Selection theorems and their applications

Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Z0rich

263 T. Parthasarathy Indian Statistical Institute, Calcutta/India

Selection Theorems and their Applications

Springer-Verlag Berlin. Heidelberg. New York 1972

A M S S u b j e c t Classifications (1970): 49 A 35, 54 C 65, 90 D 15

I S B N 3-540-05818-4 S p r i n g e r - V e r l a g B e r l i n • H e i d e l b e r g . N e w Y o r k I S B N 0-387-05818-4 S p r i n g e r - V e r l a g N e w Y o r k • H e i d e l b e r g • B e r l i n This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1972. Library of Congress Catalog Card Number 72-78192. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach/Bergstr.

PREFACE

This volume of lecture notes contains results on selection theorems and their applications.

Some of the material of this volume had been

given as seminar talks at Case Western Reserve University during ]969 - 1970. This volume contains nine sections each of them followed by selected references.

We hope this volume will be profitable to

specialists in game theory, dynamic programming,

control theory, mathe-

matical economics as well as to all mathematicians

interested in this

area of Mathematics. I wish to express my sincere thanks to the following Professors: Henry Hermes, Marc Jacobs, Ashok Maitra and Sam Nadler Jr., for several useful suggestions. A particular measure of gratitude is due to Mr. Arun Das who patiently and accurately prepared the final typescript of this volume. My wife Ranjani proof-read the manuscript.

To her I owe my heart-felt

thanks. Finally I wish to express my gratitude to the Indian Statistical Institute for providing the excellent research facilities and to Springer - Verlag for undertaking the publication of these notes.

T. Parthasarathy November 20, 1971

Indian Statistical Institute

CONTENTS

Introduction

. . . . . . . . . . . . . . . . . . . . . . . . . .

I. C o n t i n u o u s

selections

. . . . . . . . . . . . . . . . . . . .

2. C o n t i n u o u s

selections

on m e t r i c

continuum

. . . . . . . . . .

I

3

14

3. C o n t i n u o u s s e l e c t i o n s and s o l u t i o n s of g e n e r a l i s e d differential equations . . . . . . . . . . . . . . . . . . . .

24

4. M e a s u r a b l e

35

5. G e n e r a l

selections

theorems

6. T w o a p p l i c a t i o n s

7. v o n - N e u m a n n ' s

and an a p p l i c a t i o n

on s e l e c t o r s

selections

choice

theorem

8. On the u n i f o r m i z a t i o n

of sets

9.

on s e l e c t i o n

Supplementary

remarks

stochastic

games.

. . . . . . . . . . . . . . . .

of m e a s u r a b l e

measurable

to

49

. . . . . . . . . .

58

. . . . . . . . . . .

65

in t o p o l o g i c a l

theorems

spaces

.....

. . . . . . . . .

76

85

This volume

is dedicated

to my parents.

I 2~rRODU3 t'lo ~ The purpose of these notes is to prove a few selection theorems and to mention some applications of these theorems. briefly outline the contents of these notes.

We will now

These notes are divided

into nine sections. We start with a selection theorem due to Michael,

in section

one, which yields a characterlsation of paracompactness. 2 we are concerned with the following question. exists a continuous selection on subsets of some topological

Suppose there

2 )[ (= space of all nonempty closed

space X), then what can you say about X ?

In other words for what space section on 2 X

In section

X

does there exist a continuous

?

The aim of section three is to show how the existence of continuous

selection for certain set-valued ~aps leads to the

existence of classical solution of some generallsed differential equations.

In section four, we prove a selection theorem due to

Dubins and Savage and then we apply this theorem to prove the existence of optimal

stationary strategies for the two players in

zero-sum two-person stochastic games. In section five we establish the following result due to Kuratowskl

and Ryll-Nardzewskl.

metric space.

Let

F : X --> 2 Y

x " F(x) ~ G # @ ~ e S

Let

Y

be a complete separable

be a measurable map [That is,

whenever G is open in

Y

and

S

countably additive family induced by a field of subsets of Then there is a selector G is open in Y.

Y

such that

X].

f-l(G) e S whenever

We also deduce a few more selection theorems with

the help of this theorem. hypothesis,

f - X->

is the

In section six, assuming the continuum

%~ present an example (due to Orkin) of a non-analytic

subset of [0, I] which is a Blackwell space.

This example depends

on a measurable selection theorem proved in the previous section. In section seven we prove essentially the following measurable choice theorem due to von-Neumann.

Let

T = [0, I] and

arbitrary complete separable metric space. set-valued function from

t.

F

be an

is an analytic

T to X, then there is a Lebesgue measur-

able point-valued function almost all

If

X

f

: T -- X

such that

f(t) • F(t) for

We use this result while characterising extreme

points of sets of vector functions. Given a set E in the cartesian product and Y, a set and

I[X U

of

U

is said to uniformlse E, if the projections E

and

U

through

the set

Y

for each

x ~ ]IX E

above

consists of a single point.

x

X x y of two spaces X

onto

(~x) ,~Y)/] U

X

7[X E

coincide, and if,

of points of

Y

lying

In section eight, we are

concerned with the question of the existence of such uniformising sets. In the last section we mention further results on selection theory with some remarks.

I. CO~21NUOUS SELZCTIONS One of the most interesting and important problems in topology is the extention problem. T~o topolgical spaces given,

and

Y

together with a closed subset A of X, and we ~ u l d

know whether every continuous function to a continuous function U

X

A) into

Y.

f

from

g , A -~ Y

are llke to

can be extended

X (or at least from some open

Sometimes there are additional requirements on

which frequently take the following form ". for every must be an element of a preassigned

subset of

which we call the selection problem, the extention problem,

and presents

Y.

x ¢ X,

f,

f(x)

This new problem,

is clearly more general than a challenge even when

A

is

the null set or a one-point set (where the extention proble= is trivial). Let

X

and

Y

denote topological spaces and

family of non-empty subsets of Y. for

~

every

is a continuous

f : X--

If Y

$ : X -- 2 Y such that

2 Y denote the then a selection

f(x) z $(x) for

x ¢ X.

Example I.I: Let @(x) = u-l(x).

u

: Y -- X

Then

is continuous and

f

be onto.

is a selection for

f(x) ~ u-l(x) for every

Example 1.2: Let ~u." X -> 2 Y, let ~A.

Define

~ : X -" 2 Y by

@(x) = V ( x )

if

if and only if

x ~ X-A. f

Define

A < X

if and only if

and

f ". ~ -> Y

is a selection for ~

$

by f

x ¢ X.

~(x) = ~g(x)~

Then

$ ". X --~ 2Y

g if

be a selection for x s A

and

is a selection for

which extends

g.

The selection problem of the first paragraph can now be rephrased as follows : Under what conditions on " X-> ion for

2Y

can every selection for

@, or at least for

@IU

@iA

X,

A : X, Y and

be extended to a select-

for some open set

U ~ A

The purpose of this section is to give a solution to the above problem. Definition" Call ~ : X -~ 2Y lower semicontinuous l.s.c.) if

{x : $(x)~U~@}

is open in

X

(abbreviated as

for every open set U c Y .

The first step in answering the question, is provided by the following elementary but important necessary condition. Proposition I.I'.

If

~ " X -~ 2 Y has the property that, for every

Xoe X, there exists a selection for $~U (U a neighbourhood of which has a preassigned value

yo~ @(xO) at

xo)

xo, then ~ is lower

semlcontinuo us. Exsmple I.I*'. ~

is

I.s.c. if and only if

Example 1.2"" If ~ i s

u

is open.

l.s.c., A is closed in X and g

is continuous,

then ~ is l.s.c. Proof of proposition i.I: Let G = ~x : $(x) ~ V ~ ~}

V

be open in

is open in ~.

Y ~ we must show that

For each

Xo~ G, pick a

Yo ¢ #(Xo): then by assumption,

there exists a selection

for some neighbourhood

xo such that

U*o " Uo ~ x

Uo of

". fo(X) ~ V), then

is ~ontalned in

G.

Hence

f(Xo) = Yo"

fo for #~Uo, Now if

Uo* is a neighbourhood of

G

x0

which

is open and the proof is complete.

One can easily prove the following facts : (i) If

# : X -> 2 Y is l.s.c,

= ~

for every

x

where ~

of V(x) and #(x) respectively, (ii) If

# : X -~ 2Y is

with the property that the function (iii) If

Y

and if

l.s.c, #(x)NU

: X -~ 2 Y is such that and

then

(~x)

~ is l.s.c.

and if

U

is an open subset of

@(x) = U ~ $(x) is

is a topological linear space, and if ~

Y

is non-empty for every x ~ X, then

@ : X -~ 2 Y defined by

l.s.c., then the function

denote the closure

: X-~

l.s.c.

~ " X-,

~Y is

Z Y, defined by ~(x)--eonvex-hull

Now we are set to prove the following important result due to Michael which gives a characterisation of paracompact spaces. X

Recall

is paracompact if it is Hausdorff and if every open covering of

X

has an open locally finite refinement. Theorem l.l: The following properties of a

Tl-space

X

are equivalent

[i]. (a) X

is paracompact.

(b) If

Y

is a Banach space then every l.s.c, function ~'X -~ ~ ( Y )

(family of all closed convex subsets of Y) admits a selection. First we shall establish the follo~ing lemma. ~

~

If

X

semi-continuous and

is paracompact,

function from

X

Y

a normed linear space,

to the ~on-empty convex subsets of Y

r ~ O, then there exists a continuous

f(x) ~ S r ( T ( x ) ) Proof" Let

for every

~ a lower

x ~ X,

where

Uy = ~x : y ~ Sr(~Y(x)) )

f : X -~ Y Sr(Y(x))

for

y s Y.

such that

= ~y~d(y, ~y(x~)qr). Then

Uy = ~x ~ W~(x) C~ Sr(Y ) # ~}. Since y is l.s.c.,

U

is open for every

the open sphere of radius is a covering for

X

r with centre

and hence, since

open locally finite refinement if

~ # ~].

~W~ : ~ ~ A)

y]. X

W~ C V~ .

Clearly

Sr(Y) stands for ~Uy : y ~ Y) it has an

[ We assu~e V~ # V

there exists an open covering

Now we shall construct a family

~Poo : ~ ~ A) of continuous functions from interval with the following properties and

there

is paracompact,

~V~ " ~ ~ A).

By a theorem of Dieudonue, such that

y.

X

to the closed unit

" (i) p~ vanishes outside

~ p~<(x) = I for every x ~ X. By the normality of X c(a A we can find continuous functions q~ : X - ~ [0,I] such that ~ ( x ) ~ l when

(ii)

V~

x s W~

and

~(x)

= O

when

x ~ X-V .

Define

%(x) p~((x)

z

....

~e

Then a

~p~)

satisfies all our requirements.

y(~) ¢ Y

such that

fCx) =

V~ C Uy(~).

z

~e To s e e t h a t

this

neighbourhood U, f

Define

p~(x) y ( ~ ) . A

~rks,

U

Pick for each ~ ~ A~

observe

first

that

each

x ~ X

intersecting only finitely many

x • X

therefore

f

has a neighbourhood on which is continuous on

X.

f

Sr(V(x))

Ivbreover, for each

and hence

Hence

is continuous and

is a convex combination of finitely many

lie in the convex set

has a

V~ and on this

is the sum of finitely many continuous functions.

every

f(x)

"

A

x z X,

y(<), all of which

f(x) 8 Sr(~g(x)).

This

completes the proof of the lemma. Proof of theorem I.I : We will prove first

(a) ~

(b).

To prove

this it is sufficient if we construct a sequence of continuous functions (i)

fn : X -~ Y

such that for every

fn(X) z Srn.2(fn.l(X))

fn(X) ~ Srn(¢(x))

for

for

n = 2,3,...

n = 1,2, ...,

be sufficient, because then the

~fn}

where

for every

We construct fl

~fn}

and

(ii)

r n = 1/2 n.

This will

is uniformly Cauchy and

hence converges uniformly to a continuous f(x) ~ ¢(x)

x ~ X,

f

and that it follows

x z X. by induction.

The existence of a function

satisfying (ii) follows from lemma I.I.

Suppose

fl, f2,...,fn

have been constructed to satisfy (i) and (ii) and let us construct fn+l

to satisfy (i) and (ii).

For each

x ¢ X, let

Sn+l(X) = $(x) ~ S r (fn(X)).

Then

n

@n+l(X) that

is never empty by induction hypothesis.

@n+l is l.s.c.

is open if x° ~ V

U

That is, we must show

is open.

Let us check

={X.@nvl(X) O U W

To do this, we will show that every

has a neighbourhood contained in

Yo e @(x O) and a positive

k < rn

V.

iWow for

such that

W I = { x : $(x) ~ ~ ( f n ( X o ) ) ~

xo e V, pick

Yo ~ ~ ( f n ( X o ))"

Let

U ¥ @ } ,

W 2 = ~ x : fn (x) 6 Srn.k (fn (xo))) .

Clearly

W I is open since

is continuous:

and

can apply lemma I.I

xo ¢ W I N

l.s.e, and

W 2 C V.

!(fn(X))

for every and

x.

(a) ~

collection

(b) ~

P

Y = ~i(~)

(a).

E L y(U)L Ge~

Clearly

C

@(x) -- C

~ ~ y " y e Y,

Let ~ b e

Y

E

~

of

X

has a partition

to the non-negative and every

y(U) = O

i < oo }.

U a ~ZL and Y.

for every

x e X.

p e P

an open covering of

is a Banach space.

for every

for every

X

x 6 X

LM(U)

is a closed convex subset of

@(x) e ~ ( Y )

I.S.C.

This completes

To show this~ it is enough

for every

= { y i y . ~L--> R,

Plainly

is

(#(x)).

of continuous functions from

C = { y : y e Y, y(U) _> O

Clearly

such that

subordinated to it, this means that there exists a

E p(x) = I peP vanishes outside some U e ~ .

tiy Ii =

We

(b).

reals such that

Let

#n+l is l.s.c.

fn+l : X -~ Y

fn+l(X) ~ Sr

if we establish that every open covering P

fn

n+l

Now we shall prove

of unity

is open for

But then

nthe proof of

WE

Since

to find a continuous

fn+l(x) e Srn+l(¢n+l(X)) fn+l(X) e S r

¢ is

X.

Here Let

E y(U) = I }.

~ow for

x • X, let

U ¢ ~Lwlth

x ~ U }.

We will now show that

Let us first o f all show that, for every there exists a

y' a C

such that

for only finitely many pick

U1, U 2 . . . .

U ¢ ~LL.

Un ¢ ~

~jy-y'll~ ~

y'(U) -- 0

if

y(Ui~=y'(U i) for

such that

y(U i) > 0

i = 2,3,... n.

Now we will show that

@

is l.s.c.

y'

and

U I, U 2 . . . .

Since

y(U i) ~ 0

of

that

@

:

x

neighbourhood of Y' ~ ~(x')

for ~ Ui x.

y e @(x) and

Un

i

and y' e C

y'(U I) = y(U I) + 1-6 and

To show this it is

if

x' ~ U, there exists a

Suppose, therefore, that

for all

y' satisfies all our require-

then there exists a neighbourhood

that, for every

y'(U) ~ 0

It is easy to see that

enough if we establish the following ~ O,

~ ~ O,

¢/2, and then define

U ~ ( UI, U2, ... U n } ,

li y-y' Li_~ 2(I-~) ~ ~, and therefore ments.

and

and

To find such a y', we need only

Y(U I) + Y(U 2) + ... + Y(U n) = ~ ~ 1 by

y ~ C

x ~ X, U

of

x

y ~ @(x), and in

y' E @(x') with ~ > 0

be as given above.

X

such ily-y'li~¢.

are given, and let Let

u = UI ~ . . .

NUn.

i = 1,2,... n, it follows from the definition for

i = 1,2,... n

and hence

U

is indeed a

It also follows from the definition of

for every

x' ~ U, and therefore

U

~ that

satisfies all our

requirements. By assumption (b), there now exists a selection For each

U ~ ~I, define

fu " X --~ ~

by

to ~b6, since

for

@.

fu(x) = If(x)] (U).

now follows immediately from the definitions that is a partition of unity on

f

It

~ fu : U ~ ~ }

X, and this partition is subordinated

fu vanishes outside

U

for every

U ~ ~.

This

completes the proof of the theorem. Remark I.I : Even if

X

is the closed unit interval in the above

theorem, (b) becomes false if as the following example shows.

Banach is replaced by 'normed linear'

Example 1.3 • Let

X

be the closed unit interval.

exists a separable normed linear space and a $ : X -> ~ ( Y )

( = closed convex subsets of

Then there

l.s.c, Y)

function

for which there

is no selection. Proof: Let

Z

be the set of rationals in

ordered as a sequence

Zl, z2, . . . .

y(x) # 0

for only finitely many

y(x) _~ 0

for all

/

x ¢ Z }.

C

if

X, and suppose

Let

Y = {y : y E

x ¢ Z ~ .

Finally

x ~ X-

Let

Z

is

~(Z),

C =~y'y ¢ Y,

let

Z

¢(x) C ~{

y " y ~ Y, y(z n) _> I/n }

It is easy to check that no selection for Suppose each

zn z Z

¢ is

l.s.o.

if

x -- z n.

Let us show that there is

¢.

f

were a selection for

has a neighbourhood

kf(x)](z n) • I/2n

X

f

is continuous,

such that

By induction pick a sequence k of distinct integers such that znk+l ~ ~ U for all k. i=l n i

{nk} Then

whenever

U n in

$ , since

x ~ ~n"

{ ~nk } is a sequence of closed subsets of

intersection property, then

and hence there is an

[f(Xo)](znk ) • 0

for all

X

with co finite

xo ~

~ k=-I

Unk.

But

k, which is impossible.

As applications of the above theorem we cite the following two propositions " Proposition 1.2 " Let and

X

be a paracompact space,

$ : X -~ F~(y), a l.s.c, function.

re(x) = inf { ~ l.s.c,

with

m(x) > O.

y ~L : y ~ ~(x) }

p(x) Z 0

for all

Y

Let

and suppose that x,

and

Then there exists a selection

a Banach space

p : X -~ R

p(x) > re(x) whenever f

for

$

such that

is

I0 llf(x) il_~ p(x) for every

x ¢ X.

~¢(x) N{

\ {o}

(x)

Let

y : Y ¢ Y, ~ Y

--> ~ ( Y )

• X

by

is

l.s.c.

@(x) -- y(x----~then

@

Hence if we define is also

l.s.c.

by the above theorem, there now exists a selection and

f

p(x)>0

i: p(x)--0.

It is not hard to check that ~ @

~ ~ p(x)} if

f

Hence

for

@

satisfies all our requirements.

Proposition 1.3 :

A topological space

X

is paracompact if and

only if it is dominated by a collection of paracompact subsets. Before proving proposition

1.3, we ~dll start with a

definition. Def~nltion :

Let

X

closed subsets of

be a topological space and d5 a collection of

X.

