If $f:[0,1]\to [0,1]$ is given by

$$ f(x)= \begin{cases} 2x & \mbox{ if } x\in [0,1/3)\\ & \\ 2x-\frac{2}{3} & \mbox{ if } x\in [1/3,1/2)\\ 2x-\frac{1}{3} & \mbox{ if } x\in [1/2,2/3)\\ & \\ 2x-1 & \mbox{ if } x\in [2/3,1] \end{cases} $$ show that $f$ is Ergodic with respect of Lebesgue measure.

**Idea:** The technique is to show that given an invariant set of positive measure, this set will have full measure.

A very common process is to partition the domain, choose a point of density in the invariant set, use the the Lebesgue Density Theorem and the bounded distortion property

\begin{align*} \dfrac{m(f^k(E_1))}{m(f^k(E_2))}=\dfrac{m(E_1)}{m(E_2)} \end{align*}.

below is the iteration graph $f^5 (x)$. We see that the image size of each subinterval is $2/3$. The main problem that for some interval $E_1$, $f^k(E_1)$ is not $(0,1)$ integer "At first". I think that when we iterate $n$ times for very large $n$, we will have a graph virtually without breaks, with the linear arms. This would justify that the measure of each invariant set of positive measure is $1$. But it is only my intuition, I could not justify formally. Can anyone give a tip?