If $\alpha$ irrational, then $F(x,y)=(x+\alpha,x+y)\mod1$, $T^{2}\to T^{2}$ preserves Lebesgue measure and is not weak mixing

Question

Let $\alpha$ be an irrational number. How do I prove that $$F(x,y)=(x+\alpha,x+y)\mod1,\qquad T^{2}\to T^{2}$$ has the following properties:

1. Lebesgue measure invariant: $\lambda(F^{-1}([a,b)\times[c,d)))=\lambda([a,b)\times[c,d))$ for all $[a,b)\times[c,d)\subset[0,1)\times[0,1)$.
2. Is not weak mixing, i.e. does not satisfy $$\frac{1}{n}\sum_{i=0}^{n-1}|\lambda(F^{-i}(A)\cap B)-\lambda(A)\lambda(B)|\to0$$ for all Borel measurable $A,B\subset T^{2}$.

What I tried for 1: I tried to find an explicit expression for the set $F^{-1}([a,b)\times[c,d))$. So $(x,y)\in F^{-1}([a,b)\times[c,d))\iff x\in[a-\alpha,b-\alpha) \ \wedge \ y\in[c-x,d-x)$. But I don't know how to deal with the $x$-dependence of the interval $[c-x,d-x)$.

What I tried for 2: Literally no idea…

Any suggestions are greatly appreciated!

Answer 1

For 1, don't try to find an explicit expression. The point is that for each $x \in [a-\alpha,b-\alpha)$, there are exactly $d-c$ choices for $y$, so $\lambda(F^{-1}([a,b)\times[c,d))) = (b-a)(d-c)$.

For 2, the map $\mathbb{T} \to \mathbb{T}, x\mapsto x+\alpha$ is not weakly mixing, so no way this map is. To make this rigorous, consider $A = B = (0,\frac{1}{2})\times \mathbb{T}$ (say); then we'd want $$\frac{1}{N}\sum_{n=0}^{N-1} \left|\lambda\left(S^{-n}\left((0,\frac{1}{2})\right) \cap \left(0,\frac{1}{2}\right)\right)-\lambda\left((0,\frac{1}{2})\right)\lambda\left((0,\frac{1}{2})\right)\right| \to 0,$$ where $S: \mathbb{T} \to \mathbb{T}, Sx := x+\alpha$. But this is false (think about it pictorially/intuitively -- it's the reason that $S$ is not weakly mixing).