Let $V$ be a compact Riemannian manifold, $G$ the set of diffeomophisms of $V$, let $\nu$ be a probability measure in $G$. Suppose that $\exp\_{x}$ is diffeomorphism in $\mathcal{B}\_{2I}(x)\subset V$ ball of radius $2I$ for all $x\in V$, then we define
$$
\begin{array}{rl}
\delta\_{1}(T) & ={\displaystyle \sup\left\{\left.\frac{d\left(T(x),T(y)\right)}{\left\|\exp\_{x}^{-1}y\right\|} \: \right| \: (x,y)\in V^{2}, \:\: 0<d(x,y)\leq I \right\} } \\
\delta\_{2}(T) &= {\displaystyle \sup\left\{ \left. \frac{\left\| \exp\_{T(x)}^{-1}T(y)-DT(x)\exp\_{x}^{-1}y \right\|}{\left\| \exp\_{x}^{-1}y \right\|} \:\right|\: (x,y)\in V^{2},\: 0<d(x,y)\leq I,\: d(T(x),T(y))\leq I \right\} }.
\end{array}
$$
Now suppose that:
$$
(H) \qquad \int \log^{+}\left( \delta\_{1}(T)+\delta\_{2}(T) \right)\nu (dT) <\infty.
$$
Let $\Omega=G^{\mathbb{Z}}$, this is measurable space with product measure $\mathbb{P}=\nu^{\otimes \mathbb{Z}} $, and we consider the projection $T\_{i}:\Omega\rightarrow G$, that is, if $\tau$ is the right translation then $T\_{i}(\tau\omega)=T\_{i+1}(\omega)$.


Show that 
$$\lim\_{n\rightarrow \infty}\frac{1}{n}\log^{+}\left( \delta\_{1}(T\_{n})+\delta\_{2}(T\_{n}) \right)=0. \tag{1}$$


**Remark:** This question arises because I am trying to understand the article in this [link](http://archive.numdam.org/article/AIHPB_1987__23_1_111_0.pdf) (see pages 6,7), the author says that (1) is a consequence of the hypothesis (H) and the Birkhoff's ergodic theorem. I do not see how (1) you can follow these two facts, I would appreciate someone helping me to understand it. 



About the ergodic theorem of Birkhoff in the context of a compact Riemannian manifold
What this boils down to is the following:


Let $\tau$ be a measure-preserving transformation of a probability space $(\Omega,\mathbb P)$. Suppose that $f$ is a measurable function on $\Omega$ such that $\int f\,d\mathbb P(\omega)<\infty$. Then $(1/n)f(\tau^n \omega)\to 0$ a.e. 


The cheapest way to see this is from the Birkhoff ergodic theorem:
Let $A\_nf(\omega)=1/n(f(\omega)+\ldots+f(\tau^{n-1}\omega))$. 
The Birkhoff ergodic theorem states that $A\_nf(\omega)$ is a convergent sequence for a.e. $\omega$, with limit $f^\*(\omega)$, say. Then $\frac 1nf(\tau^n\omega)=\frac{n+1}nA\_{n+1}f(\omega)-A\_nf(\omega)\to f^\*(\omega)-f^\*(\omega)=0$.


There are lower-tech ways of getting the same result using the Borel-Cantelli lemmas. 


