Darboux property of non-atomic sigma-additive nonnegative measures equivalent to the AC?

Question

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.

Answer 1

Here is a proof that only uses countable choice. It is taken from Fremlin, Measure Theory, Volume 5, Number 566F. The chapter 56 of this multi-volume work is a great source on the discussion of choice and measure theory. $\renewcommand{\subset}{\subseteq}$

Let $(X,\mathcal A,\mu)$ be a finite measure space. In the following, all sets are and can be taken from $\mathcal A$.

(0) Let $A \subset X$. Then for all $\epsilon>0$ there is $B \subset A$ such that $\mu(B) \in (0,\epsilon)$. This can be proven by induction: by definition, $A$ has a subset $B$ with $\mu(B) \in (0,\mu(A))$. Then either $B$ or $B\setminus A$ have measure smaller than $\frac12 \mu(A)$.

(1) Let $A \in X$. Then there is $B \subset A$ such that $\frac13\mu(A) < \mu(B) \le \frac23 \mu(A)$. Suppose not. Then $$ \gamma := \sup \left\{ \mu(B) : \mu(B) \le \frac23 \mu(A) \right\} \le \frac13\mu(A). $$ Due to step (0), $\gamma >0$. Let $B_n$ be such that $\gamma - 2^{-n} \le \mu(B_n) \le \frac23 \mu(A)$. It follows $\mu(B_n) \le \gamma$. Define $C_n :=\bigcup_{j=1}^n B_n$. Per induction it follows $\mu(C_n) \le \frac23 \mu(A)$ and $\mu(C_n) \le \gamma$. Let $D:= \bigcup_{j=1}^\infty C_n$. Then $\mu(D) = \gamma$. By step (0), there is $E \subset X \setminus D$ such that $\mu(E) \in (0, \frac13\mu(A))$. Then $\gamma< \mu(D \cup E) \le \frac23 \mu(A) $. Contradiction.

(2) For every $n$, there are finitely many sets $A_1 \dots A_m$ with $\mu(A_i) \le \left( \frac23 \right)^n$ and $\bigcup_{j=1}^m A_j = X$. Proof by induction using step (1). Set $\mathcal C_n := (A_1 \dots A_m)$.

(3) For every $n$, construct these partitions $\mathcal C_n$ as in step (2). Uses only countable choice, as partition $\mathcal C_{n+1}$ does not depend on $\mathcal C_n$ but is constructed independently in step (2). Let $(C_n)$ be an enumeration of $ \bigcup_{n=1}^\infty \mathcal C_n$.

(4) Now let $A \subset X$, $t\in (0, \mu(A))$. Define sets $F_n$ and $G_n$ by $$ F_0 = \emptyset, \quad G_0 = A. $$ $$ F_{n+1} = \begin{cases} F_n \cup ( G_n \cap C_n) &\text{ if } \mu(F_n \cup (G_n \cap C_n)) \le t\\ F_n & \text{ otherwise } \end{cases} $$ and $$ G_{n+1} = \begin{cases} F_n \cup ( G_n \cap C_n) &\text{ if } \mu( F_n \cup ( G_n \cap C_n)) \ge t \\ G_n & \text{ otherwise}. \end{cases} $$ No choice involved. Then it follows by induction $F_n \subset F_{n+1} \subset G_{n+1} \subset G_n$. Define $F:= \bigcup_{n=0}^\infty F_n$ and $G := \bigcap_{n=0}^\infty G_n$. Then $\mu(F) \le t \le \mu(G)$.

Suppose $\mu(F) < t$. Then there is $C_n$ such that $\mu( C_n \cap (G\setminus F)) \in (0, t- \mu(F))$. This implies $$ \mu(F_n \cup ( G_n \cap C_n)) \le \mu( F\cup ( G_n \cap C_n)) \le \mu(F) + \mu(G_n \cap C_n) \le t. $$ Hence by the construction of $F_{n+1}$ we get $G_n \cap C_n \subset F_{n+1} \subset F$, which is impossible.

It follows $ \mu(F)= t$, which proves the claim.

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Actually, Fremlin proves a much stronger result. He constructs a function $F: \mathcal A \times [0,\mu(X)] \to \mathcal A$ such that $$ F(A, t) \subset A , \quad \mu( F(A,t)) \le \min(t, \mu(A)) $$ and $t \mapsto F(A,t)$ is monotonically increasing.

Answer 2

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 $\mu(Y)=a$.