A SURVEY OF
MEASURABLE SELECTION THEOREMS
S.M.SRIVASTAVA
INDIAN STATISTICAL INSTITUTE,KOLKATA
Lectures delivered at University of Dortmund,Germany
April 14,16,2010
2
1.Introduction.
Measurable selection theorems are useful as well as interesting in
their own rights.They have been used widely,so much so that many
important selection theorems have been discovered by their users.The
oretically they form a very important topic in descriptive set theory.
It won’t be much of an exaggeration to say that they form the heart
of classical descriptive set theory.In these lectures we shall give an
overview of important measurable selection theorems.
Let X and Y be arbitrary sets.A multifunction F:X →Y is a
function with domain X and values nonempty subsets of Y.
The set
{(x,y) ∈ X ×Y:y ∈ F(x)}
is called the graph of F.
For any set A ⊂ Y,we deﬁne
∃
A
(F) = {x ∈ X:∃y ∈ F(x)(y ∈ A)}
and
∀
A
(F) = {x ∈ X:∀y ∈ F(x)(y ∈ A)}.
So,
∃
A
(F) = {x ∈ X:F(x) ∩A = ∅}
and
∀
A
(F) = {x ∈ X:F(x) ⊂ A = ∅}.
Now let A be a family of subsets of X and Y a metrizable space.We
say that F is Ameasurable if ∃
U
(F) ∈ A for every open set U in Y;
F is strongly Ameasurable if ∃
C
(F) ∈ A for every closed set C in
Y.Thus,
F is Ameasurable if for every open set U in Y,
{x ∈ X:F(x) ∩U = ∅} ∈ A
;it is strongly Ameasurable if for every closed set C in Y,the set
{x ∈ X:F(x) ∩C = ∅} ∈ A.
In particular,a point map f:X → Y is called Ameasurable if
f
−1
(U) ∈ A for every open set U in Y.
A function s:X →Y is called a selector for F if s(x) ∈ F(x) for
every x ∈ X.
By axiom of choice,a multifunction always admit a selector.
But when does a measurable multifunction admit a measurable se
lector?This is a fundamental question and it crops up at several places.
In this series of talks we shall give a survey of major selection theorems.
3
In applications,usually A is a σalgebra.However,there are inter
esting cases where it is not so.
If X is a topolgical space and A is the family of all open sets,A
measurable multifunctions are called lower semicontinuous.So,F
is lower semicontinuous if and only if for every open set U in Y,
{x ∈ X:F(x) ∩U = ∅}
is open in X.
Similarly,F is called upper semicontinuous if for every closed
set C in Y,
{x ∈ X:F(x) ∩C = ∅}
is closed in X.
Note that F is upper semicontinuous if and only if for every open
set U in Y,∀
U
(F) = {x ∈ X:F(x) ⊂ U} is open in X.
Another interesting case is obtained as follows.Let X be a metrizable
space.A countable union of closed sets in X is called an F
σ
set in X
and a countable intersection of open sets in X is called a G
δ
.It is a
standard fact that every open (closed) set in a metrizable space X is a
F
σ
(G
δ
) set in X.Thus every open and every closed set in a metrizable
space is simultaneously F
σ
and G
δ
.It is also clear that the set Δ
0
2
of all
sets which are simultaneously F
σ
and G
δ
is an algebra on X.Also note
that set of all F
σ
sets coincides with the family of countable unions
of sets in Δ
0
2
.It is clear that all lower and all upper semicontinuous
multifunctions are F
σ
measurable.If X and Y are metrizable spaces,
a function f:X →Y is called a class 1 function if for every open set
U in X,f
−1
(U) is an F
σ
set.
2.Preliminaries.
A second countable,completely metrizable space is called a Polish
space.This is the category of topological spaces on which deep and
useful descriptive set theory takes place.
An algebra L on a set X is a nonempty family of subsets of X
which is closed under complementations and ﬁnite unions.Note that
if A ∈ L,X = A∪(X\A) ∈ L.So,X and ∅ belong to all algebras on
X.
If an algebra is also closed under countable unions,it called a σ
algebra.
A measurable space is a pair (X,A) with X a nonempty set and
A a σalgebra on X.
