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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).

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  • $\begingroup$ It might be easier to check that $F$ is the composition of two L-preserving maps, namely, $(x,y)\mapsto(x+\alpha,y)$ and $\mapsto(x,x+y)$. Both by a Fubinification of a proof that $x\mapsto x+\alpha$ is L-preserving. $\endgroup$ – kimchi lover Nov 17 at 18:37
  • $\begingroup$ Or show that it preserves measure of triangles. Or consider it as a map on the plane first and then... $\endgroup$ – copper.hat Nov 17 at 18:37
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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)$.

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  • $\begingroup$ Thanks for your reply! How do you conclude that $\lfloor f(A)\cap[z_{1},z_{1}+1)\times[z_{2},z_{2}+1)\rfloor$ are disjoint subsets of $[0,1)^{2}$? How do you use the bijectivity of $F$? $\endgroup$ – Jens Nov 18 at 12:19
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    $\begingroup$ @Jens: I (hopefully) simplified my answer to clarify my approach. $\endgroup$ – copper.hat Nov 18 at 15:01

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