Compactness and existence of Pareto-efficient cake partitions I am trying to understand a fundamental statement in the theory of cake-cutting.
BACKGROUND:
There is a certain "cake" $C$ (a subset of $R^n$).
The cake is divided among two agents, 0 and 1. Each agent receives a piece $A_i$, with $A_0\sqcup A_1=C$.
Each agent has a certain non-atomic value measure, $m_i$, over pieces of $C$. So for every partition of the cake to $\{A_0,ֻA_1\}$, the values enjoyed by the agents are $m_0(A_0)$ and $m_1(A_1)$. The measures are all in the range $[0,1]$.
A partition $\{B_0,B_1\}$ Pareto-dominates $\{A_0,A_1\}$ if $m_0(B_0)\geq m_0(A_0)$ and $m_1(B_1)\geq m_1(A_1)$, with at least one of the inequalities being strict.
An efficient partition is a partition that is not Pareto-dominated by any other partition. I.e., it is a partition that cannot be improved for one agent without hurting the other agent.
A famous paper in the theory of cake-cutting (Berliant et al, 1992) says that "if each agent's preferences can be represented by a non-atomic measure, then efficient partitions exist. This follows immediately from the compactness part of the Lyapunov theorem", which says that "the range of any real-valued, non-atomic vector measure is compact".
MY QUESTION: How does the existence of efficient partitions follows from the compactness of the range of vector measures?
The "vector measure" he refers to is probably the measure which assigns, to every piece of cake $A$, the valuations of the agents to that piece: $M(A)=(m_0(A),m_1(A)$. The range of this vector measure is a compact subset in $R^2$. 
An efficient partition is a maximal element of the partial order defined by "Pareto-dominates". So, my guess was that the existence of efficient partitions is somehow related to existence of maximal elements in compact sets. But, I don't know if this is true and if so, how to prove it?
EDIT: I read the main paper cited by the authors (Dubins&Spanier, 1961). That paper indeed does not rely on a compactness theorem on ranges of vector measures, but rather specifically proves the compactness of the matrix space containing the following matrices:
$$(m_i(A_j))_{i=0..1,j=0..1}$$
where the $\{A_0,ֻA_1\}$'s run over all possible partitions.
I concated Prof. Marcus Berliant for more explanations.
 A: I'm not sure what vector measure is being referenced by the authors you mention. In order for me to explain my doubts, let's establish some measure-theoretic preliminaries: The measures $m_0$ and $m_1$ "eat" subsets of $\mathbb{R}^n$ and "spit out" real numbers in $[0,1]$. The subsets they eat lie in a $\sigma$-algebra $\mathcal{M} \subset \mathcal{P}(\mathbb{R}^n)$, where $\mathcal{P}(\mathbb{R}^n)$ is the power set of $\mathbb{R}^n$. 
Note that choosing $A_0$ already determines $A_1$, since $A_1$ is the complement of $A_0$ in $C$ (i.e. $A_1= C \setminus A_0$). Thus the function you described, i.e. 
\begin{equation*}\tag{1}
\{A_0,A_1\} \mapsto (m_0(A_0),m_1(A_1)),
\end{equation*}
is equivalent to a function $A \mapsto (m_0(A),m_1(C\setminus A))$. Unfortunately, this function is not a vector measure (it doesn't satisfy finite additivity, since disjoint unions make the first entry grow and the second entry shrink). One could hope that the map in equation (1) is the restriction of some measure on a $\sigma$-algebra containing $\mathcal{M}\times \mathcal{M}$. Note that $\mathcal{M} \times \mathcal{M}$ is almost never a $\sigma$-algebra (it needs to be enlarged to be closed under unions and complements). I'm not sure what this mysterious measure would be.
My suggestion: See if you can figure out how the authors are defining their measure (and on what collection of sets the measure is defined). Then update your question with this information and see if there are any new leads. 