Lemma 2.1.Let X be a nonempty set,Y a metrizable space and
A be a family of subsets of X which is closed under countable unions.
4
Then every strongly Ameasurable multifunction F:X → Y is A
measurable.
Proof.Let U be an open set in Y.As Y is metrizable,U = ∪
n
C
n
,
C
n
’s closed in Y.
{x ∈ X:F(x) ∩U = ∅} = ∪
n
{x ∈ X:F(x) ∩C
n
= ∅}.
This shows that
∃
U
(F) = ∪
n
∃
C
n
(F).
The result now follows.
A measure µ on a measurable space (X,A) is a map µ:A →[0,∞]
such that every sequence {A
n
} of pairwise disjoint sets in A,
µ(∪
n
A
n
) =
n
µ(A
n
).
The measure µ is called ﬁnite if µ(X) < ∞;σﬁnite if there is a
sequence {A
n
} of sets in A such that X = ∪
n
A
n
and for each n,
µ(A
n
) < ∞.It is called a probability if µ(X) = 1.A probability
space (measure space) is a triple (X,A,µ) such that Ais a σalgebra
on X and µ a probability (resp.measure) on X.
A measure space (X,A,µ) is called a complete measure space if
whenever µ(N) = 0 (such sets are called µnull) and M ⊂ N,M ∈ A.
For a metric space X,the smallest σalgebra containing all open
sets is called the Borel σalgebra of X,which we shall denote by B
X
.
Unless otherwise stated,a metric space will always be equipped with
its Borel σalgebra.Sets in B
X
are called Borel sets in X.
Further,if X and Y are metric spaces,a map f:X → Y is called
Borel measurable or simply Borel if
f
−1
(U) ∈ B
X
for every open set U in Y.It follows that
f
−1
(B) ∈ B
X
whenever B is Borel in Y.
A measure µ on B
X
,X metrizable,is called continuous if µ({x}) =
0 for every x ∈ X.
A Borel subset of a Polish space is called a standard Borel space.
Standard Borel spaces satisfy many regularity properties:
Theorem 2.2.Every uncountable standard Borel space is of cardi
nanility c,the continuum.In fact,they contain homeomorphs of the
Cantor set.
5
Theorem 2.3.If B is a Borel set in a Polish space X and if µ is
a ﬁnite measure on X,then for every > 0,there is a compact K and
an open U in X such that K ⊂ B ⊂ U and µ(U\K) < .
Theorem 2.4.Every standard Borel space is a continuous image of
N
N
.
Sketch of the proof of 2.4.One ﬁrst shows that every Polish
space is a continuous image of N
N
.Now ﬁx a Polish space X.We
show that the family of subsets of X which is a continuous image of N
N
contains all open sets and is closed under countable intersections and
countable disjoint unions.Finally,one shows that B
X
is the smallest
family of subsets of X that contains all open sets and is closed under
countable intersections and countable disjoint unions.
Theorem 2.5.(Borel isomorphism theorem) If X and Y are
two uncountable standard Borel spaces,then there is a bijection f:
X →Y such that both f and f
−1
are Borel.
Borel isomorphism theorem is an important result because it allows
one to choose a standard Borel space at its own convenience.Aparticu
larly simple proof of it was obtained by B.V.Rao and S.M.Srivastava.
Theorem 2.6.(Measure isomorphism theorem) Let X be an
uncountable standard Borel space and µ a continuous probability on X.
Then there is a Borel isomorphism f:X → [0,1] such that for every
Borel set B in [0,1],λ(B) = µ(f
−1
(B)),where λ denotes the Lebesgue
measure on [0,1].
A metrizable space Y is called an analytic space if there is a stan
dard Borel space X and a continuous surjection f:X →Y.
Analytic spaces play a fundamental role in measurability.Clearly
every standard Borel space is analytic and analytic spaces are separa
ble.
It is also known that
every uncountable Polish space contains an analytic nonBorel set.
Theorem 2.7.Let X be a Polish space and A ⊂ X.The following
conditions are equivalent:
(1) A is analytic.
(2) A is a continuous image of N
N
.
(3) There is a standard Borel space Y and a Borel surjection f:
Y →A.
6
(4) There is a standard Borel space Z such that A is the projection
of a Borel set B in X ×Z.So,for every x ∈ X,
x ∈ A ⇔∃z ∈ Z((x,z) ∈ B).
