a) for every function $f \in \mathscr{K}(X)$, $f \circ \pi$ is essentially $\mu$-*integrable*.

b) for every compact set $K \subset X$, $\overline{\pi}^{-1}(K)$ is essentially $\mu$-*integrable*.

We have just observed that a) implies that the pair $(\pi, 1)$ is $\mu$-adapted. Consequently (§ 4, No. 4, Th. 2), for every compact set $K \subset X$, the function $\varphi_K \circ \pi = \varphi_A$, where $A = \overline{\pi}^{-1}(K)$, is essentially $\mu$-integrable, in other words, a) implies b).

Conversely, suppose that $\overline{\pi}^{-1}(K)$ is essentially $\mu$-integrable for every compact subset $K$ of $X$, and let us show that a) is verified. Indeed, let $S$ be the support of $f$; since $S$ is compact, setting $A = \overline{\pi}^{-1}(S)$ we have, by hypothesis,

Since $f \circ \pi$ is $\mu$-measurable (Ch. IV, § 5, No. 3, Th. 1), we see that $f \circ \pi$ is essentially $\mu$-integrable (§ 1, No. 3, Prop. 9).

Property b) is obviously equivalent to the following (which is therefore also equivalent to a)):

c) For every point $x$ of $X$, there exists a neighborhood $V$ of $x$ such that $\overline{\pi}^{-1}(V)$ is essentially $\mu$-*integrable*.

This follows from Th. 1 of §4, No. 2.

For, if $N = X - \pi(M)$ then $\pi^{-1}(N)$ does not intersect M, therefore is locally $\mu$-negligible, consequently (Cor. 2) N is locally $\nu$-negligible.

For, it follows from Cor. 3 that $\pi(\mu)$ is concentrated on $\pi(S)$, therefore if $S'$ is the support of $\pi(\mu)$ then $S' \subset \pi(S)$. On the other hand, $\pi^{-1}(X - S')$ is a locally $\mu$-negligible open set (Cor. 2), hence is $\mu$-negligible (Ch. IV, §5, No. 2, Cor. 2 of Prop. 5). Therefore $\pi^{-1}(X - S') \subset T - S$, consequently $\pi(S) \subset S'$, which proves the corollary.

This is an immediate consequence of Prop. 3 of §4, No. 3.

However, the image under $\pi$ of a $\mu$-measurable subset M of T is not necessarily $\nu$-measurable, even if $\pi$ is continuous and M is $\mu$-negligible (Exer. 7 and §8, Exer. 1).

Suppose, moreover, that $\pi$ is continuous and proper. Then, for f to be $\nu$-integrable, it is necessary and sufficient that $f \circ \pi$ be $\mu$-integrable.

It suffices to apply Th. 2 of §4, No. 4.

In particular, for every compact set $K \subset X$, $\nu(K) = \mu(\pi^{-1}(K))$. It follows from this and Cor. 3 of Prop. 2 that if $\mu$ is atomic (§ 5, No. 10) then so is $\pi(\mu) = \nu$. For, let M be the set of $t \in T$ such that $\mu(\{t\}) \neq 0$; since $\mu$ is carried by M, $\nu$ is carried by $\pi(M)$; moreover, for every $x \in \pi(M)$ we have $\nu(\{x\}) = \mu(\pi^{-1}(x)) > 0$, since $\pi^{-1}(x)$ contains at least one point of M. Therefore $\nu$ is atomic (§ 5, No. 10, Prop. 15).

a) Suppose that $\pi$ is $\mu$-proper and let $\mu' = \pi(\mu)$. For $\pi'$ to be $\mu'$-proper, it is necessary and sufficient that $\pi''$ be $\mu$-proper, in which case $\pi''(\mu) = \pi'(\pi(\mu))$ ('transitivity of the image of a measure').

b) Suppose that $\pi'$ is continuous, and that $\pi''$ is $\mu$-proper; then $\pi$ is $\mu$-proper, $\pi'$ is $\pi(\mu)$-proper and $\pi''(\mu) = \pi'(\pi(\mu))$.

Under the hypotheses of a), for $\pi''$ to be $\mu$-measurable it is necessary and sufficient that $\pi'$ be $\mu'$-measurable (No. 2, Prop. 3). On the other hand if $K$ is a compact subset of $T''$, then $\pi''(K) = \pi^{-1}(\pi'(K))$; for $\pi''(K)$ to be essentially $\mu$-integrable, it is necessary and sufficient that $\pi'(K)$ be essentially $\mu'$-integrable, by the Cor. of Th. 1. Finally, if $\pi''$ is $\mu$-proper then, setting $\mu'' = \pi''(\mu)$, we have, for every function $f \in \mathscr{H}(T'')$,

by Th. 1 of No. 2, which completes the proof of a).

