# Dubins–Spanier theorems

The Dubins–Spanier theorems are several theorems in the theory of fair cake-cutting. They were published by Lester Dubins and Edwin Spanier in 1961. Although the original motivation for these theorems is fair division, they are in fact general theorems in measure theory.

## Setting

There is a set $U$, and a set $\mathbb {U}$ which is a sigma-algebra of subsets of $U$.
There are $n$ partners. Every partner $i$ has a personal value measure $V_{i}:\mathbb {U} \to \mathbb {R}$. This function determines how much each subset of $U$ is worth to that partner.
Let $X$ a partition of $U$ to $k$ measurable sets: $U=X_{1}\sqcup \cdots \sqcup X_{k}$. Define the matrix $M_{X}$ as the following $n\times k$ matrix:
    $M_{X}[i,j]=V_{i}(X_{j})$
This matrix contains the valuations of all players to all pieces of the partition.
Let $\mathbb {M}$ be the collection of all such matrices (for the same value measures, the same $k$, and different partitions):
    $\mathbb {M} =\{M_{X}\mid X{\text{ is a }}k{\text{-partition of }}U\}$
The Dubins–Spanier theorems deal with the topological properties of $\mathbb {M}$.

## Statements

If all value measures $V_{i}$ are countably-additive and nonatomic, then:
* $\mathbb {M}$ is a compact set;
* $\mathbb {M}$ is a convex set.
This was already proved by Dvoretzky, Wald, and Wolfowitz.

## Corollaries
### Consensus partition

A cake partition $X$ to k pieces is called a consensus partition with weights $w_{1},w_{2},\ldots ,w_{k}$ (also called exact division) if:
    $\forall i\in \{1,\dots ,n\}:\forall j\in \{1,\dots ,k\}:V_{i}(X_{j})=w_{j}$
I.e, there is a consensus among all partners that the value of piece j is exactly $w_{j}$.
Suppose, from now on, that $w_{1},w_{2},\ldots ,w_{k}$ are weights whose sum is 1:
    $\sum _{j=1}^{k}w_{j}=1$
and the value measures are normalized such that each partner values the entire cake as exactly 1:
    $\forall i\in \{1,\dots ,n\}:V_{i}(U)=1$
The convexity part of the DS theorem implies that:$^{:5}$
          If all value measures are countably-additive and nonatomic,
          then a consensus partition exists.
PROOF: For every $j\in \{1,\dots ,k\}$, define a partition $X^{j}$ as follows:
    $X_{j}^{j}=U$
    $\forall r\neq j:X_{r}^{j}=\emptyset$
In the partition $X^{j}$, all partners value the $j$-th piece as 1 and all other pieces as 0. Hence, in the matrix $M_{X^{j}}$, there are ones on the $j$-th column and zeros everywhere else.
By convexity, there is a partition $X$ such that:
    $M_{X}=\sum _{j=1}^{k}w_{j}\cdot M_{X^{j}}$
In that matrix, the $j$-th column contains only the value $w_{j}$. This means that, in the partition $X$, all partners value the $j$-th piece as exactly $w_{j}$.
Note: this corollary confirms a previous assertion by Hugo Steinhaus. It also gives an affirmative answer to the problem of the Nile provided that there are only a finite number of flood heights.

### Super-proportional division

A cake partition $X$ to n pieces (one piece per partner) is called a super-proportional division with weights $w_{1},w_{2},...,w_{n}$ if:
    $\forall i\in 1\dots n:V_{i}(X_{i})>w_{i}$
I.e, the piece allotted to partner $i$ is strictly more valuable for him than what he deserves. The following statement is Dubins-Spanier Theorem on the existence of super-proportional division $^{:6}$
Theorem — With these notations, let $w_{1},w_{2},...,w_{n}$ be weights whose sum is 1, assume that the point $w:=(w_{1},w_{2},...,w_{n})$ is an interior point of the (n-1)-dimensional simplex with coordinates pairwise different, and assume that the value measures $V_{1},...,V_{n}$ are normalized such that each partner values the entire cake as exactly 1 (i.e. they are non-atomic probability measures). If at least two of the measures $V_{i},V_{j}$ are not equal ( $V_{i}\neq V_{j}$), then super-proportional divisions exist.
The hypothesis that the value measures $V_{1},...,V_{n}$ are not identical is necessary. Otherwise, the sum $\sum _{i=1}^{n}V_{i}(X_{i})=\sum _{i=1}^{n}V_{1}(X_{i})>\sum _{i=1}^{n}w_{i}=1$ leads to a contradiction.
Namely, if all value measures are countably-additive and non-atomic, and if there are two partners $i,j$ such that $V_{i}\neq V_{j}$, then a super-proportional division exists.I.e, the necessary condition is also sufficient.

