7>7<[

/y§i No. 7L//6

POLISH SPACES AND ANALYTIC SETS

THESIS

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF SCIENCE

By

Kimberly Muller, B.S.

Denton, Texas

August, 1997

Muller, Kimberly, Polish Spaces and Analytic Sets. Master of Science (Mathe-

matics), August, 1997, 81 pp., references, 4 titles.

A Polish space is a separable topological space that can be metrized by means

of a complete metric. A subset A of a Polish space X is analytic if there is a Polish

space Z and a continuous function f : Z —y X such that f(Z^ = A. The idea of

an analytic set was introduced by Souslin and Lusin and was developed mainly by

the Polish and Russian schools of mathematics. We use Polish spaces and analytic

sets to study measure spaces. We show that every Borel subset of a Polish space is

analytic and that every analytic set is universally measurable. After proving that

each uncountable Polish space contains a non-Borel analytic subset we conclude

that there exists a universally measurable non-Borel set.

7>7<[

/y§i No. 7L//6

POLISH SPACES AND ANALYTIC SETS

THESIS

Presented to the Graduate Council of the

University of North Texas in Partial

Fulfillment of the Requirements

For the Degree of

MASTER OF SCIENCE

By

Kimberly Muller, B.S.

Denton, Texas

August, 1997

TABLE OF CONTENTS

Page

Chapter

1. INTRODUCTION !

2. POLISH SPACES 3

2.1 Countablility Axioms 3

2.2 Disjoint Unions n

2.3 Product Spaces

2.4 Each Polish Space is a G$ 25

2.5 Borel Subsets of Polish Spaces and Measurable Spaces 33

3. ANALYTIC SETS 47

3.1 Borel Subsets of Polish Spaces 47

3.2 The Space Af 49

3.3 Zero-Dimensional Spaces 54

3.4 Non-Borel Analytic Sets 5g

4. THE SEPARATION THEOREM AND ITS CONSEQUENCES 66

4.1 Borel Measurable Functions 66

4.2 Borel Isomorphisms 71

5. THE MEASURABILITY OF ANALYTIC SETS 74

REFERENCES 8 1

CHAPTER I

INTRODUCTION

In both analysis and topology, many of the spaces most commonly studied are

separable and metrizable. Often, the metric is also complete. The study of such

spaces, historically called Polish spaces, has led to many interesting results. This

paper is a study of Polish spaces and analytic sets. Many definitions are needed to

begin this study. We say that a sequence (xn)n in a metric space X is a Cauchy

sequence, if for every e > 0 there exists an integer N such that d(xm,xn) < e if

m,n > N. A metric space is said to be complete, if every Cauchy sequence is

convergent. The space of real numbers with the standard metric is an example of a

complete metric space. However, the open interval (0,1) with the standard metric is

not complete because (~)n is a n example of a sequence which is Cauchy but which

does not converge to a point in (0,1). We will find later that it is possible to find

another metric which metrizes (0,1) with the standard topology which is complete.

In the early chapters we will develop several ways of producing a complete metric

from a given metric.

A space is said to be separable if it contains a countable dense subset. Again, the

space of real numbers is an example of a separable space because the set Q made

up of all rational numbers is countable and dense. A Polish space is a separable

topological space that can be metrized by means of a complete metric. From above

we see that the space of real numbers with the standard metric is an example of a

Polish space.

If X is a Polish space a subset A of X is analytic if there is a Polish space Z and

a continuous function f : Z ^ X such that f ( Z ) = A. The idea of the analytic set

was introduced by Souslin and Lusin and was developed mainly by the Polish and

Russian schools of mathematics between the First and Second World Wars. We will

use analytic sets to study measurable spaces. We will show that every Borel subset

of a Polish space is analytic and we will also prove the existence of a non-Borel

analytic set.

If (X, A) is a measurable space, a subset of X is said to be universally measurable

with respect to (X, A) if it is ^-measurable for every finite measure fx on (X, A).

In the final chapter after showing that each analytic subset of a Polish space X

is universally measurable with respect to (X ,B(X) ) where B(X) is the class of all

Borel subsets of X , we will be able to conclude that each Polish space contains a

non-Borel set which is universally measurable.

CHAPTER II

POLISH SPACES

2.1 Countability Axioms

Before we begin our study of Polish spaces we first need to define a few useful

topological properties. First recall that a space X is said to have a countable

basis at x if there is a countable collection B of neighborhoods of x such that each

neighborhood of x contains at least one of the elements of 3. A space that has a

countable basis at each of its points is said to satisfy the first countability axiom.

A topological space X is said to satisfy the second countability axiom if X has a

countable basis for its topology. Any set having the discrete topology or any set

which is metrizable would satisfy the first countability axiom. If the space X is

metrizable, an example of a countable basis at any point x £ X is the set that

consists of all balls centered at x with radius r, where r is a rational number.

Any countable set having the discrete topology would clearly satisfy the second

countability axiom. However an uncountable set having the discrete topology would

be an example of a space which satisfies the first but not the second countability

axiom.

The countability axioms are useful in the study of Polish spaces because of the

relationship between second countability and separability. In general, the property

of separability is a weaker property than second countability. The following re-

sult shows that any topological space which has a countable basis also contains a

countable dense subset.

Propos i t ion 2.1.1. If X is second countable then X is separable.

Proof. Suppose X is second countable. Let B = {B{ ji E N} be a countable basis

for X. For each i 6 N choose a point bi E Bi, and let S = [J{M- Then S is

countable because N is countable. Obviously S C X. Let x E X and U be an

open set containing x. Choose a positive integer j such that Bj C U and x E Bj.

Then bj E U and bj E S. Therefore every open set containing x intersects S and

x E S. Consequently, S = X. Since X contains a countable dense subset, X is

separable. •

Now we have that every second countable space is separable. However the con-

verse is not necessarily true. Consider the following example. Give the set of real

numbers R the lower limit topology. The set of all intervals of the form [a, b) where

a and b are real numbers form a basis for JR.. The set R with the above topology is

often called the Sorgenfrey line. Clearly, Q is a countable dense subset of R. Now

suppose B is a basis for R. Choose for each x E R an element Bx in B such that

x £ Bx C [x, x + 1).

If x then it is clear that Bx ± By because x = gib Bx and y = gib By. Conse-

quently B must be uncountable because R is uncountable. Therefore the Sorgenfrey

line is an example of a space which is separable but which is not second countable.

However using the following result we see that if a space is metrizable the properties

of separability and second countability are equivalent.

Propos i t ion 2.1.2. Let X be a metrizable space. If X is separable, then X is

second countable.

Proof. Let X be a separable, metrizable space. Let d be a metric for X, and let

S = {xi\i E N} be a countable dense subset of X. Let

B — {Bd(x i , r ) \x{ E S and r E Q}.

Then B is countable because S and Q are countable. Also B is a collection of open

sets in X with the metric topology.

Let U be an open set in X and let p G U. Because S is dense in X, p G S.

Case 1: p G S. Since U is open, q G Q can be found such that Bd(p,q) C U.

Therefore we have p G Bd(p,q) G B.

Case 2: p ^ S. Choose e > 0 so that Bd(p, e) C U. Such an e exists because U is

open. Also, Bd(p, | e ) is open. Choose j 6 N such that xj G S and xj G Bd(p, | e ) .

Since p S and xj G S, p ^ xj. Therefore 0 < d(p,xj) < | e . Choose rj G Q so

that

0 < d(p, Xj) <rj< ^e.

Hence p G Bd(xj,rj). Let y G Bd(xj,rj). Then

2 2 1 1

d(p,y) < d(p,Xj) + d(xj,y) < - e + - e = e.

Therefore p G Bd(xj,rj) C Bd(p, e) C U.

In both cases there exists an element of B containing p which is a subset of U.

Therefore B is a basis for the metric topology on X. Since B is countable, X is

second countable. •

We could conclude from the above argument that the Sorgenfrey line is not

metrizable. Consequently the Sorgenfrey line is an example of a space which is not

Polish. Now we have that each Polish space has a countable basis. The following

result is also useful.

Propos i t ion 2.1.3. If X is a second countable topological space and if Y is a

subspace of X, then Y is second countable.

Proof. Suppose X is second countable. Let B = {Bi\i £ N} be a countable basis

for X. Then C == {Bi Pi Y\Bi E B} is a countable basis for the subspace topology

on Y. Therefore Y is second countable. •

In the remainder of the paper if A is a subset of the topological space X then

Ac will be used to denote the complement of A in X . We can now use the above

theorems to establish the following theorem about Polish spaces.

Proposi t ion 2.1.4. Each closed subspace, and each open subspace of a Polish space

is Polish.

Proof. Let X be a Polish space and let Y be a subspace of X. Then X by definition

is separable and metrizable. Also, X is second countable by Proposition 2.1.2.

Therefore Y is second countable by Proposition 2.1.3. Consequently Y is separable

by Proposition 2.1.1. Let d be a complete metric for X.

Case 1: Suppose Y is closed. Since X is a Polish space and Y is a subspace of

X every Cauchy sequence in Y converges to a point in X. Since Y is closed every

Cauchy sequence in Y converges to a point in Y. Therefore, the metric inherited

by restricting d to Y is a complete metric for Y. Hence each closed subspace of X

is Polish.

Case 2: Suppose Y is open. Without loss of generality suppose that Y ^ X.

Recall that d(x,Yc) = i n f { d ( x , z ) : z £ K c } for all x £ X. Define the function

d0 : Y x Y —> R by

d0{x,y) = d(x,y) + 1

d(x,Yc) d(y,Yc)

We need to show that dQ is a metric for Y. Note that if x £ Y then d(x, Yc) > 0

since Y° is closed.

i) Since d(x,y) > 0 for all x,y £ X and absolute values are non-negative,

d0(x,y) > 0 for all x,y £ Y. Also, d(x,y) > 0 if x ^ y, thus dQ(x,y) > 0 if x ^ y. 1 1 If x = y, d(x, y) = 0 and

d(x,Yc) d(y,Yc) ii) Since d(x,y) = d(y,x),

do(x,y) — c?(a;,y) +

= 0 therefore d0(x, y) = 0.

d{x,Y°) d(y, Yc)

d(y,x) +

d0(y,x).

1

d(y,Yc) d(x, Yc)

iii) Now suppose that x,y,z £ Y. Then

d0(x,z) = d(x,z) +

< d(x,y) + d(y,z) +

= d0(x,y) + d0(y,z).

1

d(x,Yc) d(z, Yc)

1

d(x,Y°) d(y, Yc) +

d{y,Y°) d(z, Y c )

Therefore d0 is a metric in Y. Let Td0 be the topology inherited from dQ. The

next few paragraphs demonstrate that T(i0 is equivalent to the subspace topology

Td inherited from the metric topology on X.

Suppose that y £ Y and e > 0. Then B<i(y, e) is a basis element for Td containing

the element y. Consider Bda(y,e), which is a basis element for Tj0. Let x £

B<i0(y,e). Then d0(x,y) < e. It follows that d(x,y) < e. Hence x £ Bd(y,e).

Therefore, y £ Bda{y, e) C Bd(y, e) and Tda is finer than Td.

Again let y £ Y and e > 0. Then Bda (y, e) is a basis element for Tda containing y.

Since y EY and Y is open in X, there exists a neighborhood of y which is disjoint

from Yc, giving d(y, Yc) > 0. Choose S > 0 such that

4 < min{r- 2 S i '

Consider Bd{y,&). Let x £ Bd(y,S). The following argument shows that

d(x,y) > \d(y,Yc) - d(x,Yc)\.

Let z e Y c . Then

d(x,z) < d(x,y) + d(y,z)]

consequently,

inf{d(x, z) : z € Yc} < inf{d(x, y) + d(y, z) : z € F c } .

Equivalently,

and

Similarly,

thus

Hence

d(x,Yc) < d(x,y) + inf{d(y, z) : z £ Yc}

< d(x,y) + d{y,Y%

d(x,Yc) - d(y,Y°) < d{x,y).

d(y,z) < d(x,y) + d(x,z);

inf{d(y,z) : z 6 Yc} < inf{d(x,y) +d{x,z) : z 6 Yc}.

and

d{y,Yc) < d(x,y) + xni{d{x,z) : z e Yc}

< d(x,y) + d(x,Yc),

d(y,Yc) - <%,FC) < d(x,y).

Therefore \d(y,Yc) - d(x,Yc)\ < d(x,y).

Now it needs to be shown that x € Bdo(y, e). First we find that

d0(x,y) = d(x,y) -f

= d{x,y) +

d(x,Yc) d(y,Yc)

d(y,Yc) - d(x,Yc)

d(x,Yc)d(y,Yc)

< d(x,y) + d(x,y)

d{x,Y°)d(y,Y°)

Also, since

d(x,y) > |d(y,Yc) - d{x,Yc)\ > d(y,Yc) - d(x,Yc),

we have that

d(x,Yc) > d(y,Yc) - d(x,y) = \d(y,Yc) - d(x,y)|,

which yields the inequality

d0(x,y) < d(ar,j/) +

Since d{x, y) < 5,

d(x,t/)

d(x,y)

d(y,Yc)[d(y,Yc) - d(x,y)]

d(y,Yc)[d(y,Yc) - d(x,y)\ <

d(y,Yc)[d(y,Y<) - d(x,y)\

Moreover,

and

d(y,Y°) - d(x,y) > d(y,Yc) - S,

d(y,Y")[d(y,Y") - d(x,y)} > d(y,Y°)[d(y,Y°) - <S],

I d ( y , Y c ) \ d ( y , Y c ) - rf(x,!/)]| > \d(y,Yc)[d(y,Y<) - i] | .

