For, verifying that $\Lambda$ is scalarly essentially integrable for a positive measure $\eta$ on $T$ comes down to verifying that $t \mapsto \lambda_t(g)$ is $\eta$-measurable and admits a finite essential upper integral, with respect to $\eta$, for every function $g \in \mathscr{H}_+(X)$. The proposition therefore follows at once from Prop. 11 of §1, No. 4 and its Corollary 2.

It follows immediately that every scalarly essentially $\mu$-integrable mapping is also scalarly essentially $\mu'$-integrable for every measure $\mu' \leq \mu$.

In this section we shall limit ourselves to the study of scalarly essentially integrable mappings of $T$ into $\mathscr{M}_+(X)$ that have the property contemplated in the following definition:

a) If $\Lambda$ is vaguely continuous, then the mapping $t \mapsto \lambda_t^\bullet(f)$ is lower semi-continuous for every lower semi-continuous function $f \geqslant 0$ defined on $X$, $\Lambda$ is $\mu$-adequate, and we have the relation

b) If $\Lambda$ is vaguely $\mu$-measurable, then $\Lambda$ is $\mu$-adequate.

c) If the topology of $X$ admits a countable base, then $\Lambda$ is vaguely $\mu$-measurable (hence also $\mu$-adequate).

Let $f$ be a lower semi-continuous function $\geqslant 0$ defined on $X$. Let $F$ be the set, directed for the relation $\leqslant$, of functions $g \in \mathscr{K}(X)$ such that $0 \leqslant g \leqslant f$. For $g \in F$, denote by $h_g$ the function defined on $T$ by $h_g(t) = \lambda_t(g)$. Similarly, set

(§1, No. 1, Prop. 4). Let us make the following hypothesis, weaker than that of a): assume only that the restriction of $\Lambda$ to $S$ is vaguely continuous, where $S$ is a closed subset of $T$ that contains the support of $\mu$. For $g \in F$, denote by $\overline{h}_g$ the numerical function that coincides with $h_g$ on $S$ and has the value $+\infty$ on $C_S$. Set $\overline{h}_f = \sup_{g \in F} \overline{h}_g$; then $\overline{h}_f = h_f$ on $S$. For every $g \in F$, the function $\overline{h}_g$ is lower semi-continuous; $\overline{h}_f$ is therefore lower semi-continuous and, since the family $(\overline{h}_g)_{g \in F}$ is directed,

(Ch. IV, §1, No. 1, Th. 1 and §2, No. 3, Prop. 6). Since $h_f = \overline{h}_f$ on $S$, hence almost everywhere, this may also be written $\mu^*(h_f) = \nu^*(f)$, an equality identical to (5). Similarly, $f$ and $\overline{h}_f$ being lower semi-continuous, the preceding relations yield the equality $\mu^\bullet(\overline{h}_f) = \nu^\bullet(f)$ (§1, No. 1, Prop. 4); since $\overline{h}_f = h_f$ on $S$, it follows that $\mu^\bullet(h_f) = \nu^\bullet(f)$ (§1, No. 1, Prop. 1), an equality identical to (4). The mapping $\Lambda$ is therefore $\mu$-pre-adequate; but one could have replaced everywhere in this argument $\mu$ by $\mu' \leqslant \mu$, and $\nu$ by $\nu' = \int \lambda_t \, d\mu'(t)$, because $\Lambda$ is also scalarly essentially $\mu'$-integrable and $S$ contains the support of $\mu'$. It follows from this that $\Lambda$ is $\mu$-adequate.

Suppose $\Lambda$ is vaguely continuous; we may take $S = T$; then $h_f = \overline{h}_f$ is lower semi-continuous, which completes the proof of part a) of the statement.

Assume that $\Lambda$ is vaguely $\mu$-measurable and let us prove b). The set $\mathscr{K}$ of compact subsets K of T such that the restriction of $\Lambda$ to K is continuous being $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15), there exists a summable family $(\mu_\alpha)_{\alpha \in A}$ of measures on T such that $\mu = \sum_{\alpha \in A} \mu_\alpha$ and the support of each of the measures $\mu_\alpha$ belongs to $\mathscr{K}$ (§2, No. 3, Prop. 4). For every $\alpha \in A$, the mapping $\Lambda$ is scalarly essentially $\mu_\alpha$-integrable, and we set $\nu_\alpha = \int \lambda_t d\mu_\alpha(t)$; the family $(\nu_\alpha)$ is summable, and its sum is equal to $\nu$ (Cor. of Prop. 1). If $f$ is a positive lower semi-continuous function defined on X, the first part of the proof, applied to the measures $\mu_\alpha$ and the closed sets $S_\alpha$, shows that:

$1^\circ$ $h_f$ is $\mu_\alpha$-measurable for every $\alpha \in A$, hence is $\mu$-measurable (§2, No. 2, Prop. 2), and

$2^\circ$ $\int^* f(x) d\nu_\alpha(x) = \int^* d\mu_\alpha(t) \int^* f(x) d\lambda_t(x)$.