Then <~ dominates

X

if, whenever

A ~ X

has a closed intersection with every e l ~ e n t of some subcollection ~I

of

~

R _ ~

which covers .

is closed.

If

If ~ dominates Y

f

f~B

is closed.

and if

(~I C

6~ then

~/ ~ I

f ".%Jd31 -~ Y

is continuous for every

B ¢ ~I,

is continuous.

Proof of proposition 1.3 " we need only take 6 part.

X

A

is any topological space and if

is a function such that then

A, then

The " o n l y if'' part is obvious, since

= { X } .

Let us therefore prove the ''if''

By theorem I.I, it is sufficient to prove that if

Banach space, and selection.

So let

~ . X-~ Y

and

~I(y) #

is

l.s.c., then

~

where ~ ~

, and

h

is a selection for

# I U

.

is a

admits a

be given, and let us find

Consider the class J~ of all couples of the form

Y

f. (~, h)

We partially

II order o~ in the obvious manner.

From the remark made above, it

follows that every simply ordered sub-class of ~ upperbound,

and hence, by 2orn's lemma, ~

say (~o,ho).

f = h o.

Suppose ~ o # ~ :

B e ~

B duo .

Let

with

has a maximal element

We need only show that ~o = ~

simply take

selection for

~ i B~

B ~(~o

has an obvious

, for then we can

then there exists a

) = B*.

Now

and hence by theorem I.I

holB*

is a

and lemma 1.2

(stated and proved below) it can be extended to a selection for

$ ~ B.

If we now let

hi : U ~ - " if

Y

by

x z B, then

than

h l ( X ) = ho(X)

(~l,hl)

(~o,ho).

~ I = ~ B }~3 ~ o if

g

and define

x SU~o

and

hl(X)=

g(x)

is an element of ~ which is larger

This contradicts the maxi~ality of

(~o,ho)

and

thus the proof is complete. Lemma 1,2 :

If

elements of ~

~

C_ 2 Y

contains all one-point subsets of

, then the following t ~

properties of ~

are

equivalent. (a)

Every lower semicontinuous

(b)

If

(where Proo_____f: let

V

0 . X ---~ A

O " X--~ ~

admits a selection.

is l.s.c., then every selection for

0 ~ A

is a closed set) can be extended to a selection for

That (b) =~ (a) " X -~ ~

is obvious.

be l.s.c., let

a selection for ~

.

Let

To show that

A c X

@.

(a) =~ (b),

be closed and let

g

be

~ : X -~ ~f be defined as follows.

$(x)-- { g(x) } = ~(x) It is not hard to check that

for

x s A

for

x

@

is

s

X-A.

l.s.c., hence admits a selection

by assumption (a) and this selection has the required properties and the proof is thus complete.

12 For the proper understanding theorem I.I and theorem 1.2 (stated below without proof) the reader should keep in mind the following consequence of lemma 1.2, where tion space with respect to continuous If

~

c

g : A -~ Y

2Y

itself, then (a)

Every

(b)

Y

is called an exten-

if, for every closed

A < X, every

can be extended to a continuous

contains every one point subset of

Y

f " X -~ Y.

and also

Y

(a) ---->(b) below : l.s.c.

~ ". X - ~

~

admits a selection.

is an extention space with respect to

Theorem 1.2 : Y

X

Y

If

X

X.

is paracompact and zero-dimensional and if

is a complete metric space, then every lo~er semicontinuous

function

~

from

X

to the non-empty, closed subsets of

Y admits

a selection. Proof of theorem 1.2 is similar in nature to the proof of theorem I.I and hence omitted. L~ecall that if every finite open covering of refinement, in other words,

X

X

X

is zero-dimensional

has a disjoint finite open

has a base consisting of sets which

are both open and closed]. The significance of theorem I.I lies in the fact that it is the first characterisation of paracompactness which deals with continuous functions rather than coverings.

Further the theorem

Yields the fact that every paracompact space and hence every metric space has property (b) of theorem 1.l.

i3 REFERENCES

[i]

E. Michael,

Continuous selections I,

Ann. of N~th. 63

[1956], 361-382.

[~]

E. Michael,

A survey of continuous selections, Springer-

Verlag Lecture notes in Mathematics, Edited by W. M. Fleischman, on ''Set valued mappings, selections and topological properties of

54-58.

2X'' . 171 [19701,

2.

CONTINUOUS SELECTI03~30~i ~ETRIC COI~TIg~UUM

In this section we are concerned with the following problem " X

is any topological space and

closed subsets of subspace of S

if

X

2X.

f(E) e E

for every

f : S -- X E c S.

Let

X

S

be a

is called a selection on

Suppose there exists a conti-

2X, then what can you say about

words for what spaces on

is the space of all nonempty

with the Vietoris topology.

A function

nuous selection on

2X

X

? In other

does there exist a continuous selection

2X ? Throughout this section we assume

unless otherwise stated. Hausdorff. sets of

X

to be a metric continuum

By 'continuum' we mean compact, connected,

F2(X) will stand for all two-element and one element

X.

A subbase for the Vietorls topology on

X

consists of

all sets having one of the following forms : ~ F ~ F e 2 X, where

G

F ~G

~ ~ }

and ~ F • F ¢ 2X, F

is an arbitrary open set in

X.

But if

metric space, the Hausdorff metric on

2X

induces the Vietorls

topology.

The Hausdorff metric d(%B)

where

~B

= max [ sup p(x,B), A

z 2 X and

if and o~ly if

d

p

on

is a compact

ZX is defined as follows

sup ~(x,A)] B

is the metric on

2 X is compact [I].

X

~ G )

X.

Also,

X

is compact

The purpose of this section is

to prove the following theorem [2]. Theorem 2.1:

If

M

Is a metric continuum for which there is a

continuous selection on Remark 2.1:

Note that if

continuous selection on

F2(M ) X

then

M

is an arc.

is a continuum and if there is a

F2(X) , then there is also a continuous

15

selection on

2 x.

However this does not remain true in the more

general case, for example in the space of reals IS]. Definition : F£(S). to

If

An element

f

is a set and suppose a ¢ S

f

x ¢ S-{a},

Lemma 2.1-

If

S

is a selection on

is an end element of

provided that either each element

or for each

S

S

X,

S

S-~a},

is a set and

Suppose

i -- 1,2,8

f({a,x})= a

f(~a,x}) -- x. f

Is a selection on

has at most two end-elements with respect to

Proof :

with respect

al,a2, as

F~(S), then

f.

are three end-elements.

Then

for

either (i,l) ...

f(ai, x) = ai or

(i,Z) ... Suppose that (l,1) holds. so that Since

f(al, x) = x . Then for

(J,l) cannot hold and

(J,2) must.

(2,2) holds this cannot be

cannot be

as.

J = 2,3,

a2

f(al, aj) = al ' Consider

f(a2, a3).

and since (3,2) holds this

But it must be one or the other.

Suppose on the

other hand that (l,£)holds.Then by a similar argument both (2,1) and (3,1) hold.

But then

f(~a2, a3} ) is both

a2

and

as, which

is also impossible. Lemma 2.2 :

Let

S

be

continuous selection on S

Is a cut-point of

Proof" Let

H

f(~x,c~) = x that

c

and connected,

F2(S).

and

K

and suppose

f

Then each non-end element

is a c

of

S (that is, S -~c} is not connected).

be the set of all points

f(~Y,c}) = c.

otherwise

T2

x

in

S -~c}

be the set of all points Then neither

H

nor

K

y

in

is empty,

would be an end element , f~rther

H

and

such that S -~c} such since K

are

16 disjoint. lemma,

We shall now show, in order to complete the proof of the

that

we show

H

c

is a cut-point of

and

K

are open.

there is an open set

U

f(~Y,C}) = y.

U t H

Hence

S.

Let

To do this it is enough if

x ~ H.

containing

x

so that

By continuity of

such that if H

is open.

f,

y ~ U,

Similarly

K

is

open. Proposition 2.1 " suppose

S be T e (= ~lausdorff) and connected and selection is a contlnuousAon F2(S). Then S contains at most

f

Let

t~o non-cut points. Pro.of :

This follows from lemma 2.1 and 2.2.

Proposition 2.2 : Every continuum

K

of more than one point has

at least t~o non-cut points. Proposition 2.3 : If non-cut points then that is,

K

K K

is homeomorphic

2.1, 2.2 and 2.3. 2.2 and 2.3.

D

_

x

Of course, we have yet to establish propositions

The rest of this section is devoted to the proof of

~

• A simple chain connecting tw~ points X

is a sequence

a ¢ U 1 only, .

The proof is immediate from propositions

We will start with some preliminaries.

space

D

~

and y

UI,U2,...

b s U n only and

- A space in

such that

X

in

of open sets of

Ui f~ Uj # ~

X, there is a continuous

f(O) -- x

Definition : We call y

Un

and

X

X

a

whenever

and X

b

of a

such that

li-jl _~ I.

is pathwise coILuected if for any two points

f(1) = y.

range) is called a path from

and

to the unit interval I,

is an arc.

Proof of theorem 2.1 :

them.

is a metric continuum with exactly t ~

x

to

f~mction

The function

f " [0,I] -~ X f (as well as its

y.

arcwise connected, for any t ~

if there is a homeomorphi~n

points

f " [0, I] -> X

x

such

17 that

f(0) = x

and

f(1) = y.

range) is called an arc from

The function x

to

f

(as well as its

y.

At this stage we would like to make the following observations'. For a proof of these refer [4]. (i)

If

X

is connected and ~% is any open cover of

any two points of

a

and

b

of

X

X, then

can be connected by a simple

chain consisting of elements of '~. (ii)

Every pathwise connected space is connected.

(ill)

Every connected, locally path~ise connected (that is

each point has a nelghbourbood base consisting of pathwise connected sets) X

is pathwise connected.

Definition: A continuum

K (compact, c o ~ e c t e d

irreducible about a subset

A

of

X

proper subcontinuum of

K

contains

is irreduclble between

a

and

One can show that if K

Let

is a point

p ~ X

X

and

A.

If

X

is a set V

be a connected

A-- ~a,b}, we say

X-

~p}

A

p

X - ~p} LThat is, X

K

Tl-Space.

A cut point of

is not connected.

is a cut point of U

whose union is

and

A

of

using 2orn's lemma.

X, we call it a non-cut point of

empty open subsets of

is

and no

is any continuum, any subset

{p, U, V} where

disconnect

A c K

X

b.

K

such that

not a cut point of of

provided

lies in a subcontinuum irreducible about

Definition:

T2-space) in

V

If

X.

X p

is

A cutting X

and

U

are disjoint non-

}~ - ~p}.

The property of being a cut point (in fact, of being a cutting) is preserved under homeomorphism, but continuous maps can destroy cut points. [0, ~ ]

For example consider the map

onto the unit circle in

R 2.

f(x) = (Cos x, Sin x) of

18 ~

:

If

then

K

U ~/{p}

is a continuum and

and

V ~p)

~p,U,V}

defined on

K

K

onto

closed sets U ~ ~p)

Lemma 2.4: then each of

each

x

if

x~

~

P

if

x ~ V

U ~ ~p}, and

[Since

and

f

is continuous on each of the

V ~ ~p), so

f

is continuous.

U ~ ~p} = K-V, U • ~p}

If

K U

is a continuum and and

V

{x, Ux, Vx}

of

V u~ ~p}

is contained in

x e U

inclusion,

is closed in

Thus

K

and

{p,U,V} is a cutting of

contains a non-cut point of

Suppose each point

Ux

U~p}

The part of the theorem in parentheses follows].

they disconnect say

J

U ~ {p}

thus compact.

cutting

But the map

is the continuous image of a connected space and therefore

connected.

Proof:

U ~p}.

by

f(x) = carries

K

are connected (and thus are continuum).

Proof: It suffices to prove the lemma for f

is a cutting of

x

in

K.

U

If both

K.

is a cut point inducing a Ux

and

Vx

meet

V u)~p},

which is impossible by lemma 2.3. U.

by lemma 2.3.

Now

U x <~ {x}

Since

{Ux~{X}

i~ [Ux t# ~x}] xeU

K,

So one,

is a continuum for : x~

U} is directed by

is a nonempty continuum contained in

U. Pick r ~ Uq, then

q ~ x~U ~ LOx~/~x }].

Then

q ~ U r [otherwise

and disconnect it].

Then

Ur

U r ~ {r}

Uq ~ and

U

Vr

(as above) and if both meet

does not contain

Vq<3 ~q} q

which

is a contradiction. Proof of proposition 2.2" cutting (p,U,V) of

K

If

p

is a cut point of

exists, and each of

U

and

K, V

then a contains a

19 non-cut point of no cut point of

K, by the previous lemma. K

On the other hand, if

exists, certainly there are two non-cut points

and the proof is complete. Let

N

be the set of non-cut points of a continuum

suppose a proper subcontinuum

L

show that this is impossible]. {x, U, V}

of

U

say

and

V

K

exists and L ~

U.

of If

L

Then

K

contains

- ~y}

is connected,

[We will

must lie in one or the other of V ~ ~x}

being a continuum itself

and

U • {x}

y # x.

Then

is connected and these

sets meet, so their union is connected. But their union is while

y z V, hence

ion since

L

y ~/ U, hence not in L ,

K

K -~y),

this is a contradict-

contains all the noncut points of

shown that a continuum

and

x ~ K-L, then a cutting

has two noncut points and thus a ~oncut point [VU~x}]

~.

K

K.

Thus we have

is irreducible about the set of its r~n-

cut points. Definitionfrom

b

A cut point

if a cutting

p

in a connected space

{p, U, V}

The set consisting of

a,b

and

E(a,b).

b

is denoted by

defined by : Pl -< P2

if

exists with

and all points

p

Pl = P2

Lemma 2.5" (a)

If

E(a,b)

a s U

separates a and

or

Pl

b ~ V.

which separate

The separation order on

E(a,b).

is a linear order :

a

E(a,b) is

separates a from

This is easily seen to be a partial order on separation order on

X

P2"

In fact, the

(Prove this).

E(a,b) has more than t~4o polnts, its order

topology is weaker than its subspace topology. (b) a

and

b, then

topo Io gy.

If

K

is a continuum with exactly t~o non-cut points

E(a,b) = K

and the topology on

K

is the order

20 Proof." (a) %

It suffices to note that, for

(~ E(a,b)

and

Vp (b E(a,b)

Eta, b) --

p ~ E(a,b), the sets

are open in

: q

E(a,b) and

Ca, b), q < p}

Vp /-~E(a,b) = ~ q" q ~ E(a,b), p ~ q} (b)

If

any cutting

p z K

~p, U, V}

contain one of

a

and

and

p

is not one of

of

K, by lemma 2.4

b.

Thus

a J

or

b, then given

and

p s E(a,b), so

V

each

E(a,b) = K.

From (a), the order topology is weaker than the given topology on K.

Suppose conversely,

assuming that

p

that

U

is not one of

is open in a

or

K

and

meets

K-U.

then form a family of closed subsets of section property (each

But the sets K

p.

If

[r,s] (~K-U

with the finite inter-

Jr, s] is closed in

their intersection (in the compact space U

containing

U

p z (r,s), the closed interval

jr, s] = { q " r < q < s }

ps

First

b, we will show that

contains some interval (r,s) = {q " r < q < s) not, then whenever

p ~ U.

K

K)

by part (a)). is nonempty.

Thus

But

and

which leads to a contradiction.

If

p

is one of a

or

b,

the

argument is similar. :

Every continuum with exactly two non-cut points is a

totally ordered set with the order topology induced by its separation order. D _ ~ : onto

Y

Let

X

and

Y

be ordered spaces.

is an order isomorphism if

x < y <=~ f(x) < f(y).

f

is

I-I

A map and

f

of

X

21 Every order isomorphism is homeomorphism relative to the order topologies on

X

rationals in (0, I) ; form

K/2 n

for

with

Y.

that is,

n = 1,2,.o.

easy to see that p, q ~ P

and

P

Let P

P

denote the set of dyadic

consists of all numbers of the

and

K = i~2,...,2n-I.

Then it is

has no largest or snallest element and if

p < q, then for some

r ~ P,

p < r < q.

Let

D

be any countable linearly ordered set with the following property (i) It has no largest or smallest element p < q

then there exists a

order isomorphic to

with

Let

D

r ~ D

D

(b)

given

with

p ~ r < q.

a

and

b.

p,q ~ D

with

p < q

b

(i.e. f(~

D

onto

P.

But each point

Is a cut point, dividing

x < y ~

D)

D

Note that "

whenever and

x ~ Ap

and

K

is order isomorphic Let

p

into sets y ~ Bp).

K ~

be an order other than and

Bp

It follows that

f(Bp (~ D) form a Dedekind cut of the dyadic

and thus uniquely determine an element

Defining

F(a) -- 0 f

f

of

rationals

extending

and

F(p) of (0, I).

F(b) = I, we have completed the Job of

to what is obviously an order isomorphism,

a homeonorphism of

K

onto

I.

and thus

This completes the proof of

proposition 2.3. With the aid of theorem 2.1, one can prove the following theorem.

is

the~e is an element

P, the set of dyadic rationals in (0, I).

or

D

p ~ r < q.

isomorphism of a

Then

with

has no smallest or largest element

From the remarks made above it follows that to

p, q ~ D

be a countable dense subset of

not containing the non-cut points (a)

If

P.

Proof of proposition 2.3" K

r ~ D

(ii)

For a proof refer [2].

22 Theorem ~:~ %

If

M

is a locally compact separable metric space

for which there is a continuous

selection on

F2(M) , then

M

is

homeomorphic to a subset of the real llne. We close this section with a final remark, namely, if is a metrlzable continuum, exists if and only if

X

then a continuous is an arc.

selection for

X 2X

23 REFERESCES

[i]

K. Kuratowski,

Topology, Vol. I, Acad. Press and

P.W.N.,

[1966]. [2]

K. Kuratowski, S.B.Nadler, and

G.S.Young,

Continuous

selections on locally compact separable metric spaces, Bulletin De L'Academie Polonaise Des Sciences

18,

[1970], 6-11.

[3]

E.

Michael,

Topologies on spaces of subsets,

Try.

Amer.

M_aZh. soc. 71, [1951], 15~-1s2.

[4]

S.

Willard,

General Topology,

Company, [1970].

Addison-Wesley Publishing

3.

CONTINUOUS SELECTIONS

A~D SOLUTiORS OF GENERALISED

DIFFERENTI AL EQUATIONS Let

En

denote Euclidean

origin centered ball of radius

n b

dimensional in

En,

space of all nonempty compact subsets of topology.

and

Bn

a closed,

_~(Bn) the metric

B n with the Hausdorff

We consider the generalised differential equation

x(t) ~ R(x(t)), x(O) = x° ~ E n ( ~ ( t ) where

space,

R - E n --> _~(Bn)

is continuous.

is an absolutely continuous for almost all

t

function

= dx(t) dt

)

(I)

"'"

A solution of equation (I) $

such that

in some neighbourhood of zero,

O(t) ~ R(#(t)) 0(0) = x° ,

a

classical

solution of equation (I) is a continuously differentiable

function

~

with

~(0) = x0

and

O(t) e R ( ~ ( t ) ) f o r

t

in

some

neighbourhood of zero. The aim of this section is to show how the existence of continuous selection for equation (I).

