Proving a transformation of the interval is ergodic Consider the function $f(x):[0,1]\rightarrow[0,1]$ given by
$$\begin{cases}2x & 0 \leq x \leq\frac{1}{2}\\x-\frac{1}{2 } &\frac{1}{2}< x\leq 1 \end{cases}$$
I found the measure with density given by $\rho=\frac{4}{3}\chi_{[0,\frac{1}{2}]}+\frac{2}{3}\chi_{(\frac{1}{2},1]}$
(with respect to Lebsegue measure) is invariant for this transformation.
My question now is: how can I prove this system with this measure is ergodic?
I thought to use the approach with invariant functions and Fourier series, but I'm not sure on how to write Fourier expansion with a measure different than Lebesgue's. I also thought to exploit a possible conjugacy with symbolic shift, but wasn't able to prove that $[0,\frac{1}{2}]$ and $(\frac{1}{2},1]$ constitute a Markov partition of the unit interval.
Any ideas?
 A: They gave as the exact same question in the course "introduction to analysis", as an example for uses for Fourier series, so the solution is bases on Fourier series.
Let's take $f : \Bbb R \rightarrow \Bbb C $ periodic and continuous such that
$\int_{0}^{2\pi} |f(x) - f \circ T(x)|^2 = 0 $, we want to show that there exists $ c \in \Bbb C$ such that $f \equiv c$.
From Parseval's identity, we know that $\int_{0}^{2\pi} |f(x) - f \circ T(x)|^2 = \sum_{k \in Z} |\widehat {f - f \circ T} (k) |^2 \\.$
Therefore we can infer that $\widehat {f - f \circ T} (k)=0 \Rightarrow 
 (by \ linearity) \  \hat f(k) = \widehat{ f \circ T} (k) \ \forall k \in \Bbb Z$
After doing some calculation we can also infer that for any "even" $k \in \Bbb Z$ , $ \widehat {f \circ T}(k) = \hat f (\frac{k}{2}) $.
so by induction we can infer now that for any $p \neq 0$,
$ \widehat {f \circ T}(2^n\cdot  p) = \hat f (p) $
From Riemann–Lebesgue lemma, we can now infer that
$$ \lim_{n \to \infty}\widehat {f \circ T}(2^n\cdot  p) = 0 \Rightarrow \hat f(p) = 0 , \ \forall p \neq 0$$
So now we conclude that Fouriee series of $f$ is $\hat f(0)$ and therefore
$ f \equiv \hat f(0)$, so f is constant, as required.