The formula (4) follows on summing over $\alpha$ (§2, No. 2, Prop. 1). Applying the preceding argument to an arbitrary measure $\mu' \leq \mu$ (which is legitimate, since $\Lambda$ is scalarly essentially $\mu'$-integrable and vaguely $\mu'$-measurable, cf. §2, No. 2, Prop. 2), we conclude that $\Lambda$ is $\mu'$-pre-adequate, and b) is proved.

Finally, assuming that the topology of X admits a countable base, let us show that every scalarly essentially $\mu$-integrable mapping $\Lambda : t \mapsto \lambda_t$ of T into $\mathscr{M}_+(X)$ is vaguely $\mu$-measurable. This will result from the following lemma:

Let $X'$ be the Alexandroff compactification of X, which is a metrizable compact space (GT, IX, §2, No. 9, Prop. 16 and Cor.); we identify $\mathscr{K}(X)$ with a subset of $\mathscr{C}(X')$. Let $S'$ be a countable dense subset of the Banach space $\mathscr{C}(X')$ (GT, X, §3, No. 3, Th. 1); we can suppose that $S'$ contains the constant function $n$ for every $n \in \mathbf{N}$. Let $(U_n)$ be a sequence of relatively compact open sets in X, with union X, such that $\overline{U}_n \subset U_{n+1}$ for all n (GT, I, §9, No. 9, Prop. 15), and let $\varphi_n$ be a function in $\mathscr{K}_+(X)$ equal to 1 on $\overline{U}_n$. We denote by S the countable set of elements of $\mathscr{K}(X)$ of the form $\varphi_n g$ ($n \in \mathbf{N}, g \in S'$). If $f \in \mathscr{K}(X)$, let $(g_n)$ be a sequence of elements of $S'$ that converges uniformly to $f$, and let k be an integer such that the support of $f$ is contained in $U_k$. Finally, let m be an integer that is an upper bound for the norms of the functions $g_n$. The functions $f_n = \varphi_k g_n$ belong to S and satisfy the statement, with $\varphi = m \varphi_k$.

The lemma having been established, and the mapping $t \mapsto \lambda_t(g)$ being essentially $\mu$-integrable for every $g \in S$, the mapping $t \mapsto (\lambda_t(g))_{g \in S}$ of T into $\mathbf{R}^S$ is $\mu$-measurable (Ch. IV, §5, No. 3, Th. 1). The set $\mathscr{K}$, of compact subsets K of T such that the restriction of this mapping to K is continuous, is therefore $\mu$-dense, and it will suffice to show that the restriction of $\Lambda$ to every $K \in \mathscr{K}$ is continuous. Now, let $f$ be any element of $\mathscr{K}(X)$, $f_n$ and $\varphi$ elements of S satisfying the statement of the Lemma; the function $t \mapsto \lambda_t(f)$ is then the uniform limit on K of the continuous functions $t \mapsto \lambda_t(f_n)$; it is therefore continuous on K, and the proposition is proved.

Let g be a lower semi-continuous function on X such that $f \leqslant g$. For every $t \in T$,

therefore, by (4) and Prop. 4 of §1,

The first of the inequalities (6) then follows from the definition of $\int^* f(x)\, d\nu(x)$ (Ch. IV, §1, No. 3, Def. 3), and the second follows immediately from it. The inequality (7) is proved in an analogous way if $\Lambda$ is vaguely continuous, using (5) instead of (4).

