Lemma 5

Let $f\in \mathscr{F}_{\pi}(G)$ be a $\mu$-measurable function equal to zero outside $C\cdot H$ such that the function $\|f\|$ is $\nu$-integrable on $G/H$. Let $s$ be the E-valued function on G such that

$v(x)^{-1}f(x)$ if $x\in C\cdot H$

0 if $x\in G$ - $C\cdot H$.

a) The function $s$ is $\mu$-measurable; it belong to $\mathscr{F}_{\pi}(G)$ and is equal to zero outside $C\cdot H$;

b) The function $us$ belong to $\mathscr{L}^1(G)$ and $(us)^{\pi}=f$ almost everywhere modulo H.

The function $s$ is equal to zero outside $C\cdot H$ by definition; it is $\mu$-measurable since the function $f$ is $\mu$-measurable and $v(x)>0$ for every $x\in C\cdot H$ (INT, IV, p. 193, § 5, n$^o10$, prop. 16). One has $s\in \mathscr{F}_{\pi}(G)$ because $v\in \mathscr{F}_1(G)$.

The function $f$ is locally $\mu$-integrable, since $\|f\|$ is a $\nu$-integrable function (INT, VII, p. 47, § 2, No.$^o3$, prop. 6, c), noting that the measure $\nu^{\sharp}=\kappa \cdot \mu$ is equivalent to $\mu)$, hence the function $s$ is also so by formula (5). The function $us$ is measurable and zero outside a compact subset of G; since $u$ is bounded, it follows that $us$ is $\mu$-integrable.

For all $x\in G$ - $S_{us}$, we have

since $s(xh) =\pi (h)s(x)$; the last assertion follows from prop. 4, a).

When $H =\{e\}$ and $\pi$ is of dimension 1, the following proposition is none other than INT, IV, p. 184, § 5, No.$^o6$, th. 5.