The irrational rotation is ergodic. The proof should use the idea of density point. Consider $f_{\alpha}:S^{1}\rightarrow S^{1}$ the rotation of unit circle of angle  $2\pi\alpha$, and let $\mu$ the Lebesgue measure in $S^{1}$.
Let $\alpha$ irrational, show that $\left(f,\mu\right)$ is ergodic.  You should use the idea of density point.
Remark: This is a idea of the proof:


I understand the idea of proof, I do not understand is why it says: A small neighborhood of the $y_{0}$ consist basically of points of the set $B$.
 A: When the author writes that a small neighbourhood of the point of density $y_0 \in B$ consists "basically of points of the set $B$", they are just giving a more casual version of the definition of a point of density. Indeed, if $y_0 \in B$ is a point of density, then a small neighbourhood of $y_0$ may look like $(y_0 - \epsilon, y_0 + \epsilon)$ for some small $\epsilon > 0$. Now, let's consider "how much of this neighbourhood is in $B$", i.e. consider the ratio
$$
\frac{\mu((y_0 - \epsilon, y_0 + \epsilon) \cap B)}{\mu(y_0-\epsilon,y_0 + \epsilon)} = \frac{\mu((y_0-\epsilon,y_0 + \epsilon) \cap B)}{2\epsilon},
$$ 
then by definition of a point of density, this ratio is very close to 1, provided that $\epsilon > 0$ is sufficiently small. Said differently, "almost all" of the points of this small neighbourhood $(y_0-\epsilon,y_0 + \epsilon)$ of $y_0$ lie inside $B$.
A: Suppose  that exist $A$ invariant set with $0<\mu(A)<1$. Let $\epsilon>0$ take $x\in A$ and $y\in A^c$ density points.  Then  there  is $R>0$ such that
$\mu(B(x,r)\cap A)>(1-\epsilon)2r$ and $\mu(B(y,r)\cap A^c)>(1-\epsilon)2r$ for all $r\leq R$.
Take $r\leq R$, then  there 
exist $k\geq 1$ such that $d(f^k(x), y)<\frac{1}{3}r$. So
$B(y, \frac{2}{3}r)\subset B(f^kx, r)=f^k(B(x, r))$. Therefore,
\begin{eqnarray*}
2r&=&\mu(B(x,r)\cap (A\cup A^c))\\
  &=&\mu(B(x,r)\cap A)+\mu(B(x,r)\cap A^c)\\
  &=&\mu(B(x,r)\cap A)+\mu(f^k(B(x,r)\cap A^c))\\
  &=&\mu(B(x,r)\cap A)+\mu(B(f^k(x),r)\cap A^c)\\
  &\geq&\mu(B(x,\frac{2}{3}r)\cap A)+\mu(B(y,\frac{2}{3}r)\cap A^c)\\
 &>&(1-\epsilon)2(\frac{2}{3}r)+(1-\epsilon)2(\frac{2}{3}r)
\end{eqnarray*}
So
$$2r>2(1-\epsilon)\frac{4}{3}r \Rightarrow 1>(1-\epsilon)\frac{4}{3}r$$ 
for all $\epsilon>0$. Then $1\geq\frac{4}{3}$ as $\epsilon\to 0$. Contradicts!
