Invariant functions and Kac Theorem

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I am solving the following exercise (1.2.1) from Viana & Oliveira's "Foundations of Ergodic Theory".

Show that if $f: [0,1] \rightarrow[0,1]$ is a measurable transformation preserving the Lebesgue measure $m$ then $m$-almost everywhere $x \in [0,1]$ satisfies $$\liminf_n n|f^n(x)-x| \le 1.$$

Following the hint presented in the book, I will assume that the claim is false and try to reach an absurd using Kac's Theorem.

Suppose the claim is false. Then there is a set $B$ with $m(B)>0$ such that there exist a $b>1$ and a $k \in \mathbb{N}$ with $n|f^n(x)-x|>b$ for every $n \ge k$.

We can take a density point $a$ of $B$ and $r$ small enough such that the set $E=B\cap B(a; r)$ satisfies $m(E)>r$.

Now I am trying to find a lower bound to $\rho_E$, the transformation of first return to $E$. If I could show that $\rho(E)(x) >\frac{b}{r}$ for example, we would have $$\int_E \rho_E \; dm \ge m(E)\cdot\frac{b}{r}\ge b>1,$$

and that would violate Kac's Theorem, but I am having trouble to show a fit lower bound. I would appreciate any help.

Thanks!

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You are on the right track. Repeating the first part of the argument: Suppose the claim is false. Then there is a set $B$ with $m(B)>0$ such that there exist a $b>1$ and a $k \in \mathbb{N}$ with $n|f^n(x)-x|>b$ for every $n \ge k$.

Since $B$ contains no fixed points of an iterate $f^j$, the sets $$B_\ell:=\{x \in B: \, \forall j \in [1,k], \quad |f^j(x)-x|>1/\ell\}$$ satisfy $\cup_{\ell \ge 1} B_\ell=B$, so $m(B_\ell)>0$ for some $\ell$.

Choose a density point $a \ne 0,1$ of $B_\ell$ and fix $\delta>0$ such that $(2-\delta)b>2$. Then for all small enough $r$, the set $E=B\cap (a-r,a+r)$ satisfies $m(E)>(2-\delta)r$. We fix such $r <1/(2\ell)$, and also require that $b/(2r)>k$.

Fix $x \in E$. For every $n \in [k , b/(2r)] $, we have $$|f^n(x)-x|>b/n \ge 2r \,.$$ The inequality $|f^n(x)-x|> 2r$ also holds for $n \in [1,k]$, by our choice of $r$ and the definition of $B_\ell$.

Thus the return time $\rho_E$ to $E$ satisfies $\rho_E(x) \ge \frac{b}{2r}$ for all $x \in E$, so $$\int_E \rho_E \; dm \ge m(E)\cdot\frac{b}{2r}\ge (2-\delta)b/2>1 \,,$$ contradicting Kac' lemma.