Let $f: [0,1] \to [0,1]$ a measurable transformation that preserves the Lebesgue measure $\mu$. So we have that $\liminf_n n\mid{f^n(x)-x}\mid\leq1$ $\mu$-almost everywhere.
One trying to solve this problem is: Suppose that the conclusion is not true. Than $\exists $ $j \geq1, c>1$ such that $\mu(D)>0$, where $D$={$x\in[0,1];n\mid f^n(x)-x\mid>c,\forall n\geq j$}. Let $d \in D$ be a density point of $D$ (we know that it exists, by Lebesgue's theorem). Consider $E=D\cap B(d,\epsilon)$, for $\epsilon$ small. I am trying to obtain a lower bound for $R_E(x)$=min{$n\geq1;f^n(x)\in E$}, for any $x\in E$ and than use that $\int R_Ed\mu=\mu([0,1])-\mu(E_0^*)=1-\mu(E_0^*)$, where$E_0^*=${$x\in[0,1];f^n(x)\notin E,\forall n\geq1$}(Kac theorem) to get a contradiction.
Does someone have some suggestion to solve this problem?
Hint: Use the fact that the Lebesgue measure of $] f(x)^n, f(x)^{n+1}[$ is constant and apply the triangular inequality.