R

For instance,

leads to (classical) if

R

solution of

has convex values,

the existence

of classical solutions of equation (I) may be shown in many ways [2]. This may be proved as follows. for En

R(x),

Let

r(x)

be the clrcumcentre

that is : the centre of the unique smallest sphere in

containing

selection for

R(x) . R.

But

Then

[I, pp. 76-77]

~(t) = r(x(t)),

r

is a continuous

x(O) = x0

has a solution

and this solution is a classical solution of (I). We shall show in this section that if variation,

is of bounded

a classical solution of equation (I) exists, while under

a weaker condition, These conditions map

R

termed property

~

a solution is shown to exist.

are pertinent to the question of when a continuous

Q " [O,T] -~ _~(Bn) admits a continuous selection,

i.e. a

25

continuous map

q " [O, TJ -> En

is shown that if exists.

q

with

q(t) ~ Q(t)

is of bounded variation,

A special case of this is if

Q

k, then one may show

q, also with Lipschitz

k.

We assume, throughout, where

It

is Lipschitzian in the

admits a Lipschitz continuous selection

constant

t.

a continuous selection

Hausdorff metric, say with Lipschitz constant Q

for all

that

~ " E n -> _~(Bn) continuously,

B n is as in the introduction.

of continuous (n norm, and

CLO, T] will denote the space

vector valued) functions on

Cb[O,T]

the compact subset of

[O,T] with the uniform

C[O,T] defined by

Cb[O,T] = { x : x ~ C~O,T], x(O) = x°,Ix(t)-x(t')l~ blt-t'l} . If

Q ~ C(B n)

and

y a En

we use the notation

p(y, Q) = inf { iy-qL " q ~ R } ,

while for

Ql' Q2 ~ -~(Bn),

h(Ql, Q2) denotes the Hausdorff distance between these sets. For

Q " [0,T] -> _9(Bn), define the variation of

subinterval

it - ~

t], ~ > O, denoted by

Vt

•

Let

P

denote a partition of

of points

(Q)

t-~

Property A " lira a- ~0

IT vt~ 0

=

For each

sup PeO

as follows.

it-a, t], i.e. a finite collection

For the partition

P,

k vt't-~ (Q • P) = ~=OZ h(Q(t ~+i ), Q(ty ))

vt (Q) t-~

on the

'

t-~ = to • t I < ... < tk+ 1 = t, and let

set of all such partitions.

Q

~

denote the

define

and

vt (Q, P). t-q

x ¢ Cb[O'T],

(R(x(.))) dt -- 0

vtt-~ (R(x(.))) ~ LI[O,T]

uniformly for

x s

Cb[O , T].

and

26

(Assume

x(t) = x(t) = x(O)

is defined for

Vt (R(x( • ))) t-~ This convention will be assumed throughout,

t-~ < O.

for

t < ~

so that

when necessary). At this Juncture we would like to make the following observations (i)

"

It is clearly possible that the variation

unbounded for some the special case

t, Q

yet

vt~

(Q) ~ L I.

t (Q) Vt_ ~

For example,

is

consider

a point valued function in the interval

[O,1]

defined by J(l-t)

Sin(l/l-t)

~en

t # I

when

t = I.

q(t) 0

(il)

If

R

is Lipschitzian,

say

h(R(x), R(x')) _~ Klx-x'l,

clearly property A is satisfied. Indeed vtt.~ (R(x(.)) ,P) -k k ~=07 h(R(x(t +i)), R(x(t ))) _< K b ~=OZ (t +l-t ) = K b ~ for x a Cb[O,T] . (iii)

Let

R " E n --> ~(B n) continuously and let

We define the variation of Let

R

in

S, d e ~ t e d

~y:

ly-x°l_~ b T}.

V(R,S) as follows.

k

be any finite collection of points yl ,...,yk+l in S such k that Z ly +i _ y ~ _~ b T and let /k denote the set of all such ~=i k collections. Let V(R,S,k) = Z h(R(yi+l), R(yi)) and V(RpS) is i=l defined as sup V(R, Stk) . If V(R,$) < oc we say R has bounded k~A variation in S. Clearly if R has bounded variation in S, then for any

x ~ Cb[O,T] ,

is finite for all In this case,

t

V~.~ (R(x(.)))


in fact

vot(R(x(.)))

in [0,T] and continuous as a function of

Vtt-~(R(x('))) = Vot(R(x('))) - vt-~(R(x(')))o

Vtt_~ (R(x(.))) -- 0

as

~ -~ 0

uniformly in

x ~ Cb[0,T].

t.

and hence

27 (iv) For of

Q.

Q ¢ =C(Bn)

For each

let

x~

Qe

Cb[0,T]

[increasing the radius of notation family.

Fu].

Then

denote a closed we consider

B n by

~

• ~ 0

neighbourhood

Re(x(.)) . [0,T] -~C(Bn].

if necessary but retaining the

~ Re(x(.)) " x e Cb[0,T] )

is an equicontlnuous

One may note that in the proof of theorem S.I, which follows,

the essential point is to conclude that one may obtain a continuous (or measurable) selection r x . x s Cb [O,T] )

If

R~ (x(.)) such that the family

is conditionally co~pact.

Theorem S.l [4]. Let (a)

r x for each

R : En -~ _~(Bn) continuously.

R is of bounded variation in the ball ~y "ly-x°l_~ b T}

equation (I) admits a classical solution on (b)

If

solution on

R

satisfies property

The family

R(x(. ))

each

~i ~ O,

i -- 1,2,..., 6i ~ 0

let

6i -~ 0

as a map of

as

i ~

qo e R(x°).

For each

Let

-61_~ t_~ 6 i.

i x (t) = x° +

for I%

t ¢ [-61, ~i ], and choose - q~l = P(qo' R(xi(6i )))"

r i to

[-61, 26 i]

~(Bn).

For

oo.

as follows.

if

[0,T] to

We assume, without loss of generality,

r i ". [-61, T] --~ En ri(t) = qo

is equicontinuous

-~ 0 as i -~oo and for each ¢i h(R(x(t)), R(x(t'))) ~ ¢i if

be such that,

~t~t'~ { 6i' x e Cb[O,T].

Pick

equation (i) admits a

~R(x(.)) ~ x e Cb[O,T] }

where we consider each

that

4

~O, TJ.

[O,T].

Proof of (a) "

¢i' let

Then,

i = 1,2,... Let

define

ri(t) = qo

if

-61 -~ t_~ 6i"

Define

t I ri( % ) d~ 0

... (2)

ql s ~(xi(~i )) such that We define

rJ(28i ) as

qlI

as the straight llne segment Joining

and extend qo and

qi

28 on the interval

[8i' 2~i]"

We now extend

equation (2). Proceeding inductively, if interval ['~I' j ~i ] choose

xi to xi

[-~I' 2~i]

by

is define on the

q~ ~ R(xi(J~i )) such that

lqij-I - q~L = P(q~-l' R(xi(J~i )))' define rJ((J+l) ~i ) as

q~

and extend ri to [-~I' (J+l) ~i ] as the straight line segment joining qJ-I i and q~ on ~J~i,(J+l) ~i ]. Then extend xi to the interval [-~I' (J+l) 8i ] by equation (2). Continue in this manner until each function that

r i s C[-~I, T]

ri and x i while

is defined on

L-~I, T].

We note

xi ¢ Cb[O,T] , when restricted to

[O,T]. Furthermore we have (i)

iri((j+l) 8i ) _ ri(JSi)i = ~q~ _ qJ-I ~ i = P(qj-l'i R(xi(j~i))) h(R(xi((j-l) ~i ))' R(xi(J~i )))~ ~i

(ii) Given any ~J~l-ti ~ ~i"

t s [O,T] there exists an integer

j such that

Then

p(ri(t), R(xi(t))) ~ ~ri(t) - ri(J~i)l + ~(ri(JSi ), R(xi(J~i)))

+ h(a(xi(J~i)), a(xi(t))) ~ 3~ i (iii) For any

t,t' z[O,T], let

Lt-J~il ( ~i' Lt'-J'~il ~ ~i" that

J' ~ J.

j,J~ be integers such that Assume, with no loss of generality

Then

~ri(t) - ri(t,)l ~ p(r1(t), R(xi(jsi))) j'-I i + E h(R(x (~+l)~i)) , R(xi(p~i))) + p(ri(t'), R(xi(j'~i))) J'-I 8~ i + E hCR(i((~+l)~i)), R(xi(v~i))]

29 We are now in a position to prove conclusion (a) of theorem 3.1. If

R

that

is of bounded variation, ~ri(.))

is a bounded equicontlnuous

Indeed,

given any

i _~ i*

implies

0 _< ~ ~ ~,

it follows (from observation

8 • O, choose

i~

[ri(t)

ri(t')[

~,

convergent subsequauce {xik( • )}

as desired. ~rik(.)}

From (il) we see

such that for x e Cb[O,T].

i _~ i*, it-t'[ ( 6

Thus

~r i}

Then then

has a uniformly

which converges,

has a uniformly convergent

x 8 C b [O,T]

~ • 0

uniformly for

from (iii), above, it follows that if

C[O,T].

sufficiently large so that

4¢ i • 8/2 ; no,~ choose

vt.~ R(x(.)) < 8/2

sequence in

(iii)

subsequence,

say, to

r.

Then

say converging to

r(t) ¢ i~(x(t)), since

R(x(t)) is

t x(t) :x°+ S r(T)dT. 0 t , x(O) = x °, showing x is a

closed, while taking a limit in equation (2) gives Thus

~(t) = r(t) ¢ R(x(t)) for all

classical solution of (I). To obtain conclusion (b) of theorem 3.I, if property

A

i

satisfies

we find from (iii) that

T T t I [ri(t) - ri(t-~)idt • 48i2 + ~ Vt_~(R(xl(.))) 0 -0

dt.

Now given any

8 • 0 we can use property A to choose a ~ • 0 0 T such that ~ Vt (R(x i (.)))dt < ~/2 if 0 < ~ < Oo, all t-~ 0 i = 1,2, . . . . Next choose i* sufficiently large so that for i _~ i*, T 4 8iT ~ e/2. Then for i _~ i v, 0 _~ ~ < Oo, f [ri(t)-ri(t-~)Idt < ~. 0 By [5, Theorem 20, pp 298], ~ r i} is LIfO,T] conditionally compact. Thus it has an converges,

LILO,T]

say, to

subsequence,

r.

convergent subsequence, Again

convering say,

~x ik) to

~r ik} , which

has a uniformly convergent

x e Cb[O,T].

From (ii) we see

30 r(t) s R(x(t)) almost everywhere, while taking a limit in equation (2) gives

x(t) = x° + Itr(z) d~ or 0 everywhere, x(O) = x°, showing that

~(t) = r(t) s R(x(t)) almost x

is a solution (nOt necessari-

ly a classical solution) of equation (i).

This completes the proof

of theorem 3.1. Remark 3.1 :

The following question naturally arises.

equlcontinuous family

~Q~ " ~ ~ A}

of mappings

when does there exist a measurable selection that the family

~q~ : ~ ~ A}

is

Given an

Q~ " [O,T] -~ _~(Bn),

q~ for each

Q~

such

LIIO, T] conditionally compact.

sufficient condition is shown to be, essentially property (A).

A

It

would be interesting, if one could show the existence of such a family

~q~ • c( a A}

without assuming property (A).

This would then

lead to an existence theorem for a solution of equation (i) with : E n -~ C(B n) merely continuous.

The existence of a solution in

this case seems to be an open question [4]. Remark 3.2 "

We shall show in another section the existence of

measurable selections for continuous or lower-semlcontinuous mappings F : X->

2Y

where

metric space and

X 2Y

is a metric space r

Y

a complete separable

the set of nonempty closed subsets of

the Hausdorff topology.

Y

In general, one cannot expect more than a

Balre selector, indeed if

Q : [O,T] -~ _~(B~) continuously, many

examples exist to show that

Q

need not adz~it a continuous selection.

Consider the following example due to Hermes [3]. Example 3.1 :

For

t ~[0, I]

s(t)-- ~ x - - ~ s ~ , and for

t

• 0

let

with

let

y-- Sin~ : t _ ( ~ a m

}~

E2

31 I A(t) =

Cos I/t

Sin i/t 1

-Sin I/t

Cos I/t

Define

I R(t)

To show

R

show that R

=

A(t) S(t)

if t ~ 0

S(O)

if t-- O

is continuous in the Hausdorff topology, it suffices to R

is continuous at

0.

contains a continuous function

subset of the cylinder t ~ O

But

h(R(t), R(0)) _< ~t.

r, the graph of

S(O) ~ [0, i].

is a connected

But the gap in R(t) for

will disconnect a n y such arc on

continuous on

r

If

S(0) ~ 10,1].

Thus

R

is

[O,1] but there does not exist a continuous point

valued function

r

defined on

[0,I] with values

r(t) ~ R(~).

However one can prove the following theore~ [4]. Theorem 8.2 "

Let

Q ". [O,T] -~ _~(Bn) continuously. vT(Q) • G O

has bounded variation in [O,T], that is, admits a continuous selection

r

where

variation with same variation as Proof :

Choose

J = 0,...,k. (1)

For any

such that

Q

is also of bounded

k, consider the points

qo k ~ Q(O): ql E Q(T/k) and such that

lqk- qkl = p(qk, Q(T/k)), and inductively rk : [O,T] -~ E n

Then

Q

Q.

For each positive integer

O, T/k, 2T/k, ...,T.

r

Suppose

q~ e Q(JT/k)).

as the polygonal arc Joining the points

Define q~,

Then t ~ [O,T] and any

it - JT/kl • T/k.

t ~ [(J-l)T/k, JT/k].

Then

k, there exists an integer

J = J(k)

Assume with no loss of generality, that p(rk(t), Q(t))_< Irk(t) - rk(jT/k) +

p(rk(jT/k), Q(t))_~ h(Q(J-l)T/k), Q(Jr/k))+ h(Q(JT/k), q(t)).

32

(ii)

For

t, t' s LO, T]

and any

k, let J, J'

that

~t-

JT/kl < T/k,

Then

~rk(t) - rk(t')l _~ rk(t) - rk(jT/k) +

be integers such

~t' - J'T/k ~ < T/~.

rk (~T/k){

jl-I Z jrk((~+l)T/k)

-

+ Irk (J'T/k) - rk(t')l

jl- I <

We now show that

~rk}

Z

~ ----j

h(Q( ()/+l) Y/k), Q ( ~ T / k ) )+h(Q(t' ), Q(JT/k)).

is equicontinuous.

k*

sufficiently large so that for

if

Itl-t21 < T/k*.

Next, since

is continuous as a function of uniformly continuous, Vt,t (Q) < s/S

if

~/3

is of bounded variation,

Vt(Q)

on

C0,T], as is

and we can choose a

It-t'L ~ ~.

~ > 0

Then, from (ii), we have : for

Clearly the sequence

uniformly convergent subsequence, Let

t s ~O, TS

that is, set

and

J(k)T/k -> t

J(k) as

Vtt ~ (Q), hence

so that

It-t'l ~ ~, Irk(t) - rk(t')l < s/S + ~/S + a/3 = ~ ty is shown.

~ > O, choose

k > k*, h(Q(tl) , Q(t2)) < Q

t

Given any

~r k)

and equicontlnui-

is bounded hence it has a

say converging to

r ~ CLO, T].

be an integer such that It-J(k)T/kl k -> oo.

Q(t) is closed, it follows that

the desired continuous selection.

k _> k*,

< T/k,

From (i), and the fact that the r(t) ~ Q(t), that is, r

is

This completes the proof of

theo rein S. 2.

Equations of the form (i) arise naturally in many ways. Consider the implicit differential x(O) = x°.

equation

f(t, x, k) = O,

This may be reduced to an equation of the form (I) with

R(t,x) -- ~ v " f(t, x, v) = 0 }.

(Note that here R • [0, I] ×En->En).

Another way in which equation (I) appears is in the theory of control

33 systems having equations of the motion of the form ~ = f(t,x,u), 0 x(O) = x , where the control function u may be chosen as any measurable

r

vector valued function with value at time

a preassigned set

U(t)
R(t,x) =

and one may say

R

t

in

Here

~ f(t,x,u) : U ~ U(t) }

has the representation (f, U).

about control problems in another section.

We shall study

34 REFERENCES

[i]

H. C. Eggleston,

Convexity,

Cambridge Tracts in Mathematics

and Mathematical Physics No. 47, Cambridge University Press London-NewYork, [1988].

[2]

A. F. Filippov,

Classical solutions of Differential Equations

with multivalued Right Hand Sides (Eng. Trans. )

SIAM J.

Control, 5 [1967], 609-621.

[3]

H. Hermes,

The generalised Differential Equation

Advances in Math

[4]

~ ¢ R(t,x),

2 [1970], 149-169.

H. Hermes, On continuous and measurable selections and the existence of solutions of Generalised Differential Equations, Proc. Am~r. Math. Soc.

[5]

N. Dunford, and J. T. Schwartz,

29 [1971], 588-542.

Linear Operators I,

Interscience Pub. Inc., New York, [1958].

4.

MEASURABLE SELECTIONS AND AN APPLICATION TO STOCHASTIC GAMES The aim of this section is to prove a selection theorem due

to Dubins and Savage[7~.We then apply this theorem to prove the existence of optimal stationary strategies for the t~o players in stochastic games. We start with some preliminaries. space with metric

p.

Denote by

closed subsets of A .

~

Let A

be a compact metric

the collection of all nonempty

We introduce a metric

d

on

2

- the

Hausdorff metric (which we defined in section 2), that is, for any A, B e

2

,

d(4

B) = max ~

For any sequence

~ ~

sup p(x, B), x~A : n--1,2,

sup p(y, A)} y~B

... }

define

lim

{ p : p ~ ~ , there exists an increasing sequence of natural numbers such that

Pk

~ ~

and

Pk

n lira An = ~ p • p a A , ~ clear that

lira A

Pn ~ ~

and !ira ~

~

=

k I ~ k 2 < ... -- p } .

Define

n such that

are closed.

Pn -~ p }" In case

we say that the limit exists and denote it by

I~

It is ~

= lira

lira ~ .

Definition . Let X be a metric space and let F be a map from ~h X --> ~ . Call F upper semicontlnuous in the sense of Kuratowski if

xn -~ x

implies

lira F(x n) C

F(x).

Now we would like to make the following observations. (i) ( 4

,d)

is a compact metric space.

(ii) d C ~ , A ) - ~ O

if and only if

lira ~

= A.

36 z~ (iii) F

If

F

is

u.s.c,

is Borel measurable,

subset of

X

that is,

whenever

See Kuratowski

from a metric space

B

X

into

~ x : F(x) ~ B #

is a Borel subset of

2

~ }

, then is a Borel

.