Therefore,

d{y,Yc)[d(y,Yc) - d(x,y)]

Recall that S was chosen so that

< d{y,Y*)[d{y,Y*) - 5}

c\2

5 < ed(y,Yc)

2 + ed(y,Y°Y

10

Consequently,

2S + Std(y,Yc) < ed(y,Yc)2,

2S < Cd(y,Y')[d(y,Y') - <S],

and

> d(y,Yc)[d(y,Yc) - S\

Therefore, d0(x,y) < | e + \e = e. Hence x £ Bdo(y,e) and therefore x G

Bd(y,$) Q Bd0(y,e), and Td is finer t han Td0. Since Td C Td0 and Tj 0 C Td,

Td = Tdo and d0 metrizes the subspace topology on Y.

All t ha t is left to be shown is tha t d0 is a complete metric. Let (xn)n be a

Cauchy sequence under d0. Since d(xi,xj) < d0{xi,xj), the sequence (xn)n is a

Cauchy sequence under d and converges to a point x € X. Suppose x £ Y. Then

x E Yc and d(x, Yc) = 0. Let N £ N. Choose M such tha t if m, n > M then

d0(xm, xn) < 1. Choose q> M such tha t

d(xM,Yc) d(xq,x) <

Nd(xM,Y°) + 1

Since

we have tha t

Hence

d(xq,Yc) < d(xq,x) < d(xM,Yc),

1 1 <

d(xM,Yc) d(xq,Yc)

d0(xM,xq) = d(xM,xq) + 1

> 1

d(xM,Yc) d(xq,Y°)

1

d(xM, Yc) d(xq,Yc)

1 1

d(xq,Yc) d(xMiYc)

11

d(xq,x) CI(XM,YC)

Nd(xM,YC) + 1 _ 1 > D(XM,Yc) D(XM,Yc

= N.

This is a contradiction to d0(xM? xq) < 1- Consequently x £ Y, and therefore Y is

complete under d0 and Y is Polish. •

In the introduction we observed that the interval (0,1) with the standard metric

is not complete. However the above proof gives us a means of finding a metric which

metrizes (0,1) with the standard topology which is complete.

2.2 Disjoint Unions

If (X, T) and (Y, <S) are two topological spaces with X and Y disjoint, we define

a topology on the union Z = X U Y by taking as our open sets O C Z for which

OC\X eT and OC\Y 6 S. We call the space Z with this topology the disjoint union O

of X and Y and denote it by X U Y. If (X a ,Ta) is a n y indexed family of topological o

spaces with Xa Pi Xp = 0 for a 7̂ /3, we define their disjoint union Z = [jXa to

be the union of the Xa with O C Z defined to be open if O D Xa £ Ta for each

a. Each space Xa is referred to as a direct summand. If the sets Xa are not all o

disjoint, we take Ya = Xa x a and consider the disjoint union [jYa. The following

results show that many of the topological properties of a disjoint union depend only

on the topology of each of the direct summands.

o

Proposit ion 2.2.1. Let Z = {jXa. Then

a) the map f : Z —> Y is continuous if and only if each restriction f\xa i>s

continuous}

b) a set F C Z is closed if and only if F D Xa is closed for each a, and

c) Z is Hausdorff if and only if each of the spaces Xa is Hausdorff.

12

Proof, a) Suppose tha t / : Z —> Y is continuous. Let U be an open set in Y.

We assert that (f\xa D Xa for all a. First let x G ( / \ x a ) ~ 1 { U ) -

Then x G Xa and f{x) G U. Consequently, x G fl X a and ( / | x a ) - 1 ( £ 0 Q

/ - 1 ( ? 7 ) n X a . Now let X e f~l{U) n Xa. Then f{x) G U and x G Therefore

* £ ( / U J - 1 ^ ) a n d

( / k a r 1 ( ^ ) = r 1 ( ^ ) n x Q

as desired. Because / is continuous, / - 1 (Z7) is open in Z. Also, since Z is a disjoint

union / - 1 ( J 7 ) 0 is open in Z for each a. Therefore ( / | x a ) _ 1 ( ^ ) is open and

f\xa is continuous for each a.

Now suppose that f\xa is continuous for each a. Let U be open in Y. From

above, we have that ( f \ x a )_ 1 ( ^ ) = / _ 1 (U)C)Xa. Since f\xa i s continuous for each

a, we have tha t / - 1 ( { 7 ) D Xa is open for each a. Therefore by definition of disjoint

union, f~l(U) is open in Z. Hence / is continuous.

b)Let F C Z be closed. Then Fc is open and Fc fl Xa is open for each a. Since

Fc fl Xa = Xa — (Xa D F) for each a , Xa — (Xa fl F) is open in Xa for all a , and

consequently Xa D F is closed in Xa for all a.

Now suppose that Xa fl F is closed for each a. Then Xa — (Xa fl F) = Fc fl Xa

is open for each a. Therefore Fc is open in Z and F is closed in Z.

c) Suppose Z is Hausdorff. Let x and y be elements of Xa for some a. Since x

and y are also elements of Z there exist open sets U and V in Z such tha t x G U,

y G V and U fl V = 0. Because Z is a disjoint union, U Pi Xa and V D Xa are open

in Xa. Since x G U fl Xa, y G V D Xa and (U fl Xa) f l ( F n Xa) = 0, we have that

X a is Hausdorff for each a .

Now suppose for each a we have that Xa is Hausdorff. Let x,y G Z with x ^ y.

Then x G Xai and y G Xa. for some i,j. Suppose i =/=• j. Then we have that

Xai fl Xa. = 0 and both Xai and Xaj are open in Z. If i = j there exists disjoint

13

open sets U and V in Xa{ such that x <E U and y G V. Since U n Xp = 0 and

V H X/3 = 0 for (3 ̂ a i we have that U and V are open in Z. In either case we have

that x and y can be separated by open sets. Therefore Z is Hausdorff. •

Therefore, if a space is a disjoint union we can determine the properties of the

disjoint union by studying the properties of the individual summands. In the next

proposition we learn a little more about these direct summands.

P r o p o s i t i o n 2 .2 .2 .

a) If a subset Xi of a topological space X is a direct summand (in some disjoint

union), then Xx is both open and closed in X.

b) If is a subset of X which is both open and closed, then X = X\ U X2 where

X 2 = X - X 1 .

Proof, a) Suppose X = X\ UX2. Then, since X is open and X\ is a direct summand,

X fl Xi = Xi is open in X. Similarly, X2 is open in X . This implies that X - X2

is closed in X. Since Xx n X2 = 0 and X = Xx U X2 we have that X - X2 = Xx

and therefore X\ is also closed.

b) Suppose that Xx C X is both open and closed. Let X2 = X - Xx. Clearly

X = Xi U X2 and X\ H X2 = 0. Let 0 be an open set in X. Then we have that

Xx n O is open in X\ with the subspace topology and X2 fl O is open in X2 with

the subspace topology. Therefore 0 is open in Xx U X2. Now suppose 0 is an open

set in Xx U X2. Then 0 = (OflXi)U(On X2). Since O is open in Xx U X2, 0 fl Xx

and O D X2 are open in X] and X2 respectively. Since both Xx and X2 are both

open in X we have that 0 is open in X. Therefore X = Xx U X2. •

To obtain some of our results, it will often be necessary to construct a bounded

equivalent metric from a given metric. The following proposition gives us a method

for constucting such a metric.

14

Proposit ion 2.2.3. Let X be a metric space with metric d. Then

d0(x,y) = min{d!(x, y), 1}

is a bounded metric that induces the given metric topology of X, and X is complete

under d0 if and only if X is complete under d.

Proof. It first must be shown, that d0 is a metric.

i) Clearly da(x, y) — 0 if and only if d(x, y) = 0. Since d is a metric we have that

d0{x,y) = 0 if and only if x = y. If x ^ y then d0(x,y) = d(x,y) or d0(x,y) = 1.

In either case d0(x,y) > 0.

ii)If d0(x,y) = d(x,y) then d(x,y) = d(y,x) < 1. Therefore

do(y, x) = d(y, x) = d(x, y) = d0(x, y).

If d0(x,y) = 1, then d(y,x) = d(x,y) > 1. Therefore d0(y,x) = 1 = d0(x,y). In

either case d0(x,y) = d0(y,x).

iii)Suppose x, y, and z are three points in X. Either d0(x,y) + d0(y,z) < 1 or

d0(x,y) + d0(y,z) > 1. If d0(x,y) + d0(y,z) < 1 we have that d0(x,y) < 1 and

d0(y,z) < 1. This implies that d0(x,y) = d(x:y), d0(y,z) = d(y,z) and

d(x, z) < d(x, y) + d(y, z) = d0(x, y) + d0(y, z) < 1.

Therefore d0(x1zs) d^x^z^j d0(x^y^j c?0(y,z). If d0(y,z^ ^ 1 then

clearly d0(x,z) < d0(x,y) + d0(y, z).

Consequently dQ is a metric. Now let x 6 X and let e > 0. Let S = min{e, 1}.

Then clearly Bd0(x,5) C Bd(x,e). This implies that the topology on X inherited

from d0 is finer than the topology on X inherited from d. Again suppose that

x € X and e > 0. If e < 1, then Bd(x,e) = Bdo(x,e). If e > 1 then Bd(x, 1) C

X = ^d0{x,e). Consequently the topology on X inherited from d is finer than

15

the topology on X inherited from d0. Therefore the two metrics induce the same

topology. Since a sequence is Cauchy with the metric d if and only if the sequence

is Cauchy under the metric d0 we have that X is complete under the metric d if

and only if X is complete under the metric d0. •

In later proofs, we find that it is often useful to form disjoint unions from existing

Polish spaces. Above we have found that disjoint unions often preserve the proper-

ties of the individual summands. Now we find that if each of the direct summands

is Polish, then the disjoint union is also Polish.

Propos i t ion 2.2.4. The disjoint union of a finite or infinite sequence of Polish

spaces is Polish.

O

Proof. Let Xi,X2,X$,... be Polish spaces and let Z = [jXn be their disjoint union.

For each n let D„ be a countable dense subset in Xn and let dn be a complete

metric for Xn. Using Proposition 2.2.3 we can assume that dn(x, y) < 1 for each n

and each x, y G Xn. Define d such that for each x,y G Z dn(x, y) if x, y £ Xn for some n

d(x,y) 1 otherwise.

It is easily seen that d defines a metric on Z. We need to show that the topology

induced by d is the topology of Z. Let x G Z and e > 0. Then x G X j for some

positive j. Clearly if e > 1, Bd(x,e) = Z which is open in the disjoint union.

If e < 1, Bd(x,e) = Bdj(x,e) which again is clearly open in the disjoint union.

Similarly if U is open in the disjoint union, then there exists e such that 0 < e < 1

and Bdj(x,e) C U. Consequently Bj(x, e) C U. Therefore d metrizes the topology

of Z. Let be a Cauchy sequence in Z. For some N G N if m , n > N we have

that d(xm,xn) < 1. Therefore for all i > TV, Xi G X j for some j. This implies that

(xi)i^zN a Cauchy sequence in X j and therefore converges to a point of X j . Since

X j C Z, (xi)i converges to a point in Z. Now we have that Z is complete.

16

All that remains is to show that D = | J n Dn is a countable dense subset of Z.

Clearly since each of Dn is countable and D is the countable union of the sets Dn.

D is also countable. Let U 7̂ 0 be an open set in Z. Then for all i £ N, U D Xi is

open in X{. For some some j we have that U fl X j 7̂ 0. Since Dj is dense in X j

there exists an x £ U fl X j such that x £ Dj. Since Dj C D we have that x £ D.

Also since x £ U fl X j we have that x £ U. Therefore U contains a point of D and

D is dense in Z. Since Z is metrized by a complete metric and contains a countable

dense subset, Z is Polish. •

2.3 Product Spaces

Now we turn our attention to product spaces. Recall that if (X, 7~) and (7, <?)

are two topological spaces, we define a topology on the product 1 x 7 by taking

as a base the collection of all sets of the form U\ x U2 where U\ £ T and U2 £ <S.

This is called the product topology for X x Y. If (XaiTa)a is any indexed family of

topological spaces, we define the product topology on ]~|a Xa by taking as a base all

sets of the form aUa, where Ua £ Ta and Ua = Xa except for a finite number of

a . In the following proposition another method of constructing a bounded metric

is used. We see that if p is a metric for the space X then the metric p* given by

p* = is a bounded metric on X which yields the same topology as p.

Propos i t ion 2.3.1. Let (Xk,Pk)k be a sequence of metric spaces. We define their GO

direct product Z = X& by letting Z be the space of all sequences (xk)k with k=1

CO Xk G Xk and defining a metric r on Z by setting r(x,y) = ^ ^~kP\{xk-> Vk) where

k=i X = (xk)k, 'y = (yk)k and pi = Then

a) r is a metric and the sequence in Z where X{ — {xn, Xi2,...) converges to

y E Z if and only if (xij)(^zl converges to yj for each j,

b) (Z7t) is complete if each (Xk,PK)

17

are c) and if for each k the space (X\k,Pk) and {Yk,&k) are homeomorphic, then so oo oo

the spaces n xk and n Yk. k=1 k=1

Proof, a) Let Z and r be defined as above. Let x,y E Z. For each i E N,

pi{xj,yi)

1 + pi(xi,yi) < 1 therefore

OO / \ oo

r(x,y) = < y 2 - ' = i.

