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Measurable representations of preference orders


Author: R. Daniel Mauldin
Journal: Trans. Amer. Math. Soc. 275 (1983), 761-769
MSC: Primary 90A06; Secondary 04A15, 28C15, 54H05
DOI: https://doi.org/10.1090/S0002-9947-1983-0682730-4
MathSciNet review: 682730
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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

DOI: https://doi.org/10.1090/S0002-9947-1983-0682730-4
Keywords: Preference order, continuous order preserving map, universally measurable, analytic set
Article copyright: © Copyright 1983 American Mathematical Society

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