[2] for a proof of these observations.

Now we shall establish the selection theorem of Dubins and Savage.

We shall follow the proof as given in Maitra [3].

be a compact metric space, v

an -~ ao

for all

implies

A

a Borel subset of a Polish space and

a bounded, upper semicontinuous real valued function on

is,

limsup v(an) _< V(ao).

Assume

A? that

Iv(a)i _< H

a ~ A.

Lemma 4.1 : iS

S

Let

Define

v* : 2 A -~ R

by

v*(K) = max v(a). azK

Then

v*

U.S.C.

proof : exists

As

v

ao ~ K

is

u.s.c,

such that

and assume that for some subsequence

and

~ V(an,)

K

v*(K)

compact it follows that there

-----

V(ao).

an ~ Kn,

such that

v*(K n) = v ( ~ ) . v(an,) ->

compact, there exists a subsequence an,, -~ a. limsup

It follows that

Now suppose

a ¢ K

of

and since

{a n,} v

m,

K

Choose a

li--~v*(Kn).

{an,,}

v*(K n) = lim v(a n, ,) < v(a) _< v*(K)

Kn

is

As

A

is

such that u.s.c.

which proves

v*

is

u. s. c,

Le~ma 4 2 :

For each

(K, x) = ~ a : a ~ K is the domain of (K, x) ~ @ 1"

V ) Then

a compact metric space. to

£A.

K ~ 2A and

and

v(a) Z x } .

the set don

x ~ I-N, M]

V

define

Denote by

don V

~ (K,x) : (K,x) E 2 A x [ - M , is closed in

Furthermore

V

is

2A x[-M, u.s.c,

N]

from

(that M]

and

and so don

37 Proof .

Since

element of

v

2 A.

is

Next let us show that

(Kn, x n) ~ dora V, Let

u.s.c., for any real

exists a subsequence

that

dora V

n = 1,2,..., and suppose

a n ¢ ~ (Kn, Xn) , n = 1,2, . . . .

Consequently

c, ~ a " v(a) ~ c }is an

a ~ K

{ank ) and

a ~ V (K, x).

of

x = lira ~nk

A

is compact, there

such that < llmsup

Hence (K, x) z dom

Let

(Kn, x n) -> (K,x).

Since

{an}

is closed.

%

-~ a.

v(%)

_< v(a), so

~, which is, therefore

closed. Finally, in order to prove that show that a n -~ a

V

is u.s.c,

(Kn, x n) --> (K, x), an ¢ V (Kn, Xn) , a ~ V (K, x) •

imply that

Consequently,

since

Since

we have to

n = 1,2,...,

K n -~K,

a~

K.

x = lira x n _< limsup v(a n) _< v(a), a • V

(K, x).

This completes the proof of lemma 4.2. L emma 4,S "

Define

v(a) = v*(K)}.

V

Then

on V

2A

by

V(K) = ~a : a ~ K

and

is a Borel measurable map from

2A

into

2 A" Proof :

As

v

is closed. since

v

Let is

closed, a e K. Let

is

D

u.s.c.

VCK) is nonempty.

Let us show that it

an e V(K), n = 1,2,..., and suppose

an -> a.

u.s.c., v*(K) = limsup v(an) _< v(a) and as Consequently

a e V(K).

be a Borel subset of

2 A.

Hence

V

Note that

K

maps

2A

Then,

is into 2 A.

VCK) = V (K,v*(K)).

Consequently, { K : K ~ 2A

and

V(K) ~ D }

-- proJ [ {(K,x) . (K,x) ~ dora V and tV (K,x) ~ O}~{(K,x)'v*CK)=x~ .

.

.

(i)

]

38 As

v*

is u.s.c,

into L-M, M]

and hence its graph is a Borel set in

(See ~2]). map from

by lamina 4.1, it is a Borel function from

Also

V

dora V

is

into

u.s.c, 2 A.

2A ~L-M,

M]

by lemma 4.2, and so it is a Borel

Consequently,

~(K,x) : (K,x) ~ dora V, is a Borel subset of

2A

V (K,x) ¢ D }

dora V, which being closed in

the former set is a Borel subset of

2 A × [-M,M]

2A ~

~-M, MS,

as well.

Thus

the set within square brackets on the rlght-hand side of (I) is a Borel subset of

2 A × ~-M, MS.

Finally, projection being a

continuous map, and moreover, I-i in this case t it follows by a well-known theorem of Lusln K '. K ~ 2 A

and

is a Borel map. Lemma 4 4 : Define

Ecf

V(K) ¢ D)

be a bounded

u" • S ~

R

by

As

so that

u*(s) is well-deflned.

is

u~(s n) = U(Sn, an) ,

is

~an,,)

is :

U - S -~ 2 A Then

U

u.s.c, Let

by

of

lim U(Sn,,, nl

u*

V

function on

S ~A.

sn -~ s

and suppose

Choose a subsequence M~reover, ~an, )

as

A

such that

~u~(Sn,) )

is compact, an,, -~

a

.

u.s.c., it follows that,

llmsup u*(s n) =

L _ ~

Let

n = 1,2, . . . .

there is a subsequence

Hence

u.s.c,

u*(Sn,) -~ limsup u*(Sn) .

u

Hence

u'(s) = max u(s,a). Then u* is u.s.c. a~A u.s.c., for fixed s, u(s,. ) is u.s.c, in a,

Proo__..~f•

Since

2 A.

that

This completes the proof of lemma 4.8. u

such that

[21, pp 4871

is a Borel subset of

Let

u

Kuratowski

u

an ,,) <

!

u(s,a) • u~(s)

~

~

•

and the proof is complete. be a bounded

U(s) = { a " a ¢ A

is a Borel map.

u.s.c, on and

S ~A.

Define

u(s,a) = max a'¢A

u(s,a')}.

39 Proof of this lemma is similar to lemma 4.3. Next we shall state and prove the selection theorem due to Dubins and Savage. Theorem 4.1 :

Let

u

be a bounded

exists a Borel measurable map

f

u.s.c,

from

S

on

S x~

into

A

Then there

such that

u(s, f(s)) = max u(s,a) for all s. [ In this theorem A is assumed asA to be compact metric and S is a Borel subset of some Polish space]. Proof "

Choose a sequence

valued functions on

4

{Vn} , n : 1,2,... of continuous real

which separate points of

A.

one may choose a sequence of functions dense in

[For instance

C(A)].

For each

vi, define, for K ~ £A, vi(K ) : ~ a : a ~ K Then by lemma 4.3, each be as in lemma 4.5.

Define

By virtue of lemma 4.5 from Since

S

into A

2 A.

Vi

and

vi(a) : sup vi(a') ) . a'~K

a Borel map from Ul(s) = Vl(O(s))

it follows that each

Moreover, for each

is compact and

2A

s,

into

and Un

2 A.

Let

Un(S)=Vn(Un_l(S)) .

is a Borel map

U(s) D Ul(S ) ~ U2(s)

Ui(s) are closed in

U

A

...

it follows

Co

co

f~ Ui(S ) # @. Suppose now that for some s ~ S, a,a' ~ ~ Un(S) i=l n=l and a # a'. Then, for every n, as a,a' s Un(S) , it follows that Vn(a) = Vn(a') , which contradicts the separating property of the OO

sequence

~Vn}.

Hence

a singleton, say,

a = a'

and

for each

s,

(~ Un(S) n=l

is

{f(s)}.

Next, let us show that for every

s ~ S,

~f(s)} = lim

Un(S)

n-)oo

in the Hausdorff metric of Then there exists a

2 A.

~ank } with

Fix

s

and suppose that

ank z Unk(s)

and

~

-~

a ~ limUn(S). a

.

As

40 each

UmCS) is closed and

that

a ~ Urn(s) and consequently, a ~

li--~ Un(S) C ~f(s)). Hence

ank ~ Urn(s) co for all

n

S -~ ~A map from

it now follows that S

into

~A.

As each

Moreover as

f

U

iim Un(S).

is a Borel map from

n

is a Borel measurable

Finally it is Not hard to check that the 2A

is isometric to

is a Borel measurable map from

f(s) ~ U(s)

max u(s,a) for every as A

it follows

Hence

~f(s)} ~

@ " s -~ ~f(s))

class of all singletons belonging to follows that

~ Un(S). n--I

Also it is clear that

~ f(s) ) = lira U (s).

n k _~ m

for each

s s S.

s ¢ S, ~e get

S

into

A.

It

A.

u(s,f(s)) =

This completes the proof of (the

selection) theorem 4.1. Stochastic Games • S, 4

B, q, r.

A stochastic game is determined by five objects :

Here

S

is a nonempty Borel subset of a Polish

space, the set of states of a system: A

is a Nonempty Borel subset

of a Polish space, the set of actions available to Player I ,

B

is a nonempty Borel subset of a Polish space, the set of actions available to Player II ~ q

is the law of motion of the system ,

it

associates (Borel measurably) with each triple (s,a,b) a S x A x a probability measure

q(.Ls, a,b) on the Borel subsets of

S ,

the reward function, is a bounded measurable function on S ~ A

B r,

~B.

Periodically (say, once a day) Players I and II observe the current state

s of the system and choose actions

a

and

b, respectively ,

the choice of the actions is made with full knowledge of the history of the system as it has evolved to the present.

As a consequence of

the actions chosen by the players, two things happen ~ Player II pays player I new state

r(s,a,b) units of money, and the system moves to a

s' according to the distribution

whole process is repeated from the new state

q(. is, a,b). s'.

Then, the

Furthermore, there

41

is specified a discount factor today is worth

8 n at

n

8, 0 _< 8 < i, so that the unit income

days in the future.

The problem, then,

is to maximize the total expected discounted gain of Player I as the game proceeds over the infinite future and to minimize the total expected loss of Player II. A strategy v for Play I is a sequence

~I' w e , " ' ,

where

=n

specifies the action to be chosen by Player I on the n th day by associating (Borel measurably) with each history Sn_ I, an_l, bn_l, sn)

of the system a probability distribution

~n (. lh) on the Borel sets of

A.

A strategy

said to be stationary if there is a Borel map where

PA

h = (sl, al, bl,...,

~

for Player I f

from

S

to

is PA '

is the set of all probability measures on the Borel sets

of A, such that denoted by

~n = f

for each

n > I , and, in this case, ~

is

fco. Strategies and stationary strategies for Player I!

are defined analogously. A pair (~, r) of strategies for Players I and II associates with each initial state

s

an

nth-day expected gain

rn(= , C)

(s)

for Player I and a total expected discounted gain for Player I c<)

I( ~, F) (S) = n=17 8 n'l r n ( ~, F) (s).

The functions

rn(~, F) are, plainly, Borel measurable ;

consequently,

I(=, F) is a Borel function.

A strategy inf F

sup I(~, F)

~* is optimal for Player I (s)

for all r' and all

I( ~*, F') (s) _>

s ~ S , strategy

F* for

7r

Player II is optimal if ~'

if

and

and all

value if

s ~ S.

sup inf I(~, F)(s) _> I(~', F ~) (s) F

We shall say that the stochastic game has a

sup inf I(~, F)(s) = inf sup I(~, F)(s) v

for all

F

for every

s ~ S.

42

In case the stochastic game has a value, the quantity sup inf I(~, F )(s), F ?r function.

as a function on

S, is called the value

The stochastic game problem was first formulated by Shapley [6] who took

S, 4

B

to be finite, assumed that play would terminate

in a finite number of stages with probability one, and considered only what we have called stationary strategies.

Shapley was able to

prove under these conditions that the stochastic game has a value and that both players have optimal strategies.

See [8] or [61.

We

shall prove a generalisation of Shapley's result under suitable assumptions on

S, A, B, q, r.

Specifically, we shall assume that

(1) S, A, B

are compact metric spaces, (ii) r

function on

S ~

and

b n -~ bo,

A ~B,

and (ili) whenever

is a continuous sn J

So,

q(-LSn, an, bn) converges weakly to

an -~ ao,

q(. ISo, ao, bo).

These conditions will remain operative throughout the rest of this section. In order to prove the main theorem, we need some lemmas. X

is a compact metric space, we denote by

probability measures on the Borel sets of

PX X.

If

the space of all It is well-known that

PX, endowed with the weak topology, is a compact metric space. Lemma 4.6 " S ×A

xB.

Let

i

be a continuous, real-valued function on

Then, i(s, ~, k) = IJ i(s,a,b) d ~(a)dk(b), s ~ S,

~ PA' k s PB, is a continuous function on

S x PA ~(PB"

For a proof see [4]. Lemma 4.7 " where

X

Let

u

be a bounded, continuous function on

is a Borel subset of a Polish space and

metric space.

Then

u* . X -~ R

defined by

Y

X ~

Y

is a compact

u*(x) -- max u(x,y) y~ Y

45

is continuous.

Moreover

u, " X --> R

defined by

u. (x) = rain u(x,y) is continuous. y~Y Proof of this lemma is easy and hence omitted. Lemma 4.8 : where

X

u

be a bounded,

continuous function

is a Borel subset of a Polish space and

metric space. into

Let

Y

Then, there exist Borel maps

such that

U(X,g(x)) =

f

Y

y

is a compact

and

u(x, f(x)) = max u(x,y), x z X y~Y

X x

g

from

X

and

rain u(x,y), x ~ X. y~Y

This lemma is an immediate consequence of theorem 4.1. If

X

is a topological space, denote by

all bounded,

continuous functions on

X.

C(X) the family of

For each

w ~ C(X), define

Kw(S,~,k) = r(s,~,k) + ~ I w(-) dq(. {s,@,k), s ¢ S, ~ ¢ PA' k ¢ PB

where

r(s, S, k) = ~I r(s, a, b) d @(a) d k(b)

and

q(.~s, S, k) = 5S q(.is, a, b) d iz(a) d k(b). It follo~s from lemma 4.6, S ~PA

XPB"

Kw

is a continuous function on

From Sion's minimax theorem Is] we have,

sup inf ~PAXSPB

Since

Kw

Kw(S,~,k) =

is continuous on

Inf XSPB

S ~

sup Kw(S,~,k) , s e S. SaPA

PA "( PB

it follows from lemma 4.7 that

sup

by max

Thus, we have

and mln, respectively.

max ~¢PA

mln XSPB

Kw(S,U,X) =

mln XSPB

and

and inf

PA' PB

are compact,

can be replaced above

max Kw(S,~,X) , s ~ S. ~ePA

44 Lemma 4.9 : from

S

For each

into

PA

max

and

w ~ C(S), there exist Borel maps PB, respectively,

f

and

g

such that

rain ke PB

Kw(s , ~, X) = rain k~ PB

Kw(S , f(s), X)

~ e PA

min XeP B

max ~eP A

Kw(S , ~, k) = max 6eP A

Kw(s , u, g(s)), s ~ S.

and

Proof "

We prove the first assertion.

S ~ S, ~ ¢ PA" S XPA.

By lemma 4.7,

~

Let

¢(s,s) = rain Kw(S,~,k) , Xe PB is a continuous function on

Hence, by virtue of lemma 4.8, there exists a Borel map

f " S -~ PA max

rain

~eP A

kaP B

such that Kw(S , ~, X)

~(s,f(s)) = max ~(s,N) = ~zP A for all

s s S.

On the other hand,

¢(s, f(s)) =

min Kw(S , f(s), X) for all X,P B the proof of the first assertion. For each Tw(s) =

w e C(S) define

max ~ePA

min XePB

Len~na 4.10 :

The operator

P_~

Wl, w 2 e C(S).

if

Let

w ~ C(S),

check,

T

T T

~ w Ii denotes

is monotone, we get

This completes

as follows •

Kw(S , @, X) = rain XePB

It is clear from lemma 4.7,

:

Tw

s $ S.

maps

max Kw(S , @, k). ~¢PA

C(S) into

C(S).

is a contraction mapping o n Plainly,

w 1 _< w 2 + LLWl-W211,

sup ~w(s) ~. ssS

C(S). where,

Since, as is easy to

Tw 1 _< T(w 2 + li Wl-W2i~ )

-- Tw 2 + ~liwl-w 2il

45 Consequently and

w 2)

Hence,

Tw~-Tw I _< 8 ~l-W21i , T

Similarly,

Twl-TW 2 --~ 8 ~tWl-W21L

which shows

is a contraction mapping as

(interchanging

wI

II Twl-TW2il ~ ~ ~ Wl-W21L

~ < I.

This completes the

proof of the lemma. Since

C(S), when equipped with the supremum norm, is a

complete metric space,

T

has a unique fixed point in

virtue of the Banach fixed point theorem. fixed point of

T.

Let

w~

C(S), by

be the unique

Then it follows from the definition of

Kw,(S , ~, k) and lemma 4.9 that there exist Borel maps from

S to

PA

and

PB, respectively,

f* and

such that, for every

g*

s ~ S,

w*(s) = min [r(s,f'(s),k) + 8 I w*(. ) dq (.Is,f*(s),k)] kcP B = max [r(s,~,g*(s)) + ~ I w~(.) dq(.Is,N,g*(s))] ~ E PA = r(s,f*(s),

...(*)

g*(s)) + 8 I w*(.) dq(.Is, f*(s), g*(s)).

We shall show after one lemma that

w* is the value function of the

stochastic game. Our next task is to solve the above functional equation (*). To accomplish this, denote by functions on function from to

PB,

S.

With each ordered pair (f,g), where

S to

S to PA

and

we associate an operator

(L(f,~)w)(s) = r(s,f(s),g(s)) We may interpret Player I,

M(S) the f~nily of all bounded Borel

g

f

is a Borel

is a Borel function from

L(f,g) - M(S) --> M(S)

S

defined by

+ ~ S w(.) dq(.Is, f(s),~(s)),

s ¢ S.

(L(f,g)w)(s) as the expected amount Player II pays

when the initial state of the system is

and II take actions according to

f(s) and

g(s),

s,

Players

I

and the game is

46

terminated at the beginning of the second day with Player II paying Player I

w(t) units of money, where

t

is the state of the system

on the second day. Lemma 4,11 " M(S)

and

The operator

L(f,g) is a contraction mapping on

l(fOo gO~ is its unique fixed point in

The proof is straight-forward and omit it.

M(S). Now we are ready

to state our theorem [4]. Theorem 4.2 :

Let

S, A, B

be compact metric spaces, let

continuous, real-valued function on moreover, that, whenever S ,~ A '~ B,

S ~

A

,( B,

converges weakly to

be a

and assume

(Sn, an, b n) -~ (So, ao, b o)

q(. ISn, an, b n)

r

in

q(. !So, ao, be) .

Then, the stochastic game has a value, the value function is continuous, and

Players

I

and

II

have optimal stationary

strategies. Proof:

Observe that the above functional equation (*) can be

rewritten as

L(f*, g~) w* = w ~.

w ~ = l(f.(oo), g,(Oo)).

