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.