Corollary

— The continuous linear mapping $f \mapsto f^b$ of $L^1(X, \lambda^\#)$ into $L^1(X/H, \lambda)$ defined by Prop. 5 is surjective.

Suppose first that $X/H$ is paracompact and let $h$ be a function on $X$ satisfying the conditions of Prop. 8. If $k$ is a $\lambda$-integrable numerical function on $X/H$, then $h \cdot (k \circ \pi)$ is $\lambda^\#$-integrable and $(h \cdot (k \circ \pi))^b = k$ (Prop. 9).

In the general case, let $u \in L^1(X/H, \lambda)$. There exists a function $f \in \mathscr{L}^1(X/H, \lambda)$ with class $u$ and zero outside a countable union of compact sets $K_n$. Let us define recursively a sequence of relatively compact open sets $U_n$ of $X/H$ such that $U_{n+1} \supset K_n \cup \overline{U}_n$, and let $V$ be the union of the $U_n$. Then $V$ is an open subset of $X/H$, a countable union of compact subsets $\overline{U}_n$, hence is paracompact (GT, I, § 9, No. 10, Th. 5). Set $Y = \overline{\pi}^{-1}(V)$ and let $\lambda_V$ (resp. $\lambda_Y^\#$) be the measure induced by $\lambda$ (resp. $\lambda^\#$) on $V$ (resp. $Y$). It is clear that $Y/H$ may be identified with $V$ (GT, I, § 3, Prop. 10) and that $\lambda_Y^\#$ may be identified with $(\lambda_V)^\#$. Moreover, $f$ is zero outside $V$ and belongs to $\mathscr{L}^1(V, \lambda_V)$. Therefore, there exists $g \in \mathscr{L}^1(Y, \lambda_Y^\#)$ such that $g^b = f$ almost everywhere in $V$. Extending $g$ by 0 on $X - Y$, one obtains a function $g_1 \in \mathscr{L}^1(X, \lambda^\#)$, and it is clear that the class of $g_1^b$ in $L^1(X/H, \lambda)$ is none other than $u$.