It follows from lemma 4.11

that

In view of this we have,

I(f*(°°), g*(°°))(s) -- max [rCs,~, g*(s)) + asP A 8 I l(f *(°°),g*(°°))(.)dq(. Is,~,g*(s))]

= min [rCs, f*Cs),k) + ~ I ICf *(9°),g*(e('))( )dq(-Is, f*(s),k)] ke PB It follows from Blackwell

[Theorem 6, pp ~$2, [I]] or from

[Theorem 3.1 in [4]], that,

I(f *(~),g*(~))(s) = sup I (~, g*(~)) (s) 7r

=

inf l(f*(°°) C)(s) F

for

s ¢ S.

47 Consequently, l(f*(°°),g *(°°)) = sup I (~,g,(Oo)) Z inf sup I(~, C ). ?r v ?r On the other hand,

l(f*(°°),g *(°°)) = inf l(f "(°°) c ) < F

sup inf I( ~, V) ~ C

Hence, inf sup I( =, c) = sup Inf I(~, v). C ~ = f This proves that the stochastic game has a value, that the value function is I(f,(oo), g,(OO)) = w~ and so continuous, and that f, (co) g, (oo) , are optimal stationary strategies for Players I and II respectively. Remark 4.1 "

This completes the proof of theorem 4.2.

If in theorem 4.2 we allow

S

to be merely a Borel

subset of a Polish space, our proof breaks down because lemma 4.6 fails.

However, we can eliminate this difficulty by imposing

somewhat stronger conditions on

q

and

r, namely, that

should be a continuous function on

S ~

PA x p B

I w(.) dq (. Is, ~, k) for every when q

and

A

w ~ C(S). and

r,

B with

are finite]. A, B

and that

should also be continuous on [ ~te

r(s,s,k)

S ~

PA ~

that these conditions will be satisfied Then under these conditions on

compact metric spaces and

S

a Borel subset

of a Polish space, the conclusions and the proof of theorem 4.2 remain valid.

PB

48

REFERENCES

[1]

D. Blackwell, Stat

S6 [1968],

[2]

K. Kuratowski,

[3]

A. Maitra,

A. Maitra

226-235.

Topology

Vol I, Acad. Press, P.W.N. [ 1 9 6 6 ] .

Sankhya Series A. 30 [1968], 211-216.

and

T. Parthasarathy,

Jour. optimi, theory

[5]

Ann. Math.

Discounted dynamic programming on compact metric

spaces,

[4]

Discounted dynamic programming,

T. Parthasarathy

and

t~o-person games,

On stochastic games,

And its Appl , 8 [1970], 289-~00.

T.E.S. Raghavan,

Some topics in

American Elsevier Publishing Company,

New York [1971].

[6]

L. S. Shapley,

Stochastic games,

Proc. National. Acad. Sci

U.S.A., 39 [1983], i095-II00.

[7i

L.E. Dubins and L.J. Savage, Fow to gamble if you must, Xcgraw-~Ki!, New Yor~ [1965].

5.

GFz~ERAL THEORE~S ON SELECTORS

In this section we shall first prove a general theorem on selectors

due to Kuratowski

theorems on selectors Jacobs

and

[2].

Ryll-Nardze%,ski

and deduce a few

Lastly we shall state a theorem of

[3]. Let

space.

X

be a set o f arbitrary elements

Let

S

be a countably

additive

and

Y

a metric

family of subsets of

X

cO

[that is, if

An a S

for

the following statement Lemma 5.1 "

Let

f(x) = lim fn(X)

fn " X -- Y

Let

assumption Ifm

for

whenever

whenever

K = ~

c Y.

G

I/n

for

Then

G

Let

and let

is uniform.

is open in

is open in

Suppose that

Y

(I) n.

Y.

K n = ~ y " p(y, K) _<

there is a sequence

- fL _<

U An ¢ S ]. n=l

n = 1,2,...,

where the convergence

f-l(G) ~ S

Proof "

then

is true.

fnl(G) s S Then

n = 1,2,...

m I < m 2 < m 2 < ...

n = 1,2,

....

I/n }.

By

such that

We shall show that

n oo

f-l(K)

First,

let

x ~ f-l(K),

=

/h n=l

that is,

f-I mn

(Kn)"

f(x) ~ K.

As

Ifm (x) - f(x) l~ I/n, n

we have

finn(X) z Kn,

that is,

x E f-I (Kn)

for each

n.

Let

mn

fm (x) ¢ Kn, n

that is,

P(fm (x), K) _< I/n. n

As

f(x) = lim n---oo fmn(X),

it follows that (owing to the continuity of the metric) P(f(x), K) = 0 of our lemma.

and hence

f(x) ¢

K = K.

that

This completes

the proof

5o Let B

L

be a field of subsets of

are members of

Denote by

S

L, then so are

X.

A[3 B,

[In other words, if A ~B

Theorem 5.I [4] :

Let

Y

L, that is,

L.

be a complete separable metric space.

F : X -~ 2Y (the space of all closed nonvoid subsets of

be such that

~ x " F(x)/A G M @ } ~ S

Then there is a selector whenever Proof : Y.

X - A].

the countably additive family induced by

the family of countable unions of members of

Let

and

G

is open in

Let

A

f :

X->

whenever

Y

G

such that

Y)

is open in

Y.

f-l(G) c S

Y.

R = (rl, r2, ..., ri, ...) be a countable set dense in

We may suppose of course that the diameter of

shall define

f

as the limit of mappings

n = O, 1,...,

satisfying condition (I) n

Y

is

fn " ~ -~ R

< I.

We

where

and the two following

conditions.

(2) n

...

P(fn(X), F(x)) <

i/2n

(3)n

-.-

ifn(X) - fn_l(X) J <

I / ~ ~-I

for

n > O.

(The basic idea of proof is similar to the proof of theorem I.I). Let us proceed by induction. Thus

(I)°

given (2)n_ I.

and

n ~ O,

(2)0 that

Put

fo(X) = r I

are fulfilled, fn-I

for each

f4ow let us assume, for a

satisfies conditions

(1)n_ 1

Put n Ci

=

~ x "

n

D i -- { x : =

n Ci

P(ri, F(x)) < I/2n } ,

Ir - fn_1(x)! < 1/2n-z /h D n i "

x ¢ X.

},

and

51 We have,

X = ~

k9 ~

X, there is by Since

tJ . . . .

For,

(2)n_l, y ¢ F(x)

~ rl, r2, ... }

~r i - yL < 1/~ n

and

being a given point of

such that

}y - fn.l(X)i _< i/2 n-I .

is dense, we can find a iri - fn.l(X) l < 1/2 n-l.

B ni

Denote by

x

the open ball

ri Hence

such that x s ~.

{ Y " IY - rll • 1/2 n } •

It

(B~-I) . = ~ x " F(x) (h B ni # @ } and D ni = fnll n n Hence it follows that C i ¢ S coand D i e S and consequently follows that

~e

S.

~

Consequently

A~ = j=l h]

~i 'J '

where

the double sequence (i,J) in a simple sequence s = 1,2,...,

~i, J ¢ L. (ks, m s )

Arrange where

and put

s~ms

We have,

X = ~

~

~

tJ

...

~s t9 ...

This identity allows us to define a mapping fn(X) = rks fn

if

satisfies

fnl(rk

x • ~s"

(~I ~ "''wEns-l)"

(1)n, (2) n and (3) n.

) = Esn - ( ~ 1 ~

...

fn " X -~ R

~ ~s_l ).

as follows"

We shall show that

By definition

As

L

is a field,

it follows

s that

fnl(rk ) ~ L s

fn I (r i) e S

for each

Z ~ R (since (I) n

R

fn I (r i) =

i.

~J fnl(rk ), we have k s= ~ s

Consequently

is countable and

B

fnl(Z) ~ S

for each

countably additive).

Thus

is satisfied. For a given

Put

and as

k s = i.

clear that

x

let

Hence we have

s

satisfy,

x ~ Esn C

fn'S satisfy (2) n and (3) n

~

= Cin C~ D nl

and it is

Thus the sequence

52 fo' fl' "'', fn "'" (1)n, (2) n By (3) n

and

has been defined according to the conditions

(3) n.

and by the completeness of the space

converges uniformly to a mapping

f . X --> Y.

follows

is open in

f'l(G) e S

whenever

f(x) e F(x) according to

G

(2) n.

Y, this sequence By lemma 5.i, it Y.

Finally

Thus the proof of theorem S.l is

complete. Remark 5.1 : [namely by

The theorem remains true by replacing the condition

~ x " F(x) C~G # @ } E S

~ x "

F(x)~

K # ~ } e L

can be seen as follows. F

set

:

G = K IU

Y

K2•

whenever

whenever

G

is open in

K c Y

is closed.

being metric, every open ...,

where

Y ]

Kn = ~n"

This

G <

Y

is an

Hence

O0

x :

Remark 8.2 "

F(x) ~ G # ~ }

~9 { x : F(x)(AK n ~ ~ } n=l

For each complete separable metric space

a choice function f

=

f . 2Y -~ Y

Y, there is

of the first class of Baire.

may be assumed to be continuous if

dim Y = O

observations can be seen as follows. 2Y

identity mapping defined on topology,

the sets

K " K ~Q

{ @ }

is closed in

Y.

F~

in

2Y,

~ K :

2Y.

G

is

K ~ G ~ @ }

F

hence a member of

and

G8

and

X = 2 Y, F = the

According to the Vietoris

are closed in As

F~

Y

These

Put in theorem 5.1,

which are

Also

(that is, if

contains a countable base composed of closed-open sets).

L = field of subsets of

~ S.

2Y

provided

in L.

are open and the sets

Y,

so

G

is open and

~ K " K ~ G # ~ }

Q is

It is not difficult to check

(from theorem S.l) that there is a choice function of the first class of Baire, that is,

f'l(G)

is an

f "

2Y -~ Y

F~

set for

53 every open

G

we denote by

in L

Y.

to

S.

G = QI ~ Q2 ~3

...,

~ K . K ~ Qn # ~ ) ~ L

Since the members of

f-l(G)

dim

the field of closed-open subsets of

assumption we have Consequently

In the particular case where

where

and

Qn

Y = O,

2 Y.

By

is closed open.

~ K " K ~G

# ~ ) belong

S are open sets, it follows that

is open for every open set

G

in

Y

and hence

f

is

continuous. Now we shall deduce the following t ~ Theorem 8.2 ~2~ : be a

Let

X

theorems from theorem 8.1.

be a set with

~-ring

S

and let

T2-space which is the union of a family of at most ~

Y (= first

uncountable ordinal) compact metrizable subspaces in such a way that any compact subset of subfamily.

Let

Y

F :

X-~

F-I(c) = ~ x : F(x) ~ C Then there exists an x s X

and

2Y

(closed subsets of

# g ~ ~ S

f : X -~

f-l(c) ~ S

Definition :

lies in the union of an utmost countable

Y

whenever

~enever

such that C

C

Y).

Suppose

is compact in

f(x) ~ F(x)

is compact in

for every

Y.

A Lusin space is a separable metrizable space which is

the image of a complete separable metric space under a continuous I-I function. Theorem 5.~ . Lusin space.

Let Let

X

be a set with

~-algebra

S

and

Y

be a

F : X -~ 2 Y (= nonempty closed subsets of

Y).

Suppo s e

x :

F(x) ~ B

Then there exists an x

and

Y.

f-l(B) ¢ S

# ~ )

¢ S

f " X ~

Y

for every Borel set such that

for every Borel set

B

B

in

Y.

f(x) ~ F(x) for every in

Y.

54 Proof of theorem 6.2 : the hypothesis be

Let the family of compact sets described in

{ Y< ". ~ • w }

than or equal to ~L .

For each

where

w

~ • w,

is an ordinal less

define

X~ = F-I(Y~) - <J{ F-I(Y~) : 8 • ~ }.

Clearly, the

pairwise disjoint, and their union is

X.

a-ring

S

obtained by restricting

fact, a a-algebra on

X

3

(that is

and each of the sets since

X

to

S~c

for all

~ < w,

a point-closed function

S

for all

if

x e X<.

It is clear that

complete with any compatible metric. f~

such that

every closed subset

C

of

f~(x)~ Yd"

f(x) = f~(x)

Clearly,

f

is a selector for

every compact subset of

Then

Y,

and let

~ ~ ~

~

C

of

Thus for all

2Y<

by

Y.

C

of

x~

F.

S.

Noreover,

~ e w, define

F~(x) = F(x) N Y~

Y<.

~ S~

for each

Further

F~(x)

and

x~,

f " X --~ Y

Y<

S~

for

by

C

f'l(c) ~ S

for

be a compact subset

be an at most countable ordinal such that

FsI(c ) C

is

~ • w.

~urthermore,

For, let

f~l(c)¢

implies that,

fjl(c N Y s ) C

is the

Thus by theorem 5.1, for each

~ow define

if

X~

is, in

~.

F~I(c) = X< ~F-I(c)

closed (and therefore compact) subset

we have an

since

S~

the

X~, we have that

x ~ X~.

F~ : X~ -~

Each

is a n ember of

Note that, by the definition of F(x) (AY~ ~ ~

X~.

X~

{ F-I(Y<) - F'I(y~) ". ~ • ~ }

F-I(Y~), ~ _< ~,

E S, we have that

are

Assign to each

X~ ~ S<),

intersection of the countable family

X~'s

kJ { F~I(Y r) : r_~ ~ } = @,

55 since

r i ~ < B

implies that

F~l (Yr) = I[ x~ Xs : F~(x) m Y r # ~ } < { x

It follows that

: F(x)

Yr

X

F-l(Yr ) -

f-l(c) --h9 ~ f~l (C /bye) • 6 ~ ~( } •

f~l(C ~ y~) 8 S~ C S

for all

~ _~ ~ .

Hence

Plainly

f-l(c) s S

and thus

the proof is complete. Proof of theorem 8.2 :

Let

¢ " P --~ Y

be a

I-i

function from a complete separable metric space Define

F,. X --~ P

by

F, = ¢-I o F.

is continuous.

Moreover

for every

G

For let

P.

P

onto

Y.

F~(x) is closed for each

x ~ X, since $

open in

continuous

~ x : F,(x) ~ G ~ ~ ) e S G

be open in

P.

Then

G

has a compatible metric which makes it a complete metric space. Hence, the subspace conclude that

¢(G)

G

open in

for

F

Y

is a Lusin space.

¢(G) is a Borel subset of

F~ I (G) = F - I ( ¢ ( G ) ) f, : X -~ P

of

~ S.

such that P.

Define

Y.

Hence we can

It follows that

Applying theorem 5.1, we have an

f,(x) ~ F~(x)

and

f -- ¢ o f, " X -~ Y.

fil(G) e S Then

f

for every is a selector

since f(x) = ¢(f,(x)) ¢ $(F~(x)) = $ o $-ICF(x)) = F(x) To prove that

f-l(s) ~ S

for every

x s X.

B

it is sufficient to prove

set

in

Y, B

in

Y.

Thus, suppose that

is open by the continuity of

¢, and

for every Borel set

f-l(B) s S B

is open in

for every open Y.

Then

$-I(B)

f-l(B)-- f~1(¢-l(B))~ S and

thus the proof of the theorem is complete.

56 Remark 5.3 •

Theorem 8.3 generalises a theorem of Castaing [i]

where he assumes

X

to be a compact metric space and

S

the

corresponding family of Borel sets. We shall end thls section by stating one more selection theorem due to Jacobs [3]. X

For this we need some terminology.

be a complete separable metric space.

Radon measure defined on a compact metric space.

Let

continuous in

x

for every fixed that

f " T ~

X-~

T 2 space Y

for every fixed x.

Let

with t

y ". T --~ Y

y(t) ¢ f(t,X) for every

t

Let

~ T

Let

denote a positive and let

Y

be a

f(t,. ) locally uniformly

and

f(., x) measurable in

t

be a measurable mapping such

and

F

:

T -~ 2 X

denotes the

mapping defined by

g(t) = ~ x " x s X

and

f(t,x) = y(t) }

Now we are set to state our final theorem of this section. Theorem 5.4 ." Under the assumptions stated in the above paragraph, there exists a measurable function x(t) ~

V(t)

for each

t ~ T.

x :

T -~

X

such that

For a proof refer to Jacobs [3].

57 REFEREN CE3

[i]

C. Castaing,

Quelques problems de measurabilite lies a la

theorie de la commande,

C-~- Acad. Sci. Paris

262 [1966], 409-411.

[2]

C. J. Himmelberg

and

F. S. Van Vleck,

Some selection

theorems for measurable f~mctions, Canadian. Mat h

[3]

Marc.

Q.

21

[1969], 394-299.

Jacobs,

Remarks on some recent extentions of

Filippov's implicit functions le~ma,

5 [1967],

[4]

selectors,

13

SIAM. _J. _ ~

622-627.

K. Kuratowski and C. ~yll-Nardzewski,

Sciences

Jour.

Bulletin

A general theorem on

De L'aAcademl Polonaise Des

(Serie des sciences math. astr. et phys)

[1965], 39?-403.

6.

TWO APPLICATIONS OF MEASORABLE SELECTIONS

In this section we shall give two applications of measurable selection theorems (which we have proved in section S).

The first

result is an example of a non-analytic (in fact non-Lebesgue measurable) subset of Michael Orkin [3B.

[o, IB which is a Blackwell space, due to

The second result is a theorem on stochastic

[4].

games

We will start with some preliminaries. Definition " set

X

A countably generated ~-algebra ~

of subsets of a

is called a Blackwell space if every countably generated

a-algebra ~ C

~

~,

=~.

that is, ~

having the same atoms as

~

coincides with

In [I] Blackwell proved that every analytic subset of a Polish space (endowed with the relative Borel a-field) is a Blackwell space.

We shall now construct an example of a non-

analytic subset of [0, I] which is a Black,ell space. The construction • Borel sets of a set

A c

I. I

al)

Neither

a2)

If

such that exists

f

Let

I = [0, I].

be the a-field of

Using transfinite induction, we will construct

with the following properties . A

nor

Ac

contains an uncountable analytic set.

is a countable to one Borel function from

Sf = ~ x " f(x) • x }

x ~ A

Let ~

such that

f(x) ~ A

We proceed with the construction. of uncountable Borel sets in

I.

I -9 I

is uncountable then there and

f(x) ~ x.

We first well-order the class (This class has power c).

We

next well-order the class of all Borel functions which satisfy the following conditions.

59 f I)

f

is countable to one, that is

countable for every f 2)

Sf

~ f-l(x)}

x.

is uncountable where

Sf = ~ x : f(x) # x).

This class of functions also has power

c.

t~o disjoint collections of nested sets We select distinct points

We inductively construct

~,

E~

[Here

fl

fl(rl) -- s I.

We let

as follows.

Xl' Yl s B! (the first uncountable

Borel set in our ordering) and distinct points such that

~

rl, s I e I- ~xl-Yl)

-- ~x!' rl' Sl}'

is the first function in our ordering.

are reached in our inductive procedure (where we select

x~, y~

from

B~-

~[-J<< ( 4 U

such that

r<, s~

f< (r<) = s<.

from

I

-

~ BU<<

A

=

We can do this c.

