Download The Measurable Projection and Selection Theorems

The Measurable Projection and
Selection Theorems
The goal of these notes is to give a simple and straightforward proof
of the Projection and the Selection Theorems. The proof of these theorems depends on the theory of Suslin sets. The theory of Suslin sets
describes the images of measurable sets under continuous mappings1
and the theory depends on the delicate interaction of the measurable
and topological structure of complete separable metric spaces. We will
not discuss this theory in detail we are merely trying to give a simple
proof of the aforementioned theorems. The most remarkable phenomena
one should follow below is the role of the completeness of the underlying
-algebra, that is the role of the subsets of the measure zero sets.
1
The de…nition of the Suslin sets
De…nition 1 If (X; ) is a topological space then B (X) denotes the
Borel -algebra generated by (X; ) : In these notes F (X) will denote
the set of closed sets.
Lemma 2 If the topological space (X; ) is metrizable then B (X) is
the smallest family of sets containing the open sets and closed under
countable union and intersection. The same is true for the closed sets
of X:
Proof: Let A denote the family of sets in the lemma. By the de…nition
of the Borel -algebra A B (X) : As (X; ) is metrizable every closed
set is a countable intersection of open sets, so F (X) A. Let
A0 $ fA : A 2 A and Ac 2 Ag :
As A contains the closed and the open sets the open sets are in A0 . A
is closed under the countable union and intersection therefore A0 is also
1
The inverse images of open sets under continuous mappings are open. The question is what we know about the images of open sets. Are they measurable at all?
1
closed under the countable union and intersection, so by the de…nition
of A obviously A0
A; that is A0 = A: Hence A is closed under
complementation as well, so A is a -algebra. Therefore B (X)
A;
hence A = B (X) : To prove the second part of the lemma one should
remark that in a metrizable space every open set is a countable union of
closed sets.
De…nition 3 A topological space (X; ) is called a Polish space if there
is a metric d : X X ! R+ generating such that (X; d) is a complete
separable metric space.
Obviously a closed subset of a Polish space is a Polish space and the
topological product of countable many Polish spaces is a Polish space.
De…nition 4 Let (X; ) be a topological space. An S X is a Suslin set
in X if there is a Polish space P and a continuous mapping h : P ! X
such that S = h (P ). The topological space (X; ) is a Suslin space if X
is a Suslin subset of the topological space (X; ) : We shall denote the set
of Suslin subsets of X by S (X) :
Lemma 5 The following properties of the Suslin sets are obvious:
1. If X and Y are topological spaces, f : X ! Y is continuous and
S is a Suslin set in X then f (S) is a Suslin set in Y:
2. If (Sn ) are countable many Suslin spaces then the topological product n Sn is also a Suslin space.
3. Every Polish space is a Suslin space and a closed subset of a Polish
space is a Suslin set.
The next property of the Suslin sets is crucial.
Lemma 6 The Suslin sets of a topological space are closed under the
countable union and intersection2 .
Proof: Let (Sn ) be a sequence of Suslin sets in a topological space
X: Let Tn be a Polish space and let fn : Tn ! X be continuous with
f (Tn ) = Sn . Let T $ n Tn be the topological product of the Polish
spaces Tn . Obviously T is a Polish space. Observe that N with the
discrete topology is also a Polish space so N T is a Polish space. If
h (n; (t1 ; t2 ; t3 ; : : :)) $ hn (tn )
2
But not under the complementation!
2
then h is trivially continuous so [n An = h (N
Z $ f(n; (t1 ; t2 ; t3 ; : : :)) 2 N
is a closed subset of N
is a Suslin set in X.
T ) is a Suslin set in X.
T : h1 (t1 ) = h2 (t2 ) = h3 (t3 ) =
g
T hence it is a Polish space. So \n An = h (Z)
Every closed subset of a Polish space is a Suslin set. As every open set
in a metric space is a countable union of closed sets it is clear that every
open set of a Polish space is also a Suslin set. One can prove more:
Lemma 7 If X is a Polish space then B (X)
true if X is a Suslin space.
