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Hoi, ive been breaking my head on this fora few days.. Ive been trying to show that $T:[0,1)\to [0,1)$ given by

$$ T(x)= \begin{cases} 3x & \mbox{ if } x\in [0,1/3)\\ & \\ \frac{3x}{2}-\frac{1}{2} & \mbox{ if } x\in [1/3,1) \end{cases} $$ is Ergodic.

Let B be T-invariant,i.e. $T^{-1}B=B$. If $\mu(B)>0$, I want to show that $\mu(B)=1$.

Can we show that if $\mu(B)<1$ that $\mu(TB)>\mu(B)$,

is that obvious? So that it is clear that B can not T-invariant?

We know nothing of the set B, which may be little complicated set. Any idea how we can show ergodic property of this transformation. (it is measure preserving).

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Regardless of how complicated $B$ is, the behaviour of $T^{-1}$ is extremely simple. What have you tried? –  Erick Wong Nov 14 '12 at 0:58
    
Many many things...applying Knopps lemma, finding isomorphism between this system and another Ergodic system. Kind of clueless right now. Its behaviour is relatively simple to understand, but I find it hard to prove that a T-invariant set of positive measure must have measure 1. One can understand that T increases any interval $[u,v)$ in measure, so it is intuitively clear...kind of. –  DinkyDoe Nov 14 '12 at 1:19
    
I must apologize. I misinterpreted your parenthetical comment to mean you were trying to first show it is measure preserving. –  Erick Wong Nov 14 '12 at 2:05
1  
Some thoughts: First, if you identify the circle with $\mathbb{R}/\mathbb{Z}$, then your map $T$ induces a continuous map on the circle. It is a degree two map, which seems like it should be topologically conjugate to $z\mapsto z^2$. Presumably you know $z\mapsto z^2$ is ergodic. Another thought (though I have no idea if this would work) would be to show that every $T$-invariant $L^2$ function on $[0,1)$ is constant by decomposing such a function into its Fourier series $f(x) = \sum_n c_n e^{2\pi inx}$ and seeing how $T$ acts on this decomposition. –  froggie Nov 14 '12 at 15:18
    
The ergodicity of this map follows directly from the main result of "Ergodic transformations from an interval into itself" by Li and Yorke published in the transactions of the AMS in 1978. –  Cantor Nov 24 '12 at 18:15
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1 Answer 1

up vote 7 down vote accepted

For each k there are partition of $[0,1)$ in intervals $I{'s}$ such that $T^k$ restricted to each $I$ is a bijection on $[0,1)$. In each $I$

$$T^k(x)=ax+c $$

where $a$ and $c$ depends on $k$ and $I$.

Hence for every pair of measurable sets $A$ and $B$ in $I$ we have

$$ \dfrac{m(T^k(A))}{m(T^k(B))}=\dfrac{a^km(A)}{a^km(B)}=\dfrac{m(A)}{m(B)} . $$

Let $B$ be an invariant measurable set with positive measure, then,

$$m(B)\geqslant \dfrac{m(T^k(I\cap B ))}{m(T^k(I))} =\dfrac{m(I\cap B)}{m(I)}.$$

Fix a Lebesgue point $p$ in $B$ (we can choose a such point because $B$ has positive measure). For each $k$ choose a interval $I_k$ how above containing $p$. Since the diameter of the $I_k$ converge to zero and $p$ is Lebesgue point, we conclude that

$$ m(B)\geqslant \lim_{k\to\infty} \dfrac{m(I_k\cap B)}{m(I_k)}=1.$$

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Nice solution!! –  user27456 Dec 14 '12 at 22:22
    
that's just great :P thnx. –  DinkyDoe Jan 12 '13 at 12:17
1  
Nice solution!!! –  Elias Jan 12 '13 at 19:48
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