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!
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.
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.