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.
 A: 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 $W$.
Feel free to comment because I am not 100% sure of what I'm stating.
Hope that helps,
