It is a well-known fact that the irrational rotation on the unit circle$S^1$ is ergodic with respect to Lebesgue measure. But each proof I have seen uses Fourier Analysis.
Now, Can someone give a proof without Fourier Analysis?
[Math] Ergodicity of irrational rotation
ergodic-theoryirrational-numbers
Best Answer
Here I prefer to view the irrational rotation as $(\mathbb{T},S)$ where $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ and $S:x\mapsto x+\alpha$ where $\alpha$ is an irrational number. Let $\mu$ be the Lebsgue measure.
First it is not hard to deduce from that fact that for any $\epsilon>0$ there exists integers $m$, $n$ and $k$ with $m\neq n$ and $|m\alpha-n\alpha-k|<\epsilon$ that $n\alpha(mod\ \mathbb{Z})$ is dense in $\mathbb{T}$.
Now suppose $B\subset \mathbb{T}$ is a measurable set invariant under $S$. For any $\epsilon>0$ we can find a function $f\in C(\mathbb{T})$ such that $||f-\chi_B||_1\leq\epsilon$. Since $B$ is $S$-invariant, we have that $||f\circ S^n-f||_1\leq2\epsilon$ which means $||f(x+n\alpha)-f(x)||_1\leq2\epsilon$. Since $n\alpha(mod\ \mathbb{Z})$ is dense in $\mathbb{T}$ and $f$ is continuous we have that $||f(x+t)-f(x)||_1\leq2\epsilon$ for any $t\in\mathbb{T}$. So using Fubini's theorem we have
$$\left \| f-\int f(t)dt \right \|_1=\int \left |\int(f(x)-f(x+t))dt \right |dx\leq \int\int|f(x)-f(x+t)|dxdt\leq 2\epsilon. $$
And so $$||\chi_B-\mu(B)||_1\leq||\chi_B-f||_1+||f-\int f(t)dt||_1+||\int f(t)dt-\mu(B)||_1\leq 4\epsilon,$$ and since this holds for arbitrarily $\epsilon>0$ so $\chi_B$ must be equal to a constant function a.e. and so $\mu(B)=0$ or $1$. So $(\mathbb{T},S)$ is ergodic with respect to the Lebsgue measure $\mu$.