Consequently, r : Z x Z —>• R.

i) Again let E Z. Since pk(xk,yk) > 0 for all k, r(x,y) > 0. Since pk is

a metric for each k1 if x = y, xk = yk for each fc, and pk(xk,yk) = 0 for each k.

Therefore

r ( x , x ) = f ; 2 - t1

/ " ' ( l i ' X t ) , = 0 . fc_ - + Pk{xki Xk)

Now suppose r(x,y) = 0. Then

2_ f c pk(xk,yk) _ Q

1 + Pk{xk,yk)

for all k. Thus pk(xk,yk) = 0 , giving xk = yk and consequently x = y as desired.

ii)Clearly r(x,y) = r(y,x) because pk(xk,yk) — Pk(yk,xk) for all k.

iii)All that is let to show is that r satisfies the triangle inequality. Let x,y,z E Z,

then

pk(xk,yk) . Pk(yk,zk) r(X,y) + r{y,z) = £ £ 2" l + w ( w , , t )

~kPk{xk,yk)[l + pk(yk,Zk)] + pk(yk,Zk)[l + p(xk,Vk)\ = £ 2 ~

k=1 (1 + pk(xk,yk))(i + pk(yk,zk))

°° -k Pk{xk,yk)+ 2pk(xk,yk)pk(yk,zk) + Pk{yk,zk) E 2 " * t k= 1

+ pk(xk,yk)pk(yk,zk) + Pk(xk,yk) + Pk(yk,zk)

> ^ n - k pk(xk,yk) + 2pk{xk,yk)pk{yk,zk) + pk(yk,zk) E 2 " ' r k= 1

+ 2pk(xk,yk)pk(yk,zk) + pk(xk,yk) + pk(yk,zk)

18

= T . 2 - " — , — k=i pk(xkiVk j+i pk(%k ,yk) pk(yk >zk)+pk(yk ,zk) oo -

> E 2 - ' , ' , + 1

k= 1 pk(Xk^k) Pk{xk^k)

k=1 l + Pk(xk,Zk)

= r(x7z).

Hence r is a metric.

Let (xi)i be a sequence in Z which converges to y, where Xi = (xn, Xii,...) for

each i and y = (yi , j/25 L ( ,t j 6 N and let e > 0. F ind 8 > 0 such tha t 8 < ,

giving tha t 8 8e < e and Choose N <E N such tha t n > iV implies tha t

r ( « n , y ) < Then

, . _ 0 - f c PkjXnkiVk)

r { x n , y ) - ^ l l + p ^ . y f c ) < 2> /C= 1

for all n > N. This implies t ha t if n > N, then

2 -i Pj(Xnj->yj) ^ 1 + pj{xnj,yj) 2-?'

P j ( X n j , V j ) ^ ^

1 + pj(xnj,yj)

P j { x n j 1 V j ) < $ + ^Pj{xnj') Vj)i

Pj(xnj,yj)( 1 -5) < ^

(J Pj{Xnj->Vj) <

1 - < T

and

P j ( x n j i V j ) < e*

Hence converges to y j for all j G N.

Now suppose (#m)n?=i converges to yj for all i £ N. Let e > 0 and choose N <E N

such tha t for all n > TV oo 1

£ 2 - ' < r . k=n-\-1

19

For each i < N, Mi can be found such tha t pi(xni,yi) < for all n > Mi. Let

M = max{iV, Mi , M 2 , M j v } - Let n > M. Then

w / t ( X . , , , ) = £ 2

1 +pi(a ;ni ,y i ) 2=1 N

Pifanii Vi)

4" Pii^xni^ y>t

__ o~?' ^ g - t

j=1 1 Pi(xnif Hi) i—N+i ^ P^Xn^V 0

N oo / x

^ / - >k i o - i Pi\xni,yi)

;=»+! 1 + * (*« . , ! / , )

N oo

^ E 2 _ i | + E 2 _ i

i-1 i = i V + l

^ 1 1

~ 2 2

= e.

Therefore (arn)ra converges to y in Z .

b)Let {xi)i be a Cauchy sequence in Z. Let e > 0 and let j £ N. Choose N such

tha t r{xn, xm) < 2~- 7 (^y) for all n , m > N. Then for each n,m > N we have tha t

E0 - t W t \ „ 0 - j 6

2 — _ = r (a ; B ) a ; m ) < 2 J — — j- ^ 1 "t~ Pi\Xni 5 ^ 1

This implies tha t

2~i Pj(xnj) amj) ^ 2~j e

1 -)- P j{xnj ^Xmj^ 6 1

and therefore P j ( x n j , x m j ) < e. Consequently is a Cauchy sequence in

Xj for all j. By hypothesis ( x n j ) 1 converges to the point yj and by par t (a)

this implies tha t (x,), converges to the point y = (y1,y2,...) G Z. Therefore Z is

complete.

c) Assume tha t for each k the spaces (Xk, pk) and (Yk, crfc) are homeomorphic. For O O

all A; let fk : Xk —> Yk be a homeomorphism from Xk onto Yk. Define / : Xk —> k=i

20

oo

n Yk by f(x) = (fi(xi), f2(x2), fs{x3)...) where x = (xi,x2,x$,...). Since each fk k=1 is a bijection, it is clear that / is also a bijection. Suppose (xi)(-Z1 is a convergent

oo sequence in Yl Xk• By part (a), x is a convergent sequence in Xj for each j.

k=1 Since f j is a homeomorphism we have that is a convergent sequence in

oo Yj. Again by part (a) we have that ( f (x i ) ) f l 1 is a convergent sequence in Yk.

k=1 Therefore / is continuous. The proof that / _ 1 is continuous is similar. Therefore

oo oo oo the spaces -^k and ]~| Yk are homemorphic. The proof is similar that if n xt

k= 1 fc=l k= 1 oo

and Yk are homeomorphic then for each k the spaces (Xk,pk) and (Yk,crk) are k= 1

also homeomorphic. •

Proposit ion 2.3.2. If (X,p) and (F, a) are two metric spaces then the product

topology on X x Y is the same as the topology induced by the product metric.

Proof. Let T represent the product topology on X x Y, define r on X x Y by

r({x1,yl),(x2,y2)) = \J p{x i,x2)2 + o"(yi,y2)

2,

and let S represent the topology induced by the metric r . Let (a, b) 6 l x F and let

A x 5 b e a basis element of (X x Y, T) which contains (a, 6). By definition of product

topology, A is open in X and B is open in Y. Consequently there exist > 0 and

eg > 0 such that Bp(a, e^) C A and Ba(b, eg) Q B. Let e = m i n l e ^ e s } ; then

(a, b) £ Bp(a, e) x Ba(b, e) C A x B. Let (c, d) £ Br((a, 6), e); then

\/V(a,c)2 +<j(6, d)2 = r((a, 6), (c, d)) < e.

This implies that p(a, c) < e and <r(6, d) < e. Therefore (c, G?) £ A X B and

Br((a, b), e) C 4 x 5 . Hence TCS.

Now let (a,b) e X x Y and let e > 0. Let S = ^ and let (c,d) € Bp(a,S) x

5^(6,5). Then

-((a,6),(c,d)) = yjp{a,c)2 + <t(6,o?)5

21

< v ^ 2 + £2

" V ¥ + 2 "

= e.

Hence (c,d) G Bp(a,8) x B(r(b,S) C _Br((a, 6), e). Consequently S CT. Therefore

the two topologies are equal. •

Now we return to our study of Polish spaces.

Proposit ion 2.3.3. The product of a finite or infinite sequence of Polish spaces is

Polish.

Proof. Let Xi,X2,X$,... be a finite or infinite sequence of Polish spaces. Assume

that Xi ^ 0 for all i. For each n let dn be a complete metric in Xn such that

dn(x,y) < 1 for all x,y G Xn. Also define d such that

d{X) y) = ^ ^ 2 d n (x n , y n ^ n

where x = (xux2,...) and y = (t/i,t/2, •••)• Let X->V € \[Xn. n

i) If d(x,y) = 0 it must be that dn(xn,yn) = 0 for all n. Since each dn is a

metric, it must be that xn = yn for all n which implies that x = y. Also if x = y it

is obviously true that d(x,y) = 0.

ii) Clearly since dn(xn,yn) = dn(ynixn) for all n we have that d(x,y) = d(y,x).

iii) Let x,y,z G \\Xn. Then n

y) -|- z) ^ ^ 2 dn[xn^ yn) ^ ^ 2 dn{yn^ zn) n n

^ ^ 2 [dn[xn^ yn) 4~ dn(yn^ zn)~\ n

^ 2 dn[xn^zn) n

= d(x, z).

22

Therefore d is a metric for the space JJ Xn. n

Next we show that the topology Td inherited from d is the same as the product

topology, Tp. Let x £ J ] Xn. Let U be a basic open set in Xn, Tp) which contains n n

x. Then

U = Ui X U2 X U3 x ... X Ui x Xi+1 x Xi+2...

for some choice of i £ N and U& open in Xk, k < i. For each Uk choose an such

that 0 < €k < 1 and Bdk(xk, efc) Q Uk• Let e = minje i , e2, e3,..., e;}, and let

V = Bdl(xi,e) x Bd2(x2,e) x ... x Bd{(xi,e) x x X i + 2 x ....

Then V C U. Consider Bd(x, 2 V) which is open in X n with the topology Td. n

Let y £ Bd{x, 2~le). Then

d(xj t/) — ̂ ^ 2 j/n) 2 e. n

Consequently for all n < i,

2 dn(^n, ) < 2 €

and dn(^n5J/n) < < e. Therefore y G V C £7, 2~~*e) C 17, and C T^.

Again let x G J ] and let 0 < e < 1. Note that for all x, y G f ] we have n n

d(x,y) = < 1. n n

Consider Bd(x, e) which is a basis element for Td which contains x. Since the sum

^ 2 - r a < oo we can choose N such that 2~n < | e . Let « n>iV

Z7 = 2^:) X Bd2(x2, X ... X BdN(xN, ^~r) X Xjy+i X ...

and let y £ {/. Then

?/) ^ ^2 dn(xn,yn^

23

^ ^ 2 dn[xn,yn^ -f- ^ ^ 2 dn(xn, y n ) n<iV n>N

^ ^ ^ (^n ? Un ) H~ ^ ^ 2 dn (xn, yn) n<.N n>iV

^ Ne 1 < |_ —6 ~ 2 i V 2

= 6.

Therefore {7 C e), C Tp , and we have that d metrizes the product topology

on l\Xn. n

All that is let to show is that the metric d is complete and that the product space

is separable. Let ( x n ) n be a Cauchy sequence under <i, where for each n, the point

%n = (^ni j xn2-> Xn3,...). Fix j and let e > 0. Consider the sequence in X j .

Choose N G N such that m , n > N implies that d(xn,xm) < 2-«?e. Then

d^XfijXffi^j — ̂ ^ 2 dk{^Xjik, ) <C 2 ^6. k

Consequently

2 ^d j (x T l j , xmj) <C 2 *̂ 6,

and

dj , %mj ) "\ ۥ

Since this is t rue for all m, n > N we have that is a Cauchy sequence in X j .

Since X j is Polish, the sequence converges to a point of X j . Since this is true for all

J5 the sequence ( z n ) n converges to a point of \\Xn. Therefore ]\Xn is complete n n

under d.

For each n choose a countable base Bn for Xn. Let

$ — {Ui x U2 x ... x Un x Xjv+i x ...\Ui G Bi and N £ N}.

24

Then B is a base for J~[ Xn and B is countable. Since ]^| Xn is a metrizable space with n n

a countable base, by a previous theorem Xn is separable and therefore Polish. • n

Now we have that the product of a sequence of Polish spaces is Polish. The next

result gives a rather explicit way of obtaining a countable dense subset.

L e m m a 2 ,3 .4 , Suppose that (Xnjdn)n is a sequence of non-empty separable met-

rizable spaces and that

Dn —• ^ G

is a countable dense subset of Xn for each n. If an is chosen arbitrarily in Xn for

each n G N; then the set

D = {(xi i 1 ?x 2 i 2 , ...,XiVijV,ajv+i,aiv+2? •••) : N £ G Dj for 1 < j < N}

is a countable dense subset ofY[Xn. n

Proof. By the proof of Proposition 2.1.2 the set

Bn = {Bdn(xni,r)\xni G Dn and r G Q}

is a countable base for (Xn,dn). Let U be the collection of sets of the form

Bd1{xul,rl) x Bd2(x2i27r2) x ... x BdN (xNiN, rN) x XN+t x XN+2 x ...,

for some choice xjii G Dj and r ? G Q, and for some N G N. The set U is a countable

base for n nXn. For each Xn choose an element an G Xn. Let D be the set of

elements of the form

for some choice xjij G Dj and N G N. Then D has a one-to-one relationship with

U. Consequently D is countable.

25

Let A be an open set in J}n Xn. Then there exists a set

B = Bd1 (x i^ , r i ) x Bd2(x2i2,r2) x ... x BdN(xNiN,rN) x XN+1 x XN+2 x

which is a subset of A. Then

Hi 5 ®Nijv ? aiV+2 j ^ ^

Therefore D is dense. •

2.4 Each Polish Subspace is a

We now have that open subsets and closed subsets of Polish spaces are Polish

and that disjoint unions and products of Polish spaces are Polish. The purpose of

this section is to show that a subspace of a Polish space X is Polish if and only if it

is a G$ in X. Recall that a subset Y of X is a Gs in X if Y can be written in the

form Y = P|n Un where each Un is open in X. Also, a subset Y of X is an in X

if Y can be written in the form Y = \JnVn where each Vn is closed in X. Before

we can show that a subspace of a Polish space X is Polish if and only if it is a Gs

in X, we first need the following results.

