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A result commonly, and probably erroneously, 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.

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    $\begingroup$ I haven't written down a full proof but I don't see why this result would use any more than dependent choice. $\endgroup$ Dec 9, 2015 at 17:08
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    $\begingroup$ I don't know. The proof I know is essentially the same as the one worded here: math.stackexchange.com/a/1398683/20748, and I haven't tried to make it work with the DC only. $\endgroup$ Dec 9, 2015 at 17:20
  • $\begingroup$ Story: I first came across this theorem in an undergrad measure theory course. It was from a list of 10 challenge problems that we could turn in at the end of the semester to get an A. (Some were quite difficult. For example, one of the problems was to show every Borel set was countable or size continuum.) I figured out the Zorn's lemma proof eventually and turned it in. The professor didn't like it, and kept telling me I could do it without ZL. Eventually he took it (and for some naive reason, I thought "without ZL" meant using transfinite induction). (Continued in next comment...) $\endgroup$
    – Jason Rute
    Dec 10, 2015 at 2:04
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    $\begingroup$ Fast forward a few years into grad school, I was helping a friend with their measure theory homework. This problem came up and I suggested that they should use Zorn's Lemma. They became quite angry insisting that no problem (in this class) should ever require AC. We had a fight over it. Eventually, it turned out I was wrong and the professor "of course" wanted them to do it without the axiom of choice. (Now, I realize dependent choice was used in the proof, but non-logicians generally use that without notice.) I don't know if this has a moral, but I am much more careful about using AC now. $\endgroup$
    – Jason Rute
    Dec 10, 2015 at 2:13

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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 $\mu(Y)=a$.

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    $\begingroup$ The next step is to prove that you need at least some choice. For example, we might show that the conclusion fails in one of the standard models of $\neg\text{AC}$. Better, would be a reversal of the argument you give from DC. $\endgroup$ Dec 9, 2015 at 22:22

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