Theorem 1

— Let $f$ be a function with values in a Banach space $F$ or in $\overline{\mathbf{R}}$, and let $H$ be the set of $t \in T$ for which $f$ is not $\lambda_t$-integrable.

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.