S (X). The same is
Proof: S (X) is closed under the countable union and intersection. If
X is a Polish space then F (X)
S (X), so by Lemma 2 B (X)
S:
If X is a Suslin space then X = h (T ) ; where T is a Polish space. If
A 2 B (X) then h 1 (A) 2 B (T )
S (T ) ; hence as h is continuous
A = h (h 1 (A)) 2 S (X) by Lemma 5.
Example 8 If X is a separable Banach space and if E; F 2 B (X) ; then
E + F 2 S (X) :
Y $ X X is a Polish space and G $ E F 2 B (Y )
function h (x; y) $ x + y is continuous so h (G) 2 S (X) :
S (Y ) : The
Example 9 If X and Y are Suslin spaces and B 2 B (X
prX (B) 2 S (X) :
It is su¢ cient to recall that the projection is continuous.
2
The measurability of Suslin sets
Lemma 10 Let ( ; B; ) be a measure space and let
(A) $ inf f (B) : A
B 2 Bg
be the outer measure generated by : If An % A then
lim
n!+1
(An ) =
3
(A) :
Y ) then
Proof: If
(An ) = +1 for some n then the relation is obvious,
therefore one can assume that
(An ) < +1 for every n: As B is a
-algebra by the de…nition of
for every n there is a Bn 2 B such
that An
Bn and
(An ) = (Bn ) : For every 1
k
n obviously
Ak Bk \ Bn Bk hence
(Bk ) =
(Bk \ Bn )
(Ak )
(Bk ) :
Therefore
(Bk \ Bn ) =
As
is additive
(Bk ) =
and as
(Bk \ Bn ) + (Bk n Bn )
(Bk ) < +1 obviously
(Bk n Bn ) = 0: Hence for every n
([nk=1 Bk ) =
As
(Bk ) :
(Bn ) :
is increasing
lim
n!+1
(Bn ) = lim
n!+1
(An )
([n An )
([n Bn ) = lim
n!+1
([nk=1 Bk ) = lim
n!+1
(Bn ) :
Hence the second relation above is an equality which proves the lemma.
From now on let X be a metric space and let (X; B (X) ; ) be a …nite
measure space. Let
: P (X) ! R+ be the outer measure generated
by : The following proposition shows the importance of Polish spaces
in the theory of Suslin sets.
De…nition 11 Let ( ; B; ) be a measure space. By de…nition the measurability of a set S
means that there are B1 ; B2 2 B such that
B1 S B2 and (B2 nB1 ) = 0:
Proposition 12 If X is a metric space then every S 2 S (X) is
measurable.
-
Proof: Let f (T ) = S where T is a Polish space and f : T ! X is
continuous. As T is a Polish space T = [j Tj ; where (Tj )
1 and
every Tj is closed. (One can use the closed balls with radius 1=2 around
points of a dense, countable subset of T:) Let
An $ f [nj=1 Tj :
4
As every Tj is a Polish space it has a decomposition Tj $ [k Tjk , where
Tjk are closed and (Tjk ) 1=2: Let
Ajn $ f ([nk=1 Tjk ) :
One can continue the procedure and construct a sequence
An1 ;n2 ;:::;n $ f ([nk=1 Tn1 ;n2 ;:::;k )
1=2k . Obviously for every k and (n1 ; n2 ; : : : ; nk ) 2
where (Tn1 ;n2 ;:::;nk )
Nk
An1 ;n2 ;:::;nk ;n % An1 ;n2 ;:::;nk :
Let " > 0. As An % S by the previous lemma there is an index n1 2 N
such that
"
:
(An1 ) > (S)
2
Now assume that we have already de…ned the indexes n1 ; n2 ; : : : ; nk . As
An1 ;n2 ;:::;nk ;n % An1 ;n2 ;:::;nk there is an index nk+1 2 N such that
An1 ;n2 ;:::;nk ;nk+1 >
"
(An1 ;n2 ;:::;nk )
2k+1
:
The sequence k 7! An1 ;n2 ;:::;nk is obviously decreasing and for every k 2 N
(An1 ;n2 ;:::;nk ) >
(S)
k
X
"
>
j
2
j=1
(S)
":
For every k 2 N let