Let us pass to the proof of (8). The mapping $t \mapsto \lambda_t^*(1)$ is measurable, and finite locally $\mu$-almost everywhere. The set $\mathscr{K}$ of compact subsets of $T$ such the restriction of $t \mapsto \lambda_t^*(1)$ to $K$ is finite and continuous is therefore $\mu$-dense, and Prop. 4 of §2, No. 3 implies the existence of a summable family $(\mu_\alpha)_{\alpha \in A}$ of positive measures, with supports belonging to $\mathscr{K}$, such that $\mu = \sum_{\alpha \in A} \mu_\alpha$. The mapping $\Lambda$ is $\mu_\alpha$-adequate for every $\alpha \in A$; set $\nu_\alpha = \int \lambda_t d\mu_\alpha(t)$. Prop. 1 shows that $\nu = \sum_{\alpha \in A} \nu_\alpha$, and the relation (4), applied to the measure $\mu_\alpha$ and the function 1, shows that $\nu_\alpha$ is a bounded measure (because $\lambda_t^*(1)$ is bounded on $\operatorname{Supp}(\mu_\alpha)$). Let us then write the formula (6) for the measure $\mu_\alpha$, replacing the symbol $\int^*$ in the first member by $\int^\bullet$, which is legitimate by Prop. 7 of §1; then

(the last equality due to the fact that $\lambda_t$ is bounded locally almost everywhere, and Prop. 7 of § 1). The inequality in (8) is then obtained by summing on $\alpha$ (§2, No. 2, Prop. 1).

If no hypothesis analogous to that of c) is made, the inequality (8) may fail (Exer. 2).

For, $f$ is the sum of a sequence of functions $f_n \geq 0$ such that $f_n$ is zero outside a compact set $K_n$ for $n \geq 1$, and $f_0$ is $\nu$-negligible (§1, No. 2, Prop. 6); $f_0$ is then $\lambda_t$-negligible except for $t$ forming a set that is locally $\mu$-negligible (and even $\mu$-negligible, if $\Lambda$ is vaguely continuous) by Cor. 1, and the statement then follows at once.

b) Suppose that f is constant on the complement of a ν-moderated subset of X, and that Λ is vaguely continuous; then M is μ-negligible.

c) Suppose that $\lambda_t^\bullet(1) < +\infty$ locally μ-almost everywhere; then M is locally μ-negligible.

Let us first prove a) (resp. b)). Since every ν-integrable set is contained in a ν-integrable open set, the function f is constant on the complement B of a countable union of ν-integrable open sets. There exists a partition of X − B formed by a ν-negligible set N and a sequence $(K_n)$ of compact sets such that the restriction of f to each $K_n$ is continuous. Let S be the set of $t \in T$ such that N is not $\lambda_t$-negligible: S is locally μ-negligible (resp. μ-negligible) by Cor. 1 of Prop. 3. The sets $K_n, B, N$ are measurable for every measure on X, and the restriction of f to each of them is $\lambda_t$-measurable for every $t \notin S$. The function f is therefore $\lambda_t$-measurable for every $t \notin S$ (Ch. IV, §5, No. 10, Prop. 16).

To establish c), let us take up again the notations in the proof of Prop. 3; since f is ν-measurable, it is measurable for each of the measures $\nu_\alpha \leq \nu$. Now, these measures are bounded, hence moderated, and it follows from a) that M is locally $\mu_\alpha$-negligible for every $\alpha \in A$. This implies that M is locally μ-negligible (§2, No. 2, Cor. 2 of Prop. 1).

a) Suppose that f is ν-moderated. The set N is then locally μ-negligible, the function $t \mapsto \int^\bullet f(x) d\lambda_t(x)$ is μ-measurable, and

b) Suppose that f is ν-moderated, and that Λ is vaguely continuous. The set N is then μ-negligible, the function $t \mapsto \int^* f(x) d\lambda_t(x)$ is μ-measurable and μ-moderated, and

c) Suppose that $\lambda_t^\bullet(1) < +\infty$ locally μ-almost everywhere. The set N is then locally μ-negligible, the function $t \mapsto \int^\bullet f(x) d\lambda_t(x)$ is μ-measurable, and (9) holds.

Let us first prove a) (resp. b)), assuming that f is ν-moderated. The assertions concerning the set N have already been established (Prop. 4, and Cor. 2 of Prop. 3). By Prop. 6 of §1, No. 2, we may limit ourselves to proving a) (resp. b)) in each of the following special cases:

1) The function $f$ is $\nu$-negligible. 2) There exists a compact set $K$ such that $f$ is zero outside $K$ and the restriction of $f$ to $K$ is continuous.

