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Sierpiński's theorem states that nonatomic probability measures take a continuum of values. What if I assume that $\mu$ is a countably additive probability measure on $(X,2^X)$ and further that $\mu(\{x\})=0$ for all $x\in X$ (a weaker assumption than non-atomicity). Does it follow that $\mu$ takes on every value in $[0,1]$?

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    $\begingroup$ If $|X|$ is a measurable cardinal, then there is a $\sigma$-complete ultrafilter $\mathcal U$ on $X$. You may define a countably additive probability measure $\mu$ on $(X,2^X)$ by setting $\mu(Y) = 1$ if $Y \in \mathcal U$ and $\mu(Y) = 0$ if $Y \notin \mathcal U$. This measure takes only two values and assigns measure $0$ to every singleton. Thus a partial answer to your question is "assuming large cardinals, no it doesn't follow." I'm posting this as a comment rather than an answer in the hopes that someone more knowledgable can find a counterexample without using large cardinals. $\endgroup$ – Will Brian Jan 17 at 14:14
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If $|X|$ is a measurable cardinal, then there is a $\sigma$-complete ultrafilter $\mathcal U$ on $X$. You may define a countably additive probability measure $\mu$ on $(X,2^X)$ by setting $\mu(Y) = 1$ if $Y \in \mathcal U$ and $\mu(Y) = 0$ if $Y \notin \mathcal U$. This measure takes only two values (hence is atomic) and assigns measure $0$ to every singleton. Thus the answer to your question is "no it does not follow" (assuming the existence of a measurable cardinal).

Furthermore, we can show that the measurable cardinal is necessary, in the sense that if there is an example $\mu$ of an atomic measure having the properties you describe, then there is a measurable cardinal. To see this, first note that any such measure $\mu$ must be atomic, by the theorem you quoted in your post. Fix $Y \subseteq X$ such that $\mu(Y) = c > 0$ and if $Z \subseteq Y$ then either $\mu(Z) = 0$ or $\mu(Z) = c$. (This is what it means for a measure to be atomic.) Letting $\mathcal U = \{Z \subseteq Y : \mu(Z) = c\}$, it is not difficult to show that $\mathcal U$ is a $\sigma$-complete ultrafilter on $Y$. This shows that $|Y|$ is $\geq$ the least measurable cardinal.

Thus a fuller answer to your question is "no it doesn't follow . . . or if it does, then the existence of measurable cardinals is inconsistent."

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