a) For every $x \in X$, the function $\xi \mapsto f(x\xi)$ on H belongs to $\mathscr{K}(H)$; one sets

b) The function $f^\chi$ is continuous, is zero outside SH, and satisfies $f^\chi(x\xi) = \chi(\xi)f^\chi(x)$.

c) If g is a continuous real-valued function on X and satisfies $g(x\xi) = \chi(\xi)g(x)$, then $(fg)^\chi = f^1g$ ($f^1$ being given by the formula (1) with $\chi$ replaced by the representation $\xi \mapsto 1$ of H in $\mathbf{R}_+^*$).

d) If $\eta \in \mathbf{H}$, then $(\delta(\eta)f)^{\chi} = \chi(\eta)\Delta_{\mathbf{H}}(\eta)^{-1}f^{\chi}$.

Let $x_0 \in X$ and let $V$ be a compact neighborhood of $x_0$ in $X$. The set of $\xi \in \mathbf{H}$ such that $V\xi$ intersects $S$ is also the set of $\xi \in \mathbf{H}$ such that $V\xi$ intersects $S \cap VH$, hence is compact in $\mathbf{H}$ since $S \cap VH$ is compact and $\mathbf{H}$ operates properly in $X$ (GT, III, §4, No. 5, Th. 1); then Lemma 1 of §1, No. 1 proves a) and the continuity of $f^{\chi}$. The rest of b) is obvious. Finally, c) and d) result from the following calculations:

Linearity is immediate. It is clear that $f^{\chi} \geqslant 0$ for $f \geqslant 0$. It then suffices to apply the following lemma:

Let I be a relatively bounded linear form (Ch. II, §2, No. 2) on $\mathscr{K}^{\chi}(X)$. Then $f \mapsto I(f^{\chi})$ is a relatively bounded linear form on $\mathscr{K}(X)$, that is, a measure $\mu_{I}$ on $X$. The mapping $I \mapsto \mu_{I}$ is injective by Prop. 2. The measures $\mu_{I}$ on $X$ so obtained may be characterized as follows:

a) There exists a relatively bounded linear form I on $\mathscr{K}^\chi(X)$ such that $I(f^\chi) = \mu(f)$ for all $f \in \mathscr{K}(X)$. b) $\delta(\xi)\mu = \chi(\xi)^{-1} \Delta_H(\xi)\mu$ for all $\xi \in H$. c) For all $f, g$ in $\mathscr{K}(X)$,

(2)

d) If $f \in \mathscr{K}(X)$ is such that $f^\chi = 0$, then $\mu(f) = 0$.

a) $\Rightarrow$ b): If $\mu(f) = I(f^\chi)$ then, taking into account Prop. 1 d),

whence $\delta(\xi)\mu = \chi(\xi)^{-1} \Delta_H(\xi)\mu$.

b) $\Rightarrow$ c): Suppose hypothesis b) is satisfied. Note that the functions $(x, \xi) \mapsto f(x)g(x\xi)$ and $(x, \xi) \mapsto f(x\xi)g(x)$ on $X \times H$ are continuous with compact support (because $H$ operates properly in $X$); this established, Th. 2 of Ch. III, §4, No. 1 permits us to write:

which proves c).

c) $\Rightarrow$ d): If c) is verified and if $f^\chi = 0$, then $\mu(f \cdot g^1) = 0$ for all $g \in \mathscr{K}(X)$, thus $\mu(f) = 0$ on choosing $g \in \mathscr{K}(X)$ such that $g^1 = 1$ on $\mathrm{Supp}\, f$ (which is possible by Prop. 2 applied with $\chi = 1$).

d) $\Rightarrow$ a): If condition d) is satisfied, there exists a linear form I on $\mathscr{K}^\chi(X)$ such that $\mu(f) = I(f^\chi)$ for $f \in \mathscr{K}(X)$, and this form is relatively bounded by virtue of Prop. 2.

b) Conversely, let $\mu$ be a measure on $X$ such that $\delta(\xi)\mu = \Delta_H(\xi)\mu$ for all $\xi \in H$. There exists one and only one measure $\lambda$ on $X/H$ such that $\mu = \lambda^\#$.

This is a special case of No. 1, Prop. 3.

a) If f is a numerical function $\geq 0$ then, for $x \in X$,

b) For f to be $\beta_{\dot{x}}$-measurable, it is necessary and sufficient that the function $\xi \mapsto f(x\xi)$ on H be $\beta$-measurable.

c) Suppose that f is a function on X, with values in a Banach space or in $\overline{\mathbf{R}}$; then, for f to be $\beta_{\dot{x}}$-integrable (resp. essentially $\beta_{\dot{x}}$-integrable), it is necessary and sufficient that the function $\xi \mapsto f(x\xi)$ on H be $\beta$-integrable (resp. essentially $\beta$-integrable), in which case $\int_X f(y)\, d\beta_{\dot{x}}(y) = \int_H f(x\xi)\, d\beta(\xi)$.

