and consider this time the situation where $A\colon X \to {\mathbb{G}}$ assumes a *finite* number of values. Such $A$ cannot satisfy the richness condition, so the previous results do not apply. Nevertheless we can prove that measurable dichotomy holds generically.

\[d.unif hyp\] A bounded set $\Sigma \subset {\mathbb{G}}$ is called *uniformly hyperbolic* if there exists $\lambda>1$ such that $$\|A_n \cdots A_1\| > \lambda^n
\quad \text{for all $A_1,\ldots,A_n\in\Sigma$.}$$

(Notice that the first option in definition \[d.measdich\] just amounts to saying that $A(X)$ is a uniformly hyperbolic set.)

Given an $N$-tuple of matrices $\Sigma=(A_1,\ldots,A_N)$, for simplicity we also write $\Sigma$ for the set $\{A_1,\ldots,A_N\}$.

\[t.discrete\] Let $N \ge 2$ be an integer. There exists a residual set ${\mathcal{R}}\subset {\mathbb{G}}^N$ such that for every $\Sigma \in {\mathcal{R}}$:

-   either $\Sigma$ is uniformly hyperbolic;

-   or for every measurable map $A \colon X \to \Sigma$ which assumes every value in $\Sigma$ on a set of positive measure, there is a residual set ${\mathcal{R}}_A \subset {\mathit{Aut}}(X,\mu)$ such that ${\mathit{LE}}(A,T)=0$ for every $T \in {\mathcal{R}}_A$.

For the reader’s information, we mention a characterization of uniform hyperbolicity obtained in [@ABY]: A compact set $\Sigma\subset {\mathbb{G}}$ is uniformly hyperbolic iff there exists an open set $U \subset {\mathbb{P}}^1$ 