Theorem 3 (Lebesgue)

— Every complex measure $\theta$ on T may be written in one and only one way in the form $\theta = g \cdot \mu + \theta'$, where $g$ is locally $\mu$-integrable and $\theta'$ is a measure alien to $\mu$. Then $|\theta| = |g| \cdot \mu + |\theta'|$.

When $\theta$ is positive, this follows at once from the theorem of F. Riesz (Ch. II, §1, No. 5, Th. 1) applied to the fully lattice-ordered space $\mathscr{M}(T)$ of real measures on T, and to the band generated by $\mu$ in this space, on taking into account Cor. 2 of No. 5, Th. 2; moreover, $\theta'$ and $g \cdot \mu$ are then positive, which implies that $g$ is positive locally $\mu$-almost everywhere (Cor. 3 of Prop. 3). To treat the case that $\theta$ is not positive, set $\nu = |\theta|$, $\nu = f \cdot \mu + \nu'$ (where $f$ is positive and where $\nu'$ and $\mu$ are alien to each other), and $\theta = v \cdot \nu$, where $v$ is a universally measurable function of absolute value 1 (Cor. 3 of Th. 2). We then have (Prop. 8) $\theta = g \cdot \mu + \theta'$, with $g = v f$ (so that $|g| = f$) and $\theta' = v \cdot \nu'$ (so that $|\theta'| = \nu'$ by Prop. 2); the measures $\theta'$ and $\mu$ are alien to each other by Cor. 2 of Prop. 13. It only remains to establish the uniqueness of the decomposition. Thus, suppose that $\theta = g \cdot \mu + \theta' = g_1 \cdot \mu + \theta'_1$, where $\theta'$ and $\theta'_1$ are alien to $\mu$; $|\theta' - \theta'_1|$ is bounded above by $|\theta'| + |\theta'_1|$, therefore $\theta' - \theta'_1$ is alien to $\mu$, hence also to $(g_1 - g) \cdot \mu$. The relation $\theta' - \theta'_1 = (g_1 - g) \cdot \mu$ then implies that the two members are zero, which proves uniqueness.

Recall (No. 5, Th. 2 and Scholium) that the space $L^1_{loc}(T, \mu; \mathbf{C})$ may be identified (by means of the mapping $g \mapsto g \cdot \mu$) with a subspace of $\mathscr{M}_\mathbf{C}(T)$. With this convention, Theorem 3 takes the following form: