For $\alpha$ irrational, prove that $F(x,y):=(x+\alpha\mod1,x+y\mod1)$, $T^{2}\to T^{2}$ preserves the Lebesgue measure. 
Technically, this is semi-duplicate of one of my previous questions, but since it hasn't been answered yet, I figured that I should reformulate and focus on just one single question.

Suppose that $\alpha\in\mathbb{R}$ is an irrational number. How do I prove that $$F(x,y):=(x+\alpha\mod1,x+y\mod1),\quad T^{2}\to T^{2},$$ preserves the Lebesgue measure (on $[0,1[\times[0,1[$)? Recall that measure preserving means that $\text{Leb}(F^{-1}(A))=\text{Leb}(A)$ for all measurable subsets $A\subset[0,1[\times[0,1[$. I tried to prove it directly and I tried to use Fourier analysis (i.e. that $\forall f\in L^{2}$, $\int_{T^{2}}f\circ F=\int_{T^{2}}f$), but I didn't succeed. I spent 3 days on this problem without progress, so any suggestions are greatly appreciated!
EDIT: I'm not sure, but I suspect that this result is also true for rational $\alpha$ and that irrationality is required in the proof of non-weak-mixing (see link above).
 A: I am sure there is a less messy way, but it escapes me.
Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be $f(x) = (\alpha+x_1, x_1+x_2)$.
Note that $\det {\partial f(x) \over \partial x} = 1$ and so $f$ is measure preserving as a map on the plane.
Let $g(y) = (y_1-\alpha, y_2-y_1+\alpha)$ and note that $g = f^{-1}$ (as a map on the plane).
Let $\lfloor x \rfloor = (x_1 \operatorname{mod} 1, ..., x_n \operatorname{mod} 1)$.
Note that if $z \in \mathbb{Z}^2$then $\lfloor f(x+z) \rfloor = \lfloor f(x) \rfloor$ and similarly for $g$. It is straightforward to see that we can define $F,G$ on $T^2$ such that $F(\lfloor x \rfloor) = \lfloor f(x) \rfloor$
and similarly for $G$.
Furthermore, it is straightforward to show that $G$ is the inverse of $F$ (hence $F$
is a bijection).
To reduce clutter, let $I_z = [z_1,z_1+1) \times [z_2,z_2+1)$,
$I_0 = [0,1)^2$.
Now take $A \subset I_0$, then $f(A) = \cup_{z \in \mathbb{Z}^2}(f(A) \cap I_z )$ where the last union is clearly a disjoint union.
If we let $A_z = f^{-1} (f(A) \cap I_z)$, then the $A_z$ are disjoint and
$A = \cup_{z \in \mathbb{Z}^2} A_z$.
Since $F$ is a bijection, the sets $F(A_z)$ are disjoint.
Furthermore, if $x \in A_z$, then 
$F(x) = f(x)-z$, so we have
$m A_z = m f(A_z) = m F(A_z)$.
Hence $ m A = \sum_{z \in \mathbb{Z}^2} m A_z = \sum_{z \in \mathbb{Z}^2} m f(A_z) = \sum_{z \in \mathbb{Z}^2} m F(A_z) =m F(A)$.
