Proposition 4

— Let $f$ be a $\nu$-measurable function defined on $X$, with values in a topological space $G$, and let $M$ be the set of $t \in T$ such that $f$ is not $\lambda_t$-measurable. a) Suppose that $f$ is constant on the complement of a $\nu$-moderated subset of $X$; then $M$ is locally $\mu$-negligible.

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).