there exists $\eta>0$ with the following properties: If $A$, $B\colon (X,\mu)\to {\mathbb{G}}$ are measurable maps with $\|A\|_\infty$, $\|B\|_\infty \le C$ and $$\|A-B\|_1 = \int_X \|A-B\| \; d\mu < \eta$$ then $$\left| \Lambda_k(A) - \Lambda_k(B) \right| < \delta.$$

Take $\eta = C^{-k+1}\delta$. Fix any $T \in {\mathit{Aut}}(X,\mu)$. We can estimate pointwise $$\left| \log \|A^k_T \| - \log \|B^k_T \|\right| \le
\left| \|A^k_T \| - \|B^k_T \|\right| \le
\| A^k_T  - B^k_T \| \le
C^{k-1} \sum_{i=0}^{k-1} \|B \circ T^i - A \circ T^i\| \, .$$ Dividing by $k$ and integrating over $X$ we obtain $\left| \Lambda_k(A,T) - \Lambda_k(B,T) \right| < \delta$.

\[rem.isomorphism\] If $A\colon X \to {\mathbb{G}}$ is measurable and bounded, and $S\colon  (X,\mu) \rightarrow (X',\mu')$ is an isomorphism, it is clear that $\Lambda_k (A\circ S) = \Lambda_k(A)$, because $\Lambda_k(A \circ S, T) = \Lambda_k (A,S \circ T \circ S^{-1})$ for every $T \in {\mathit{Aut}}(X,\mu)$.

In fact, we will prove in lemma \[l.push\] a stronger result: *$\Lambda_k(A)$ depends only on the push-forward measure $A_* \mu$.* To prove that, we will need lemma \[l.square\] below.

In what follows ${\mathbb{I}}$ denotes the unit interval $[0,1]$ and $m$ denotes Lebesgue measure on ${\mathbb{I}}$ or (by abuse of notation) on the square ${\mathbb{I}}^2$.

\[l.square\] Let $A\colon 