Proof.Since every standard Borel space is a continuous image of
N
N
(2.4),(2) follows from (1) by the deﬁnition of analytic sets.Since
the graph of a Borel map is Borel,(3) implies (4).Since projection is
a continuous map,(4) implies (1).
Let Z be any set.Asystemof sets in Z is a family {A(n
0
,...,n
k−1
)}
of subsets of Z indexed by the set of all ﬁnite sequences (n
0
,...,n
k−1
)
(including the empty sequence e) of natural numbers.The set
∪
α∈N
N ∩
k∈N
A(αk),
where N denotes the set of all natural numbers and for α ∈ N
N
,αk =
(α(0),...,α(k−1)),k ∈ N,the restriction of α to k,is called the result
of Souslin operation on {A(n
0
,...,n
k−1
)}.
Theorem 2.8.Let Z be a Polish space and Y ⊂ Z.The following
conditions are equivalent:
(1) Y is analytic.
(2) Y is the result of Souslin operation on a system of closed sets
in Z.
(3) Y is the result of Souslin operation on a system of Borel sets in
Z.
Analytic spaces enjoy many regularity properties.
Theorem 2.9.Let X be a Polish space and Y ⊂ X analytic.
(1) If Y is uncountable,it contains a homeomorph of the Cantor
set.In particular,every uncountable analytic space is of cardi
nality c.
(2) Y is measurable with respect every σﬁnite measure on X.In
other words,every analytic space is universally measurable.
(3) Y has the propery of Baire,i.e.,there is an open set U and
a set N of ﬁrst category in X such that Y = UΔN,where Δ
denotes the symmetric diﬀerence.
The following result is truly a ﬁne and a very useful one.Descriptive
set theory got started with it.It is due to Souslin.
Theorem 2.10.(Souslin) Let Z be a Polish space and A and B two
disjoint analytic subsets of Z.Then there is a Borel set C in Z such
7
that
A ⊂ C & B ∩C = ∅.
In particular,a subset Y of Z is Borel if and only if both Y and Z\Y
are analytic.
The following generalization of this theorem plays key role in selec
tion theory.
Theorem 2.11.Let Z be a Polish space and {A
n
} a sequence of
analytic sets in X such that ∩
n
A
n
= ∅.Then there exist Borel sets
B
n
⊃ A
n
such that ∩
n
B
n
= ∅.
A particularly elegant proof of 2.11 was obtained by Mokobodzki.
Theorem 2.12.Let Z be a Polish space and {A
n
} a sequence of
pairwise disjoint analytic sets in Z.Then there is a sequence {C
n
} of
pairwise disjoint Borel sets such that ∀n(C
n
⊃ A
n
).
Proof of 2.12.Using Souslin’s theorem (2.10),for each n,get a
Borel set C
n
such that C
n
⊃ A
n
and C
n
∩ ∪
m=n
A
m
= ∅.Now take
B
n
= C
n
\∪
m=n
C
m
.
Using 2.4 and 2.12,one proves the following important result.
Theorem 2.13 Let X be a standard Borel space,Y a Polish space
and f:X → Y a Borel map which is onetoone.Then f(X) is
standard Borel.It follows that f:X →f(X) is a Borel isomorphism.
Theorem 2.14.Let X,Y be Polish spaces,A ⊂ X and f:A →Y.
The following conditions are equivalent:
(1) A is Borel and f is Borel measurable.
(2) The graph of f is Borel in X ×Y.
We close this section by stating a corollary of 2.11 in the form that
we shall use.Let X be a Polish space.A subset C of X is called
coanalytic if X\C is analytic.The following simple result is fairly
useful.
Proposition 2.15.Let X be Polish and C ⊂ X.The following are
equivalent:
(1) C is coanalytic.
8
(2) There is a standard Borel space Z and a Borel B ⊂ X×Z such
that for every x ∈ X,
x ∈ C ⇔∀z ∈ Z((x,z) ∈ B).
2.15 is not at all hard to prove.
Theorem 2.16.Let X be a standard Borel space and {C
n
} a
sequence of coanalytic subsets with union X.Then there is a se
quence {B
n
} of pairwise disjoint Borel sets such that ∀n(B
n
⊂ C
n
)
and ∪
n
B
n
= X.