Hk $ cl (An1 ;n2 ;:::;nk ) 2 F (X)
and let
H (") $ \k Hk 2 F (X)
The sequence (Hk ) is decreasing and as
H (") = lim
k!+1
(Hk )
lim
k!+1
B (X) :
is …nite
(An1 ;n2 ;:::;nk )
(S)
":
We show that H (") S: Let x 2 H (") : This means that x 2 Hk for every
k. By the de…nition of H1
B (x; 1) \ An1 6= ;:
This implies that
f (Tn ) \ B (x; 1) 6= ;
5
for certain sets Tn with n
n1 : Now by the de…nition of H2
B (x; 1=2) \ An1 n2 6= ;
which implies that for certain indexes n
n1 and m
n2
f (Tnm ) \ B (x; 1=2) 6= ;:
One can continue the procedure for all n1 ; n2 : : : ; nk . As the number of
sets Tm1 ;m2 ;:::;mk in the construction is in…nite there is an index m1 n1
such that Tm1 contains in…nitely many of the selected T -sets. As the
number of the sets Tm1 n is also …nite there is an index m2
n2 such
that Tm1 m2 contains in…nitely many T -sets. Therefore one can construct
a sequence sk 2 Tm1 ;m2 ;:::;mk such that
d (x; f (sk ))
1=2k :
As Tm1 ;m2 ;:::;mk+1
Tm1 ;m2 ;:::;mk by the construction d (sk ; sk+1 ) < 1=2k ;
therefore (sk )
T is a Cauchy sequence. As T is complete there is
an s 2 T such that f (sn ) ! f (s) which means that x = f (s) 2 S:
Therefore H (") S: Let now
H $ [n H1=n :
Obviously H 2 B (X) and H
(H )
S and also for every n
H (")
(S)
1
n
which means that (H ) = (S) : On the other hand by the de…nition
of the outer measure there is a set K 2 B (X) such that S K and
(K ) =
(S) =
(H ) :
Hence K n H has -measure zero and it contains the set S n H . This
implies that S n H has -measure zero and since H is -measurable
S will be also -measurable.
De…nition 13 The measurable space ( ; A) is called complete if for
every E
;E2
= A there is a …nite measure : A ! R+ such that E
is not -measurable.
Example 14 If ( ; A; ) is a …nite or -…nite, complete measure space3
then ( ; A) is a complete measurable space. Rn with the Lebesgue measurable sets is a complete measure space.
3
The completeness of a measure space means that all subsets of measure zero sets
are measurable.
6
Proposition 15 Let ( ; A) be a complete measurable space, X be a
metric space and let f : ! X be an A-measurable mapping4 . Then
f
1
(S) 2 A
for every S 2 S (X) that is the inverse images of the Suslin sets are
measurable.
Proof: Assume that f 1 (S) 2
= A: Then by de…nition there is a …nite
measure : A ! R+ such that f 1 (S) is not -measurable. If
(A) $
f
1
(A)
then (X; B (X) ; ) is a …nite measure space. Let be the outer measure
of . By the previous proposition S is -measurable, hence S has the
decomposition S $ K [ L where K 2 B (X) and
(L) = 0: Hence
there is an F 2 B (X) such that L F and (F ) = 0: As f measurable
f 1 (K) 2 A and by the de…nition of f 1 (F ) 2 A has -measure zero:
Hence f 1 (L) f 1 (F ) is -measurable. Therefore
f
is
1
(S) = f
1
(K [ L) = f
1
(K) [ f
-measurable. This is a contradiction therefore f
1
(L)
1
(S) 2 A:
Example 16 Every S 2 S (R) is Lebesgue measurable.
With f (x) $ x it is trivial from the just proved proposition.
Example 17 Let X be a Suslin space and let B 2 B (R
prR (B) is Lebesgue measurable.