We also select

~x<, y~} }

< c, and by noting the

f<, including the fact that

k~# ~ < c

c c),

We can do this by observing that the set of

able and Borel, must also have power

We then let

B~, f~,

is an ordinal

(4 U EB) ~

points we have thus far removed has power conditions imposed on

E1 = ~ Y! }"

When

~

E~).

because every uncountable Borel set has power distinct points

is at most

c.

Sf , being uncount-

We then let

~.

By the method of construction,

4

Ae

(complement of

A)

do

not contain uncountable Borel sets, hence no uncountable analytic sets, so condition (a I) is satisfied. satisfying

Also any Borel function

(f I) and (f 2) appeared somewhere in the induction

process so (a 2) is satisfied.

f

60 We now consider the pair ( 4 Borel

G-field on

A, that is,

~)

where ~ *

is the relative

~3" -- sets of the f o ~

B ~

A,

B~. Theorem 6.1 : ~(on

A)

If ~ *

is a countably generated subfield of

whose atoms (on A) are the points of

4

then

~*

= ~

*.

[Here we assume the continuum Hypothesis]. We need .the following selection theorem due to Proposition 6.1 "

Let

f " Z -~ W

to one function where spaces. (i)

and

W

be a Borel measurable countable are Bore! subsets of some Polish

Then there exists a Borel subset

f~E

Proof:

Z

is

I-I and

Note that if

(ii) f

E. A. Stchegolkow.

f(E)

E

of

= range of

Z

such that

f.

is a countable to one continuous function

defined on a complete separable metric space taking values in a Polish space then the forward image of every Borel set is also Borel - for a proof of this see Kuratows~i, It is er~ugh to prove proposition 6.1 countable to one and

Topology Vol I. when

f

is continuous,

Z = N N ( = the set of all sequences of positive

integers. Let

X

subsets of Plainly

= range of Z),

f.

such that,

Define F:X--> ~Z (nonempty closed F(x) = f-l(x).

F(x) is clo~ed for each

x

since

f

is continuous.

F

is also measurable since, the set,

iS Borel (by the remark made above) for every open set

U

in

Hence by Karatowski --Ryll-Nardzewski's

F

has a

selection

g.

That is,

g " X -~ Z

theorem

(5.1).

is Borel measurable and

Z.

61

g(x) z F(x) = f-l(x) for every E = g(X).

Then

proposition

E

Plainly

g

is

I-I.

Take

has all required properties stated in

6.1.

Proof of theorem 6.1 : class for

x.

~

by the sets

Let

Fn = B n ~ A

be a countable generating

and consider the sub-~-field of ~ Bn.

Call this

~-field ~ .

on

I

generated

We claim the atoms of

are countable sets. If T

T

is an atom of ~ ,

since

contains at most one point in

( ~

separates points on

A,

Since

T

is a Borel set

being countably generated) and because

T

contains at most

one point of set for

Ac

A,

T ~ Ac

A.

~

is Borel.

Thus

T

#I Ac

contains no uncountable Borel set.

is a countable

Thus

T

is a

countable set. We now use the Borel selection theorem, that is proposition 6.1. g

This proposition implies the existence of a Borel function

from

I -~ I

which maps each atom of ~ onto a point in itself.

Since the atoms of one.

~

are countable sets, g

[See the remark 6.1 below].

is countable.

Indeed if

S

We clai~

must be countable to Sg = ~ x : g(x) # x }

is uncountable, then

g

g

the class of functions we used in the construction of must exist distinct points impossible since of

~

g

x,y ~ A

~ith

belongs to 4

g(x) = y.

and there

This is

maps each atom of ~ into itself and no atom

contains more than one

A

point.

But the fact that

S

g countable implies that all but a countable n~Eber of the atoms of ~

are singletons.

Thus ~

separates the points of

exception of at most a countable set. =

.

I

with the

From this it follows that

Thus the proof of theorem 6.1 is complete.

is

62

Remark 6.1 .

Our use of the selection theorem was accomplished by

using the well-kno~n fact that if ~ a-field on

is a countably generated

I, say, then there is a Borel function

f

which maps

distinct atoms into distinct points (not necessarily in the same atom) and from wkich ~ through

can be got by mapping the Borel sets of

f-I (for example, if

FI, F2,... , generate

~

,

I

let

Oo

f

=

Z n=l

2fn 3n "

where

fn(X) = I

If we consider the graph of

f,

if

x e Fn

and=O

if x ~ Fn).

which is a Borel set in the

corresponding product space, the horizontal sections of the graph correspond to the atoms of

~.

Since in the case, we were

considering, these sections were then countable sets, the conditions for the selection theorem were satisfied and the selection function so obtained mapped each atom of %

onto a point in itself.

Before stating our next result on stochastic games, we shall write down our assumptions. stochastic games]. A

and

B

S

will be a complete separable metric space.

are finite sets

function of

s

[See section4 for details regarding

for fixed

r(s, a, b) is a bounded measurable a, b.

Now we shall state our theorem [4].

TheoreE 6.~ : Under the assumptions stated above, the stochastic game has a value and the value function is Borel measurable. more Players I and II have optimal stationary strategies. R

~

:

Let

rw

:

S-->

PA ~

PB

where

rwCs) = ~[ (~',x') : max [r(s,s,~,') + ~ f w(-) dq(. Is,~,X')] = rain [r(s,t~',X) + ~ I w(.) dq(. Is,#',X)} k

Further-

63 and

w

is a bounded real-valued Borel measurable

These set valued functions Olech [2]. theorem 6.2.

rw

function on

S.

are measurable by a result of

Now one can use (the selection)

theorem 5.3 to prove

The rest of the proof follows along similar lines

to that of theorem 4.2, -with some minor ,~difications the proof of theorem 6.2 is omitted.

and hence

64 REFERENCES

[1]

D. Blackwell,

On a class of probability spaces, Proceedings

of the third Berkeley Symposium on Mathematical Statistics and Probability Vol 2, p 1-6, edited by J. Neyman, University of California Press [1956].

[2]

C. Olech,

A note concerning set valued measurable functions,

Bulletin De L'Academie Polonaise Des Sciences, S _ ~ ~e~ S i e ~ ,

[3]

M. Orkin,

~.ath, ~s~ro et Pl~vs. 13 [1965], 317-321.

A Blackwell space which is not analytic,

To appear.

[4]

T. Parthasarathy, Discounted and positive stochastic games, Bull. Amer, Math. Soc

77 [1971], iS4-136.

7.

VON-NEUMANN'S

MEASURABLE CHOICE THEOREM

The purpose of this section is to prove a measurable theorem due to von-Neumann.

choice

After giving the proof, we shall

mention a few consequences of this theorem. Choice Theorem 7.1 [4] ". Let

S

be a complete separable metric

space,

S

and

A

an analytic set in

defined and continuous for all values.

Let

assumes.

K

Let

A

and has real numerical

be the set of all values, which ~ (k)

is a monotone,

a ~

F(a) a function which is

be an arbitrary

nondecreasing,

F(a), a c A

N-function,

rlght-continuous,

that is,

T (k)

bounded function.

Under these assumptions we have (i)

K

is a

T (k)

measurable set.

(ii)

there exists a function

X ~ K

and assumes values in (a)

k ~ K

If

with (b)

Proof :

0

S

such that

is an open subset of

f(k) c 0 For every

is

S, then the set of all

"[ (X)-measurable.

k ~ K,

F(f(k)) = ~.

We shall prove the theorem when

and nowhere constant

~-i~anction,

a topological mapping of a ~ ~ • b

f(k), which is defined for all

where

a

=

T (k)

therefore

is a continuous

k -~ % (k) = ~

is

- co ~ k ~ oo on the finite interval T~(- co),

b

= p~(+ co).

The general

situation can be reduced to this case [4]. Let us now replace and then write again

X

F(a), f(k) by for

our assertions unaffected, ~.

The domain of

k

~.

(F(a)),

f("C-I(~)),

This leaves the structure of

but there are these changes regarding

is now

a

~ k • b

and

not

- oc ~ k • + co.

66 In this new domain we have to use the notion of common (Legesgue)measurability and not that one of T(k)-measurability. -9 I~(k)

[Our mapping

transformed the latter into the former].

We now introduce the '0 space of Baire', So, which is the set of all sequences

4 = (n!,n 2 .... ),

nl, n 2 ... = 1,2, . . . .

It is metrized by the definition Dist [(ml,m2,...) , (nl,n2,...)] = u

if

mv = nv

= _I_ if

vo

vo

for every

is the smallest

integer for which

Thus

SO

is a complete, separable metric space.

set of all those

~ = (nl, n2, ...), for which

a finite number of

vo

exists such that

mvo • %

establishes an ordering of all

~ ~ 4

we have sphere in 41, 42) . then

is open.

~ < 4 So

[The countable nv W i

occurs for

If

So.

(ml, m2,...) • (hi,n2,...) and If

p ~ 4, and

in the entire sphere is a set of the form

mv ~

= nv

for

P -- (nl, n2,...)

if a

is fixed, then the set of vo

is defined as above, then

Dist (~, B) • I/v o. ~l -< ~ ~ 42

Every

(with fixed ¢ ~ O,

belongs to it, if and only if,

Dist ~ (ml,m2,...) ' (nl, n2,... ) } < ~, where

So] .

v < vo ,

If its center is (ml, m2, ...) and its radius

v = 1,2, ..., v I

m v W nv.

v's only is obviously everywhere dense in

On the other hand the definition,

v.

vI

may be written in the form

that is

m v = nv

is the greatest integer

for

_< i/~.

This

~I -< p ~ c(2 if we put

~I = (ml'''',mVl-l, mvl, I, I, ...), ~2 =(ml,"',mVl-l, mvl+l,l,l, --~-

67 As

A

Is an analytic set In a complete, separable metric

space, therefore there exists a function and continuous for every

~ ~ So

and ~Lich maps

[The mapping may be many to one]. the set of all

There As

F(~(~)), ~ s So.

continuous for every

~ z So

~(~) which is defined

K

So

onto

A.

Is clearly

F(@(~)) also Is defined and

and the real numbers form a complete,

separable metric space, we can conclude that

K

is an analytic

set and consequently measurable. Consider now a wlth

F(@(~)) = X

k ~ K.

by

Denote by the set of all

T(A).

as

F(@(~)) Is continuous,

nI

with

As T(X)

(n I, n2, ...) ~ T(A)

the smallest

n2

(us, n4, ...)

by

with ~

k ¢ K,

Is closed. for auy

etc.

Let

for every

<(~)=

with

~(A) ~ ~o"

F(@(~)) = ~.

{ B : ~ < ~o ) M( ~ o ) If

~,

... ).

It

and therefore <(X) ~

has a first element,

In

~o

and consider the set

That is, ~ ~ ~o

k e M(~o) wi~h

in other words So

under

M(~o) for all

is clearly equlvalen~

8 ~ T(~), that Is of a ~i(~o)

F o ~ .

is any subset of

Hence

wlth

~(A) ¢ T

by

i~(T),

if

form

~l -< ~ ~ ~2, therefore

result proves therefore, that

So , d e ~ t e T

~ < ~o

is the image of open set we can conclude

is an analytic set and hence measurable. ~

n~,

~(X).

to the existence of a with

T(A)

by

T(%) for ~ y

(n° ' ~ ,

T(X)

In other ~ r d s

Choose now a fixed X's

Denote the smallest

besides its definition guarantees that

~ ~ T(A).

and thls Is

is not empty and

(n2, n3, ... )

(nl, n2, n S . . . ) ~

is clearly a condensation polnt of ~(A) ~ T(A),

T(X)

~ ~ SO

the set of all

A's

Is a sphere, then it has the

i~(T) = M(~2) - M(~l)" ~(f) is measurable.

Our above If

T

is

68 merely an open set, then, as

So

is separable, we can write

T

CO

as the sum of a sequence of spheres : T = ~ Ti. NOw co i--I N(T) -- ~_# N(TI) , and as ~N(Ti) } are all measurable, so is i=l

Let finally

O

be an open subset of

the set of all

c( ~ SO

is continuous,

therefore

measurable.

That is,

with

~(~) ~ O.

S,

As

0

T(O) is open too.

X s N(T(O))

means

and denote by is open and Thus

~(T(O))

~(X) z T(O)

T(O) #(c() is

and hence

~(~(~)) ~ 0. Observe finally, that as

The measurability of

K

~(X) ~ T(A) (assuming

A ~ K),

being already established, we have

proved statement (1) of our theorem.

And in order to prove

statement (ll) it suffices to define

f(A) = #(~(A)).

Thus the

proof of theorem 7.I is complete. For the statement of our next theoreE we need some definitions. We shall need the concept of standard measurable space.

This is

a measurable space that is isomorphic to a cartesian product of {0,I} points

countably many times. O

and

1

The space

~U,I}

consists of the

only and all subsets are measurable.

An isomor-

phism between two measurable spaces is a one-one function from one onto the other, that takes measurable sets onto measurable sets in both directions.

The unit interval, any Euclidean space, and in

fact any uncountable Borel subset of shy separable complete metric space, with the usual Borel a-field, is standard, to state a theorem due to As/mann [i].

i~ow we are ready

69 Theorem 7.._._..______"_~2 Let (T, ~)

be a ~-flnite measure space, let

a standard measurable space, T ~X

whose projection on

and let £

is

(bebesgue) measurable function for almost all

t

in

set is measurable Remark 7,1"

be a measurable subset of

all o f

T.

Then there is a

g " T --~ X, such that

(t,g(t)) ~ G

~-measure

OJ.

This theorem follows from the previous theorem --for

requirement that

Also, X

shows.

the theorem is false without the

be standard, Let

the Borel sets, and let subspace Qf

be

['ALmost all' means that the exceptional

and of

a proof refer L1].

Lindenstrauss

T.

G

X

as the following example due to

T be the half open interval ~

be Lebesgue measure.

Let

[O,1 ) with X

be the

[O,1 ) that one usually uses to show that there are

non-Lebesgue measurable sets, that is, the sets mutually disjoint for rational (addition is modulo 1).

bet

r, D

and

X + r

are

k J ( X + r) = LU, l> r

be the subset of

T ~X

defined

by

D=

((t,

•

Here

D

is a sort of diagonal of

that

D

is measurable ; one builds a fi~zlte n~mber of s~all

rectangles whose union covers and the intersection is X

exhausts

X,

D.

T ~X°

D, lets the rectangles get smaller, Note that the projection of

but the projection on

in fact, the projection is precisely a subset of Let

~. G

It is not hard to check

= k_3 (D + (r, 0)). r

D

I' does not exhaust X,

on T ,

which may be considered

70 Clearly

D + (r, O)

The projection of projection of

G

is measurable, D + (r, O) on

exhausts

follows that for each such that

t

(t,g(t)) ~ G.

T.

and hence

~

is

G

Is measurable.

X + r, and hence the

Since the

X + r

are disjoint,

there is precisely one point

0

S + r -- S

If

S

for all rational

r, then

~(S)

> O.

which

1 :

hence for every

has density

ly small interval

is, t)

around

~(S ~ Is, t)) ~ ~-~) (t-s).

and

to, ~

or

s

1.

to

at

and every sufficient-

will have

In particular,

~(S ~ Is, t))

~(S) = o

T

Then we can find a point

if

m

large positive integer, we can find an interval t - s -- i/m

on a set

is a measurable subset of

£o prove thls suppose S

g

only.

We first prove a lemma : such that

g(t) ~ X

Clearly it is sufficient to prove that

there is no measurable function that differs from of Lebesgue measure

It

• (1-~)/m.

is a sufficiently ~s,t) such that

i~en

m

~(s)

= ~ ( s ,u Co, l } )

= ~ ( s r~ z (Ls, t ) + J/m)) J=l

m

=

z

~(s~([s,t)

+ J/m))

J=l =

m 7. ~ ( ( S + J/m)

(h (Is, t) + j/m))

J=l m

=

Z

~ ( ( S rh [s,t))

+ J/m)

J=l m

=

Z

~z(S ~ is,t))

> m(l-~)/m

= 1 - e.

J=l

Since

~

may be chosen arbitrarily small,

the lemma follows.

thls lemma we deduce the following Corollary.

If

S

is a not

From

71 necessarily measurable subset of all rational

r,

Indeed, let

take

S'

then the outer measure of

S'

~(S') -- ~+(S),

where

~+

by

and includes

S,

denotes outer measure.

S + r

=

= s+(S). I

S.

S ;

Now

or

Then

S' + r

S'' + r

S"

S'' C =

0

or

{ Si } Si

= ~S'

S'

S"

I.

(For example, of measurdiffers

is measurable for all

Therefore

since

for

such that

such that the measure of

i/i).

able, and includes

~(S'') = O

S,

S + r = S

is either

to be the intersection of a sequence

~+ (S)

~(S")

S

be a measurable set containing

able sets containing from

[o,i) such that

+ r

r,

is measur-

it follows that

for all rational

r,

hence

and the Corollary is proved.

Obviously the outer measure of a measurable subset of

T

such that

X

cannot vanish; Y ~ X

and

let

Y

be

%z(Y) = N+(X) • 0.

Using density considerations as in the proof of the lemma~ we can find t%~ dlsJoin5 intervals such that

~(Y ChIl) , 0,

~+(X C~ 12) • O.

I 1 = [Sl, ti) ~(y c3 I2) • u.

and

12 =

Then also

Is2, to) ~+(X ~ I I) ~ 0,

Let xI

--

<3

( ( x ~ I l) + r)

r

LJ ((X fhi2) + r). r

Then XID

X1 + rI X rill,

-- X 1 and

and

X 2 + r -- X 2

~+(X l) _ ~ + ( X

~ I I) • 0 ~

So from the Corollary it follows that and since

~

where

denotes the inner measure.

~-

~

X 2 = g,

for all rational

Iz+(X1 ) = i,

it follows that

N'(~)

But

follows that there is no measurable function differs from example.

g'

similarly

on a set of measure zero.

r.

But

~+(X 23 ~ O.

~+(X 2) = l,

= N-(X2) = O,

X 1 = g-!(ll) , and it g'

such that

g

l~lis completeshhe

72 We shall present another version of von-Neumann's choice theorem.

Let

X

The function

F

will be defined on

be subsets of X

be an arbitrary complete separable metric space.

X.

T = [O,1]

Recall a point valued function

will be called Lebesgue measurable if

measurable subset of subset of

X.

T

X.

f-l(u)

f

from

T

to

Is a Lebesgue

for every open (or equivalently, Borel)

An analytic subset of

a Borel subset of

but its values will

X

is the continuous image of

The set-valued function

F

will be called

analytic if its graph, that is,

(t, u ) "

is an analytic subset of

Proposition 7.1 ; to

T x X.

Now we have the following

[2].

proposition

T

u ~ F(t) )

If

F

is an analytic set-valued function from

X, then there is a Lebesgue measurable point-valued function

f " T -~ X

such that

f(t) ~ F(t)

for almost all

t.

