Knopp's Lemma - Show T is ergodic

Be $\beta > 1$ non-integer.

$T_{\beta}: [0,1)\rightarrow[0,1)$ with $T_{\beta}x = \beta x$ mod$(1) = \beta x-\lfloor\beta x\rfloor$.

Show with Knopp's Lemma that $T_{\beta}$ is ergodic with respect to $\lambda$ Lebesgue measure.
(If $T_{\beta}^{-1}A = A$, then $\lambda(A)=0$ or $1$).

$\underline{\textrm{Knopp's Lemma:}}$ $B$ Lebesgue set. $\mathscr{C}$ is class of subintervals of $[0,1)$ with

a) $\forall$ open subinterval of $[0,1)$ is at most a countable union of disjoint elements from $\mathscr{C}$

b) $\forall A\in\mathscr{C}$: $\lambda(A\cap B)\geq \gamma\lambda(A)$ with $\gamma>0$ independent of A.

Then $\lambda(B)=1$.

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Are you really assuming $\beta$ to be non-integer? The Lebesgue measure does not look like an invariant measure in that case. – Le Curious Apr 19 '14 at 18:34

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

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

Can you see the class of subintervals of the Knopp's Lema ?

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