Probability space with countable subset such that every subset of positive measure meets the subset

Let $$(X, \mathcal F, P)$$ be a probability space.

Question

What kind of condition is this: there exists a sequence $$(a_n)_n \subseteq X$$ such that

$$\forall$$ measurable $$A \subseteq X$$, $$P(A) > 0 \implies a_n \in A$$ for some $$n$$.

Context. I'm reviewing a paper for a conference and the authors assume this condition on the probability space in order to prove a result. I was wondering if this condition has a technical name, etc. in the greater math literature.

• The measure is a countable combination of atoms. – Fedor Petrov Aug 9 at 8:25
• @FedorPetrov Thanks, I should have thought of this. It's rather obvious. Indeed, such a measure must be discrete, and so the condition is extremely restrictive. – dohmatob Aug 9 at 8:41
• Proof of Fedor's claim. We show that the set compliment $X\setminus (a_n)_n$ has zero measure. Indeed, $X \setminus (a_n)_n$ (assumed measurable!) contains no $a_n$, thus by the contrapositive of condition, we must have $P(X\setminus (a_n)_n) = 0$. Thus the support of $P$ is a contained in the sequence $(a_n)_n$, and so the former must be a countable combination of atoms $\Box$. I should probably close the question... – dohmatob Aug 9 at 9:41
• Actually the point $\{a_n\}$ may be non-measurable. But we may denote by $c_n$ the infimum of measures of all measurable sets containing $a_n$, this infimum is realized (since the countable intersection of the sequence of minimizing sets is measurable itself), and the set $A_n$ which realizes it is an atom. – Fedor Petrov Aug 9 at 9:58
• Thanks for the construction. I suspected measurability limitation of my argument; that's why I put the "(assumed measurable!)" in brackets :) – dohmatob Aug 9 at 10:04