This follows from Ch. V, §4, Prop. 2, Prop. 3 and Th. 2.

Since $f^\flat \in \mathscr{K}(X/H)$ for $f \in \mathscr{K}(X)$, formula (8) proves that the mapping $u \mapsto \beta_u$ of $X/H$ into $\mathscr{M}(X)$ is vaguely continuous, that the family $(\beta_u)$ is $\lambda$-adequate$^1$ for any positive measure $\lambda$ on $X/H$, and that

which furnishes a new interpretation of $\lambda^\#$.

$^1$ In the sense of the first edition of Ch. V, hence a fortiori in the sense of the second (cf. the footnote to the Example of Ch. VI, §1, No. 1).

a) Let $f$ be a $\lambda^\#$-measurable function on $X$, with values in a topological space, constant outside a countable union of $\lambda^\#$-integrable sets. Then, the set of $\dot{x} \in X/H$ such that the function $\xi \mapsto f(x\xi)$ is not $\beta$-measurable is locally $\lambda$-negligible.

b) Let $f$ be a $\lambda^\#$-measurable function $\geq 0$ on $X$, zero outside a countable union of $\lambda^\#$-integrable sets. Then, the function $\dot{x} \mapsto \int^* f(x\xi) d\beta(\xi)$ on $X/H$ is $\lambda$-measurable, and

c) Let $f$ be a $\lambda^\#$-integrable function on $X$, with values in a Banach space or in $\overline{\mathbf{R}}$. Then, the set of $\dot{x} \in X/H$ such that $\xi \mapsto f(x\xi)$ is not $\beta$-integrable is $\lambda$-negligible; the function $f^b$ on $X/H$ defined almost everywhere by the formula

is $\lambda$-integrable, and

and

d) Let $f$ be a $\lambda^\#$-measurable function on $X$, with values in a Banach space or in $\overline{\mathbf{R}}$, and zero outside a countable union of $\lambda^\#$-integrable sets. Then, for $f$ to be $\lambda^\#$-integrable, it is necessary and sufficient that

Taking into account Lemma 2, the assertions a), b) and c) follow from Ch. V, §3, Prop. 4, Prop. 5 and Th. 1 (with the exception of (13), which follows from (12) because it is clear that $|f^b| \leq |f|^b$); d) follows from b).

a) Let $N$ be a subset of $X/H$. For $N$ to be locally $\lambda$-negligible, it is necessary and sufficient that $\pi^{-1}(N)$ be locally $\lambda^\#$-negligible.

b) Let g be a function on X/H, with values in a topological space. For g to be $\lambda$-measurable, it is necessary and sufficient that $g \circ \pi$ be $\lambda^\#$-measurable.

c) Let h be a function on X/H, with values in a Banach space or in $\overline{\mathbf{R}}$. For h to be locally $\lambda$-integrable, it is necessary and sufficient that $h \circ \pi$ be locally $\lambda^\#$-integrable, in which case $(h \cdot \lambda)^\# = (h \circ \pi) \cdot \lambda^\#$.

Suppose $h \circ \pi$ is locally $\lambda^\#$-integrable. For every $f \in \mathscr{K}(X)$, $f \cdot (h \circ \pi)$ is $\lambda^\#$-integrable, therefore (Prop. 5) the function $(f \cdot (h \circ \pi))^b = f^b \cdot h$ is $\lambda$-integrable and

Since $f \mapsto f^b$ is a surjective mapping of $\mathscr{K}(X)$ onto $\mathscr{K}(X/H)$, this shows that h is locally $\lambda$-integrable and that

In particular, if $\pi^{-1}(N)$ is locally $\lambda^\#$-negligible, then $\varphi_N \circ \pi$ is locally $\lambda^\#$-negligible, therefore $(\varphi_N \cdot \lambda)^\# = (\varphi_N \circ \pi) \cdot \lambda^\# = 0$, consequently $\varphi_N \cdot \lambda = 0$ and N is locally $\lambda$-negligible. Now suppose that $g \circ \pi$ is $\lambda^\#$-measurable. Let K' be a compact subset of X/H. Let $f \in \mathscr{K}_+(X)$ be such that $f^b = 1$ on K' (No. 1, Prop. 2), and let $K = \operatorname{Supp} f$; one has $\pi(K) \supset K'$. There exists a partition of K formed of a $\lambda^\#$-negligible set M and a sequence $(K_n)$ of compact sets such that $(g \circ \pi)|K_n$ is continuous for every n. Then $g|_{\pi(K_n)}$ is continuous. Let P be the set of points of K' not belonging to $\pi(K_1) \cup \pi(K_2) \cup \ldots$; then $\pi^{-1}(P) \cap K$ is contained in M, hence is $\lambda^\#$-negligible; therefore $f \cdot \varphi_{\pi^{-1}(P)}$ is $\lambda^\#$-negligible; it follows (Prop. 5) that

therefore P is $\lambda$-negligible, and g is $\lambda$-measurable.

