2
$\begingroup$

I want to proof Lemma 6.7.2 in C.E. Silva's Book "Invitation to Ergodic Theory" (the proof is left as an excercise).

A finite measure-preserving transformation $ T $ is said to be rigid if for all measurable sets $ A $ and for every $ \varepsilon > 0 $, there exists an integer $n>0$, such that $\mu(T^{-n}(A)\triangle A) <\varepsilon$.

The lemma I'm trying to proof states:

$ T $ is rigid $\iff$ there is a sequence $n_k\rightarrow\infty$ such that $ \lim\limits_{k\to\infty}\mu(T^{-n_k}A\triangle A)=0 $, for all measurable sets $A$.

So, I think $"\Leftarrow"$ is clear. The other implication seems to be a little more tricky to me.

I found in another paper on the topic, that it's sufficient to prove that for every measurable $A$ there is a rigidity sequence, i.e. a sequence $ n_k = n_k(A)$, such that $ \lim\limits_{k\to\infty}\mu(T^{-n_k}A\triangle A)=0 $.

At this point I got lost. If I take any measurable set $A$, I'm not sure how I can construct such a sequence, only knowing, that for every $\varepsilon>0$ there exists some $n$, such that $\mu(T^{-n}(A)\triangle A) <\varepsilon$.

Any help would be appreciated.

$\endgroup$
2
  • $\begingroup$ Can the sequence $n_k$ depend on $A$, or the same sequence should work for all measurable $A$? $\endgroup$ – Berci Jan 21 at 16:42
  • $\begingroup$ the sequence can depend on $A$ $\endgroup$ – mixer Jan 21 at 16:47
2
$\begingroup$

You can construct $\{n_k\}$ recursively. For each $k\ge 1$, let $$ n_k= \min\left\{n>n_{k-1}:\mu(T^{-n}(A)\Delta A)<2^{-k}\right\}, $$ with $n_0\equiv 0$. Then $n_k\to\infty$ and $\mu(T^{-n_k}(A)\Delta A)\to 0$ as $k\to\infty$.

It remains to show that $n_k$ is well defined for all $k$, i.e., for a given $n_{k-1}$, there exists $n>n_{k-1}$ s.t. $T^{-n}(A)$ approximates $A$. Suppose not. Then $\mu(T^{-n}(A)\Delta A)\ge 2^{-k}$ for all $n>n_{k-1}$. Therefore, by the rigidity of $T$, it must be true that $\mu(T^{-n_{k-1}}(A)\Delta A)=0$. Set $B:=T^{-n_{k-1}}(A)$. Since $T$ is rigid, there is $n>0$ s.t. $\mu(T^{-n}(B)\Delta A)=\mu(T^{-n}(B)\Delta B)<2^{-k}$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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