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.
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