$T: [0,1)^{2}\rightarrow[0,1)^{2}$ by
$T(x,y) = (2x,\frac{y}{2})$, with $0 \leq x < \frac{1}{2}$
and
$T(x,y) = (2x-1, \frac{y+1}{2})$, with $\frac{1}{2} \leq x < 1$
(i) invertible
(ii) measurable
(iii) measure preserving
$\underline{\textrm{To (i):}}$
I found out, the inverse of T is $T^{-1}(x,y)=(\frac{x}{2},2y)$ for $0\leq y<\frac{1}{2}$
And $=(\frac{x+1}{2},2y−1)$ for $\frac{1}{2}\leq y<1$
Is that enough to (i)?
$\underline{\textrm{To (ii):}}$
I do not remember how to do this. Well, I know the definition, but maybe could someone show me with another example how to show a function is measurable? Because although I can explain what T is doing:
Taking a rectangle, strech it, fold it and turn it a quarter. Hence if I name the two half $A$ and $B$, then $A$ has $\frac{1}{2}$ of the area and $T(A)$ sends it again to $\frac{1}{2}$ of the area. And then $T(A)$ intersects $A$ and $B$ in a square of area $\frac{1}{2}$. Well so informally, if we take the measure of $A$, and go the streching/folding. Back, than we still have the same area for $A$? But I do not remember how to put this into formal language.
Same with (iii).
I'd be happy for any help :)
Best Luca