Proposition 9

Let $\theta$ be a bounded complex measure; for $\theta$ to be a positive measure, it is necessary and sufficient that $\| \theta \| = \theta(1)$..

The condition is obviously necessary. Conversely, suppose that $\| \theta \| = \int d\theta$, and denote by $v$ a $|\theta|$-measurable function of absolute value 1 such that $\theta = v \cdot |\theta|$. Since $\| \theta \| = \int d|\theta|$ (Ch. IV, §4, No. 7, Prop. 12) and $\int d\theta = \int v \cdot d|\theta|$ (Th. 1), the hypothesis implies that $\int (1 - v) \, d|\theta| = 0$ and therefore $\int \mathscr{R}(1 - v) \, d|\theta| = 0$. The function $\mathscr{R}(1 - v)$, being positive, is therefore zero almost everywhere, which implies that $v = 1$ almost everywhere and completes the proof.