= \pi^{-1}(K) \cap P_n^{-1}(K)$; then $m(G_n^c)< 2{\varepsilon}$. Thus $$\int_{{\mathbb{I}}^2} \|A \circ P_n - A \circ \pi \| \, dm =
\int_{G_n} (\cdots) + \int_{G_n^c} (\cdots) <
{\varepsilon}+ 2 {\varepsilon}\|A\|_\infty.$$

By lemma \[l.cont\], if $n$ is sufficiently large then $\Lambda_k(A \circ P_n, T) < \delta + \Lambda_k(A \circ \pi)$. Let $T' = P_n \circ T \circ P_n^{-1}$; then $\Lambda_k(A, T') = \Lambda_k(A \circ P_n, T)$. This shows that $\Lambda_k(A) < \delta + \Lambda_k(A\circ\pi)$. Since $\delta>0$ is arbitrary, the lemma follows.

Let us record for later use something we have proved:

\[l.pn\] There exists a sequence of isomorphisms $P_n:({\mathbb{I}}^2,m)\to ({\mathbb{I}},m)$ such that for any measurable bounded $A:{\mathbb{I}}\to {\mathbb{G}}$ we have $\| A \circ P_n - A \circ \pi \|_1 \to 0$ as $n \to \infty$.

Now we can state and prove the:

\[l.push\] Let $A, B\colon (X,\mu) \to {\mathbb{G}}$ be such that $A_* \mu = B_* \mu$. Then $\Lambda_k(A) = \Lambda_k(B)$.

We can assume that $\mu(X)=1$. By remark \[rem.isomorphism\], we can assume that $A$ and $B$ are defined over $(X,\mu)=({\mathbb{I}},m)$. Since $A_* m = B_* m$, by lemma \[l.thouvenot\] there is an automorphism $S$ of the square ${\mathbb{I}}^2$ such that $A \circ \pi = B \circ \pi \circ S$. In particular, $\Lambda_k(A 