Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\lambda(B)=1$.

share|cite|improve this question
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 ?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.