Definition of strong mixing and definition of measure-preserving

Question

I have a few questions regarding measure-preserving dynamical systems $(X,\mathcal{A},\mu,T)$.

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1) The definition of measure preserving is always stated as $$\mu(T^{-1}B)=\mu(B),$$ for all $B$.

I believe this neither implies nor is implied by the similar statement $$\mu(TB)=\mu(B).$$

Is this correct? The problem lies in the fact that $TT^{-1}B \subseteq B \subseteq T^{-1}TB$ may be strict inclusions, so the following implications may not hold: $$\mu(T^{-1} TB) = \mu(TB) \implies \mu(B)=\mu(TB)$$ and $$\mu(TT^{-1}B)= \mu(T^{-1}B) \implies \mu(B)=\mu(T^{-1}B).$$

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2) Is there some "reason" why the above definition involves $T^{-1}$ rather than $T$?

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3) The definition of strong mixing is $$\lim_{n \to \infty} \mu(A \cap T^{-n}B) = \mu(A) \mu(B)$$ for all $A$ and $B$. Again, the exponent of $T$ is negative. However, in the wine-water example below the definition, they have $T^n$. Is this an issue?

Thank you in advance

Answer 1

$1$) This is correct, there are examples of transformations which change the measure going forward.

Example Let $T:\Bbb T^2\to \Bbb T$ be a map from the $2$-torus to the $1$-torus given by projection along the diagonal. This is a continuous, surjective homomorphism on a compact group. It is easy to see any such map is measure preserving by defining the pullback measure on the domain and showing that it is a Haar measure which assigns the whole group measure $1$, and using uniqueness). However, taking the diagonal subgroup $\Delta \Bbb T\subseteq \Bbb T^2$ which has Haar measure $0$ (it is of infinite index) in the domain but full measure in the target gives us that $\mu_2(T(\Delta \Bbb T))\ne \mu_1(\Bbb \Delta \Bbb T)$.

Example Let $T:\Bbb T\to\Bbb T$ be a map from the $1$-torus to itself given by $$T: z\mapsto z^2$$ Then we have that $T$ is a continuous, surjective homomorphism on a compact group so that it is measure preserving, just as in the first example, however the direct image of some sets actually increase. This is easy to see by taking, for example, $S=\{z\in\Bbb T : -{\pi\over 2}\le\text{Arg}\; z\le {\pi\over 2}\}$ which has measure $\mu(S)={1\over 2}$, however clearly $T(S)=\Bbb T$ so that $\mu(T(S))=1$.

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$2$) Inverses commute with basic set operations like complement, union, and intersection, so it is handy when studying things to be able to do that. This is the foremost reason for making the definition as it is invaluable. For bijections, there is no need to distinguish, of course, and this is the most common case, but the theorem on continuous homomorphisms from the example above is another great example of why we choose inverses.

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$3$) Because $T$ is measure preserving and bijective (you're not losing any molecules in this particular system), $\mu(T^nA\cap B)=\mu(A\cap T^{-n}B)$ which is an equivalent formulation since

$$T^{-n}(T^nA\cap B)=A\cap T^{-n}B.$$

More generally this is not so, and indeed if you recall the definition of a measurable function is that it sends measurable sets to measurable sets under pre-images, so the image may not even be in the relevant $\sigma$-algebra. (h/t m.g. for making the implicit explicit)