Proof.Let Y be a Polish space containing X.Set A
n
= X\C
n
.
Then {A
n
} is a sequence of analytic sets with ∩
n
A
n
= ∅.By 2.11,get
Borel sets D
n
such that X ⊃ D
n
⊃ A
n
and ∩
n
D
n
= ∅.Set E
n
= X\D
n
.
Then E
n
is Borel,E
n
⊂ C
n
and ∪
n
E
n
= X.Now take
B
n
= E
n
\∪
m<n
E
m
.
Here is a beautiful consequence of 2.15 that we shall use in selection
theory.
Theorem 2.17.(Dellacherie) Let X,Y be Polish spaces and B ⊂
X ×Y Borel with sections B
x
open in Y.Then we can write
B = ∪
n
(B
n
×U
n
),
B
n
’s Borel and U
n
’s open.
3.Kuratowski—Ryll Nardzewski Selection Theorem.
Kuratowski and Ryll Nardzewski proved a very general selection the
orem.It encompasses a lot of major selection theorems.Its proof is
simple at the same time beautiful.
Let X be a nonempty set and L an algebra of sets on X.Let L
σ
denote the smallest family of subsets of X containing L and closed
under countable unions.The following basic facts about L
σ
will be
used without proof.
Proposition 3.1.If {A
m
} is a sequence in L
σ
,then there is a
sequence {B
m
} of pairwise disjoint sets in L
σ
such that ∀m(B
m
⊂ A
m
)
and ∪
m
B
m
= ∪
m
A
m
.
Sketch of the proof.Write A
m
= ∪
n
A
mn
,A
mn
∈ L.Enumerate
the double sequence {A
mn
} in a single sequence,say {D
k
}.Set C
k
=
9
D
k
\∪
l<k
D
l
.Now take
B
n
= ∪{C
k
:C
k
⊂ A
n
∧ ∀m< nC
k
⊂ A
m
}.
The proof of the following proposition is quite routine.
Proposition 3.2.Let Y be a Polish space and {s
n
} a sequence of
L
σ
measurable functions from X to Y.If {s
n
} converges uniformly to
s:X →Y,then s is L
σ
measurable.
Theorem 3.3.(Kuratowski and Ryll Nardzewski) Let X and L
σ
be
as above and Y a Polish space.Then every closed valued L
σ
measurable
multifunction F:X →Y admits a L
σ
measurable selector.
Proof.Fix a complete metric d < 1 on Y inducing its topology and
a countable dense set {r
m
} in Y.
We claim that there is a sequence {s
n
} of L
σ
measurable functions
from X to Y satisfying the following conditions for every n and every
x ∈ X:
(1) d(s
n
(x),F(x)) < 2
−n
.
(2) d(s
n
(x),s
n+1
(x)) < 2
−n
.
Assuming this,we complete the proof ﬁrst.Since (Y,d) is a complete
metric space and {s
n
} uniformly Cauchy,{s
n
} is uniformly convergent,
say to s.By 3.2,s is L
σ
measurable.Since F(x) is closed,by (1),
s(x) ∈ F(x).
We construct the sequence {s
n
} satisfying (1) and (2) by induction
on n.Set s
0
≡ r
0
.Note that (1) is satisﬁed for n = 0.
Suppose s
n
is L
σ
measurable and satisﬁes (1).For each m,deﬁne
A
m
= {x ∈ X:d(s
n
(x),r
m
) < 2
−n
}∩{x ∈ X:F(x)∩B(r
m
,2
−(n+1)
) = ∅}.
Since s
n
and F are L
σ
measurable,each A
m
∈ L
σ
:A
m
is the intersec
tion of following two sets:
A = {x ∈ X:d(s
n
(x),r
m
) < 2
−n
}
and
B = {x ∈ X:F(x) ∩B(r
m
,2
−(n+1)
) = ∅}.
Since {r
m
} is dense in Y,∪
m
A
m
= X.
Now get paiwise disjoint B
m
⊂ A
m
in L
σ
as in 3.1.In particular,
∪
m
B
m
= X.
Deﬁne s
n+1
:X →Y by
s
n+1
(x) = r
m
,if x ∈ B
m
.
