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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.

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    $\begingroup$ The measure is a countable combination of atoms. $\endgroup$
    – Fedor Petrov
    Commented Aug 9, 2019 at 8:25
  • $\begingroup$ @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. $\endgroup$
    – dohmatob
    Commented Aug 9, 2019 at 8:41
  • $\begingroup$ 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... $\endgroup$
    – dohmatob
    Commented Aug 9, 2019 at 9:41
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    $\begingroup$ 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. $\endgroup$
    – Fedor Petrov
    Commented Aug 9, 2019 at 9:58
  • $\begingroup$ Thanks for the construction. I suspected measurability limitation of my argument; that's why I put the "(assumed measurable!)" in brackets :) $\endgroup$
    – dohmatob
    Commented Aug 9, 2019 at 10:04

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