Measurable representations of preference orders

Abstract

A continuous preference order on a topological space Y Y is a binary relation ≼ \preccurlyeq which is reflexive, transitive and complete and such that for each x , { y : x ≼ y } x,\{y:x \preccurlyeq y\} and { y : y ≼ x } \{y:y \preccurlyeq x\} are closed. Let T T and X X be complete separable metric spaces. For each t t in T T , let B t {B_t} be a nonempty subset of X X , let ≼ t { \preccurlyeq _t} be a continuous preference order on B t {B_t} and suppose E = { ( t , x , y ) : x ≼ t y } E = \{(t,x,y): x{ \preccurlyeq _t}y\} is a Borel set. Let B = { ( t , x ) : x ∈ B t } B = \{(t,x):x \in {B_t}\} . Theorem 1. There is an S ( T ) ⊗ B ( X ) \mathcal {S}(T) \otimes \mathcal {B}(X) -measurable map g g from B B into R R so that for each t , g ( t , ⋅ ) t,g(t,\cdot ) is a continuous map of B t {B_t} into R R and g ( t , x ) ⩽ g ( t , y ) g(t,x) \leqslant g(t,y) if and only if x ≼ t y x{ \preccurlyeq _t}y . (Here S ( T ) \mathcal {S}(T) forms the C C -sets of Selivanovskii and B ( X ) \mathcal {B}(X) is a Borel field on X X .) Theorem 2. If for each t , B t t,{B_t} is a σ \sigma -compact subset of Y Y , then the map g g of the preceding theorem may be chosen to be Borel measurable. The following improvement of a theorem of Wesley is proved using classical methods. Theorem 3. Let g g be the map constructed in Theorem 1. If μ \mu is a probability measure defined on the Borel subsets of T T , then there is a Borel set N N such that μ ( N ) = 0 \mu (N) = 0 and such that the restriction of g g to B ∩ ( ( T − N ) × X ) B \cap ((T - N) \times X) is Borel measurable.