# Transactions of the American Mathematical Society

Published by the American Mathematical Society, the Transactions of the American Mathematical Society (TRAN) is devoted to research articles of the highest quality in all areas of pure and applied mathematics.

ISSN 1088-6850 (online) ISSN 0002-9947 (print)

The 2020 MCQ for Transactions of the American Mathematical Society is 1.43.

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by R. Daniel Mauldin
Trans. Amer. Math. Soc. 275 (1983), 761-769 Request permission

## 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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