Under the hypotheses of b), let $K'$ be a compact subset of $T'$. Then $K'' = \pi'(K')$ is compact, therefore $\pi''(K'')$ is essentially $\mu$-integrable, therefore $\pi^{-1}(K') \subset \pi''(K'')$ is essentially $\mu$-integrable (Ch. IV, §5, No. 5, Prop. 7), so that $\pi$ is $\mu$-proper. The proof is then concluded by applying part a) of the statement.

Set $\nu = \pi(\mu)$. For $g$ to be locally $\nu$-integrable, it is necessary and sufficient that $gf$ be $\nu$-integrable for every function $f \in \mathscr{H}(X)$; since $gf$ has compact support, it comes to the same to say that $gf$ is essentially $\nu$-integrable, and this is equivalent to saying that $(g \circ \pi)(f \circ \pi)$ is essentially $\mu$-integrable (Th. 1). But, by Th. 1 of §5, No. 3, this signifies that $f \circ \pi$ is essentially integrable for $\rho = (g \circ \pi) \cdot \mu$, and, by definition, this says that $\pi$ is $\rho$-proper (since $\pi$ is obviously $\rho$-measurable). Moreover,

(No. 2, Th. 1 and §5, Th. 1), which proves the relation (4).

For $\pi$ to be $\mu$-measurable, it is necessary and sufficient that $\pi$ be $\lambda_\alpha$-measurable for all $\alpha \in A$ (§ 1, No. 4, Cor. 2 of Prop. 11). Suppose that this condition is satisfied; to say that $\pi$ is $\mu$-proper is then equivalent to saying that, for every function $f \in \mathscr{K}_+(X)$,

Now,

(§ 1, No. 4, Prop. 11); the first member is thus finite for every $f \in \mathscr{K}_+(X)$ if and only if the family $(\pi(\lambda_\alpha))$ admits a supremum $\theta$ in $\mathscr{M}(X)$, in which case $\int (f \circ \pi)\, d\mu = \int f\, d\theta$, a relation equivalent to (5).

c) Let $f$ be a function defined on $X$, with values in a Banach space $F$; if $f \circ \pi$ is essentially $\theta$-integrable, then $f$ is essentially $\nu$-integrable and

Taking into account formula (8), these results may be deduced from Cor. 2 of Prop. 2, from Prop. 3, and from Th. 1 of No. 2.

It suffices to prove that, for every function $f \in \mathscr{K}(G(I))$,

Now, this formula has already been proved for $g \in \mathscr{K}(I)$ (FRV, II, §2, No. 1, formula (1)). Let us pass to the general case; there exists a sequence $(g_n)$ of functions in $\mathscr{K}(I)$ such that: $1^\circ$ the sequence $(g_n(t))$ tends to $g(t)$ almost everywhere in $I$; $2^\circ$ there exists a $\mu$-integrable function $h \geq 0$ such that $|g_n| \leq h$ for all $n$ (Ch. IV, §3, No. 4, Th. 3). It follows at once from Lebesgue’s theorem that, setting $G_n(x) = c + \int_{x_0}^x g_n(t)\,dt$, the sequence $(G_n)$ converges uniformly to $G$ on $I$, and that the numbers $G_n(a+)$ and $G_n(b-)$ tend respectively to $G(a+)$ and $G(b-)$. Let $f'$ be a function in $\mathscr{K}(\mathbf{R})$ that extends $f$; the foregoing proves that $f'(G_n(t))$ tends to $f'(G(t)) = f(G(t))$ for all $t \in I$; applying Lebesgue’s theorem, one sees that the formula (12) results from the formula

by passage to the limit.

(formula for change of variable in the Lebesgue integral).

For, $\mathbf{f}(G(t))$ is integrable for the measure $|g|\cdot\mu$, hence also for the measures $g^+\cdot\mu$ and $g^-\cdot\mu$; it follows from Th. 1 (No. 2) that $\mathbf{f}$ is integrable for the image measures $G(g^+\cdot\mu)$ and $G(g^-\cdot\mu)$, hence also for the measure $\varphi_J \cdot \nu$, and that (13) holds, on taking into account Prop. 8 and formula (9).

It can happen that $\mathbf{f}$ is integrable on $J$ for Lebesgue measure, but that $t \mapsto \mathbf{f}(G(t))g(t)$ is not integrable on $I$ for Lebesgue measure (Exer. 10).