Proposition 2.4.1. Each closed subset of a metric space is a Gs-

Proof. Let C be a non-empty closed subset of a metric space X. For all n G N, let

1 1 An = jx € X

If x G C,

d(x, C) < n

d(x, C) = inf{<i(x, z)\z G C} = 0 < —,

n

and x G An. Since this was not dependent on the choice of n we have C C P|n An.

Now suppose x G p|n An. Then d(x, C) < ^ for all n. This implies that each open

set containing x contains a point of C and therefore x G C = C. Consequently,

26

n „ A , C C. Since we have containment in both directions, p | n An = C. We finish

the proof by showing that for each n the set An is open. Fix n and let x G An.

Then d(x,C) < Let 7 = d(x,C) and let 8 < ^ — 7. Note that 8 > 0. Let

y G B(x, 5); then

d(y, C) = in f{d (y , z)\z G C}

< in f {d (x , y) + d(x, z)\z G C}

= d(x,y) + inf{d(x, z)\z G C}

< 8 + 7

1 < n

Therefore y G An. Hence for each x G An an open neighborhood of x can be found

which is a subset of An. Therefore An is open for all n. Since C = f ] n An, the set

C is a Gs. •

T h e o r e m 2.4 .2 (Cantor's N e s t e d Set Theorem) . Let X be a complete metric

space. If (An)n is a decreasing sequence of non-empty closed subsets of X such that

limn diam(An) = 0; then P l n = = i An contains exactly one point.

Proof. Let X be a complete metric space. Let (An)n be a decreasing sequence of

non-empty closed subsets of X such that limn diam(j4n) = 0 . For each n G N

choose xn G An, and let e > 0. Since limn diam( J4n) = 0, a positive integer N can

be found such that diam(An) < e for all n > N. Let m,n > N. Since (An)n is a

decreasing sequence, xm,xn G Ajv; therefore

d(xm,xn) < diam(Aiv) < e.

This implies that (xn)^L1 is a Cauchy sequence. Because X is complete, (a;n)^i_1

converges to x G X. Let i G N. Then also converges to x. Since (An)n

27

is a decreasing sequence, each element of [xn)'^>

=i is an element of At which is

closed. Therefore x £ A{ for all i, and x £ HnLi An. Since limn diam(An) = 0,

if y £ 0^=1 it must be that x = y. Therefore An contains exactly one

point. •

Lemma 2.4.3. Suppose (X,d) is a metric space and that {An)n is a sequence of

sets such that limn diam(An) = 0 . Then limn diam(An) = 0.

Proof. Suppose that the hypotheses are satisfied. Let e > 0. Choose TV £ N such

that diam(Ara) < | for all n > TV. Let n> TV and x,y £ An. Since Bd(x, | ) is an

open set containing x we can choose w £ An such that w £ Bd(x, | ) . Similarly we

can choose z £ An such that z £ Bd(y, f )• Then

d{x, y) < d(x, w) + d(w, z) + d(z, y) < e.

Hence diam(An) < e. Since e was arbitrary limn diam(Are) = 0 . •

The following lemma is important in this section, but it will also be used often

throughout the remainder of this paper.

Lemma 2.4.4. Let (X, T) and (V, S) be topological spaces and let f •. X Y be a

homeomorphism. If (V, 5*) is a Polish space then (X, T) is also a Polish space.

Proof. Suppose the hypotheses are satisfied. Let d be a complete metric for (Y, S).

Define the function a : X x X —»• K. by

<r(x,y) = d(f(x),f(y)).

It needs to be shown that a is a metric for X. Let x,y,z £ X.

i) Since d is a metric d(f(x),f(y)) > 0; consequently, a(x,y) > 0 . If x — y,

f(x) = f ( y ) which implies that

a (x,y) = d(f(x)J(y)) = 0.

29

implies that ( / ( # i ) ) ^ i is a Cauchy sequence in Y. Since Y is complete

converges to some point y £ Y. Since / is surjective y = f(x) for some x £ X.

Again let e > 0. Choose N £ N such that d ( / (#n) , / (# ) ) < e for all n> N. Then

a (xn,x) = d(/(®«),/(s)) <

for all n > N. Hence converges to x £ X. Consequently cr is a complete

metric.

Let 5 be a countable dense subset of Y. Consider the set f~1(S) C X. Clearly,

since / is one-to-one, / - 1 (S) is countable. Let U be an open set in X and note that

f(U) is an open set in Y. Choose s € S such that s £ f(U). Then / - 1 ( s ) £ U.

Since is also in f~1(S) we have that / _ 1 ( 5 ' ) is dense in X. Since X is

separable and cr is a complete metric for X, we have that X is Polish. •

If Y is a subspace of X and A is a subset of Y it is possible that the closure of

A in Y is a proper subset of the closure of A in X. In the following proof A is used

Y

to denote the closure of A in X and A is used to denote the closure of A in Y.

Now we are ready to prove the main result of this section.

Proposit ion 2.4.5. A subset of a Polish space X is Polish if and only if it is a

G$ in X.

Proof. Let X be a Polish space and suppose that Y is a Gj-subset of X. Let (Un)n

be a sequence of open sets in X such that Y = (\nUn. By Proposition 2.1.4, Un

is Polish for all n € N. Also, by Proposition 2.3.3, Y\Un is Polish. Let A be the n

subset of IJ Un defined by n

A = < u = (ui , u2 , . . .) G Un Uj = u/c for all j,k >. n ^

Suppose x = (x\, #2,-••) is a limit point of A. For each i £ N choose U{ £ A

such that U{ = (un,Ui2,...) and the sequence ( u j ) ^ 1 converges to x. Note that for

30

all j G N, (uij)^ converges to xj. Since m G A, mj = uik for all j,k. Hence

x j — %k for all j,k and x G A. Consequently, A contains all of its limit points and

is therefore closed. By Proposition 2.1.4, A is Polish.

Let f : Y —y A be such that f ( y ) = (y,y,y,. . .). Note that since Y = f]nUn,

f ( y ) e A for all y e Y. Similarly if (x,x,x,...) e A, x € Ui for all i, giving x € Y.

L e t (yi)U 1 b e convergent sequence in Y. Since /(?/;) = (un, ui2, «i3, •••) where

uij = Vi f ° r a h ji we have ( ^ i = and therefore converges for all

j. Thus (f(yi))il1 converges. Similarly if (f(yi))il1 is a convergent sequence, the

sequence also converges. Because / and / - 1 are both continuous and / is

one-to-one and onto, / is a homeomorphism of Y onto A. By Lemma 2.4.4, Y is

Polish.

Now suppose the subspace Y in X is Polish. Let d be a complete metric for the

topology of X and let da be a complete metric for the topology of Y. For each n £ N

let Vn be the union of those open subsets W of X for which W D Y is non-empty and

WHY has a diameter of at most J under d0. Let us show that Y = Y D ( f | n K) -

First suppose y £ Y. Obviously y (E Y. Let n £ N. Then Bd0(jji ~) is open in Y

and is therefore equal to W fl Y for some W open in X. Since W f)Y = Bd0(y, -)

has a diameter of at most ^ in Y, W C Vn. Also since y 6 Bdo(y, ^) , we have

yew CVn. This is true for all such n, hence y E f | n vn a n d Y C Y n ( f ) n Vn).

Now suppose x € Y fl ( f | n Vn). This implies that x e Vn for all n. Since Vn

is the union of open sets, for each n choose Wn open in X such that x G Wn and

diamd0(Wn fl Y) < Notice that because x is also an element of Y, Wn Pi Y / 0.

By replacing Wn with a smaller open neighborhood of x we can assume without loss

of generality that is a decreasing sequence and that diamd{W n) < By

Lemma 2.4.3, we have that limn[diam(/o(VFra D Y )] = 0. Also since n was arbitrary

limn[diam(i(l/Fn)] = 0. Since Y is complete under d0 the Cantor Nested Set Theorem

31

implies that y E F can be chosen such that y E f)n{Wn fl Y Y ) . Since

wnc\Y c wn n F c wn n F

for all n, we have y E P|n Wn. Also x E Wn C Q n Wn. Therefore again applying

the Cantor Nested Set Theorem x = y. From above we have that y E F , therefore

x E Y. This gives that Y D (p|n Vn) C Y. Consequently Y = Y fl (P|n Vn). By

Proposition 2.4.1, F is a Gj , therefore Y is a G&. •

The Cantor set C is the set which consists of all those real numbers of [0,1]

that have ternary expansion (anJn for which an is never 1. If x has two ternary

expansions, we put x in the Cantor set if one of the expansions has no term equal

to one. The set C can be obtained by first removing the middle third (§, §) from

[0,1], then removing the middle thirds ( | , | ) and ( | , | ) of the remaining intervals,

and inductively continuing this process. Since the complement of the Cantor set is

the union of open intervals, the Cantor set is closed. This implies that the Cantor

set is a Gs in [0,1] (Proposition 2.4.1). Since the set [0,1] is clearly a Polish space,

by the above proposition the Cantor set is also Polish. The following proposition

gives us another way of representing the Cantor set that will be useful later in the

study of analytic sets.

Proposit ion 2.4.6. The Cantor set is homeomorphic to the set {0,1}N .

Proof. Define / : C -» {0,1}N by = (&;), where bi = 0 if a{ = 0 and

bi = 1 if a, = 2. Suppose (cn)n / (dn)n then for some i we have a ^ di which

implies / ( ( c n ) n ) ^ f((dn)n). Therefore / is one-to-one. Now suppose (bn)n E

{0,1}N. Define (an)n such that an = bn if bn = 0 and an = 2 if bn = 1. Then

/((«n)n) = (bn)n and / is onto. Let ( a ; ) -^ be a convergent sequence in the Cantor

set, where cii = (ajj, a^, . . . ) . Then, for each j , ( a ^ ) ^ 1 is a convergent sequence.

This implies that is eventually constant. Let (a{j )^ 1 converge to x j for

32

each j. Then clearly ( / ( a ; ) ?^ converges to f(x), where x = (xj)j. Therefore /

is continuous. Since C is compact and {0,1}^ is Hausdorff we have that / is a

homeomorphism. •

Another approach to finding a homeomorphism from {0,1}N to the Cantor set

is to view the Cantor set as the set of all elements of the form where each

rik takes on the value zero or one. Then map each element (rik)k in {0,1}N to

Efc n h - T o see that this map is one-to-one suppose that (Tik)k 7̂ ( m k ) k and choose

the smallest integer i such that m Without loss of generality suppose that

rii = 1 and m4- = 0. Then

0 0 0 0

rik ~ mk nk — mk Erik ~ m k _ y -

zk ~ ^ W k—1 k—i -j CXJ

h+ ? k=i-\-

1 00

OO

nk - m k

3k k=i+l

Therefore OO

°° 1 > - - V —

3« ^ 3* k=i-\-1

- 1 -1(1. ~ 3 i ~ 2

> 0.

2mk 3fc 2-J 3k •

k=1 k= 1

Therefore the map is one-to-one. The proof that this map is also continuous and

onto is similar to the proof above.

By Lemma 2.4.4 we can now conclude that the set {0,1}N is a Polish space.

The space N is another example of a Polish space because it is a closed subset of

the Polish space R. By Proposition 2.3.3, is also a Polish space. This space is

extremely useful in the study of analytic sets and will often be denoted by Af. Also

the space X of irrational numbers in the interval (0,1), together with the topology

33

it inherits from JR., is Polish because it is a Gg in (0,1). Now let us use Proposition

2.4.5 to show that Q is not Polish.

Proposition 2.4.7. The set Q of rational numbers with the topology it inherits as

a subspace of R is not Polish.

Proof Suppose that { x is a sequence of open subsets of R such that Q =

C\nUn. Then R — Q = R — C\nUn = Un(^ — Un). This implies that the irrationals

are an T^. Clearly for each n the set R — Un has an empty interior because every

open set would contain a point of Q and consequently would not be a subset of

R — Un. Since <Q> is countable and therefore also the union of nowhere dense sets

we have that R is the union of nowhere dense sets. This contradicts the Baire

Category Theorem. Consequently Q is not a Gs• By Proposition 2.4.5, Q is not a

Polish space. •

2.5 Borel Subsets of Polish Spaces

and Measurable Spaces

We turn to some basic facts about the Borel subsets of Polish spaces. First we

need a few definitions. A collection A of subsets of X is called an algebra if X £ A

and

i) A U B G A whenever A, B € A,

ii) A° 6 A if A 6 A,

Hi) A n B g A if A, B e A.

