn it also solves the case of 2.1.1. So not only do we have the measure space isomorphism, but also we have the isomorphism of measure-preserving systems if we set iterated map $T$ well.

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