Example

— Lebesgue measure on $\mathbf{R}$ is diffuse (Ch. IV, §1, No. 3, Remark 1).

To say that $\theta$ is a diffuse measure on $T$ amounts to saying that every set with finite complement carries $|\theta|$, or again that $|\theta|$ is alien to every point measure. The diffuse measures therefore form a band in $\mathscr{M}(T)$ (Ch. II, §1, No. 5, Th. 1).

Recall (Ch. III, §1, No. 3) that a complex measure $\rho$ on $T$ is said to be atomic if it is of the form $\sum_{t \in T} \alpha(t) \varepsilon_t$, where $\alpha$ is a complex function on $T$ such that $\sum_{t \in K} |\alpha(t)| < +\infty$ for every compact subset $K$ of $T$, which expresses that the family $(\alpha(t) \varepsilon_t)_{t \in T}$ is summable (§2, No. 1, Remark 2). It then follows from the remark following Cor. 3 of Th. 2 of No. 5 that $|\rho| = \sum_{t \in T} |\alpha(t)| \varepsilon_t$. The function $\alpha$ that occurs in these formulas is uniquely determined, because $\alpha(t) = \rho(\{t\})$. An atomic measure and a diffuse measure are alien to each other.