In fact if a collection A of sets satisfies (i) and (ii) then De Morgan's law would

imply that it also satisfies (iii). An algebra A of sets is called a a-algebra if every

union of a countable subcollection of sets in A is in A. Again using De Morgan's

law this would also be true if every intersection of a countable subcollection of sets

34

in A is in A. The collection B of Borel sets in a set X is the smallest u-algebra

which contains all the open sets of X. This set is often denoted B(X).

Let (Xi,A\) and (X2 , ,42) be measurable spaces. Recall that a function / :

X1 ->• X2 is said to be measurable with respect to A\ and A2 if for each B in A2 the

se^ f {B) belongs to A\. A function f '• X —y Y is said to be Borel measurable if

for each Borel set B in Y the set is a Borel set in X.

Let (X1,Ai), (X2,A2),... be measurable spaces. The product of these measur-

able spaces is the measurable space ( f j n Xn, An) where An is the <r-algebra

on n n X n generated by the sets that have the form

A\ x A2 x ... x An x Xjv+i x ...

for some positive integer N and some choice of An € An, when n = 1,2,... , JV. In

order to show that the class of Borel sets in a measure space has certain properties,

it is often useful to show that the sets which have those properties are a er-algebra

which contain the open sets. Since the collection of all Borel sets is the smallest such

er algebra, this would give us that the class of Borel sets also has those properties.

The following proposition gives us a better representation of the class of Borel sets

in product spaces.

Propos i t ion 2.5.1. Let Xi,X2,Xz,... be a finite or infinite sequence of separable

metrizable spaces. Then B{Y[n Xn) = J]n B{Xn).

Proof. By definition \[nB{Xn) is the (T-algebra on ]JnXn generated by the sets

that have the form

x B2 x ... x JBJV x A'Jv+1 x Xn+2 X ...

for some positive integer N and some choice of Bn £ B(Xn). Recall that the

projection function ^ : Y [ nX

n ^ X i defined by

~ CLi

35

is continuous. If A is a Borel set in Xi then

7T- 1(A) = Xi x X2 X . . . X X i - i X A x X i i+1 x •••

which is an element of Y\nB{Xn). Hence iri is measurable with respect to B ( X i )

and B { X [ n ( X n ) ) for each i. Note that if N € N and Bi G Xi for i < N then

N

Bi x B2 x . . . x Bn x Xn+i... = 7 r t- 1 ( J 5 i ) ;

i=i

therefore B ( X n j is the smallest a-algebra on Xn that makes each projection

7Vi measurable. The continuity of irl for each i implies tha t if A is open (closed)

in Xi, then (A) is open (closed) in X . Thus 7 1 ( A ) £ B ( Y l n X n ) . Since in-

verses preserve unions, for any Borel set B in X i , t t ~ 1 { B ) e B ( ] J n X n ) . Hence nt

is measurable with respect to $ ( r i n - X n ) and B(Xi). Since B(Xn) is the small-

est cr-algebra on J | „ X n tha t makes these projections measurable, it follows tha t

U n B ( X n ) C B ( U n X n ) .

For each n choose a countable base Un for Xn. Let U be the collection of sets

tha t have the form

U\ X U2 X . . . X UN X XN + i X Xp*j+ 2 X . . .

for some positive integer N and some choice Un <G Un. Then U is a countable

base for Y l n X n and U C Y [ n B ( X n ) . Since each open set of X [ n X n is a union of

elements of U and U is countable it follows tha t B ( \ \ n X n ) C [ \ n B ( X n ) . Thus

^ (11 n^-n) = \ [ n B { X n ) as desired. •

If X and Y are sets and / is a function from X to Y then the graph of / (denoted

by g r ( / ) ) is defined by

g r ( f ) = { ( x , y ) e X x Y : y = f ( x ) } .

The following results deal with measurable functions.

36

Proposit ion 2.5.2. Let X and Y be separable metrizable spaces and let f : X —>• Y

be Borel measurable. Then the graph of f is a Borel subset of X xY.

Proof. Let F : X x Y -4 Y x Y be the map that takes (x, y) to ( f ( x ) , y). The Borel

measurability of / implies that if A,B £ B(Y) then / _ 1 (A) and f~l ( B ) are in B(X).

This implies that F 1(A x B) G B(X) x B(Y)~, hence F is measurable with respect

to B(X) x B(Y) and B(Y) x B{Y). Therefore by Proposition 2.5.1, F is measurable

with respect to B(X x Y) and B(Y x Y). Let A = {(yx,y2) G Y x Y\yi = y2}. By

the proof of Proposition 2.4.5, A is a closed subset of F x Y .

We now need to show that gr( / ) = F ^ A ) . Let (x,y) 6 gr( / ) . By definition

f{x) = V- This implies that F((x,y)) = (f(x),y) g A. Consequently, gr( / ) C

^ " ^ A ) . Now suppose that (u,v) G F _ 1 ( A ) . Then F((u,v)) = (/(«), v) e A.

Hence f{u) — v, (u,v) G gr(/) , and gr( / ) = i ^ - 1 (A) . Since i*1 is Borel measurable

it follows that gr( / ) is a Borel set in I x F . •

Proposi t ion 2.5.3. Let (X,A), (Y,B ) , and (Z,C) be measurable spaces and let f :

('Y,B) (Z,C) andg : (X,A) (F ,0 ) 6e measurable. Then fog : (X , ^ l ) (Z,C)

is also measurable.

Proof. Suppose that C G C. Then / - 1 ( C ) G B and <7 - 1 ( / - 1 (C)) G «4. Since

( f 0 9) 1 (C) = S , - 1 ( / _ 1 (C)), the measurability of / o g follows. •

Let X be a set, and let J- be a family of subsets of X. Then the smallest <x-

algebra on X that includes T is clearly unique; it is called the cr-algebra generated

by T , and is often denoted by o{T).

Proposit ion 2.5.4. Let (X,A) and (Y,B) be measurable spaces, and let B0 be

a collection of subsets of Y such that <r(B0) = B. Then a function f : X Y

ts measurable with respect to A and B if and only if f~l{B) G A holds for each

B^B0.

37

Proof. If / : X —» Y is measurable with respect to A and B then f~x(B) G A

for each B € B0 because B0 C B. Now assume that G A holds for each

B e B0. Let T be the collection of all subsets B of Y such that f~1(B) <E A. Since

f f 1(B°) = (f~1(B))°, and / - 1 ( | J n B n ) = ( J w e have that

J~ is a cr-algebra on Y. Since J~ includes B0 it must include B. Thus f is measurable

with respect to A and B. •

Proposition 2.5.5. Let X and Y be Hausdorff topological spaces, and let f : X —>

Y be continuous. Then f is Borel measurable.

Proof. Since f is continuous, f 1 (V ) is an open subset of X for each open subset

U of Y. Since the collection of open subsets of Y generates B(Y^j, the measurability

of / follows from Proposition 2.5.4. •

Proposition 2.5.6. Let (X,A) be a measurable space, and let Y be metrizable

topological space. Then a function f : X ~^Y is measurable with respect to A and

B{Y) if and only if for each continuous function g : Y —y R the function g o f is

A-measurable.

Proof. If f is measurable with respect to A and B(Y), then the measurability of

g o / for each continuous function g follows from Proposition 2.5.5 and Proposition

2.5.3.

Now assume that for each continuous function g \ Y —y 1R the function g o f is

A-measurable, and let d be a metric that metrizes Y. Suppose that U is an open

subset of Y. There is a continuous function gu : Y —> R such that

U = {y e Y\gu(y) > 0}.

For instance if U / Y define gu by gu(y) = d(y, Uc), and if U = Y let gu be such

that gu{y) — 1. The set / - 1 (U) is equal to

{x e X : (gu o f)(x) > 0}

38

and therefore belongs to A because gu o f is ,4-measurable. Therefore / is measur-

able. •

Propos i t ion 2.5.7. Let (X,A) be a measurable space, letY be a metrizable topo-

logical space and for each positive integer n let fn : X —> Y be measurable with

respect to A and B(Y). If limn fn{%) exists for each x 6 X then the function

f '• X -» Y given by f(x) — limn fn(x) is measurable with respect to A and B(Y).

Proof. Note that if g : Y -»• R is continuous, then g(f(x)) = limn g(fn(x)) holds

for each x in X. For each n, g(fn(x)) is measurable by Proposition 2.5.6. Also

since each g(fn{x)) is real valued and measurable, lim„ g(fn{

x)) is also measurable.

Since this implies g(f(x)) is measurable, again applying Proposition 2.5.6 gives us

that / is measurable. •

If (X, A) is a measure space and C is a subset of X, then

Ac = {A fl C where A £ A}

is a a-algebra of subsets of C.

Proposi t ion 2.5.8. Let (X,A) be a measurable space, let Y be a Polish space, and

for each positive integer n let fn : X —>• Y be measurable with respect to A and

B(Y). Let

C = {x G X\\imfn(x) exists}.

n

Then C £ A. Furthermore, the map f : C —$• Y defined by f ( x ) = limra fn{x) is

measurable with respect to Ac and B(Y).

Proof Let J be a complete metric for Y. Then C is the set of those x G X for

which (fn(%))'^L1 is a Cauchy sequence in Y. We need to show that for each positive

integer n the set

1 1 An = [(yi,y2)€Y xY d(yi,y2) < n

39

is open in Y xY. Let n G N and let (x,y) G An. Choose e > 0 such that

d{x, y) + e < Let (p, 5) 6 f e) x Bd{y, |e) . Then

9) < <*(p,«) + d(x, y) + d(y, 2)

1 u ^ 1

< -e + d(x,y) + - e

1 < n

Consequently (x,y) G \e) x Bd(y,\e) C and is open. Therefore

An G H(F) x B(y) by Proposition 2.5.1. For each i,j,n define the set C(i,j, n) by

C(i,3,n) = {* e X d(fi(x), f j ( x ) ) < i } .

Define the function faj : X ->• Y x Y such that

(jM^) = CftO), /?(»)•

Suppose A and B are open in Y. Then

4>~HAxB) = { - ' ( A ) n f j \ B ) .

Since both / j and f j are measurable <f>ij is also measurable. We now show that

C{i,j,n) = An). First let x G C(i,j,n); then

Vi(x),fj(x)) < -•

n

Therefore (/,- {x), f j ( x ) ) G An. This implies that z G <t>J(An) and C(i,j,n) C

<i>ij(An)• Now let z G ^{An). Then <j>ij{z) = ( f i ( z ) J j ( z ) ) G An. This implies

that d(fi(z),fj(z)) < ± and 2 G C(i,j,n). Therefore C(i,j,n) = (f)^1 (An). Since

4>ij is measurable and An is open C(i,j,n) G A.

The fact that C(i,j, n) G A will be used to show that C G A. First we need to

show that c = n u n n n k i>kj>k

40

First let x G C. From above we have that ( / n (^) )^L 1 is a Cauchy sequence in Y. Let

n G N. There exists a positive integer k such that if z, j > k then d(fi(x), f j ( x ) ) <

Since this is true for all such n

Now suppose

* 6 nu n n c(^»-n k i>kj>k

x g nu n n c^^n)-n k i>kj>k

For each n € N there exists a k such that if i,j > k then d(fi(x), f j ( x ) ) <

Therefore ( / n (x))^L 1 is a Cauchy sequence. This implies that x € C and

*=nunn<™,™). n k i>kj>k

Consequently since each C(i,j,n) G A and A is a cr-algebra, C e A.

Let / : C —> Y be defined by f ( x ) — limn fn(x). Then by Proposition 2.5.7 we

have that f ( x ) is measurable with respect to Ac and B(Y). •

Since Ac is a subset of A it is also true that / is measurable with respect to A

and B{Y). A similar argument to that above is made in the following proof.

Propos i t ion 2.5.9. If (X, A) is a metrizable space, Y is a separable metrizable

space, and f,g\X-^Y are measurable with respect to A and B(Y), then

{x G X\f(x) = g(x)}

belongs to A.

Proof. Let d be a metric for Y. For each positive integer n let

d(yi,y2) < -n

= <(yi,y2) eY xY

41

The set An is open in Y x F . By Proposition 2.5.1, An G B(Y) x B(Y). Define the

set

1 1 d{f(x),g(x)) <

n Cn = G X

Define the function 4> : X —$Y x Y by

<A(®) = (f(x),g(x))-

Suppose A and B are open in Y; then,

r 1 ( i x B ) = r ' ( X ) n 9 - ' m .

Since / and g are measurable, <f> is also measurable. We assert that

Cn = 4>~l d(yuy2) < An).

Let x G Cn. Then d(f(x),g(x)) < i . Therefore ( / (a) , </(x)) G A n , and x G

4* l{An)- Now let a; G 4> 1(An). T h e n <f>(x) = (f(x),g(x)) G An; consequent ly ,

d(f ( x ) jg(x) ) < ^ and x G C*n. Therefore we have C n = <^ -1(An) as desired.

Since is in B(Y) x # ( F ) and 4> is measurable, Cn G A. Let C = Cn. If

x G C then d(f(x),g(x)) < i for all n, and / ( s ) = g(x). Therefore

c = {x € X\f(x) = g{x)}.