The special case 1) has already been treated (Cor. 1 of Prop. 3). To treat the second, we denote by $G$ a $\nu$-integrable open set containing $K$, by $M$ a constant upper bound for $f$, by $h$ the lower semi-continuous function $M \varphi_G$, and by $g$ the function $h - f$. Since the function $f$ is upper semi-continuous on $X$, $g$ is lower semi-continuous and positive. Moreover, $f, g, h$ are $\nu$-integrable.

Let us then apply formula (4) (resp. (5)) to the lower semi-continuous functions $h$ and $g$. By subtraction, we see that the function

is $\mu$-measurable and that the formula (9) (resp. (10)) holds. Finally, under the hypothesis of b), the function $t \mapsto \int^* f(x)\, d\lambda_t(x)$ has finite upper integral, hence is indeed $\mu$-moderated.

To prove c), let us take up again the measures $\mu_\alpha$ and $\nu_\alpha$ of the proof of Prop. 3; since $f$ is $\nu_\alpha$-measurable and $\nu_\alpha$-moderated, the assertion a) implies that $t \mapsto \int^\bullet f(x)\, d\lambda_t(x)$ is $\mu_\alpha$-measurable and that

It remains only to sum on $\alpha$, applying Props. 1 and 2 of §2, No. 2.

If $f$ is not assumed to be $\nu$-moderated, and if one does not make the assumption in c), then the relation (9) may not hold (Exer. 3).

This follows at once from Prop. 5 and the criterion for integrability (Ch. IV, §5, No. 6, Th. 5).

a) If $f$ is $\nu$-integrable, then $H$ is locally $\mu$-negligible, the function $t \mapsto \int f(x)\, d\lambda_t(x)$ (defined for $t \notin H$) is essentially $\mu$-integrable, and

b) If $f$ is $\nu$-integrable and $\Lambda$ is vaguely continuous, then $H$ is moreover $\mu$-negligible and the function $t \mapsto \int f(x)\, d\lambda_t(x)$ (defined for $t \notin H$) is $\mu$-integrable.

c) If $\lambda_t^*(1) < +\infty$ locally $\mu$-almost everywhere, then the conclusions of a) remain true for $f$ an essentially $\nu$-integrable function.

We are first going to establish a) (resp. b)). This statement is true when $f$ is a positive numerical function (Prop. 5); if $f$ is an integrable function with values in $\overline{\mathbf{R}}$, this result may be applied to the positive functions $f^+$ and $f^-$, and therefore extends at once to $f$ by subtraction. It remains to treat the case of functions with values in $F$. Let $\mathscr{H}$ be the subspace of $\mathscr{L}_F^1(\nu)$ formed by the linear combinations, with coefficients in $F$, of the functions in $\mathscr{K}(X)$; the result pertaining to real functions implies at once the validity of the statement for the elements of $\mathscr{H}$. Now, $\mathscr{H}$ is dense in $\mathscr{L}_F^1(\nu)$; therefore, for every $f \in \mathscr{L}_F^1(\nu)$, there exists a sequence $(f_n)$ of elements of $\mathscr{H}$ that has the following properties:

1) the sequence $(f_n)$ converges to $f$ in mean in $\mathscr{L}_F^1(\nu)$, and $\nu$-almost everywhere; 2) the function $g = |f_0| + \sum_{n \in \mathbf{N}} |f_{n+1} - f_n|$ is such that $\nu^*(g) < +\infty$ (Ch. IV, §3, No. 4, Th. 3).

Let $N_1$ be the set of $t \in T$ such that $\lambda_t^*(g) = +\infty$; $N_1$ is locally $\mu$-negligible (resp. $\mu$-negligible) by formula (6) (resp. (7)). For $t \notin N_1$, the $f_n$ belong to $\mathscr{L}_F^1(\lambda_t)$, the sequence $(f_n)$ converges $\lambda_t$-almost everywhere, as well as for the topology of convergence in mean in $\mathscr{L}_F^1(\lambda_t)$ (Ch. IV, §3, No. 3, Prop. 6). Let $M$ be set of $x \in X$ such that $f_n(x)$ does not converge to $f(x)$: since $M$ is $\nu$-negligible, the set $N_2$ of $t \in T$ such that $M$ is not $\lambda_t$-negligible is locally $\mu$-negligible (resp. $\mu$-negligible) by Cor. 1 of Prop. 3.

