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Let $\mu$ be a non-negative finitely additive measure. Let m be a countably additive measure.

Definition0: All singletons are $\mu-$measure zero.

Definition1: For each $A \in \mathscr{B}$, $\mu(A)> 0$, there exists $ B \subset A$ such that $0 < \mu(B) < \mu(A)$.

For countably additive measure these two seems to be equivalent. How about finitely additive measures?

My guess: Definition 1 implies 0, but 0 cannot imply 1.

High GPA
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  • Does this help? https://math.stackexchange.com/questions/204842/example-for-finitely-additive-but-not-countably-additive-probability-measure – Ethan Bolker Mar 04 '25 at 19:44
  • What is $\mathscr{B}$? – Michael Greinecker Mar 04 '25 at 22:31
  • Those definitions are not equivalent, even for countably additive measure. Let $X$ be an uncountable set, let $\Sigma$ be the countable-cocountable $\sigma$-algebra and let $\mu$ be defined on $\Sigma$ by $\mu(A)=0$ if $A$ is countable and $\mu(A)=1$ is $A$ is cocountable (that is $X \setminus A$ is countable). It is easy to see that $\mu$ is a countably additive measure and it satisfies definition0 but not definition1. – Ramiro Mar 05 '25 at 05:15
  • @Ramiro Is it true that definition 1 implies definition 0 then? – High GPA Mar 06 '25 at 06:33
  • @HighGPA In the example, singletons are not even measurable- which brings me to my unanswered question. – Michael Greinecker Mar 06 '25 at 07:29
  • @HighGPA, In the example I gave, the singletons are measurable (because they are countable subsets of $X$). Now, regarding your question, the answer is yes, definition $1$ implies definition $0$ IF the singletons are in $\mathscr{B}$. – Ramiro Mar 06 '25 at 13:23
  • @HighGPA Here is the proof. Suppose the measure space satisfies definition $1$ and all singletons are in $\mathscr{B}$. Then if any singleton $A$ would have $\mu(A) > 0$ it would contradict definition $1$ because any singleton $A$ has only two subsets: the empty set and itself. So for any $ B \subset A$, either $B= \emptyset$ and $\mu(B)=0$ or $B=A$ and $\mu(B)=\mu(A)$. – Ramiro Mar 06 '25 at 13:35

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