# Variant of Sierpiński's result on non-atomic measures

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 $$[0,1]$$?

• 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. – Will Brian Jan 17 at 14:14

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.