Now suppose that $g$ maintains a constant sign almost everywhere (and is locally $\mu$-integrable); one may suppose for example that $g(t) \geq 0$ almost everywhere in $I$. Then $G$ is an increasing continuous function on $I$, therefore $G(a+)$ and $G(b-)$ exist (but may be infinite). Moreover, $G$ is a $\lambda$-proper mapping of $I$ into $G(I)$: for, if $G(b-) \in G(I)$, there is an $x_1 \geq x_0$ such that $G$ is constant for $x \geq x_1$, and then the inverse image under $G$ of the compact interval $[G(x_0), G(b-)]$ is $\lambda$-integrable; if, on the contrary, $G(b-) \notin G(I)$, then the inverse image under $G$ of every compact interval with left end-point $G(x_0)$, contained in $G(I)$, differs from a compact interval by at most a $\lambda$-negligible interval. One argues similarly for the compact intervals with right end-point $G(x_0)$, whence our assertion. Moreover:

The first part of the statement follows from the fact that the formula (12) is valid for every function $f \in \mathscr{K}(G(I))$; for, the support of the function $t \mapsto f(G(t))$ is contained in an interval $K \subset I$ on which $g$ is integrable, by virtue of the above remarks, and it suffices to apply Prop. 8 to $K$. The second part is a consequence of Th. 1 of No. 2.

We begin by arguing on the assumption that $g$ is finite locally $\mu$-almost everywhere; we will rid ourselves of this auxiliary hypothesis at the end of the proof. Since $\pi$ is by hypothesis $\nu$-measurable, to say that $\pi$ is $\nu$-proper is equivalent to saying that $\nu^*(f \circ \pi) < +\infty$ for every function $f \in \mathscr{K}_+(T)$; $g$ being finite locally almost everywhere, we are under the conditions for applying assertion c) of Prop. 5 of §3, No. 2. Therefore

from the fact that $\lambda_t$ is concentrated on $\overline{\pi}^{-1}(t)$. We know that $g$ is $\mu$-measurable, since $\Lambda$ is $\mu$-adequate (§ 3, No. 1, Def. 1). To say that the first member is finite for every $f \in \mathscr{K}_+(T)$ is therefore equivalent to saying that $g$ is locally $\mu$-integrable (§ 5, Prop. 1), and in this case (14) follows at once from the above relations.

It therefore only remains to eliminate the auxiliary hypothesis. If $g$ is locally $\mu$-integrable, then $g$ is finite locally $\mu$-almost everywhere, and the hypothesis is indeed satisfied. Let us assume that $\pi$ is $\nu$-proper, and let us show that $g$ is finite locally almost everywhere. Let $\mathfrak{K}$ be the $\mu$-dense set of compact sets $K$ such that $\Lambda|K$ is vaguely continuous; since $g$ is measurable, we are reduced to showing that every compact set $K \in \mathfrak{K}$ such that $g|K = +\infty$ is $\mu$-negligible. Now, let $\mathscr{H}$ be the set of functions $h \in \mathscr{K}_+(X)$ such that $h \leq 1$; set $g_h(t) = \lambda_t(h)$, denote by $\Lambda_h$ the $\mu$-adequate mapping $t \mapsto h \cdot \lambda_t$, by $\nu_h$ the integral of $\Lambda_h$, and by $f$ an element of $\mathscr{K}_+(T)$ such that $f \geq \varphi_K$. Applying formula (14) to $\Lambda_h$, which does satisfy the auxiliary hypothesis, we obtain:

But the functions $fg_h|K$ form an increasing directed set of continuous functions on $K$, whose upper envelope has the value $+\infty$; by Dini’s theorem (GT, X, §4, No. 1, Th. 1), one can choose $h$ so that $fg_h|K$ is greater than or equal to an arbitrary positive number $n$, and it follows that $\int (f \circ \pi)\, d\nu \geq n\, \mu(K)$. Since the first member is finite because $\pi$ is $\nu$-proper, it then follows that $\mu(K) = 0$.

a) If $N \subset T$ is locally $\mu$-negligible, then $\overline{\pi}^{-1}(N)$ is locally $\nu$-negligible. b) If $f$ is a $\mu$-measurable mapping of $T$ into a topological space $G$, then $f \circ \pi$ is $\nu$-measurable.

We take up again the notations $\Lambda_h$, $\nu_h$, $g_h$ of the end of the preceding proof: $\nu_h$ being a bounded measure for every $h \in \mathscr{H}$, $\pi$ is $\nu_h$-proper, $g_h$ is locally $\mu$-integrable, and $\pi(\nu_h) = g_h \cdot \mu$, a measure with base $\mu$. It follows that $N$ is locally negligible (resp. that $f$ is measurable) for the measure $\pi(\nu_h)$ (§ 5, No. 3, Cor. 1 of Prop. 3 and Prop. 4). Consequently $\overline{\pi}^{-1}(N)$ is locally negligible (resp. $f \circ \pi$ is measurable) for the measure $\nu_h$ (Cor. 2 of

Prop. 2, resp. Prop. 3). Finally, one notes that the measures $\nu_h$ form an increasing directed family of positive measures whose supremum is $\nu$, and one applies Cor. 1 (resp. Cor. 2) of Prop. 11 of §1, No. 4.

Taking into account Prop. 10, this follows immediately from Th. 1 of §5, No. 3 and Th. 1 of No. 2.