An equivalence statement for measure preserving invertible transformations

Question

I am trying to prove the claim in Remark 1.2.13. : which I can if I can prove the two following which I don't know how to prove them:

1- If $T$ is a measurable transformation and B a measurable set then $T[B]$ is a measurable set?

2- If $T$ is an invertible measurable transformation $T^{-1}$ is a measurable transformation too?

If any one of the two questions wrong, then I would need another proof for the mentioned remark in the text.

Please help. Thanks!

Answer 1

In all of the Ergodic books I've seen, it is assumed that the definition of invertible means that the inverse is measurable.

In general, this need not be the case. Let $T(x)=2x\,\text{mod} 1$ and $X=[0,1)$ equipped with Lebesgue measure. This map is measure preserving, but it's inverse is not. To see that $T(x)$ is measure preserving, let $B\subset [0,1)$ be some measurable set, which can be assumed to be an interval of the form $(a,b)$. To see that $T(x)$ is a measure preserving transformation, we have the following calculation: \begin{equation} \begin{split} &\mu(T^{-1}\big{(}(a,b)\big{)}\\ &=\mu\Big{(}\big{(}\frac{a}{2}, \frac{b}{2}\big{)}\bigcup\big{(}\frac{a+1}{2}, \frac{b+1}{2}\big{)}\Big{)}\\ &=\mu((a,b)) \end{split} \end{equation} If the above calculation isn't obvious to you. Here are some concrete examples.

Suppose that $(a,b)=(0,1)$: \begin{equation} \begin{split} T^{-1}\big{(}(a,b)\big{)}&=T^{-1}\big{(}(0,1)\big{)}\\ &=\big{(}0,\frac{1}{2}\big{)}\cup\big{(}\frac{1}{2},1\big{)}. \end{split} \end{equation}
Additionally, suppose that $(a,b)=(0, \frac{1}{2})$: \begin{equation} \begin{split} T^{-1}\big{(}(a,b)\big{)}&=T^{-1}\big{(}(0,\frac{1}{2})\big{)}\\ &=\big{(}0,\frac{1}{4}\big{)}\cup\big{(}\frac{1}{2},\frac{3}{4}\big{)}. \end{split} \end{equation} Hence, $T^{-1}\big{(}(a,b)\big{)}$ is a pair of two non-overlapping intervals, which preserves measure.

Conversely for the inverse, note that if $B=[0,\frac{1}{2})$, then ${(T^{-1})}^{-1}(B)=T(B)=[0,1)$, so the measures are different. Also note that the above example generalizes to $T(x)=nx\,\text{mod} 1$ for $n$ being any positive natural number.

Answer 2

In general it is not true that bijective measurable maps are bimeasurable, see Measurability of the inverse of a measurable function. There it is shown that bijective measurable absolutely continuous maps are bimeasurable.

So let us let an invertible measure preserving transformation $f$ of a probability space $(X,\mu)$ stand for a bimeasurable bijection that preserves $\mu$ (which then implies the inverse also preserves $\mu$, as the Remark remarks).

Let me also mention that it is not uncommon to have all this be applicable $\mu$-almost everywhere, so that there are $\mu$-full measure sets $X_1,X_2\subseteq X$ such that $f:X_1\to X$ is measurable and measure preserving, $g:X_2\to X$ is measurable and measure preserving, $g\circ f = \operatorname{id}_{X_1}$, $f\circ g = \operatorname{id}_{X_2}$ (Here $g$ is $f^{-1}$ measure theoretically). A standard example is the isomorphism (in the category of measure-spaces and measure-preserving measurable maps) between the sequence space $F(\mathbb{Z}_{\geq0},\{0,1\})$ and the standard Cantor set.