Lemma 2

Let $f$ be a $\nu$-measurable positive function on $S \times T$; for every compact subset $L$ of $T$, let $I_f^L$ be the function $s \mapsto \mu^\bullet(f_s \varphi_L)$ on $S$. Then the function $I_f^L$ is $\lambda$-measurable, and

where $L$ runs over the set of compact subsets of $T$.

We first note that the inclusion $L \subset L'$ implies $I_f^L \leq I_f^{L'}$; on the other hand, $I_f^L(s) = \mu_L^\bullet((f_s)_L)$ for all $s \in S$. Formula (5) is therefore an immediate consequence of the definition of the encumbrance $\mu^\bullet$ given in

§1, No. 2. If K is a compact subset of S, and L a compact subset of T, then $\nu_{K \times L} = \lambda_K \otimes \mu_L$ by construction, and Prop. 7 of Ch. V, §8, No. 3 implies the relation

Moreover, every compact subset of $S \times T$ is contained in a compact set of the form $K \times L$; passing to the upper envelope over all K and L in the preceding formula, we therefore obtain

namely (6).

Finally, Prop. 7 of Ch. V, §8, No. 3 implies that the restriction of $I_f^L$ to every compact subset K of S is $\lambda_K$-measurable; this is equivalent to saying that $I_f^L$ is $\lambda$-measurable.