# rigid measure preserving transformation

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.

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

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.