10
Thus,for each m,s
−1
n+1
(r
m
) = B
m
.This implies that for every U ⊂ Y,
s
−1
n+1
(U) = ∪
{m:r
m
∈U}
s
−1
n+1
(r
m
) ∈ L
σ
.
In particular,s
n+1
is L
σ
measurable.
There are many special cases of this selection theorem that occur in
applications.
Theorem 3.4.Let (X,A) be a measurable space and Y a Polish
space.Then every closedvalued Ameasurable multifunction F:X →
Y admits an Ameasurable selector.
(Take L = A.)
Theorem 3.5.Let X be a metrizable space,Y a Polish space and
F:X → Y a closedvalued lower semicontinuous or upper semi
continuous multifunction.Then F admits a class 1 selector.
Proof.Let L be the family of all subsets of X which are simultane
ously F
σ
and G
δ
.Then L is an algebra on X.If F is either lower or
upper semicontinuous,F is L
σ
measurable.Hence,the result follows
from Kuratowski and Ryll Nardzewski selection theorem.
Let Π be a partition of a set X.A subset S of X is called a cross
section of Π if S ∩ C is a singleton for every C ∈ Π.For any A ⊂ X,
set
A
∗
= ∪{C ∈ Π:C ∩A = ∅}.
Thus,A
∗
is the smallest Πinvariant set containing A.
Call A Πinvariant if A = A
∗
.
Now suppose X is Polish.
We call Π Borel measurable if U
∗
is Borel for every open U ∈ X.
Similarly Π is lower semicontinuous if U
∗
is open for every open
in X;
upper semicontinuous if C
∗
is closed for every closed C in X.
Theorem3.6.(Eﬀros) Every Borel measurable partition of a Polish
space into closed sets admits a Borel crosssection.
Proof.Take A to be the set of all Πinvariant Borel sets in X.
Consider the multifunction F:X →X deﬁned by
F(x) = {x}
∗
,x ∈ X.
Then F is closed valued and Ameasurable.
11
By Kuratowski and Ryll Nardzewski selection theorem,let s:X →
X be an Ameasurable selector for F.Note that sC is constant on
every C ∈ Π and s(x) ∈ C whenever x ∈ C.
It is easy to see that
S = {x ∈ X:x = s(x)}
is a Borel crosssection for Π.
Theorem 3.7.Every lower and every upper semicontinuous parti
tion of a Polish space into closed sets admits a G
δ
crosssection.
Proof.The proof is similar to 3.6 by taking L to be the set of all
invariant sets which are simultaneously F
σ
and G
δ
.
Srivastava and Sreela Gangopadhyay (n`ee Bhattacharya) used a beau
tiful idea of changing topology and got a serious generalization of Ku
ratowski and Ryll Nardzewski selection theorem.
Proposition 3.8.Let Y be a Polish space and {C
k
} a sequence
of closed sets in Y.Then there is a ﬁner Polish topology on Y that
generate the same Borel σalgebra and in which each C
k
is open.
Proof.Let us deﬁne f:Y →Y ×2
N
by
f(y) = (y,I
C
0
(y),I
C
1
(y),...),y ∈ Y,
where for any C ⊂ Y,I
C
denotes the indicator function of C.
Then f is onetoone and Borel.Let Z = f(Y ).
By 2.13,f:Y →Z is a Borel isomorphism.
Since each C
k
is closed,Z is a G
δ
in Y × 2
N
and so,Polish.Also
note that
C
k
= {y ∈ Y:∃α ∈ 2
N
(f(y) = (y,α)) ∧ α(k) = 1}.
Now take the smallest topology on Y making f continuous.This sat
isﬁes all the conditions.
Motivated by some problems in descriptive set theory,we generalized
the selection theorem of Kuratowski and Ryll Nardzewski as follows.
Theorem 3.9.(Bhattacharya and Srivastava) Let L be an algebra
on a set X,Y a Polish space and F:X →Y a closed valued strongly
L
σ
measurable multifunction.Supose Z is a second countable space and
g:Y →Z a class 1 function.Then F admits a L
σ
measurable selector
s:X →Z such that g ◦ s is also L
σ
measurable.
