Proposition 13

In order that two complex measures $\rho$ and $\sigma$ on $T$ be alien to each other, it is necessary and sufficient that there exist in $T$ two disjoint sets $R$ and $S$ such that $\rho$ is concentrated on $R$ and $\sigma$ on $S$; $R$ and $S$ may be taken to be universally measurable.

Set $\mu = |\rho|$, $\nu = |\sigma|$, $\lambda = \mu + \nu$; since $\mu$ and $\nu$ are bounded above by $\lambda$, there exist two locally $\lambda$-integrable functions $u$ and $v$ (which one can suppose to be universally measurable by §3, No. 4, Prop. 7) such that $\mu = u \cdot \lambda$, $\nu = v \cdot \lambda$. Then

(No. 2, Cor. of Prop. 2). Let $A$ (resp. $B$) be the set of $t \in T$ such that $u(t) > 0$ and $v(t) = 0$ (resp. $u(t) = 0$ and $v(t) > 0$). If $\rho$ and $\sigma$ are alien, then $\inf(u, v) = 0$ locally $\lambda$-almost everywhere (No. 3, Cor. 2 of Prop. 3), so that the disjoint universally measurable sets $A$ and $B$ carry $\mu$ and $\nu$, respectively. Conversely, suppose that $\mu$ and $\nu$ are carried, respectively, by disjoint sets $R$ and $S$; $\varphi_R$ is measurable for the measure $\mu = u \cdot \lambda$, and $\mu = \varphi_R \cdot \mu$. By Prop. 8 of No. 4, the function $u' = u \varphi_R$ is $\lambda$-measurable, and $\mu = u' \cdot \lambda$. Similarly, if $v' = v \varphi_S$ then $\nu = v' \cdot \lambda$; one concludes by remarking that $\inf(u', v') = 0$ (No. 2, Cor. of Prop. 2).