# 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).

• 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. – kimchi lover Nov 17 at 18:37
• Or show that it preserves measure of triangles. Or consider it as a map on the plane first and then... – copper.hat Nov 17 at 18:37

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 $$m A = \sum_{z \in \mathbb{Z}^2} m A_z = \sum_{z \in \mathbb{Z}^2} m f(A_z) = \sum_{z \in \mathbb{Z}^2} m F(A_z) =m F(A)$$.

• 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 $F$? – Jens Nov 18 at 12:19
• @Jens: I (hopefully) simplified my answer to clarify my approach. – copper.hat Nov 18 at 15:01