Proof follows from theorem 7.1. Next we give an application of proposition 7.1

in the

characterisation of extreme points of sets of vector functions [2,3]. Let

A

be a compact convex subset of

extreme points,

~

the closure of

B.

Let

En,

B

MA,

MB

the set of its and

M

be B

the sets of all measurable functions from B

respectively.

T

to

En

T = [0,I]

to

A,

B

Since the space of all measurable ftuuctions from

has a linear structure, it is meaningful to talk about

the extreme points of Proposition 7.2 "

and

MA.

The set of extreme points of

MA

is precisely

73 Proof :

Clearly every point of

Conversely, let MB,

f

t

t,

f(t) = 12 g(t) + 12 h(t), t.

g(t)

and

and

g(t)

h(t)

and

and

f

If

in

h(t)

f

A

= ~2 g + 21 h,

M A-

is not in A.

Hence

so that

differ from f(t ) for

Because of proposition 7.1,

chosen so as to be measurable as well. MA

N A.

f(t) is not an extreme point of

we may choose

at least some

is an extreme point of

be an extreme point of

then for some

for each

PB

Then

g g

and and

h

h

may be are in

contradicting ~he extreme property of

f.

This proves the proposition. Another situation treated by Karlin [8] is the following. Let

IZl' ~2, " " , ~m

on

T,

and

B

and let

be a set of non-atomic totally finite measures

aI, ..., am

be as above.

that

Let

b i _~ ~ f d Ni --~ ai

the extreme points of

G

and

bl,...,b m

PrOof :

for

i = !,~,...,E.

are contained in

The extreme points of

G

and suppose

in the previous proof, and let f + e

=

g e MA

and

~(S) = ~ S~[ e d ~!' of dimension

m.

for

on

T

t @ S.

by

e = !2

g

= h ~ M A.

f

be an

Construct

g

and

For N

such

We shall now

are contained in

h

Then

f

Let

-

A

Karlin proved that

N . B

G

f g MB.

... ' SI e d N m }.

M B.

h

so that S C T

set

is a vector measure

Applying Lyapunov's theoreE on vector measures,

we obtain a subset e'

f - e

Let

MA, consiting of functions

We follow the proof due to Aumann [2].

extreme point of

En.

G

prove the following proposition. Proposition 7.~ :

be in

S

e'(t) Define

of

T

= e(t)

such that for

fl = f + e',

t s

~(S) = ! ~(T), S,

f2 = f-e'.

and

Define

e'(t) = -e(t)

Then

as

~D

o~ O

~J

0

I-*

~D

~D

II

I-b

P.

H~

H

7~

~D

!

tu,

÷

!

7~

i~ .

II

i-J

~o

r

I!

75 REFERENCES

[i]

R. J. Anmann, theorem,

Measurable utility and the measurable choice Proc. Int. Colloq., La Decision, C.N.R.S,

Aix- e n - Provence [1967],

[2]

R. J. 2amann,

15-26.

Integrals of set-valued functions,

Math. Analy. Appl. 12 [1965],

[3]

S. Karlin, F~th. Soc.

[4]

M~th.

1-!2.

Extreme points of vector functions,

~.

~er.

4 [1958], 603-610.

J. yon Net,nann, ~.

Jour.

On rings of operators, Reduction theory,

50 [1949],

pp

448-451.

8.

ON THE UNIFORMISATIO~ OF SEfS IN TOPOLOGICAL SPACES

Given a set spaces

X

and

projections X

E

in the cartesian product

Y , a set

~X E

and

U

~X U

of

{(x) × Y } U

E

and

x ~ ~X E

if

E E

through

Y

x

But, If

if the onto

the set

...

(1)

consists of a single point.

existence of such a uniformising set

and

U

E,

~

lying above

the axiom of choice.

of two

is said to uniformlse

coincide, and if, for each

of points of

X ~Y

X

and

U Y

The

follows immediately from are topological spaces

is, in some sense topologically respectable, for example belongs to some Borel class, it is natural to ask a

uniformising set not much ~ r s e .

U

that is equally respectable or at any rate

Usually there is no way of controlling the

topological respectability of sets obtained by the use of axiom of choice and quite different methods have to be used in obtaining topologically respectable uniformising sets. The earlier work of Lusin [3] on problems of this nah~re was confined to the case when for each points of

E

lying above

x

x

in

~ X E, the set (I) of

is at most countable.

The first

general result seems to have been the result obtained independently by Lusin [4] and Sierpinskl [7] showing that, are Euclidean spaces and unlformising set analytic set.

U

E

Is a Borel set in

when

X

X ~< Y,

and

Y

the

can be taken to be the complement of an

Following work by Lusin and Novikoff [6]

on the

effective choice of a point from a complement of an analytic set

77

defined by a given sieve, Kondo [6] sho~,~ed that in the Euclidean case, the complement of an analytic set could be unlformlsed by a complement of an analytic set.

Since any Borel set in a

Euclidean space is the complement of an analytic set this provided a most satisfactory generallsation of the result of Lusin and Sierpinski. closed set

Braun L2] showed in the ~uclidean plane that any E

can be unlformised by a

set can be uniformised by a

G~

G~

set and that any

F~

set.

It is easy to extend the results of Lusin and Sierpinski and of Kondo when

X

and

Y

are complete separable metric

spaces by the following mapping technique.

If

X

is a complete

separable metric space there will be a continuous function that maps a relatively closed subset irrational numbers in onto

X.

[0, I]

[0, I]

J

o

of the set

Y.

Then the product map

continuously and one-one onto

f-I ~ g - I

of ~,

X ~Y.

Borel sets and coanalytic sets in Lusin and Sierpinski or of

Io x

R 1 ~<S I.

back to

X ~Y

there results in

g

that

S1

of real maps

Hence the inverse in

X ~

Y

J o, which remain So the results of

Kondo can be applied in

when the unlformising set is intersected with

one-one

of irrational

f X g

maps Borel sets and coanalytic sets

into Borel sets and coanalytlc sets ia

f Xg

J

regarded as a subset of the set

numbers, one-one onto

by

I

Similarly there will be a continuous function

numbers in

map

of the set

regarded as a subset of

maps a relatively closed subset

Io x J 0

I0

f

R 1 ~( S 1

I0 x J 0 X xy

and mapped

a uniformising

set that is the complement of an analytic set as required.

We shall

now prove the following theorem due to Rogers &nd Willmott [6].

78

Theo~

8.1 ". Let

X

be any topological space.

q-compact metric space.

Let

~

can be unlformised by a

can be uniformised by a

X ~Y,

be any

D6

Then a closed set in

set and a

Fq set in

X ,~Y

D~-set.

We will now introduce some notations we study a cartesian product ~y

Y

be the family of finite unions

of differences of closed sets in X ~Y

Let

X xy

and definitions.

When

of two spaces we use

to denote the projection operators onto

X

and

Y

~X'

respective-

ly, so that ~x(X ~ y ) If

C

= x,

is any set in in

~y(X ~ y )

X ~Y

Y

= y

we use

with

for

x ~ X, y ¢ Y.

C (x), for

x "
and

x ~ X, to denote

the set of

y

C (y) is defined

similarly.

We also define the cylinder on a set

E

to be the

set ~xEX and use A set U C x

in

cy E U

E,

Y

to denote the cylinder.

will be said to tuliformlse a set ~X U = ~X E

and

U (x)

E

in

X ~(Y

if

contains a single point for each

~ X E.

A function

f

is defined on

is said to uniformise ~X E

into

x ~f(x) We will call a set it is of the form

Z

~ E in

Z -- P ~ Y

Y

a set

E

in

X ~

Y

if

f

so that

for every X ~Y

x ¢ ~ X E.

a cylinder parallel to

for some subset

P

of

X.

Y

if

79 Lemma 8.1 :

Let

Y*

be a compact subset of a

J

be a set of the form

Z

a closed cylinder parallel to

j

o

Jl

F - Z

n

Let

Y.

=

J - cy Jo

(x

J

~

closed

and

and

@

~ X Jo k] ~ X Jl

=

~ X J"

F

closed and

Z

a closed cylinder

Then

Y-}

is of the required form, as

F

: {Fn(X×Y*)}-z

f~ (X ~

it follows easily that

closed set in

F

Y~)

=

with

with

Let

Y.

Then the sets

V X Jo C~ ~ X Jl

Jo:{F-z}n{x

is compact,

Y.

J

J = F - Z

parallel to

X "< Y

=

are both of the same form as

Proof "

in

T 2 space

X.

~ X { F C~ (X x

Y*)]-- [ ~ X {

is a closed cylinder parallel to

J1-- J - c y

is closed. y*)}

As

Y~

is a

Hence the set

zI -- cy [ F ~ ( X ×

°YJo--°Y[

Y*)

{Fn(×~

Jo

Y.

F:~(X~

Further

y-)}. z] -- ~ -

: {F'z}-{Zl'Z} =

F-

(Z<2

Y*)] ] × Y

~)

z

8O and

Jl

has the required form.

The formulae for the intersection and union of the projections of

Jo

and

J1

a subset of

J

follow immediately from the facts that and that

J! = J - cy Jo"

Proof of theorem 8.1 :

Let

E

Y

be a closed in

with the properties

(a)

the diameter of

(b)

each point of

Yi Y

is

This proves the lemma. X x

is a-compact and metric, we can choose a sequence of compact subsets of

Jo

y.

As

Y

YI' Y2 "'"

:

tends to zero as

i -~

co

and

belongs to infinitely many sets of

the sequence. We define sets

Do, DI, D2,

Dn+ I = ~ D n n

...

inductively by tsking

(X "( Y n + l ) } k # ~

Dn - c y ( D n ~

Do = E,

(X ~ Y n + l ) ) }

...

for

n = 0,1,2,

lemma that with

F

Dn

....

It follows inductively,

is the union of

closed and

Z

2n

sets,

by the use of the

each of the form

a closed cylinder parallel to

disjoint projections with union

(2)

Y,

F - Z having

~ X E.

Co

Write

U

Then each set

=

Dn ~ ~

~X U

~ n--O

Dn .

and

U

is a

?TX Do

~ I ) ~ set.

= ?TX E

it will suffice to show that given any point

Yx

with

x "~ Yx ~

As

U.

x e ~X E, there is a unique

81

If

x c ~X E,

the set

D (x)

=

o

E (x)

Our aim is to show that the sets

is closed and nonempty.

D n(x) ,

n = 0 , _ , _ ~. ,l

closed, decreasing, nonempty, compact for and w i t h d i a m e t e r

set

u(x) =

tending

~ n=o

Dn(x)

to zero.

n

..., are

sufficiently large

This ~'~ll e n s u r e t h a t

the

consists of a single point as required.

It follows from (2) that,

Dn+l( X=) D(X) n =

for

Dn(X)

n = 0,1,2,

n = 0,1,2, ...,

integer

D(X)/hn Yn+l # ¢

D(X) n f~

if

Yn+l

=

It follows immediately that the sets • ""

sequence

if

Yn+l

are closed, decreasing and non-empty.

YI' Y2' "'" n(x),

Dn(x) (x) ~

is compact for

~(x) I)n t

"

covers

Y,

Yn(x)+l

n _~ n(x) + I.

As the

it follows that for some first # @'

This implies that

As each point of

infinitely many sets of the sequence

Y

YI, Y2' ...,

D (x) n

lies in and as the

diameters of these sets tend to zero, it follows that the diameter of E

D(X)n also

-90.

This completes the proof of the case when

is closed. ~Dw consider the case when

~-compact we can express each set

En

compact set in

E

E

is s/1 F

in the form

is closed ~nd each set Y.

Then each set

E

~ y En

~ X En

set. ~s Y is co = i_J ~ , where n=l is a subset of a

is closed (being

effectively the projection of a closed set through a compact space). Let

Un

be a

D~

set unlformlslng

~,

for

n = !, 2, . . . .

82 oo

Let

J

=

n

/-] m=l

Dnm

D

where each set

~ ~

.

It is clear

that the set

U

uniformises

--

E.

u

L # [ Un - cY ( U n=l v
)]

But oO

u

As

~X

E

n

ey(

-

u

u

-~
)

m= I

is closed for

Dnm

--

~

- cy

(

~
n, the set

<

U

E

)

w
belongs to ~ .

Remark 8.1 :

Hence

U

is a

_D~

If each open set in

~

-set as required.

is a

F

ising set for a closed set can be taken to be a for a

F~

set can be taken to be a

as follows.

and

V

in

X xy

is an

is a

G~

-set.

In this case each open set in

union of open rectangles X

G~

X

and

F~-set.

set in

U ~ V

in

V. Y

~hen, is an

X ~

set, the uniformG~

set and that

This can be seen Y

is a countable

as each open set F~-set,

in

each open set in

So each closed set and hence each set

X "~Y.

U

im

The result follows.

We shall close this section after stating one more result due to Rogers and Willmott, Theorem 8.2 : X

a proof of which can be found in [6].

Suppose that each open set in a topological space

is a Souslin set.

Let

E

be the complement of a Souslin set

83 in

X ~

I

where

I = [0,I].

Then there is a set

the complement of a Souslln set in

(a)

u

(b)

~X U

(c)

for each

X ~(I

U, that is

and that satisfies,

~n E

=

~X E

and

x ~ ~X E, the set (~x} ~

I)

f~ U

contains a single point.

Remark 8 . 2

;

It seems that Robert ~olovay has proved the

following result

LSee L8], Remark 4, pp 2]

existence of measurable cardinals.

assuming the

Every complement of an

analytic set in t~o-dimensional Euclidean plane can be unlformised by a Lebesgue measurable set of

~I] in this connexion.

:

See also

pp

24-25,

84 REFERENCES

[i]

R.J.

;mmann,

Measurable utility and the measurable

choice theorem,

Proc. Int. Colloq., La Decision,

C.N.R.S., Aix-en-Provence [1957], 15-26. [2]

S. Braun, Sur l'uniformisation des ensembles fermes, Fund. Math, 28 [1937], 214-218.

[3]

N. Lusin, Lecons sur les ensembles analytlques et leurs applications.

[4]

N. Lusin,

Paris, [19:30].

Sur le probleme de

tlon des ensembles. [5]

N. Lusln and

M.J. Hadamard d'uniformisa-

Nathematiea, 4 [19S0], 54-66.

P. Novikoff,

Choix effectlf d'un point

dans un complementaire analytic arbitraire, donne par un erible, [6]

Fund. M_aJ~, 25 [1935], 559-560.

C. A. Rogers and R. C. Wi!Imott, sets in topological spaces,

On the uniformisation of ~.

~

120 [1968],

1-52. [7]

W. Sierpinskl,

Sur l'uniformisation des ensembles

measurables (B), [8]

R. Solovay,

Fund. Math

16, [1930],

136-139.

A model of set theorey in which every set

of reals is Lebesgue measurable,

[1970], pp 2.

Ann. ~

92,

9.

SUPPLE~I~NTARY REMARKS ON SELECTION THEOREMS

The purpose of this section is to make a few remarks o n selection

theorems

covered so far.

and to

mention

applications

w h i c h we h a v e

r~t

We may not give detailed proof to any of the

theorems to be stated in this section but we give references where one can find the complete proof. (a) @

n-selections

Suppose

be a function from

2Y X

:

X

and

Y

each

2Y

~(x)

points and

~ :

is a subset of

• ~(x) x~X

a topology of when

2Y

to

X -~ 2 Y

Definition :

@(x)

is closed in X

A function

kff ~(x). x~ X

is a lower semicontinuous

A carrier

~

is coEpact for each

Ly

from

if for n

In order to relate Y

to that of

function,

X

it is

have the following properties.

@ : X-~

2Y

is said to be point-compact

=

~ x . y ~ @(X) }

y ~ Y.

A carrier will be called small if for each O,

q

Ly (defined below)

Ly

set

~

(Here

containing at most

or the topology of

convenient to require that

if the set

Y).

is said to be an at most n-selection for

x s X,

Let

X -~ 2 Y, sometimes called a carrier.

denotes the set of closed subsets of to

are metric spaces.

c

0 ~

X,

y ~ Y

there is an open neighbourhood

for which

¢)

o.

and open V

of

y

86 Remark 9.1 : $ : X-~

2Y

If

f : Y -• X

defined by

is an open and onto mapping, then

O(x) = f-l(x) is a lower semicontinuous,

small and point compact carrier. semlcontinuous,

X

Also if

is compact and

small and point-compact.

Y

$ " X -~ 2Y is upper

is regular then

$

is

The following theorem has been proved

in [6] by McAuley and Addis. Theorem 9.1 : di= X < n

Suppose

and that

X

and

Y

~ : X -9 2Y

are metric spaces with is a lo~er semlcontinuous, s=all

and point-compact carrier for which to the metric on B

c X

Y

for every

$(x) is complete with respect

x ~ LK.

Furthermore, suppose that

is a closed set (perhaps empty)

map so that

g(x) ~ ~<x)

most (n+l)-selection Re~ark 9 , 2 :

~

for each for

~

and

g : B -9 Y

x ~ B.

is a

Then there is an at

such that

~(x) = g(x) for

x ~ B.

The proof depends on the fact that certain project-

ions are closed mappings (which is not true in general) but our assumptions on (b)

~

ensures that they are closed.

Continuous carriers :

We will now briefly mention about

continuous carriers discussed in [8]. (metric • O, every

P), then a carrier every

xo ~ X

~ : X -~ 2 Y

has a neighbourhood

Y

is a metric space

is continuous if, given U

such that, for

x • U,

$(xo) ~ N~(~(x)),

where

If

~(×) c ~ (~(xo))

N ( $ C x ) ) = ~ y : pCy, #(x)) < ~ }

etc.

It is clear that

is continuous in this sense, if and only if it is continuous,

87 in the usual sense, with respect to the following topology 2Y.

This on

~

2Y

is the topology generated by the uniform structure which is obtained by taking, as a basis for the

entourages of the diagonal in

2Y ~ 2 Y, all classes of the form

= ~(A,B) ° A,B ¢ 2 Y,

~

with

~ ~ O.

A Ci~(B),

~(Y)

a selection.

Y

( = nonempty, closed, convex subsets of If

X

is a topological space, and

open, convex subsets of

Y

~

theorem when

Y

x a X

X

X

Y

Y) admits

normed linear

to the nonempty In these two

except that it is a

X

$

as well as the sets

~(x).

is finite dimensional (here we mean Lebesgue

covering dimension, that is, dim

every

X ~ n

if every finite open

has a finite open refinement

belongs to at most

n + i

~4/ such that

elements of ~L/), it is

possible to place purely topological conditions on the sets

O(x)

which are not only sufficient but, in a sense, also necessary. See [7] in this connexion. (c)

Part metric in convex sets ". ~1ow ~,zewill briefly discuss

some results on the part metric and its relation to selection theorem due to Bauer and Bear. nari es.

is

is an arbitrary metric - of course one has to

X

of

X

It ~ould be nice if one could prove a selection

have strong assumptions on the map For example if

from

admits a selection.

results no restriction is placed on topological space.