12
Proof.Let {U
n
} be a countable base for the topology of Z.For
each n,write
g
−1
(U
n
) = ∪
m
C
nm
,
C
nm
’s closed.
Let {C
k
} be an enumeration of the double sequence {C
nm
} into a
single sequence.Take a ﬁner topology on Y as in 3.8.
Now note that still each F(x) is closed,F is L
σ
measurable and g
is continuous with respect to the new topology.Take a L
σ
measurable
selector s:X →Y for F with respect to the new topology on Y.
Since g is continuous with respect to the new topology,g ◦ s is L
σ

measurable and our proof is complete.
We got another generalization of the selection theoremof Kuratowski
and Ryll Nardzewski which we have recently used to show the existence
of a continuous stochastic relation with support contained in a given
set.
Theorem 3.10.(Srivastava) Let X,Y,L
σ
and F:X → Y be as
in 3.3.Then there is a map f:X ×N
N
→Y such that
(1) for every x ∈ X,α → f(x,α) is a continuous map from N
N
ontoF(x),and
(2) for each α ∈ N
N
,x →f(x,α) is L
σ
measurable.
4.Selection Theorems for G
δ
valued multifunctions.
Following a problem raised for some representations of C
∗
algebras,
a signiﬁcant generalization of 3.6 was obtained by Srivastava.
Before we state this,let us recall the Vitali partition of the real line
R.
Example.Call two real numbers x and y equivalent if x −y ∈ Q.
Then every member of the induced partition is a translate of Q.Since
they are countable,they are F
σ
.Since they are homeomorphic to Q,
by Baire category theorem,they are not G
δ
in R.It is wellknown
that this partition of R does not admit even a Lebesgue measurable
crosssection.It is also pertinent to note that for any open set U in R,
U
∗
= ∪
r∈Q
[U +r]
is open in R.Thus,the Vitali partition is lower semicontinuous.
Using some deep results in descriptive set theory,particularly a sep
aration theorem of SaintRaymond for analytic sets with σcompact
sections,we obtained the following result.The proof of this result is
hard and therefore omitted.
13
Theorem 4.1.(Srivastava) Every Borel measurable partition of a
Polish space into G
δ
sets admits a Borel crosssection.
This result was obtained by proving the following selection theorem
for G
δ
valued multifunctions.
Theorem 4.2.(Srivastava) Let X be an analytic space,Y Polish
and F:X →Y a G
δ
valued Borel measurable multifunction with graph
Borel in X ×Y.Then F admits a Borel selection.
The following result on lowersemicontinuous partitions of Polish
spaces into G
δ
sets is worth mentioning.
Theorem 4.3.(D.Miller) Every lower and every upper semi
continuous partition of a Polish space into G
δ
sets admits a G
δ
cross
section.
5.Von Neumann Selection Theorem.
It is very wellknown that there is a Borel set B ⊂ [0,1] ×[0,1] with
all xsections nonempty,yet there is no Borel map s:[0,1] → [0,1]
whose graph is contained in B.There are several beautiful and useful
(all quite deep) results showing the existence of such s provided the
xsections satisfy some conditions.In the last section we shall discuss
such results.
Question arises:how good a selector one can have in general?A
very ﬁne and quite useful result was proved by Von Neumann.
Theorem5.1.(Von Neumann) Let X be an analytic space,Y Polish
and B ⊂ X × Y analytic with each xsection B
x
nonempty.Then
there is a map s:X → Y whose graph is contained in B and which
is measurable with respect to the σalgebra generated by all analytic
subsets of X.
Proof.Since B is analytic,there is a continuous function f from
N
N
onto B.Now deﬁne a multifunction F:X →N
N
by
F(x) = f
−1
({x} ×B
x
),x ∈ X.
Then F is a closed valued multifunction whose graph equals
{(x,α) ∈ X ×N
N
:x = π
X
(f(α))},
and so closed in X ×N
N
.
14
Since X is analytic,in particular the graph of F is analytic.It
follows that F is measurable with respect to the σalgebra generated
by analytic sets.
By Kuratowski—Ryll Nardzewski selection theorem,it has a selector,
say s
,measurable with respect to the σalgebra generated by analytic
sets.
Now take s = π
Y
◦ f ◦ s
.Since π
Y
and f are continuous,s is
measurable with respect to the σalgebra generated by analytic sets.
