Proposition 11

For a positive measure $\mu$ on $T$ to belong to the domain of $\Lambda$, it is necessary and sufficient that $\Lambda$ be scalarly essentially $\mu$-integrable.

This condition is obviously necessary. Conversely, suppose it is satisfied, and let $f$ be a lower semi-continuous positive function defined on $X$. The function $\Lambda f$ is universally measurable, hence $\mu$-measurable. We are going to prove that $⟨\mu, \Lambda f⟩ = ⟨\mu\Lambda, f⟩$; since this equality will also be valid for every positive measure $\mu' \leq \mu$, because $\Lambda$ is also scalarly essentially $\mu'$-integrable, it will follow that $\Lambda$ is $\mu$-adequate.

Let $(\mu_i)_{i\in I}$ be a summable family of positive measures with compact support, such that $\mu = \sum_{i\in I} \mu_i$ (§2, No. 3, Prop. 4); the family of measures $\mu_i \Lambda$ is then summable, and $\mu \Lambda = \sum_{i \in I} \mu_i \Lambda$ (No. 1, Cor. of Prop. 1). Consequently $\langle \mu \Lambda, f \rangle = \sum_{i \in I} \langle \mu_i \Lambda, f \rangle$ (§ 2, No. 2, Prop. 1); but $\Lambda$ is $\mu_i$-adequate, thus $\langle \mu_i \Lambda, f \rangle = \langle \mu_i, \Lambda f \rangle$. Again applying Prop. 1 of § 2, we obtain the sought-for equality: