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Suppose that I have the transformation $$T(x,y)=(x+a,y+b), a,b \in \mathbb{R}$$ from the 2-dimensional torus $\mathbb{R}^2/\mathbb{Z}^2$ to itself. We know that this transformation in ergodic with respect to the Lebesgue 2-d measure on the torus if and only if $a,b $ are rationally independent which means $na+mb \in \mathbb{Z}$ if and only if $n=m=0$.

Now according to Birkhoff's ergodic theorem for any function $f\in C(\mathbb{R}^2/\mathbb{Z}^2,\mathbb{R})$ we have that $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}f(T^j(x,y))=\int f dm$$ for almost all $(x,y) \in \mathbb{R}^2/\mathbb{Z}^2$, where $T^j$ are the iterates of $T$ and $m$ is the 2-d Lebesgue measure and the integral is over $[0,1]\times[0,1]$.

Let's say that we choose the function $f(x,y)=\cos (2\pi y)$ which is continuous in $[0,1]\times[0,1]$ and we choose $a\not \in \mathbb{Q}$ and $b=0$. Then $T$ is ergodic and $T^j(x,y)=(x+ja,y)$, so we have that $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\cos(2\pi y)=\int_{0}^{1}\int_{0}^{1} \cos (2\pi y) dydx=0$$ But this cannot be true almost everywhere since the sum on the left is equal with $\cos (2\pi y)$.

Now my Question is what am I doing wrong?

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2 Answers 2

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Your choice for $(a,b)$ does not give rationally independent elements, because $0\cdot a+1\cdot b=0$ (with $n=0$, $m=1$) hence the transformation is not ergodic.

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The only problem is with rational independence of $a$ and $b$. Since $(0)(a)+(1)(b)=0$ your $a$ and $b$ are not rationally independent. So $T$ is not ergodic.

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