# set of values of finite measure, exhaustion method

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 $0 < \mu(F) < \mu(E)$". – Patrick Da Silva Aug 6 '11 at 12:43
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. – user83827 Aug 6 '11 at 15:07

It probably has to do with Zorn's Lemma. The reason why I think this argument is using it is because he starts with a set $E_1$, and then adds a disjoint set $E_2$ such that the measure of $E_2$ is bounded from above, but not from below, which does not ensure the convergence of $$\sum_{n=1}^{\infty} \mu(E_n) = \mu\left( \sum_{n=1}^{\infty} E_n \right)$$ towards $\alpha$ (which is what we want in the end). The reason why I would use Zorn's Lemma is because you can consider the set $$W = \left\{ \{E_n : n \in \mathbb N\} | E_i \cap E_j = \varnothing, E_n \in \mathcal S, \mu \left( \sum_{n=1}^{\infty} E_n \right) \le \alpha. \right\}$$ equipped with the following partial order : say that $\{E_n\} \le \{F_n\}$ as elements of $W$ if $E_n \subseteq F_n$ for all $n \in \mathbb N$. I am not sure about this part, but I believe that in $W$ every chain has an upper bound, because if $K$ is a set of indexes for the elements of the chain $\{E_n\}_k$, you could consider the following element of $W$ : $$\left\{ \bigcup_{k \in K} E_n \right\} = F_n$$ If you define a total order on $K$ relatively to the total order in the chain, I believe something like this can hold : clearly $\{F_n\}$ is an upper bound for every element in the chain, the sets $F_n$ are pairwise disjoint, and to see that they're in $\mathcal S$, it suffices to take a countable sequence of indexes in $K$, call it $k_n$, such that for every $k \in K$, there exists $n \in \mathbb N$ such that $k_n \le k \le k_{n+1}$. This allows us to write $F_n$ as a countable and increasing (in the sense of set inclusion) union instead of an arbitrary union, so that a countable union of elements of a $\sigma$-field remains in the $\sigma$-field. (I think this is possible, but I am not sure... If someone could clear up this part I'd be pleased, I am not sure I can take a countable sequence out of a total order and go to "infinity" in the order, i.e. go as "high" as one could need. That's what I need in the argument... if this doesn't hold then I don't know where to go from here, sorry).
It remains to see that $\{F_n \} \in W$, i.e. that $$\sum_{n=1}^{\infty} \mu(F_n) = \sum_{n=1}^{\infty} \mu \left( \bigcup_{k \in K} E_n \right)= \mu \left( \sum_{n=1}^{\infty} \bigcup_{k \in K} E_n \right) = \mu \left( \bigcup_{k \in K} \sum_{n=1}^{\infty} E_n \right) = \sup_{k \in K} \, \mu\left( \sum_{n=1}^{\infty} E_n \right) \le \alpha$$ This implies every chain in $W$ has an upper bound in $W$, so that you can use Zorn's Lemma to find a maximal element in $W$, call it $\{H_n\}$. If $$\mu \left( \sum_{n=1}^{\infty} H_n \right) = \beta < \alpha,$$ it means you can find some set $J$ in $Z \backslash \left( \bigcup_{n=1}^{\infty} H_n \right)$ such that $0 < \mu(J) \le \alpha-\beta$, and then if you add the elements of $J$ to, say $H_1$, to form a new element of $W$, this contradicts the fact that $\{H_n\}$ is a maximal element in $W$.
I've posted a question about my problem over the fact that $F_n$ is required to be in the $\sigma$-field. It seems my construction cannot hold. – Patrick Da Silva Aug 6 '11 at 14:37