Remark 5.2.Since analytic sets are measurable with respect to
every σﬁnite measures (2.9),it follows that s obtained in 5.1 is uni
versally measurable.
We can generlize Von Neumann selection theorem a bit more,
Theorem 5.3.Let (X,A) be a measurable space with A closed
under Souslin operations and Y a Polish space.Suppose B ∈ A⊗B
Y
(the product σalgebra) with each xsection B
x
= ∅.Then there is an
Ameasurable s:X →Y whose graph is contained in B.
Proof.Get a sequence {A
n
} in A and a sequence {B
n
} in B
Y
such
that B belongs to the σalgebra generated by the sequence {A
n
×B
n
}.
Deﬁne I:X →2
N
by
I(x) = (I
A
0
(x),I
A
1
(x),I
A
2
(x),...),x ∈ X.
Set Z = I(X).Then a set A ⊂ X is in the σalgebra generated by
{A
n
} if and only if I(A) is Borel in Z.
It follows that C = (I ×1)(B) is Borel in Z×Y.Now get a Borel set
D in 2
N
×Y such that C = D∩ (Z ×Y ).By Von Neumann selection
theorem,there is a map s
:π
2
N(D) → Y with graph contained in D
and measurable with respect to the σalgebra generated by analytic
sets.Now deﬁne
s(x) = s
(I(x)),x ∈ X.
Since A is closed under Souslin operations,using 2.8,we show easily
that s is Ameasurable.Clearly s is a selector of B.
6.Borel Uniformizations.
Let X,Y be Polish spaces and B ⊂ X × Y Borel.A Borel uni
formization of B is a Borel subset G of B which is the graph of a
function,say f,and π
X
(G) = π
X
(B).
By 2.14,π
X
(G) is Borel and f is Borel measurable.Thus,if a Borel
set admits a Borel uniformization,its projection must be Borel.
15
Since there are analytic nonBorel sets,it follows that there exist
Borel sets that do not admit a Borel uniformization.
It is also known that not every Borel B with projection Borel admits
a Borel uniformization.
The study of the existence of Borel uniformizations lies in the heart
of classical descriptive set theory.In this section,we shall present
major uniformization theorems.Since they require deep results from
descriptive set theory,their proofs will essentially be omitted.
6.1.Borel sets with countable sections.
Lusin,the founder of descriptive set theory,proved a highly non
trivial result on Borel sets with countable sections.If one is prepared
to use recursion theory,one gets a much simpler proof than the classical
proof of Lusin.
Theorem 6.1.(Lusin) Let X,Y be Polish and B ⊂ X ×Y Borel
with sections B
x
,x ∈ X,countable.Then B admits a Borel uni
formization.
In fact,Lusin proved much more.
Theorem 6.2.(Lusin) Let X,Y be Polish and B ⊂ X ×Y Borel
with sections B
x
,x ∈ X,countable.Then B is a countable union of
Borel graphs.
6.2.Borel sets with compact sections.
The results (but not the proofs) in this section are due to Novikov.
Theorem 6.3.(Novikov) Let X,Y be Polish and B ⊂ X×Y Borel
with sections B
x
,x ∈ X,compact.Then π
X
(B) is Borel.
Proof.(Srivastava) We imbed Y into a compact metric space Z.
Since Y is a G
δ
in Z,B is Borel in X ×Z.Further,its xsections are
closed in Z.
By Dellacherie’s theorem (2.17),there exist Borel sets B
n
⊂ X and
open sets U
n
⊂ Z such that
(X ×Z)\B = ∪
n
(B
n
×U
n
).
Now observe that
B
x
= ∅ ⇔∃n
0
,...n
k
(Z = ∪
k
i=0
U
n
i
& x ∈ ∩
k
i=0
B
n
i
).
16
So,
X\π
X
(B) = ∪
{(n
0
,...,n
k
):Z=∪
k
i=0
U
n
i
}
∩
k
i=0
B
n
i
.
Corollary 6.4.Let X,Y be Polish and B ⊂ X × Y Borel with
sections B
x
,x ∈ X,compact.Then for every open set U in Y,
{x ∈ X:B
x
∩U = ∅} ∈ B
X
.