If

a Banach space then every continuous

space, then every continuous carrier

covering ~

B • It(A) } ,

Now one can prove the follo~dng results.

a topological space and @ : X -~

~c on

We will start with some prelimi-

88 Consider

a real linear space

which contains no whole llne. is denoted by in

C

by

write

~x, y].

r(> O)

x .~ y

If

If

classes of ~

The closed segment from

x, y a C

~;e say that

x + r(x-y) ~ C

if Ix, y]

that .~ defines

L, and a convex set

and

extends by some

an equivalence

There is a metric on each part of

C, C

x

is not equivalent

satisfies

r • O.

y,

firdte.

write

metric on

C. :

(i)

This metric

A mettle

d

the mapping

[0, I] ~ C ~< C (il)

into

the set

~ ~ 0

on

d

C

C.

~he equivalence

are clearly also convex.

is continuous

and each convex subset

in a position

paracompact ?r

~

of

Let

space

~ T

of

and S) < ~ ~

S

of C

is convex for

C. without any line is Now we are

theorem.

be a lower semicontlnuous

mapping

from a

to the nonempty closed convex subsets o f

of the cone

a locally compact

d

except that it is

C -- for a proof see [2].

to state the following

Theorem 9 , 2 -

C,

Then

will be called convex if

The part metric of a convex set

one part

d(x, y) = + oo.

r }.

will be called the part

~ y : y ¢ C, d(y,

convex on each part

We

It can be shown

(k, x, y) -~ k x + (l-k)y C

y

defined by

all the axioms of a metric on

not always

each

to

to

L

Ix, y] extends

d(x,y) = inf ~ log(l + I) r : L~x ,~J "~ extends by

If

x

in

y + r(y-x) ~ C.

relation in

, called the parts o f

C

space

M+(X) of nonnegative X.

Then there exists

Radon measures on a continuous

89 selection for f

#.

has the form

mapping from

T

Let

f

be any such function and

f(t) -- gt f~ ' where into

t -9 gt

L°°(~) ( = space of

t~ s ~.

Then

is a continuous

N-essentially bounded

u-measurable functions]. The proof of this theorem depends on the following observation.

The part metric

d

In each part

topologically equivalent to the functions from

vN

(containing

~) is

L~°(u)-metric on the corresponding

P~ = ~ g " g ¢ L°°(s),

g

is positive and bounded a~Tay

0 }.

For a proof see

pp

27 in

[2].

Michael's theorem in section l.

Compare this theorem with Here " t h e Banach space

Y''

is replaced by a convex set with a complete metric which is convex in the sense described above. (d)

Fixed point theorem and the selection problem :

has proved the following result. subset of the Euclidean

Let

n-space and let

continuous set valued function from for every

x ~ M, the image set

convex.

Then

a point

x°

T with

M

M

Kakutani

be a compact convex T

be an upper semi-

into itself such that

T(x) is nonempty, closed and

admits a fixed point, that is, there exists xo ~ T(Xo).

We will presently show how this

theorem can be viewed as a selection problem.

For this we need

some preli~iniaries. Let vertices

S = [al, a2, ... an+l ]

stand for an

a l, a2 . . . . ,an÷ I. Given a simplex

the point this simplex. O, I, 2, ...

[a I + a2 + ... + aj]

n-simplex

with

[al, a2, ..., aj],

Ls called the barycentre of

The barycentres of all the faces of orders of a simplex

S

determine a family

S (I)

of

n

90 simplexes,

called the barycentric

the properties (i)

n

simplex of

a face of dimension if an

O,

n

S (I)

and

~ Jl' J2 .... 'Jk} ~

S, with

contains the barycentre of

S (I) contains the barycentres

Jail , ai2, ..., ~ p ] ,

where

k < p,

~ il, i2, "'., ip } .

If we divide each n-slmplex of

S (I)

obtain a new family of n-slmplexes,

in a similar manner, we called the barycentric

division of order 2, which we denote by process,

of

of a face of dimension i, etc.

simplex of

of [aJl , aj2 , ... ajk] then

I

"

each

(ii)

division of order

S (2).

Continuing thls

we can define the barycentric division

S (k)

of order

k. Let

T "

S -> S

be an upper semicontint~us

f~inctlon such that for each subset of

S

where

S

of the kth barycentric bk

be a point of

x e S,

T(x) is a nonempty closed

is an n-simplex. division

S (k)

T(a k) and write

x ¢ S, we consider a n-slmplex

set valued

For each vertex

of the simplex

b k = ~k(ak).

S~

ak let

For each

[ k I[ k ] ~ s(k) ai I, ai~, ..., a~n+ 1

which contains it and we put

~k(X) = ¢k(Pl aikI + P2 ~ 2

The function of

S (k)

Sk

+ ... ) =

k k PlSk(ai )+P2@k(al )+ .... i 2

so defined is linear in the interior of a simplex

and is therefore continuous ~

moreover it is uniquely

determined,

even at the points which belong to several n-simplexes

of

consequently

S (k) ,•

Ok

is continuous in

S.

Hence by

9i

Brouwer's fixed point theorem (which asserts that every singlevalued continuous function from there exists a point S

xk ~

S

is compact, the sequence

k

k

~ +i ]

La I , a2, ... the point

Xk,

S

such that ~Xk}

is an

where

xk = Ok(Xk).

has a cluster point

s(k)

n-slmplex of

Since Xo.

If

which contains

we can write

...)=

Xk : $k(Xk ) = Ck (pk ak + pk a k +

k k k k Pl bl + P9 b2

=

into itself has a fixed point),

bk ~ T(a~)

and

+

"'"

+

bk ~ T ( ~ ) ,

p]k Sk(a~) + pk ~ k ( ~ ) +

k Pn+l

...

k bn+l

etc.

Without any loss of generality we shall assume that, XXk k --~ --~ Xo, Xo, a~

Since

b

k

o

k

-~ bi' k_~ bi

-> Xo,

continuous, we have

o

Pi -~ Pi o bi,

as

k -->co

k k b i ¢ T ai

o

b i ~ T xo

for

for

and

T

i -- 1,2,... n+l. is upper seEi-

i = 1,2, ... n+l.

Now we have the following proposition. Proposition 9. I "

Let

S

be an

functions defined as above and continuous with given a

¢ • 0

n-simplex.

T "

S -> S

Let

N

be

be upper semi-

T(x) nonempty and closed for each there exists an integer

Sk

x e S.

Then

such that for all

k~N, d($k(X),

Remark

9.3

"

W(x))

• ~ .

If the slngle-valued functions

Ck

(as defined

above) converges uniformly to a sing!e-valued function

¢o

then

92 ~o

is a selection for

T.

This can be seen as follows.

the definition of the functions of the functions

Sk

to

Sk

and the uniform convergence

'9o, we have,

d($o(x)0 T(x)) 2 d($o(X) , Ok(X)) + d(~(x), for

From

T(x)) ~ ~ + ~ = ~

n ~ i~(¢). Now we are ready to state the following theorem

(which is a slight extention of Kakutani's fixed point theorem) due to Patnaik " with

[9]. Let

T

be an upper semicontinuous map from

T(x) nonempty and closed.

Let

Then there exists a fixed point under

Sk

converge uniformly.

T.

The truth of this theorem can be seen as follows. Sk(Xk) = xk

and

xk -~ xo

uniformly and since that

xo ~ T(Xo).

$o

as

S -~ S

k --~co.

Since

is a selection for

T,

We have @k "~ $o it follows

In Kakutani's theorem we assume each

T(x)

is nonempty convex and closed but here

T(x) need not be convex

for each

~

x.

Of course here we assume

converges uniformly

to ensure the existence of a fixed point. (e)

Control problems of contingent equation •

~ow we shall

consider the control problem and sketch a proof due to Kikuchl for the existence of optimal control [4].

We shall make the

following assumptions. (I) defined in (2)

F(t, x, u) I xR m XR r

is a compact set (in where

I = ~to, to + a] C

F(t, x, u) is measurable in

~t ". t ~ T, F(t,x,u) c

C }

R m) valued function

t

R'.

[that is,

is measurable for every compact set

93 C

in

Rm]

for each fixed

in (x, u) for each fixed (S)

(x, u) a R m

(4)

Q(t, x)

defined in

t s I.

I x Rm

is a compact set

(in

and measurable in x

Rm ~

Rr

Q(t,x)

t

R r) valued function for each fixed

[that is, for every

we can find some nelghbourhood

N~ (Q(t, Xo))

I x

Rm.

and upper semlcontlnuous in

ts

and continuous

F(t, x, u) carries every bounded set in

into a bounded set in

E ~ 0

×

for every

V

x s V]

of

xo

xo

x ~ Rm, and

such that,

for each fixed

I. (8)

Q(t, x) carries every bounded set in

bounded set in

(6)

Rm

into a

R r.

R(t,x) = F(t,x,Q(t,x)) = ~ y • y ~ F(t,x,u),u z Q(t,x)}

is a compact and convex set (in (7)

I ~

For every

holds for every

y

t

and

such that

x

Rm) for each (t,x) s I x and

R m.

u ~ Q(t,x), x.y _< C(Ixl2+l)

y ¢ F(t,x,u), where the dot denotes

the scalar product. (8) (in in

K

is a compact set in

Rm.

Rm) valued function defined in

I

K(t) is a compact-set and upper semlcontinuous

t. (9)

f(t,x) is a real-valued defined in

measurable in

t

for each fixed If

x ~

R m, and is

R m, and continuous in

x

t e I, and is bounded from below.

u(t)

measurable in

for each fixed

I ~

is a measurable function in t

for each fixed

R r,

F(t,x,u(t)) is

x z R m, and is continuous in x

94 for each fixed

t ~ I.

Therefore

for each measurable function

u(t) the system of equations,

x(t) ¢ F(t, x(t), u(t))

for almost all

t ~ I

x(t o) -- x °

has an absolutely continuous solution if

x(t) for every

x° ~ Rm,

F(t, x, u) satisfies the assumptions stated above. We say that

control

x(t) is the trajectory corresponding to a

u(t) (measurable in

x(t) is an

m-dimensional,

t

and

~

Q(t, x(t)) on

I) if

absolutely continuous function

satisfying the above system of equations. We say that a control ~ I, transfers corresponding to

K

to

u(t)

u(t), defined for

K(t)

to _< t_< q ,

if one of the trajectories

satisfies the relations

x(t o) ¢ K

x(t) and

x( ~ ) s K ( ~ ) . We shall consider the problem of finding a control function u(t) which transfers

K

to

K(t) and wFzLch minimises the cost

func rio nal

J(x)

:

S

f(t, x(t)) dt

to where

x(t)

is one of the solutions corresponding to

represents a value of Theorem 9 . 4 • satisfied.

t

such that

u(t), and

x(t) s K(t).

Suppose that the conditions stated above are

Also suppose that there e ~ s t s

u(t) which transfers

K

t~ K(t) on

I.

at least one control

Then there exists an

95 optimal control, that is, a measurable function one of the corresponding solutions, x*(t o) s K,

attains

u*(t)

for which

x*(t), with initial condition

K(t ~) for some

t* e I, and

t*

inf JCx) = J(x*) = S fCt, x~(t)) dr, t0 where, in addition,

u*(t) ¢ Q(t, x*(t)).

We will indicate the proof of this theorem. all the

Consider the set of

x(t) satisfying

~(t) s F(t, x(t), u(t)) almost everywhere on x(t O) ~ K

and

x( ~ ) s K( ~ )

for some

addition,

u(t) s Q(t, x(t)) for some control

such solution exists by hypothesis,

~ ~ I, where, in u(t).

Since one

this set is r~t empty.

Consequently we can select a sequence of trajectories on

I,

I,

~x n (t)}

with tn

J(x n) =

I t

f(t, Xn(t)) dt

0

decreasing monotonically to value of

t

such that

Inf J(x),

Xn(t) ~ K(t).

~Jhere

tn

represents a

N~q(t) satisfy the following

relations ~n(t) ~ R(t, Xn(t)) almost everywhere on

I

and

Xn(t o) ~ K.

solutions of the contingent equation

By the compactness of

[See pp 98-96

in [4]

a proof of the compactness of solutions] we conclude that

for

96 ~*(t) s R(t, x*(t)), where

x~(t)

x~Cto) ~ K,

is a limit function of a subsequence of

~Xn(t) } .

Also we can select a further subsequence (~ithout changing the notation) such that

~tn}

is a compact interval. of

~Xn(t) }

that

converges to some

K

and

measurable function

since

I

Further, making use of the equi-continuity

and the upper semi-continuity of

x~(t o) s

t* ~ I

x*(t*) ~ K(t*).

K(t), we conclude

~ow one can select a

u*(t) such that

x*(t) ~ F(t, x*(t), u*(t))

almost everywhere on

i

and

u*(t) e Q(t, x*(t)) on

I.

Finally

tn J(xn)

=

I to

f(t, xn(t)) dt

approaches

t*

I

f(t, x*(t)) dt

to as

n -~ oo and hence

tO _< t _< t*

inf J(x) : J(x*).

Thus

x*(t)

on

is optimal.

One could state theorem 9.4 more generally in view of the results in section three but we will not be doing that now. results in this connexion refer [5] where the set

For further R(t, x) is not

necessarily convex. (f)

Set-valued measures "

We shall now discuss about set-valued

measures which have many interesting applications in Mathematical economics.

See Debreu and Schmeldler [5]

and

Vind [i0].

97 Let

(T, T)

be a measurable space, and

dimensional real vector space. (T, T)

is a function from

S

be a finite

A set-valued measure

T

@

to the ton-empty subsets of co

which is countably additive, that is,

_T.

S,

CO

U Ej) = 2 $(Ej) J=l J=l El,oc~, ..., of mutually disjoint elements

for every sequence of

on

Here the sum

Y Aj J=l

@ (

of the subsets

~,

~,

..., of

cO

S, consists of all the vectors

Z where the series is j=l aj, aj ~ Aj for every j = 1,2, . . . .

absolutely convergent, and ~

.

~

"

Let

valued measure ~(E) ~ ~(E)

~

~

a =

be a set-valued measure on (T, =T).

on (T, T)

for every

is a selector of

@

A vector-

if

E ¢ __T.

We investigate now the existence of selectors, and their oroperties.

In particular, the following problem is interesting •

Give conditions implying that for everj

E ~ T

x ~ @(E) there exists a selector

~

Recall that measure

k

@ if

t~ of

~ud every

such that

~(E) = x.

is absolutely continuous ~ith respect to the k(E) = O

implies

@(E) = ~ 0 }.

i~ext we state

a theorem due to Artstein LI]. Theorem 9.~ :

Let (T, T, k)

and non-negative.

Let

~

k.

Then for every

@

k

is finite

be a set-~alued measure, with convex

values (that is, for each S), and such that

be a measure space,

E ~ T, @(E)

is a convex subset of

is absolutely continuous with respect to x ~ @(E)

there is a selector

~

of

~

such

that ~ ( E ) -- x~ We will indicate the main steps in the proof. proof refer to Artstein's paper.

For a detailed

98 Note that it is enough to consider the case a singleton then and

$(E) -- ~ ~(E)}

%z is a measure.

Thus

~

x ¢ $(T).

certain

If

$(I'), ~ d

-- sup

p.y:

inf

Now define for every

Sp(E) # @

and

suppose dim ~(T) • O. ~(T)

then for a

~

of

x ¢

Indeed, since E.

Obviously,

Sp

Then one can show that x ~ Sp(T) it follows ~p(T) has dimension

such that

tz(r) = x.

It is clear that

~.

relative interior, one can construct a suitable

set valued function

G

such that

I G(t) dX(t) = relative interior of $(E) for every E

I G(t) dX(t)

g

t]. of

X

and

f(t) z G(t)

for

X-almost

In particular, there exists an integrable G

~(E) = I g(t) dk(t). E

E s T. =

stands for the set of all integrable functions

with respect to the measure

all values of

and

According to the induction hypothesis there is

is also a selector of When

$(T) }

p.x = U p ( E ) } .

for every

$(T).

a selector

selection

Proceed

}

p.y " y

is a set valued measure.

less than

f

E ~ T ,

E z 2, ~ p ( E ) = sup ~ p.x " x ~ O(E)}

@p(E) = ~ x " x ~ $(E)

[Here

is

p ¢ S,

•

that

$(T)

is the required selector.

x ~ relative interior of

p.x

Sp

If

is a singleton for every

by induction on the dimension of Let

E = T.

such that Then

~

I g(t) dX(t) = x. T

Define

is the required selector.

99 One can give examples to show that theorem 9.5

is false

without the convexity assumption or without the condition that @

is absolutely continuous with respect to the measure

k.

We will close this section after ~entioning the results obtained by and

Y

L. D. Brown (Personal communication).

be a non-metrizable compact space.

a-fields of subsets of

X

and

Y

Let

Let

=X and

respectively.

X = Lo, I] =Y be

Of course

is the usual Borel G-fleld containing all closed sets of Let

C

stand for the product G-field generated by means of

rectangles

A xB,

Baire a-field on

A ~ ~, B ~ Y. Y

Let

then projection of

analytic but the result is false if Y.

If

X.

Y

is the Baire ~-field on

Y Y

S ~ C. S

If

onto

__Y is the [O,I] is

is the Borel

G-field on

and if the continuum

hypothesis is assumed to be valid then there exists a measurable selection of -~ Y.

$, that is, a measurable ft!nction

L I also understand from

f " Projection LS]

L.D.~rown's Co,~munication that

he is writing a lecture motes on Statistical decision theory ,~hich will contain a chapter on applications of selections theorems to decision problems with new proofs of so~e of the selection theoreEs].

100

REFERENCES

Ill

Z. Artstein,

Set-valued measures, Research Memorandum

No. 66, [1971], Department of ~thenatics,

Hebrew

University, Jerusalem, ISRA~Iu.

[2]

H. Bauer and .H.S.Bear, The part metric in convex

sets,

Pac. Jour. l~ath. 30 [19A~9j, 15-83.

[3]

G. Debreu and

D. Schmeidler,

lhe Radon-Nikody~ derivative

of a correspondence, to appear. [4]

N. Kikuchi, Control problems of contingent equation, Pub!.

[5]

RIMS

Kyoto Univ.

Set. ~I 3 [1967], 85-99.

N. Kikuchi, On contingent eouations, Japan- U.S. seminar on ordinary differential and functional equations, [1971], preprlnt.

[6]

L. F. McAuley and D. F. Addis,

Sections and Selections,

to appear.

[?]

E. Nichael, Continuous selections If,

Ann. Math

64

[1956], 562-580. [8]

E. Michael, Continuous selections iii, 2ran. M~th 64

[1957], [9]

375-~90.

S. N. Patnalk, On the K~kutani fixed point theorem and its relationship with the selection problem I Springer Verlag Lecture notes in Nathe~atics, edited by W.M.Fleischman, on 'Set-valued mappings, selections and topological properties of 70-73.

2X ' , No .171, [1970] ,

101

[10]

K. Vind, Edge~orth-allocations in an exchange economy with many traders,

Int. Econ. Rev

8 [1964], 168-177.