Since C = P|n C n , a countable intersection of elements from A, we have C G A. •

A positive measure on A is regular if

i) each compact subset K of X satisfies

fi(K) < oo,

ii) each set A in A satisfies

= inf{^/(Z7)|A C U and U is open},

42

iii) each open subset U of X satisfies

n(U) — sup{/u(i!i")|K C U and K is compact}.

We would, like to show that each finite Borel measure on a Polish space is regular.

To do this we need the following lemmas.

L e m m a 2.5.10. Let be a measure space. If (Ak)k is an increasing se-

quence of sets that belong to A, then fi({Jk Ak) — lim^ ji(Ak).

Proof. Define a sequence {Bn)n of sets by letting B\ = A\ and Bj = Aj — Aj-1 if

j > 1- Then the sets are disjoint and measurable. Note that ( J n A n =

U„ Bn and therefore

MlM-) = E ' W n n

n = lim y jjiBi

n < ^ n i—1

lim/u(l J Bi) n '-s

i= 1 ]im/j,(An)

as desired. •

L e m m a 2.5 .11. Let X be a Hausdorff space in which each open set is an J-a, and

let fj, be a finite Borel measure on X. Then each Borel subset A of X satisfies:

(1) fJ^(A) = i n f : A C U and U is open},

and

(2) fi(A) = sup{^t(.F) : F C A and F is closed}.

Proof. Let C be the collection of all subsets A of X that satisfy (1) and (2). We

begin by showing that C contains all open subsets of X. Let V be open in X. Then

obviously V satisfies

/^(V) — inf{/n(U) : V C U and U is open}.

43

By hypothesis V is an FA. Let (CN)N be a sequence of closed subsets of X such that

V = (Jn CN. We can assume without loss of generality that (CN)N is an increasing

sequence, because if necessary we could replace each CN with (Jj=i CJ • Then by

Lemma 2.5.10, /J,(V) = limny«(Cn). It follows that

pt-iy) = sup{FJ-(F) : F C y and F is closed}.

Consequently C contains all the open subsets of X. Now let A be a, Borel set

such that for each e > 0 there exists an open set U and a closed set C such that

C C A C U and /J(U — C) < e. Since A — C C U — C and U — A C U — C we have

FJ-(A — C) < N(U — C) < e and (x{XJ — A) < /J,(U — C) < e. Therefore

n{A) < fi(A n C) + v{A n CC) = n(C) + v{A -C)< /J(C) + e,

and

fi(U) < n{U n A) + fi(U n Ac) = n(A) + - A) < fi(A) + e.

Consequently A satisfies (1) and (2) and therefore is in C. Now we can show that

C is a er-algebra. Clearly C contains X because X is open. If A 6 C and e > 0

there exist C and U which are respectively closed and open such that C C A C U

and FJ,{U — C) < e. Then UC and CC are respectively closed and open and satisfy

U° C Ac C Cc and n(Cc — U°) < e. Therefore it follows from above that Ac £ C.

Now let (Ak)k be a sequence of sets in C and let e be a positive number. For each

k choose a closed set Ck and an open set Uk such that Ck C Ak C Uk and

TIVU - CK) <

Let U = \JKUK and C = |Jfc CK. Then U and C satisfy the relations C <Z\JKAK CU

and

fi{u ~C)< ,x(U(Uk - Ck)) < - ck) < 6.

44

The set U is open but the set C can fail to be closed. However for each n the

set U L i is closed and ((Jfc=i Ck)n is an increasing sequence. Therefore using a

similar argument to that in Lemma 2.5.10 we have

n

v(U-C) = limv(U- (J Ck). n

k=1

Consequently we can find an N G N such that fJ.{U — UfcLi @k) < e- This gives that

UfcLi Ak G C, and it follows that C is a <r-algebra. Since C contains all the open

sets, B(X) C C. •

P r o p o s i t i o n 2 .5 .12 . Every finite Borel measure on a Polish space is regular.

Proof. Let X be a Polish space, let d be a complete metric on X, and let /i be a

finite Borel measure on X. Since X is metrizable, X is Hausdorff. Let U be an

open set in X. By Proposition 2.4.1 the set X — U is a Gs. Therefore U is a

By Lemma 2.5.11, each Borel set A satisfies

/i(A) = i n f { n ( U ) : A C U and U is open},

and

jJ.{A) = sup{|«(i?1) : F C A and F is closed}.

Therefore we only have left to show that

yu(̂ 4) = sup{/i(liQ : K C A and K is compact}.

First consider the case where A — X. Let (xk)k be a sequence whose terms form a

dense subset of X. Let e > 0. Note that for each n, X = U ^ ! Bd(xk, J ) . For each

n G N choose kn such that

k

M U B J ( x l , h ) > / J ( x ) - ^ . k=1

45

Define the set K by oo k IVn

« = n u z > -n=1fc=l U

The set K is closed and because d is complete in X, K is also complete. Now we

need that K is totally bounded. Let e > 0 and choose n such that ^ < | e . Clearly

{Bd{xki \)}\=i is a finite covering of K. For each fc, Bd{xk, ^ Bd(xk, je) . Let

y be a limit point of Bd{xk^~). Choose S > 0 such that ^ + S < | e . There exists

z G Bd{xk, ^ ) such that z G Bd(y,S). Then

d(xk,y) < d(xk,z) + d(y,z) < - + S < ]-e. H/ ZJ

Therefore Bd(xk, C Bd(xk, | e ) . Now we have that {Bd(xk, is an open

covering of K. For each k = 1,2, ...kn choose, if possible, yk G K such that yk G

Bd(xk, | e ) . Let

A = {k : y}~ was chosen}.

Clearly A is finite. Consider {Bd(yk,e)}kea- Let z £ K and fix k G {1,2,. . . , kn}

such that z G Bd(xk, | e ) . Since z £ K and z G Bd(xk, | e ) it was possible to find

yk G K such that yk G Bd{xk,\^)- Consequently, k G A. Therefore

d(Vk,z) < d(yk,xk) + d(xklz) < e.

Consequently z G Bd(yk,e)- Now we have that {Bd(yk,e)}k£a is a finite covering

of ET. Since each yk G K for all k G A, we have that K is totally bounded. Now we

have that K is compact. Furthermore,

oo kn ^ oo

M X - K ) < - ((J -))) < £ ^ = e, n= 1 k=1 rc=l

and therefore n{K) > / i (X) — e. Since e is arbitrary we have

fJ.{X) = sup{yu(isf) : K C X a n d K is compact}.

46

Now let A be an arbitrary Borel subset of X and let e be a positive number.

Choose a compact set K such that (J,(X — K) < t. We have that

fj-(A) = sup{/-t(F) : F C A and F is closed}.

Choose a closed subset F of A such that fJ,(F) > /J,(A) — e. Then K C\F is a compact

subset of A and

H {K DF)> n{F) - n(F n {X - K))

> fi(F) - n{X - K)

> — e — e

> jji(A) — 2e.

Since e is arbitrary

^{A) = sup{^(K) : K C A and K is compact}.

Thus fi is regular. •

CHAPTER III

ANALYTIC SETS

3.1 Borel Subsets of Polish Spaces

Now we turn our attention to analytic sets. Recall from the introduction that

a subset A of X is analytic if there is a Polish space Z and a continuous function

/ : Z —> X such that f ( Z ) = A. The purpose of this section is to show that every

Borel subset of a Polish space is analytic. Later we will see that there are analytic

sets which are not Borel.

Propos i t ion 3.1.1. Let X be a Polish space. Then each open subset and each

closed subset of X is analytic.

Proof. By Proposition 2.1.4 each closed subset and each open subset of X is Polish.

Let Y be a closed (or open) subset of X. Define / : Y —> X such that f ( y ) = y

for all y E Y. Let U be an open set in X. Then y E f~l{U) if y E U and y E F .

Also if z E U fl Y, f ( z ) = z E U] therefore / - 1 (?7) = U Pi Y which is open in Y.

Therefore / is continuous. Clearly f ( Y ) = Y. Consequently Y is analytic. •

Propos i t ion 3.1.2. Let X be a Polish space, and let A i ,A2 , ... be analytic subsets

of X. Then (Jft Ak and Ak are analytic.

Proof. For each k choose a Polish space Zk and a continuous function fk '• Z^ X

such that fk(Zk) = Ak- Let Z be the disjoint union of the spaces Z\, Z2, Z3,... and

define / : Z —»• X so that for each k it agrees with fk on Zk- Then Z is Polish by

Proposition 2.2.4. Let U be an open set in X. Then

r\u) = f-\u n AO u r\u n A2) u ... u f-\u n An)...

48

= / f 1 {u n Aj) u f ; \ u n A2) U ... u / " ' ( £ / n An)...

= / f 1 ( C f ) u / 2 - , ( t / ) u . . . u

Therefore / _ 1 (C / ) is open and / is continuous. Let y G f{Z). Then there exists a

positive integer k such that y G f(Zk) — A^. Therefore y G (Jk-^-k- Now suppose

z € Ufc Ak- Then z G Ak for some k, giving 2; G f(Zk) and z G f(Z). Therefore we

have that f ( Z ) = (Jfc Ak and \Jk Ak is analytic.

Next form the product space Zk and let A consists of those elements (zk)k i n

Zk for which fi(zi) = f j ( z j ) holds for all i and j. Let (pk)k be a limit point of A.

Let (yi)i be a sequence from Zk which converges to (pk)k, where yi = (yii, 2/i2, - --)

for each i. Then (yij)f^i converges to p j for all j. Since f j ( y i j ) = fk{lfik) f ° r all j and

k, we have f j ( p j ) = fk{Pk) f ° r all k. Thus (pk)k € A and A is closed. By Proposition

2.3.3, Zk is Polish and by Proposition 2.1.4, A is Polish. Define g : A —> X such

that g((zk)k) = /1 (z\ )• Then the continuity of f\ implies that g is also continuous.

Let x G g(A). Then there exists (yk)k G A such that g((yk)k) = fi{yi) — x-

Therefore x G Uk for all k because fk{yk) — f\{y\) = x and fk{Zk) = Uk for all k.

Consequently g(A) C C\kUk- Now suppose y G HkUk- Then for each k there exists

Zk G Zk such that fk(zk) — %• Consequently (Zk)k G A and g((zk)k) = x. Now we

have that (/(A) = ClkUk and that f\kUk is analytic. •

L e m m a 3.1.3. Let X be a Hausdorff topological space. Then B(X) is the smallest

family of subsets of X that

(a) contains the open and the closed subsets of X,

(b) is closed under the formation of countable intersections,

(c) is closed under the formation of countable disjoint unions.

Proof. Let S be the smallest collection of subsets of X that satisfy (a), (b), and

(c). Note that we can find S by taking the intersection of all such collections. Let

49

S0 = € S and AC £ <S}. It is clear that S0 C S C B(X). Thus we need to

show that S0 is a er-algebra that contains each open subset of X and it will follow

that S0 = = B{X).

It is obvious that S0 contains the open subsets of X and that SQ is closed under

complementation. Now suppose that (AN)N is a sequence of sets in S0. Then (Jn AN

is the union of the sets

AI,A\ n AI,A\ n AC2 n A$, ...

Since these sets are disjoint and belong to S, (JN AN must also belong to S. Further-

m o r e (Un AN)° i s the intersection of a sequence of sets in S, and must belong to <5.

Consequently | J n AN belongs to S0. It follows that S0 is closed under the formation

of countable disjoint unions. Therefore S0 is a cr-algebra, and S0 = S = B(X). •

Proposit ion 3.1.4. If X is a Polish space, then each Borel subset of X is analytic.

Proof. By Propositions 3.1.1 and 3.1.2 the class of analytic subsets of X satisfy

conditions (a), (b), and (c) of Lemma 3.1.3. Therefore B(X) is a subset of the

analytic sets and each Borel subset of X is analytic. •

3.2 The Space Af

Recall from an earlier example that M is the space Nn . This space is very

important in the study of analytic sets. In fact, we will show in this section that

each analytic set is the image of Af under some continuous function. This section

also provides many unique methods of producing the continuous functions needed

to show that a set is analytic. First we need to determine some more properties of

analytic sets, and we will also find that each Polish space is the image of M under

some continuous function.

50

Proposit ion 3.2*1. Let Xi,X2,... be a finite or infinite sequence of Polish spaces;

and for each k let Ak be an analytic subset of Xk- Then ]Jk A* is an analytic subset

ofUk^k.

Proof, For each k choose a Polish space Zk and a continuous function fk : Zk —> Xk

such that fk(Zk) — Ak. Define a function / : zk -)• n k X k by f{(zk)k) =

(fk(zk))k- We have that Zk is Polish by Proposition 2.3.3, and / is continuous

by Proposition 2.3.1. Clearly Zk) = Ilfc Ak. Therefore Ak is analytic. •

Lemma 3.2.2. Let X be a Hausdorff topological space, and let Y be a subspace of

X. Then

B(Y) — {A\ there is a set B in B(X) such that A = B f ) Y } .

Proof Let B{X)y denote the collection of subsets of Y that have the form B D Y

for some B in B(X). We need to show that B(Y) = B(X)Y. Let / : Y -> X be

a function such that f ( y ) = y for each y 6 Y. Then / is continuous and hence

measurable with respect to B{Y) and B(X). Since each subset B of X satisfies

/ 1(-®) —BOY, we have that B(X)y C B(Y). Now let us show that B(X)y is

a a-algebra containing the open subsets of Y. If U is open in Y by definition of

subspace topology U = V (1 X for some V open in X. Therefore B{X)y contains

all the open sets in Y. Suppose A € B(X)y. Then A = B fl Y for some B £ B(X).