Proof.Since every open set in a metrizable space is a F
σ
,there is a
sequence {F
n
} of closed sets in Y such that U = ∪
n
F
n
.Now note that
{x ∈ X:B
x
∩U = ∅} = ∪
n
π
X
(B ∩(X ×F
n
)).
Clearly,xsections of B∩(X×F
n
) are compact.So,the result follows
from 6.3.
6.4 says that the closed valued multifunction x →B
x
from π
X
(B) to
Y is Borel measurable.By 6.3,π
X
(B) is Borel.So,by Kuratowski—
Ryll Nardzewski selection theorem,we get the following theorem.
Theorem 6.5.(Novikov) Let X,Y be Polish and B ⊂ X × Y
Borel with sections B
x
,x ∈ X,compact.Then B admits a Borel
uniformization.
6.3.Borel sets with σcompact sections.
Theorem 6.6.(Kunugui) Let X,Y be Polish and B ⊂ X × Y
Borel with sections B
x
,x ∈ X,σcompact.Then B admits a Borel
uniformization.
This is a fairly hard result to prove.We would like to deduce this
from the following very beautiful result of SaintRaymond.
Theorem6.7.(SaintRaymond) Let X,Y be Polish and B ⊂ X×Y
Borel with sections B
x
,x ∈ X,σcompact.Then there is a sequence
{B
n
} of Borel sets in X×Y with compact sections such that B = ∪
n
B
n
.
SaintRaymond’s result together with 6.3 easily implies that that
B contains a Borel set C with compact sections such that π
X
(B) =
π
X
(C).Then Kunugui’s uniformization theoremfollows fromNovikov’s
result.
6.4.Blackwell—Ryll Nardzewski selection theorem.
17
Let (X,A) be a measurable space and Y a Polish space.A sto
chastic relation P:X → Y is a map P:X × B
Y
→ [0,1] such
that
(1) for every x ∈ X,P(x,∙) is a probability on Y,and
(2) for every Borel B in Y,x →P(x,B) is Ameasurable.
Theorem 6.8.(Blackwell and Ryll Nardzewski) Let (X,A) be a
measurable space,Y Polish and B ∈ A ⊗ B
Y
.Suppose there is a
stochastic relation P:X →Y such that whenever B
x
= ∅,P(x,B
x
) >
0.Then π
X
(B) ∈ A.Further,the multifunction x →B
x
from π
X
(B)
to Y admits an Ameasurable selection.
Sketch of the proof.Using familiar good set argument (or π −λ
theorem),one shows that for every A ∈ A ⊗ B
Y
,x → P(x,A
x
) is
Ameasurable.This implies that π
X
(B) ∈ A.
Next one shows that for every ,δ > 0,there is a C ⊂ B in A⊗B
Y
with sections C
x
,x ∈ X,compact and of diameter < δ.Further,for
every x ∈ X,P(x,C
x
) ≥ ∙ P(x,B
x
).Using this repeatedly,one gets
a sequence {C
n
} of subsets of B in A ⊗ B
Y
with section (C
n
)
x
non
empty,compact and with diameter(C
n
) converging to zero.This yields
the desired selection.
Theorem6.9.(Blackwell and Ryll Nardzewski) Let X,Y be Polish
and B ⊂ X×Y Borel.Suppose there is a stochastic relation P:X →Y
such that whenever B
x
= ∅,P(x,B
x
) > 0.Then B admits a Borel
uniformization.
6.5.Borel sets with nonmeager sections.
The following result was independently proved by Kechris and Sar
badhikari.Kechris gave a proof using recursion theory whereas Sar
badhikari gave a classical proof.
Theorem 6.10.(Kechris,Sarbadhikari) Let X,Y be Polish spaces
and B ⊂ X ×Y Borel.Suppose whenever B
x
= ∅,it is nonmeager in
Y.Then B admits a Borel uniformization.
Enter the password to open this PDF file:
File name:

File size:

Title:

Author:

Subject:

Keywords:

Creation Date:

Modification Date:

Creator:

PDF Producer:

PDF Version:

Page Count:

Preparing document for printing…
0%
Σχόλια 0
Συνδεθείτε για να κοινοποιήσετε σχόλιο