Suppose that $t$ does not belong to $N_1 \cup N_2$; the sequence $(f_n)$ converges in mean in $\mathscr{L}_F^1(\lambda_t)$, and converges $\lambda_t$-almost everywhere to $f$. Therefore $f \in \mathscr{L}_F^1(\lambda_t)$ and $\int f\, d\lambda_t = \lim_{n \to \infty} \int f_n\, d\lambda_t$ (Ch. IV, §4, No. 1). The set $H$ of the statement is thus contained in $N_1 \cup N_2$; it is therefore locally $\mu$-negligible (resp. $\mu$-negligible). On the other hand, the function $t \mapsto \int f\, d\lambda_t$ is equal locally $\mu$-almost everywhere to the limit of a sequence of $\mu$-measurable functions; it is therefore $\mu$-measurable. Finally, for every $t \notin N_1 \cup N_2$ and every $n$, we have

by virtue of the inequality $|f_n| \leq g$ and Prop. 2 of Ch. IV, §4, No. 2. Now, the function $t \mapsto \int^* g(x)\, d\lambda_t(x)$ is essentially $\mu$-integrable (resp. $\mu$-integrable) by Prop. 5. We may therefore apply Lebesgue’s theorem, which yields Since $\int f_n(x)\, d\nu(x)$ tends to $\int f(x)\, d\nu(x)$ as $n$ tends to $\infty$, by the hypotheses made on the sequence $(f_n)$, the relation (11) follows and we have proved a) (resp. b)).

Now suppose that $\lambda_t^\bullet(1) < +\infty$ locally $\mu$-almost everywhere, and that $g$ is an essentially $\nu$-integrable function. Let $f$ be a $\nu$-integrable function such that $g = f$ locally $\nu$-almost everywhere (§ 1, No. 3). Then $g = f$ almost everywhere for $\lambda_t$, except for $t$ forming a locally $\mu$-negligible set $P$ (Cor. 1 c) of Prop. 3). Therefore $\int g\, d\lambda_t = \int f\, d\lambda_t$ for all $t \notin P \cup H$, and this completes the proof.

The condition is obviously necessary; on the other hand it is sufficient, because every positive measure $\mu$ is the sum of a family of measures with compact support (§2, No. 3, Prop. 4): the statement then follows from Prop. 2 of §2, No. 2.

Let $\mathscr{K}$ be the set of compact sets in T such that the restriction of $f$ to K is continuous; since $\mathscr{K}$ is $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15), there exists a locally countable family $(K_i)_{i \in I}$ of pairwise disjoint elements of $\mathscr{K}$ such that the set $N = T - \bigcup_{i \in I} K_i$ is locally $\mu$-negligible (Ch. IV, §5, No. 9, Prop. 14). Let $x$ be an element of F; set

The functions $f$ and $f'$ are equal locally $\mu$-almost everywhere. On the other hand, $N \cap K$ is a Borel subset of K for every compact set K in T, since the family $(K_i)$ is locally countable. It follows that N is a universally measurable set, and that $f'$ is a universally measurable function (Ch. IV, §5, No. 10, Prop. 16).

Suppose that $\Lambda$ is a diffusion. The condition 1) is then satisfied by the definition of adequate mappings (No. 1, Def. 1) and Prop. 6; the condition 3) is satisfied by formula (4), since $\Lambda$ is $\mu$-adequate. Let $g \in \mathscr{K}_+(X)$ and let $u$ be the function $t \mapsto \lambda_t(g)$ (universally measurable, by 1)); suppose that $u$ is not locally bounded. There would then exist a compact set K such that $u$ is not bounded on K, hence there would exist a sequence $(t_n)$ of elements of K such that $u(t_n) \geqslant n^2$ for all $n \geqslant 1$; then $u$ is not integrable for the measure $\mu = \sum_{n \geqslant 1} \frac{1}{n^2} \varepsilon_{t_n}$ with compact support, contrary to the hypothesis on $\Lambda$, which implies that $t \mapsto \lambda_t(g)$ is integrable for every positive measure with compact support. The above three conditions are thus necessary. Conversely, the conditions 1) and 2) imply that $\Lambda$ is scalarly essentially $\mu$-integrable for every measure $\mu$ with compact support. Since every measure $\mu' \geqslant 0$ that is bounded above by a measure $\mu$ with compact support also has compact support, the conditions 1) and 3) express that $\Lambda$ is $\mu$-adequate for every positive measure with compact support, which is indeed the sought-for result.