Also

Y - A = Y ~{Bf\Y)

= Y n (BDY)C

= Yr\(BcU Yc)

= F n Bc.

51

Since B E B(X), B° E B(X). Hence Y — A E B(X)y• Now suppose is a

sequence of elements from B(X)y. For each i choose Bi such that At = BtC\Y and

Bi EB{X). Then

U = U(5i n r ) = ( | J 5 0 n r .

i i i

Since (J iBi E B(X) we have (J^ Ai E B(X)y. Therefore B(X)y is a cr-algebra

containing all open subsets of Y. This implies that B(Y) C B(X)y. We now have

shown that B(Y) = B{X)y. •

Proposition 3.2.3. Let X and Y be Polish spaces, let A be an analytic subset

of X, and let f : A —>• Y be Borel measurable (that is measurable with respect to

B{A) and B(Y)). If A\ and A2 are analytic subsets of X and Y respectively, then

f ( A f ) A i ) and f~1(A2) are analytic subsets ofY and X respectively.

Proof. Suppose the hypotheses are satisfied. Proposition 2.5.2 implies that gr(/) E

B(A x Y) and Lemma 3.2.2 then implies that there is a Borel subset B of X xY such

that gr( / ) = B D {A x Y"). We have that Y is analytic by Proposition 3.1.4. Also

gr( / )n(A 1 x F ) is analytic by Proposition 3.1.2 and 3.2.1. Similarly, g r ( / )n(A 1 x Y)

is an analytic subset o f X x F and therefore is the image of a Polish space Z under

a continuous map, say h. Let Try be the projection o f l x F onto Y. Then tt y is

continuous and 7Ty o h is a continuous map from Z into Y.

We want to show that (7xy o h)(Z) = f(A Pi A\). We will show this by inclusion

in both directions. Let z E {TCy o h)(Z). Then there exists an x E X such that

{x,z) E h(Z) = g r ( / ) D x Y). Then x E Ax and f{x) = z E Y. Since

the domain of / is A we have that x E A. Consequently z E f(A f| Ai) and

(Try o h)(Z) C f(A n Ax). Now suppose y E f{A fl Ax). Then there exists an

x E AD Ax such that f(x) = y. Then (x,y) E gr ( / ) fl (Ax x Y). This implies

that there exists z E A such that h(z) = (x,y) and (Try o h)(z) = y. Therefore

52

f(A n i l ) C (7Ty o h)(Z). Now we have that f(A fl A\) = (7ry 0 h)(Z), ixy o h is

continuous, and Z is a Polish space. Consequently / ( A fl Aj) is analytic.

The argument is similar to the above that gr( /) fl (X X A2) is also an analytic

subset of X xY. Now let h be a continuous function and Z be a Polish space such

that h maps Z onto gr( /) fl (X x A2). We need to show that / - 1 {A2) = ( f f x oh)(Z).

Let x G f~1(A2). Then x G A and f(x) = y for some y G A2. This implies that

(x>y) € gr ( / ) fl (X x A2). Then there exists z G Z such that h(z) = (x,y), which

implies (nxoh){z) = x. Therefore C ( t txo/i)(Z). Now let x G (nx°h)(Z).

Then there exists y G Y such that (x,y) G h(Z). This implies that f(x) = y and

y G A2. Consequently x G f~1(A2) and / _ 1 ( A 2 ) = (nx o h)(Z). Since ( irx o h) is

continuous and Z is Polish, we have that f~1(A2) is analytic. •

The following proposition uses a method of constructing a continuous function

that will be useful again later in this study. The function is constructed by first

constructing a net of nested closed sets and applying to each chain the Cantor

Nested Set theorem to find the unique point in the intersection of sets in the chain.

The continuous function sends a sequence used to index the closed sets to that

unique point. In the following proof, if A is a subset of the topological space X

denote the interior of A by A0.

Proposit ion 3.2.4. Each non-empty Polish space is the image of Af under a con-

tinuous function.

Proof. Let X be a non-empty Polish space and let d be a complete metric for X.

We shall construct a family {C{nx, n2 , . . . , nk)} of subsets of X, indexed by the set

of all finite sequences (n\, n2,..., n^) of positive integers. We do this by induction

on k. First suppose k = 1, and let be a sequence whose terms form a

53

dense subset of X. Then for each n j we can define

C(n\) = {x G X\ d(x,xni) < 1}.

Now suppose k = 2 and let (£n2)n^=i be a sequence whose terms form a dense

subset of Let

C (n i , n 2 ) = | x G C(ni) d(x,xn2) < ^

Then C(ni) — [Jn2 C(n\, 712). If we continue this process inductively, then the

following conditions hold:

(a) C(ni , n 2 , r i f c ) is closed and non-empty,

(b) the diameter of C(n\,n2, ...njt) is at most £,

(c) C(ni , n2 , . . . , nfc-i) = (Jnfc C(n 1 ; n2 , ...njt), and

(d) ^ = U „ 1 ^ ( » i ) .

We turn to the construction of a continuous function that maps J\f onto X.

Let n = (n/c)'^L1 be an element of Af. It follows from (a), (b), and (c) above

that C(n1),C(n1,n2),C(n1 , n2, n3),... is a decreasing sequence of non-empty closed

subsets of X whose diameters approach zero. Thus there is a unique element in the

intersection of these sets by Theorem 2.4.2 and we can define a function / : Af —> X

by letting / ( n ) be the unique member of p|fc C(n j , ..., n^). Note that if m and n

are elements of Af such that m; = m holds for i = 1,2,..., k then d(f( m), /(n)) < \.

It follows that / is continuous. Finally (c) and (d) above imply that for each x in X

there is an element n e Af such that x G D* C(nun2,...,nk) and hence x = /(n);

thus / is surjective. •

Corollary 3.2.5. Every non-empty analytic subset of a Polish space is the image

of Af under some continuous function.

54

Proof. If A is the image of the Polish space Z under the continuous function / , and

Z is the image of Af under the continuous function g by Proposition 3.2.4, then A

is the image of Af under the continuous function f o g . •

The following is a similar result that will be useful in a later section.

Propos i t ion 3.2.6. Let X be a Polish space. A subset A of X is analytic if and

only if there is a closed subset of Af X X whose projection on X is A.

Proof, Since Af is Polish, Af x X is Polish by Proposition 2.3.3. Any closed subset

of Af x X is Polish by Proposition 2.1.4. If A is the projection of a closed subset of

Af x X onto X , then A is analytic because the projection function is continuous.

Now suppose that A is an analytic subset of X . If A is empty, then it is the

projection of the empty subset of Af x X onto X. Otherwise by Corollary 3.2.5

there is a continuous function / : Af -> X such that f ( A f ) = A. Recall that

g r ( / ) = {(Xjy) e Af x X\y = f(x)}

is closed. Since

7 r x(gr( / ) ) = {y\y = f ( x ) for some x £ Af}

we have 7rx(gr(/)) = A. •

3.3 Zero Dimensional Spaces

A topological space is zero-dimensional if its topology has a basis that consists

of sets that are both open and closed. Among the zero-dimensional spaces are the

space of all rational numbers, the space of all irrational numbers, and each space that

has the discrete topology. Note that a product of zero-dimensional spaces is zero-

dimensional, that a subspace of a zero-dimensional space is zero-dimensional, and

that the disjoint union of a collection of zero-dimensional spaces is zero-dimensional.

55

In particular, the spaces J\f and {0,1}N are products of zero-dimensional spaces and

so are zero-dimensional. We need to show that each Borel subset of a Polish space is

the image under a continuous injective map of some zero-dimensional Polish space.

To do this we need the following lemma.

L e m m a 3 .3 .1 . Every separable metrizable space is homeomorphic to a subspace of

[0,1]N, and every Polish space is homeomorphic to a Gs in [0,1]N.

Proof. Let d be a metric for the separable metrizable space X. Let (xn)n be a

sequence whose terms form a dense subset of X. Consider the map / from X to

[0,1]N that takes the point x to the sequence whose nth term is min{l , d(x, xn)}.

Let A be the image of X under / . Clearly f : X —>• A is onto. Let y and z be

distinct points in X. Let

6 = min j £ < % * ) , 1 J .

Choose n £ N such that xn E Then d(y,xn) < 8. However,

d(z,xn) + d(xn,y) > d(y,z),

and

d(z, xn) > d(y, z) - d(xn,y) >2S- d(xn,y) >26 - S = 6.

This implies that the coordinate of / ( y ) is not equal to the n c o o r d i n a t e of

f ( z ) and f ( y ) ^ f ( z ) . Now we have that / is one-to-one.

Let (zi)i be a sequence in X which converges to y. Clearly f ( y ) e A. Let n G N

and e > 0. There exists N e N such that d(Zi,y) < e for all i > N. Suppose i > N.

Then

d(yZi, xn) ^ d(y, xnJ -(-

and

56

d(zi,xn) - d(y,xn) < d(y,Zi).

Similarly,

d(y, xn) < d(zi,xn) + d(y, Z{),

and

d(y,xn) - d(zi,xn) < d(y,Zi).

This implies that \d(y,xn) — d{zi,xn)\ < d(zi,y) < e. Therefore the nth coordinate

of ( f ( z i ) ) i converges to the nth coordinate of / (y) . Since n was arbitrary, / is

continuous.

Now suppose (f(zi)) converges to f ( y ) in A. Let e > 0, and choose n such that

d(y,xn) < e. Choose N such that for all i > N we have

\d{xn,Zi) - d(xn,y)| < e.

Then

d(xn,Zi) - d(xn, y) < e,

and

d(xn, Zi) < e + d(xn,y).

This implies that

d{zi, y) < d(xn, z^ + d(xn, y) < e + 2 d ( x n , y ) < 3e.

Therefore (zj)j converges to y, and / is a homeomorphism. In addition, if we

suppose that X is a Polish space the fact that / is a homeomorphism along with

Proposition 2.4.4 implies that A is a Polish space. Since A is Polish it must also be

a Gs in [0,1]N. •

57

Proposit ion 3.3.2. Each Borel subset of a Polish space is the image under a

continuous injective map of some zero-dimensional Polish space.

Proof. We begin by showing that each Polish space is the image under a continuous

injective map of some zero-dimensional Polish space. First consider the interval

[0,1]. It is the image of the space {0,1}N under the map F : {0,1}N —> [0,1] that

takes the sequence (xk)k to the number | £ . Each number in [0,1) that has two

binary expansions (that is, each number in (0,1) that is of the form ^ for some

m and n) is the image under F of two elements of {0,1 }N: the remaining members

of [0,1] are images of only one element of {0,1}N. Thus if we remove a suitable

countably infinite subset from {0,1}N, the remaining points form a space Z such

that the restriction of F to Z is a bijection of Z onto [0,1]. A similar argument to

that which showed that the map from {0,1}N into the Cantor set is continuous can

be made to show that F is continuous. Since {0,1}N is zero-dimensional, we have Z

is zero-dimensional. Since {0,1}N — Z is countable and consequently an we have

that Z is a G& and therefore Polish. Hence [0,1] is the image under a continuous

injective map of a zero-dimensional Polish space. It follows that [0,1]N is the image

of the zero-dimensional Polish space under a continuous injective map.

Now suppose that X is an arbitrary Polish space. By Lemma 3.3.1 there is a

homeomorphism G of X onto a subspace of [0,1]N such that G(X) is Polish and a

Gs• Let H be a continuous injective map from ZN onto [0,1]N. Then H^1(G(X)) is

a Gs in and hence is also Polish. Let H0 be the restriction of H to H~1(G(X)).

Then X is the image of the zero-dimensional Polish space H~l(G(X)) under the

continuous map G ' o H0.

We turn to the Borel subsets of X. Let F consists of those Borel subsets B of X

for which there is a zero-dimensional Polish space Y and a continuous injective map

f '• Y —$• X such that f(Y) = B. The first part of this proof along with Proposition

58

2.1.4 implies that T contains all open and closed subsets. A modification of the

proof of Proposition 3.1.2 shows that T is closed under the formation of countable

intersections and countable disjoint unions. Thus Lemma 3.1.3 implies that T —

B{X). •

Propos i t ion 3.3.3. Let X be a zero-dimensional separable metric space, let U be

an open and non-compact subset of X, and let e be a positive number. Then U is

the union of a countably infinite family of disjoint sets, each of which is non-empty,

open, closed, and of diameter at most e.

Proof. Since U is open and non-compact there is a family A that consists of open

sets and whose union is U, but A has no finite subfamily whose union is U. Let

V be the collection of all subsets of X that are both open, closed, of diameter at

most e, and included in some member of A. Since X is zero-dimensional the set U

is the union of the family V. Since X is second countable there exists a countable

subfamily of V, call it V0, whose union is U. List the sets in V0 in the sequence

Vi, V2,... and consider the non-empty sets that appear in the sequence

vu v{ n v2, vf n v2c n y3,...

It is clear that these sets are open, closed, disjoint and of diameter at most e, and

that their union is U. If all but a finite number are empty a finite sub collection of

A would cover U. Therefore there must be a countably infinite number. •

3.4 Non-Borel Analytic Sets

In this section we will show that there does exists a non-Borel analytic set. To do

this we first need to study some properties of uncountable Borel subsets of Polish

spaces. A point x of X is a condensation point of A if every open neighborhood of x

59

contains uncountably many points of A. The following lemma about condensation

points will be useful in our study of uncountable Borel subsets.

