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Let $\cal{S}$ be a $\sigma$-field of subsets of a set $Z$, and $\mu$ be a positive finite measure on $\cal{S}$ which does not contains atoms. (An atom in $(X, \cal{S}, \mu)$ is a measurable set $E$ such that $\mu(E)>0$ and for every measurable subset $F\subset E$ either $\mu(F)=0$ or $\mu(F)=\mu(E)$.) P. Halmos ["On the set of values of finite measure", Bull. Amer. Math. Soc. 53, No 2,(1947), 138-141] proved that the set of values of $\mu$ such as above, is the closed interval $[0, \mu(Z)]$. The proof consists of two parts. First he showed that every measurable set $E \subset Z$ of positive measure contains measurable subset of arbitrary small positive measure. ($E$ is not an atom, hence there exists $E \in \cal{S}$ such that $0<\mu(F)<\mu(E)$.Write $E_1$ for that one of two sets $F$, $E\setminus F$ whose measure is not greater than $\frac{\mu(E)}{2}$. Similarly there exists $E_2 \subset E_1$ such that $0<\mu(E_2)\leq \frac{\mu(E_1)}{2}$, and proceed by induction.) The second part go in the following way: If $\mu(Z)=0$ is OK. If $0<\alpha <\mu(Z)$ we may find a measurable set $E_1 \subset Z$ such that $0<\mu(E_1) \leq \alpha$. If is equality is OK, if not we may find a measurable set $E_2 \subset Z \setminus E_1$ such that $0<\mu(E_2)\leq \alpha -\mu(E_1)$. I don't understand the part below. Halmos said that by transfinite induction, if necessary, we can obtain a countable sequence of pairwise disjoint measurable sets the union of which has measure $\alpha$. The method used in the last fragment is called "method of exhaustion".

My questions are: what is the "method of exhaustion" and how to do the last part of the above proof, maybe with the Lemma of Zorn.

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    I think you meant to say "$E$ is not an atom, hence there exists $\underline F$ (not $E$) in $\mathcal S$ such that $0 < \mu(F) < \mu(E)$".2011-08-06
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    Do you know what transfinite induction is? The idea is to inductively build a sequence of pairwise disjoint sets of positive measure, indexing by ordinals as you go (at limits subtract the union of what you've built). Since you can't have uncountably many pairwise disjoint subsets of positive measure, this process terminates at some countable ordinal. Then you can just union up what you've built to get the desired measurable set.2011-08-06

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