## Sketch of Proof

Suppose w.l.o.g. that $V_{1}\neq V_{2}$. Then there is some piece of the cake, $Z\subseteq U$, such that $V_{1}(Z)>V_{2}(Z)$. Let ${\overline {Z}}$ be the complement of $Z$; then $V_{2}({\overline {Z}})>V_{1}({\overline {Z}})$. This means that $V_{1}(Z)+V_{2}({\overline {Z}})>1$. However, $w_{1}+w_{2}\leq 1$. Hence, either $V_{1}(Z)>w_{1}$ or $V_{2}({\overline {Z}})>w_{2}$. Suppose w.l.o.g. that $V_{1}(Z)>V_{2}(Z)$ and $V_{1}(Z)>w_{1}$ are true.
Define the following partitions:
* $X^{1}$: the partition that gives $Z$ to partner 1, ${\overline {Z}}$ to partner 2, and nothing to all others.
* $X^{i}$ (for $i\geq 2$): the partition that gives the entire cake to partner $i$ and nothing to all others.
Here, we are interested only in the diagonals of the matrices $M_{X^{j}}$, which represent the valuations of the partners to their own pieces:
* In ${\textrm {diag}}(M_{X^{1}})$, entry 1 is $V_{1}(Z)$, entry 2 is $V_{2}({\overline {Z}})$, and the other entries are 0.
* In ${\textrm {diag}}(M_{X^{i}})$ (for $i\geq 2$), entry $i$ is 1 and the other entires are 0.
By convexity, for every set of weights $z_{1},z_{2},...,z_{n}$ there is a partition $X$ such that:
    $M_{X}=\sum _{j=1}^{n}{z_{j}\cdot M_{X^{j}}}.$
It is possible to select the weights $z_{i}$ such that, in the diagonal of $M_{X}$, the entries are in the same ratios as the weights $w_{i}$. Since we assumed that $V_{1}(Z)>w_{1}$, it is possible to prove that $\forall i\in 1\dots n:V_{i}(X_{i})>w_{i}$, so $X$ is a super-proportional division.

### Utilitarian-optimal division

A cake partition $X$ to n pieces (one piece per partner) is called utilitarian-optimal if it maximizes the sum of values. I.e, it maximizes:
    $\sum _{i=1}^{n}{V_{i}(X_{i})}$
Utilitarian-optimal divisions do not always exist. For example, suppose $U$ is the set of positive integers. There are two partners. Both value the entire set $U$ as 1. Partner 1 assigns a positive value to every integer and partner 2 assigns zero value to every finite subset. From a utilitarian point of view, it is best to give partner 1 a large finite subset and give the remainder to partner 2. When the set given to partner 1 becomes larger and larger, the sum-of-values becomes closer and closer to 2, but it never approaches 2. So there is no utilitarian-optimal division.
The problem with the above example is that the value measure of partner 2 is finitely-additive but not countably-additive.
The compactness part of the DS theorem immediately implies that:$^{:7}$
          If all value measures are countably-additive and nonatomic,
          then a utilitarian-optimal division exists.
In this special case, non-atomicity is not required: if all value measures are countably-additive, then a utilitarian-optimal partition exists.$^{:7}$

### Leximin-optimal division

A cake partition $X$ to n pieces (one piece per partner) is called leximin-optimal with weights $w_{1},w_{2},...,w_{n}$ if it maximizes the lexicographically-ordered vector of relative values. I.e, it maximizes the following vector:
    ${V_{1}(X_{1}) \over w_{1}},{V_{2}(X_{2}) \over w_{2}},\dots ,{V_{n}(X_{n}) \over w_{n}}$
where the partners are indexed such that:
    ${V_{1}(X_{1}) \over w_{1}}\leq {V_{2}(X_{2}) \over w_{2}}\leq \dots \leq {V_{n}(X_{n}) \over w_{n}}$
A leximin-optimal partition maximizes the value of the poorest partner (relative to his weight); subject to that, it maximizes the value of the next-poorest partner (relative to his weight); etc.
The compactness part of the DS theorem immediately implies that:$^{:8}$
          If all value measures are countably-additive and nonatomic,
          then a leximin-optimal division exists.

## Further developments

* The leximin-optimality criterion, introduced by Dubins and Spanier, has been studied extensively later. In particular, in the problem of cake-cutting, it was studied by Marco Dall'Aglio.