\circ \pi) = \Lambda_k(B \circ \pi)$. So, by lemma \[l.square\], $\Lambda_k(A) = \Lambda_k(B)$.

Based on lemma \[l.push\], we can introduce the following notation: If $\nu$ is a finite measure in ${\mathbb{G}}$ with bounded support, and $k\in{\mathbb{N}}$, we write $$\Lambda_k (\nu) = \Lambda_k(A),$$ where $A\colon (X,\mu) \to {\mathbb{G}}$ is some map, defined on some non-atomic Lebesgue space, such that $A_* \mu = \nu$. (Notice the existence of such $A$.)

The map $\Lambda_k$ has the following convexity properties:

\[l.convex\] Let $\nu$, $\nu'$ be finite measures in ${\mathbb{G}}$, with bounded supports, and let $t>0$. Then

1.  $\Lambda_k(t\nu) = t\Lambda_k(\nu)$;

2.  $\Lambda_k(\nu + \nu') \le \Lambda_k(\nu) + \Lambda_k(\nu')$.

The first part is obvious, so let us show the second one. Let $I$, $I' \subset {\mathbb{R}}$ be disjoint intervals with lengths $\nu({\mathbb{G}})$, $\nu'({\mathbb{G}})$, respectively. Take a map $A: I \sqcup I' \to {\mathbb{G}}$ such that $A_* m|_I = \nu$, $A_* m|_{I'} = \nu'$ (where $m$ is Lebesgue measure). Then (1) $\Lambda_k(\nu + \nu')$, (2) $\Lambda_k(\nu) + \Lambda_k(\nu')$, are the infimum of $\Lambda_k(A,T)$ where $T$ runs over (1) all automorphisms $T \in {\mathit{Aut}}(I \sqcup I', m)$, (2) the automorphisms $T$ such that $T(I)=I$, $T(I')=I'$, respectively.

An existence result {#ss.existence}
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If $\nu$ is a finite measure 