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This is a property of $\mu$, not that of $\mathcal A$, and it is called being atomless. It is equivalent to not having sets $A \in \mathcal A$ of positive measure such that for all $B \in \mathcal A$, $B \subseteq A$ the measure $\mu(B)$ is either 0 or $\mu(A)$.

edit: Wikipedia article, complete with the proof of the property you describe from atomlessness.

edit: yup, the comments are right and I'm wrong. The precise condition for finite measures composed entirely of atoms to have full range is $a_n \leq \sum_{j>n} a_j$ - it is clearly necessary as $a_n-\varepsilon$ has to be produced somehow, and the greedy algorithm shows sufficiency.

This is a property of $\mu$, not that of $\mathcal A$, and it is called being atomless. It is equivalent to not having sets $A \in \mathcal A$ of positive measure such that for all $B \in \mathcal A$, $B \subseteq A$ the measure $\mu(B)$ is either 0 or $\mu(A)$.
This is a property of $\mu$, not that of $\mathcal A$, and it is called being atomless. It is equivalent to not having sets $A \in \mathcal A$ of positive measure such that for all $B \in \mathcal A$, $B \subseteq A$ the measure $\mu(B)$ is either 0 or $\mu(A)$.