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.
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.