Measurable representations of preference orders

Abstract:

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