A result commonly, and [probably erroneously](https://mathoverflow.net/a/223245/16537), attributed to W. Sierpiński is that every non-atomic, countably additive, nonnegative measure $\mu: \Sigma \to \bf R$, where $\Sigma$ is a sigma-algebra on a set $S$, has the weak, and hence the strong, Darboux property, which means that for every $X \in \Sigma$ and $a \in [0, \mu(X)]$ there exists $Y \in \Sigma$ such that $Y \subseteq X$ and $\mu(Y) = a$.


The theorem is provable in ZFC, and I'm just curious to know if it is actually equivalent to the axiom of choice.


In either case, I would very much appreciate a reference where the question is answered: Before asking, I browsed through Howard and Rubin's *Consequences of the Axiom of Choice*, but couldn't find anything close to what I'm looking for.



Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?
This theorem follows from Dependent Choice, and thus is strictly weaker than the Axiom of Choice. Here is a proof using only DC. Fix $X\in\Sigma$ such that $\mu(X)>0$ and let $a\in(0,\mu(X))$. We will use DC to inductively construct a sequence of disjoint measurable subsets $Y\_n$ of $X$ such that $\mu(\bigcup Y\_n)=a$.


Having defined $Y\_1,\dots,Y\_{n-1}$ such that $\mu(Y\_1)+\dots+\mu(Y\_{n-1})<a$, define $Y\_n$ as follows. Let $Z=X\setminus(Y\_1\cup\dots\cup Y\_{n-1})$. Since $\mu$ is atomless, there exists a subset $W\subset Z$ such that $0<\mu(W)<\mu(Z)$. It is also clear that there exists such $W$ with $\mu(W)$ arbitrarily small (we can get $\mu(W)\leq \mu(Z)/2$ by replacing $W$ with $Z\setminus W$ if necessary, and now iterate); in particular, there exist such $W$ with $\mu(Y\_1)+\dots+\mu(Y\_{n-1})+\mu(W)<a$. Now choose $Y\_n$ to be such a $W$ which additionally has the property that for any other such $W'$, $\mu(Y\_n)\geq \mu(W')/2$ (that is, among the possible measures of such sets $W$, $Y\_n$ is in the upper half).


Let $Y=\bigcup Y\_n$. By construction, the $Y\_n$ are disjoint and satisfy $\sum \mu(Y\_n)\leq a$, so $\mu(Y)\leq a$. Suppose $\mu(Y)<a$. As above, there exists a subset $W\subset X\setminus Y$ such that $\mu(W)>0$ and $\mu(W)+\mu(Y)<a$. Now let $n$ be such that $\mu(Y\_n)<\mu(W)/2$ (such $n$ exists since $\sum \mu(Y\_n)$ is finite). Then $W$ is a subset of $X\setminus(Y\_1\cup\dots\cup Y\_{n-1})$ such that $\mu(Y\_1)+\dots+\mu(Y\_{n-1})+\mu(W)<a$, so by our choice of $Y\_n$, we must have $\mu(Y\_n)\geq \mu(W)/2$. This is a contradiction, so we must have $\mu(Y)=a$.