X) : Then
prR (B) is a Suslin subset of R so it is Lebesgue measurable.
Example 18 If E; F 2 B (R) ; then E + F is Lebesgue measurable.
4
That is f
1
(B (X))
A:
7
3
Measurable Projection Theorem
Lemma 19 For every n 2 N
( +1
)
X a (k)
Hn $
: a 2 f0; 1gN and a (n) = 1
k
4
k=1
is a compact subset of R. For every a 2 f0; 1gN
+1
X
a (k)
k=1
4k
2 Hn
if and only if a (n) = 1:
Proof: The topological product f0; 1gN is compact and
n
o
Mn $ a 2 f0; 1gN : a (n) = 1
is a closed subset of this product. The function f (a) $
+1
P
k=1
a(k)
4k
is a
uniformly convergent limit of continuous functions so it is continuous.
Hence Hn = f (Mn ) is compact. f is injective5 so f (a) 2 Hn = f (Mn )
is equivalent to the condition a 2 Mn .
Lemma 20 Let ( ; A) be a measurable space and let X be a topological
space. For every C 2 A B (X) there is a set D 2 B (R X) and an
A-measurable function g : ! R such that
C = f(!; x) : (g (!) ; x) 2 Dg :
Proof: As C is in the -algebra generated by the measurable rectangles
of A and B (X) it is also in a -algebra generated by countable many
measurable rectangle, therefore there are sets (An )
A and (Bn )
B (X) such that
C2G$
Let
(fAn
+1
X
1
g (!) $
4k
k=1
5
Bn : n 2 Ng) :
Ak
(!) :
The simplest way to show this just to observe that if a sequence x is lexicographically smaller than y then f (x) < f (y) : To show this one should use that in the
de…nition of f one has 4 and not 2.
8
By the previous lemma for every n 2 N and ! 2
g (!) 2 Hn ()
(!) = 1 () ! 2 An ;
An
that g 1 (Hn ) = An ; where Hn R are the compact sets in the previous
lemma. As g is a sum of A-measurable functions it is A-measurable.
Let
h (!; x) $ (g (!) ; x) ;
and let
H$ h
1
H is obviously a -algebra in
the de…nition of h
h
1
(Hn
As obviously Dn $ Hn
(D) : D 2 B (R
Bn ) = g
X. As g
1
(Hn )
Bn 2 B (R
An
1
X) :
(Hn ) = An for every n by
Bn = An
Bn :
X) by the de…nition of H
Bn 2 H:
H is a -algebra and G is the smallest -algebra generated by the rectangles An Bn therefore G
H: Hence C 2 G
H; that is by the
de…nition of H there is a set D 2 B (R X) with C = h 1 (D) : By the
de…nition of h this implies the lemma.
Theorem 21 (Measurable Projection Theorem) Let ( ; A) be a complete measurable space and let X be a Suslin space. If C 2 A B (X)
then
pr (C) 2 A:
Proof: By the previous lemma there is a D 2 B (R
A-measurable function g : ! R such that
X) and an
C = f(!; x) : (g (!) ; x) 2 Dg :
From this trivially
1
pr (C) = g
(prR (D)) :
As R X is a Suslin space D 2 B (R X) is a Suslin set6 so S $ prR (D)
is a continuous image of a Suslin set so S is a Suslin set in R. But as g
is A-measurable7
pr (C) = g 1 (S) 2 A:
6
7
See: Lemma 7.
See: Proposition 15.
