Measurable representations of preference orders
Author:
R. Daniel Mauldin
Journal:
Trans. Amer. Math. Soc. 275 (1983), 761769
MSC:
Primary 90A06; Secondary 04A15, 28C15, 54H05
DOI:
https://doi.org/10.1090/S00029947198306827304
MathSciNet review:
682730
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Abstract  References  Similar Articles  Additional Information
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.

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Additional Information
Keywords:
Preference order,
continuous order preserving map,
universally measurable,
analytic set
Article copyright:
© Copyright 1983
American Mathematical Society