If N is locally $\lambda$-negligible, then $\pi^{-1}(N)$ is locally $\lambda^\#$-negligible (Ch. V, §6, No. 6, Cor. 1 of Prop. 10). If g is $\lambda$-measurable, then $g \circ \pi$ is $\lambda^\#$-measurable (ibid.). Finally, suppose h is locally $\lambda$-integrable. Then we already know that $h \circ \pi$ is $\lambda^\#$-measurable. For every $f \in \mathscr{K}_+(X)$ we have, by Prop. 5,

therefore $h \circ \pi$ is locally $\lambda^\#$-integrable.

The first assertion follows from Prop. 6, a) and c). The second follows from the first.

Replacing $f$ by $f/(1+|f|)$, one reduces to the case that $f$ is bounded, hence locally $\lambda^\#$-integrable. Let $\mu = f \cdot \lambda^\#$. The hypothesis on $f$ implies that $\delta(\xi)\mu = f \cdot \delta(\xi)\lambda^\# = \Delta_H(\xi)\mu$ for all $\xi \in H$. There then exists (Prop. 4) a measure $\lambda'$ on $X/H$ such that $\mu = {\lambda'}^\#$. By Cor. 1, there exists a locally $\lambda$-integrable function $g$ on $X/H$ such that $\lambda' = g \cdot \lambda$. By Prop. 6, $f \cdot \lambda^\# = {\lambda'}^\# = (g \circ \pi) \cdot \lambda^\#$, whence $f = g \circ \pi$ locally $\lambda^\#$-almost everywhere.

c) Let $\lambda$ be a complex measure on $X/H$. Then $|\lambda|^\# = |\lambda^\#|$.

Assume the family $(\lambda_\iota)$ to be bounded above and let $\mu = \sup \lambda_\iota$. Since $\lambda \geq 0$ implies $\lambda^\# \geq 0$, we have $\mu^\# \geq \lambda_\iota^\#$ for all $\iota$, which shows that the family $(\lambda_\iota^\#)$ is bounded above and that

Conversely, assume the family $(\lambda_\iota^\#)$ to be bounded above and let $\nu = \sup(\lambda_\iota^\#)$. Since $\delta(\xi)\lambda_\iota^\# = \Delta_H(\xi)\lambda_\iota^\#$ for all $\xi \in H$, obviously $\delta(\xi)\nu = \Delta_H(\xi)\nu$, therefore there exists a measure $\mu' \in \mathscr{M}(X/H)$ such that $\nu = {\mu'}^\#$. Since $\lambda^\# \geq 0$ implies $\lambda \geq 0$, we have $\mu' \geq \lambda_\iota$ for all $\iota$, which shows that the family $(\lambda_\iota)$ is bounded above and that $\nu = {\mu'}^\# \geq (\sup \lambda_\iota)^\#$, whence

which completes the proof of a). The assertion b) then follows at once since, for example, $\lambda^+$ is none other than $\sup(\lambda, 0)$. To prove c), it suffices to note that $|\lambda| = \sup \mathscr{R}(\alpha \lambda)$ over the complex numbers $\alpha$ of absolute value 1, and on the other hand that $\mathscr{R}(\mu^\#) = (\mathscr{R}\mu)^\#$ for every $\mu \in \mathscr{M}(X/H)$.

a) There exists on $X$ a continuous function $r$, with values $> 0$, such that $r(x\xi) = \chi(\xi)r(x)$ for all $x \in X$ and $\xi \in H$.

b) The mapping $g \mapsto g/r$ is an isomorphism of the vector space $\mathscr{K}^\chi(X)$ onto the vector space $\mathscr{K}^1(X)$.

Let us apply Prop. 1 of No. 1 on taking $f$ to be a function $\geqslant 0$ that is not identically zero on any orbit (this is possible by Lemma 1 of Appendix 1); then $r = f^\chi$ satisfies the properties of a). The assertion b) is obvious.

Let us apply Prop. 1 of No. 1, with $\chi = 1$, on taking for $f$ a function $\geqslant 0$ that is not identically zero on any orbit. We have $f^1(x) > 0$ at every point $x$ of $X$. Set $h = f/f^1$. Then $h^1 = f^1/f^1 = 1$, therefore $h^b = 1$. It follows that if $g$ is a continuous function on $X/H$, then $(h \cdot (g \circ \pi))^b = h^b \cdot g = g$.