9
4
Measurable Selection Theorem
The topological version of the Measurable Selection Theorem is the following:
Lemma 22 ( Kuratowski–Ryll-Nardzewski) Let ( ; A) be a measurable space and let X be a Polish space. Let F : ! F (X) be a point
to set mapping and assume that
1. F (!) 6= ; for every ! 2
2. for every open set G
and
X
F (G) $ f! 2
: F (!) \ G 6= ;g 2 A:
Then F has a measurable selection f :
! X that is there is an Ameasurable function f such that f (!) 2 F (!) for every ! 2 :
Proof: In order to simplify the notation one may assume that (X)
1: As X is complete for the proof of the the theorem it is su¢ cient to
construct a sequence of measurable mappings (fn ) with the properties
1. d(fn (!) ; F (!)) < 1=2n
2. d(fn (!) ; fn
1
(!)) < 1=2n 1 :
Let Q = (qk ) be a countable dense subset of X: Let be choosen f0 q 2
Q arbitrarily. As (X) 1 conditions 1 and 2 hold. Assume that fn 1
has already been constructed. Let
Ckn $ ! : d (qk ; F (!)) <
Dkn $ ! : d (qk ; fn
1
1
2n
(!)) <
;
1
2n 1
;
Ank $ Ckn \ Dkn :
Choose ! 2 : By the construction of fn 1 there is an x 2 F (!) such
that d (x; fn 1 (!)) < 1=2n 1 : As Q is dense in X there is a qk 2 Q such
that d (qk ; x) < 1=2n and
d (qk ; fn
1
(!))
d (qk ; x) + d (x; fn
This means that ! 2 Ank : Therefore
Bkn $
(!)) <
1
2n 1
= [k Ank for any n: If
x : d (qk ; x) <
10
1
1
2n
:
then
Ckn = f! : F (!) \ Bkn 6= ;g ;
Dkn = fn 11 Bkn 1 :
As fn 1 is measurable Dkn 2 A: By the conditions of the lemma Ckn 2 A
as well, so Ank 2 A: Let
fn (!) $ qk ;
! 2 Ank n [kr=11 Anr 2 A:
For any open set G
fn 1 (G) = [qk 2G fn 1 (qk ) 2 A
so fn is measurable. It is also clear from the construction that 1: and 2:
hold. (fn (!)) is a Cauchy sequence and as X is complete fn (!) ! f (!)
for some f (!) and as X is a separable metric space f is also measurable.
As F (!) is closed it is obvious that
f (!) 2 cl (F (!)) = F (!)
for every !: Hence f is a measurable selection of F:
Theorem 23 (Measurable Selection Theorem) Let X be a Suslin
space and let ( ; A) be a complete measurable space. Let F :
!
P (X) be a point to set mapping and let assume that
1. F (!) 6= ; for every ! 2
2. graph F 2 A
and
B (X) :
Then F has a measurable selection f :
! X that is there is an Ameasurable function f such that f (!) 2 F (!) for every ! 2 :
Proof: As in the proof of the Projection Theorem there is a set D 2
B (R X) and an A-measurable function g : ! R such that
graph F = f(!; x) : (g (!) ; x) 2 Dg :
As D 2 B (R X) ; it is a Suslin set in R X, hence there is a Polish
space Y and a continuous function h : Y ! R X such that h (Y ) = D:
Let
(!) $ h 1 (fg (!)g X) :
11
As F (!) 6= ; there is an x such that (!; x) 2 graph F; which implies
that
(g (!) ; x) 2 D = h (Y ) :
Hence for every ! there is an y 2 Y such that h (y) = (g (!) ; x) : By the
de…nition of
this means that y 2 (!) : Hence (!) 6= ; for every
! 2 : As h is continuous (!) 2 F (Y ). As Y is a Polish space every
open set G Y is a Suslin set. As h is continuos prR (h (G)) is a Suslin
set in R. As g is A-measurable
(G) = g
Hence
Let ' :
1
(prR (h (G))) 2 A:
satis…es the Kuratowski–Ryll-Nardzewski selection theorem.
! Y be the A-measurable selection of . For very ! 2
h (' (!)) 2 (fg (!)g
X) \ D:
Hence prR (h (' (!))) = g (!) for every !: Let f (!) $ prX (h (' (!))) :
Then f : ! X is A-measurable and for every ! 2
(g (!) ; f (!)) = (prR (h (' (!))) ; prX (h (' (!)))) = h (' (!)) 2 D:
By the construction of D and g this implies that (!; f (!)) 2 graph F;
that is f is an A-measurable selection of F .
12