measurable, measure preserving $T: [0,1)^{2}\rightarrow[0,1)^{2}$ 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
 A: (i) Yes, enough. (If true.) Have you drawn it? It maps the left half of the unit square to the bottom half of again the unit square, and the right half to the upper one. Nice mapping.
(ii) $T$ is measurable means here (I guess) that for any Borel subset $W$ of $\Bbb R^2$ (now $[0,1]^2$ is enough), we have 
$$T^{-1}(W)\ \text{ is also a Borel set}$$
Where Borel sets of $\Bbb R^2$ are generated by the filled rectangles with countably union and intersection, and complement.
(iii) is the same backwards: $V$ Borel $\implies\ T(V)$ Borel.
Can you do that? Just cut them at the appropriate halves..
A: To prove $(i)$, notice that, $T(v)=Av$, then $T$ is invertible iff $A$ is a non-singular matrix, where $T$ is a linar transformation between two vector spaces and $A$ is the representation matrix of the linear transformation $T$. In your case $A$ is given by
$$ A = \begin{bmatrix}
       2 &  0           \\[0.3em]
       0           & \frac{1}{2} \\[0.3em]
\end{bmatrix}\,.$$
You can see that the above matrix is non-singular, since the determinant is not zero. You can do the same with the other linear transformation.