Lemma 3.4.1. Let X be a separable metrizable space, and let C be the set of

condensation points in X. Then C is closed and Cc is countable.

Proof. Let W be a countable basis for X. Then x (Jz C if and only if there is a

countable open set that belongs to U and contains x. Hence C° is the countable

union of countable open sets and is therefore open and countable. This implies that

C is closed. •

Proposition 3.4.2. Let X be a Polish space, and let B be an uncountable Borel

subset of X. Then there is a continuous injective map f : Af —> X such that

/O^O B and such that B — f(M) is countable.

Proof. There is, according to Proposition 3.3.2, a zero-dimensional Polish space Z

and a continuous injective map g : Z —» X such that g(Z) = B. Thus it will suffice

to construct a continuous injective map h : J\f Z such that Z — h(Af) is countable,

and then to define / to be g o h.

Let Z0 be the collection of all points in Z that are condensation points of Z. Then

by Lemma 3.4.1 and by Proposition 2.1.4, ZQ is zero-dimensional. Also, by Lemma

3.4.1, Zr0 is countable. If x 6 Z0 every neighborhood of x contains uncountably many

points of Z. Since Z'0 is countable every neighborhood of x contains uncountably

many points of Z0. Therefore every point of Z0 is a condensation point of Z0.

Suppose that d is a complete metric on Z0. For each k we construct a family of

sets indexed by as follows. Choose a0 £ Z0 and let U = ZQ — {a0}. Since aQ

is a condensation point of ZQ and therefore a limit point of Z0 we have that U is

not closed and consequently is not compact. Let e = 1. Then by Proposition 3.3.3

we can find a sequence of sets ( A ( n i ) ) ^ = 1 which are disjoint, non-empty, open,

60

closed, and of diameter at most 1, such that | J ^ _ 1 A(ni) = U. Also each A{rt\)

consists entirely of condensation points of itself. We can repeat this construction

producing for each k and n\, n2, sets A{n\, n 2 , n - f c ) , rik — 1,2,..., that are

disjoint, non-empty, open, closed and of diameter at most and are such that

U ^ = i ^(n ~~ ^ -~nk) A(rii, ...,rik-i) less some point a(ni , . . . , n ^ - i ) . Since each

point in A{n\, n 2 , •••> i s a limit point and

a(ni,.. . ,njfe_i) £ |J A(n1,n2, ...nk), nk = 1

the set U ~ = i ^ ( n i ? n 2 ) is n o t closed and therefore not compact.

Define h : M —> Z by letting h(n) be the unique point in D^. 1A(ni ,n2, . . . ,nfe) .

Note that for each m and n in M such that m; = rti hold for i = 1,2,.. .k , we have

d(h{m)i h(n)) < j. It follows that h is continuous. Also, since A(ni,n2, . . . ,n.fc-i ,p)

and A(ni, n 2 , . . . , n j t - i , <?) are disjoint if p ± q it follows that h is injective. Suppose

that x is not a point removed in the above construction. Then x G A(n\) for

some n\. Since x ^ a (n i ) we have x G A(ni ,n2) for some n-2- Inductively for

each k there exist some choices n\, 712,such that x G A(ni , n 2 , . . . , ^fe). This

implies that there exists an n G J\f such that h(n) = x. Consequently Z0 — h(N)

is the countably infinite set consisting of points removed during the construction

of the sets A(ni,ri2, ...,njt). Since Z% is countable it follows that Z — h(Af) is also

countable. Since g(Z) = B, if we define f = g o h, f(Af) C B. Suppose 2; is in the

set B — f{M). Then z G g(Z) and 2: ^ g(h(Af)). Consequently there exists x G Z

such that g(x) = z. Then x G Z — h(Af). Since g is injective and Z — h(M) is

countable, B — f(Af) must also be countable. •

Corollary 3.4.3. Each uncountable Borel subset of a Polish space includes a subset

that is homeomorphic to {0,1}N.

Proof. Let X be a Polish space, and let A be an uncountable Borel subset of X.

61

Proposition 3.4.2 provides a continuous injective map / : M —• X such that / ( f i f ) C

A. Regard {0,1}N as a subspace of M and restrict / to {0,1}N. Then since {0,1}N

is compact this restriction is a homeomorphism of {0,1}N onto the subset / ({0,1}N)

of A. •

Since we found earlier that {0,1}' ' is homeomorphic to the Cantor set, the above

corollary also implies that each uncountable Borel subset of a Polish space includes a

subset that is homeomorphic to the Cantor set. Let us introduce some terminology

and notation. Suppose that X and Y are sets and that E is a subset of X x V.

Then for each i £ l and each y € F the sections Ex and Ey are the subsets of Y

and X given by

Ex = {yeY : (x,y) e E},

and

Ey = {x e X : {x,y) € E}.

Now let X be a set and let ZF be a family of subsets of X. A subset A of J\f x X is

universal for T if the collection of sections {j4n |n € N}, is equal to F.

Proposi t ion 3.4.4. Let A be an uncountable analytic subset of the Polish space X.

Then A has a subset that is homeomorphic to {0,1}N. Also, A has the cardinality

of the continuum.

Proof. Let A be an uncountable analytic subset of the Polish space X and let d

be a complete metric for X. By Corollary 3.2.5 there exists a continuous function

/ : M -> X such that f ( N ) = A. Choose a subset S of M such that the restriction

of / to S is a bijection of S onto A. We can do this by choosing for each a G A

exactly one n 6 Af such that / ( n ) = a. Let S0 consist of the points in S that are

condensation points of S. By Lemma 3.4.1 the set S — S0 is countable. Therefore

S0 is uncountable and we can choose «(0) and ®(1) from SQ such that x(O) ^ x(l).

62

Let 5 ( 0 ) = m i n { 1, |c?(a;(0), x ( l ) ) } . L e t

5 ( 0 ) = 5 ^ ( 0 ) , 5 ( 0 ) ) ,

a n d

5(1) = Bd(x(l), 5(0)).

T h e n 5 ( 0 ) n B{ 1) = 0. C h o o s e x ( 0 , 0 ) a n d x ( 0 , 1 ) in B{0) n S0 s u c h t h a t x ( 0 , 0 ) ±

x ( 0 , 1 ) a n d let

5 ( 0 , 0 ) = m i n | ^ ^ c ? ( a : ( 0 , 0 ) , a ; ( 0 , 1 ) ) | .

A l s o c h o o s e a?(l, 0 ) a n d x(\, 1) i n 5 ( 1 ) fl S0 s u c h t h a t a : ( l , 0 ) ^ x ( l , 1) a n d le t

5 ( 1 , 0 ) = m i n j i , ^ d ( z ( l , 0 ) , a ; ( l , 1 ) )

T h e n le t

B(0,0) = Bd(x(0,0), 5(0,0)) n £ ( 0 ) ,

B{0,1) = Bd(x{0,1), 5(0,0)) n £ ( 0 ) ,

5 ( 1 , 0 ) = Bd(x(l, 0 ) , 5 ( 1 , 0 ) ) n 5 ( 1 ) ,

a n d

5 ( 1 , 1 ) = 5 ^ ( 1 , 1 ) , 5 ( 1 , 0 p B ( l ) .

S i m i l a r l y for e a c h ( n i , r i 2 , r i k ) G { 0 , l } k c h o o s e x ( n i , « 2 , • ••«*;, 0 ) a n d

x(r i ] , t t .2 , • • • , « £ , 1) in So D B(ni,ri2, ...,rc.fc) s u c h t h a t

s ( n i , U2, ...nfc, 0 ) ^ x ( n i , n 2 , n * , 1)

a n d c h o o s e

5 ( n i , n 2 , . . . , n j t , 0 ) = m i n | ^ ^ ^, ^ < / ( « ( n i , n 2 , . . .n f c , 0 ) , x(ni, re2, n * , 1) j -

63

Let

B ( m , n 2 , ...,nfc,0) = Bd(x(n1,n2,...nk,0),S{n1,n2,...,nk,0)) n B(nun2, ...,nfc)

and let

B(nun2,...,nk,l) = , n2, ...nk, 1), 8(m, n2,..., n*, 0)) n B{nx, n 2 , "*)•

Define p : {0,1}N such that flf(n) is the unique element of n ^ = 1 B ( n 1 , n 2 ,

Note that if m and n are elements of {0,1}N such that mi = n j holds for i —

1,2,. . . ,k then % ( m ) , # ( n ) ) < It follows that g is continuous and / o g is

continuous. Again suppose that m, n € {0,1}N with m ^ n. Choose the smallest

positive integer i such that rn, / nt. Then without loss of generality we can

suppose that </(m) G jB(n i ,n 2 , . . . ,n i_ i , l ) and p(n) G B ( n i , n 2 , Since

_B(m,n2 , . . . ,n;_i ,0) n B(n i ,n 2 , . . . ,n , -_ i , l ) = 0 and / is injective on S we have

that ( / o <j)(m) 7̂ ( / o g){n) and that f o g is injective. Consequently {0,1}N is

homeomorphic to the image of {0,1}N under the map f o g .

Since / is a one-to-one function such that / { A f ) — A, we have that the cardinality

of A must be less than or equal to the cardinality of M. Since Af has the cardinality

of the continuum, the cardinality of A is less than that of the continuum.. By

the above, {0,1}N is homeomorphic to a subset of A. Since {0,1}N also has the

cardinality of the continuum, A must also have the cardinality of the continuum. •

Lemma 3.4.5. Let X be a separable metrizable space. Then there is an open subset

of M x X that is universal for the collection of open subsets of X, and there is a

closed subset of M x X that is universal for the collection of closed subsets of X.

Proof. Let V be a countable basis for X and let (Vn)n be a sequence whose terms

are the sets in V together with the empty set. Define a subset U of Af x X by

U = {(n,x) | x G Vnk for some k}.

64

Then U is open because it is the union of the open sets U(k,j), for some positive

integers k and j, defined by

U(k,j) = {n G Af\ nk = j} x Vj.

For each n in Af the section Un is given by

Un = {x G X\ (n ,x) G U} = k

Since each open set is the countable union of elements from V, we have that {?7n}n

is equal to the set of all open sets.

The complement of U is a closed subset of Af x X and is universal for the class

of closed subsets of X. •

Proposition 3.4.6. Let X be a Polish space. Then there is an analytic subset of

Af x X that is universal for the collection of analytic subsets of X.

Proof. Use Lemma 3.4.5, applied to the space Af x X, to choose a closed subset F

of Af x Af x X that is universal for the collection of closed subsets of Af x X. Let

A be the image of F under the map h that takes (m, n, x) (m, x). Then A is an

analytic subset of Af x X. Let m G Af. Then

Am = {x E X|(m, x) G A}.

Also,

Fm = {(n, x) G Af x X\ (m, n, x) G F}.

If (n, x) G Fm, then (m, n, a;) G F and (m, x) G A. This implies that x G Am.

Consequently Am is the projection on X of the corresponding section Fm of F.

Since F is universal for the closed subsets of Af x X, Proposition 3.2.6 implies that

the analytic subsets of X are exactly the projections on X of the sections Fm. Thus

A is universal for the class of analytic subsets of X. •

65

Corollary 3.4.7. There is an analytic subset of M that is not Borel.

Proof. According to Proposition 3.4.6, there is an analytic subset A of M x M that is

universal for the collection of analytic subsets of Af. Let S = {n G Af\ (n, n) G A}.

Recall that C = { (m,n) G Af X Af\ m = n} is a closed subset of Af x Af and is

consequently analytic. Also A fl C is analytic because it is the intersection of two

analytic sets. Since S is the projection of A fl C onto Af, S is also analytic. Now

suppose that S is a Borel set. Then S c is also a Borel set and is therefore analytic

by Proposition 3.1.4. Thus, since A is universal, there is an element n0 of M such

that Sc = A n o . Let us consider whether or not n0 belongs to Sc. If n0 G Sc then

the definition of S implies that (n0, n0) $ A, contradicting n0 G Sc = AUo. Likewise

if n0 ^ Sc, then (n0, n0) G A, contradicting n0 ^ Sc = Ano. In either case we have

a contradiction, therefore it must be that S is not a Borel set. •

CHAPTER IV

THE SEPARATION THEOREM AND ITS CONSEQUENCES

4.1 Borel Measurable Functions

Let X be a Polish space, and let and A2 be subsets of X. Then Ai and A2

can be separated by Borel sets if there are disjoint Borel subsets Bx and B2 of X

such that A\ C B\ and A 2 Q B2.

Theorem 4.1.1. Let X be a Polish space, and let Ai and A2 be disjoint analytic

subsets of X. Then A\ and A2 can be separated by Borel sets.

Proof. Let us begin by showing that:

a) if Ci, C 2 , a n d D are subsets of X such that for each n the sets Cn and D can

be separated by Borel sets, then U n C * a n d D c a n b e s e P a r a t e d B o r e l s e t s '

and

b) if E1,E2,..., and FUF2,... are subsets of X such that for each m and n the

sets Em and Fn can be separated by Borel sets, then jjm Em and [Jn Fn can be

separated by Borel sets.

First consider assertion (a). For each n choose disjoint Borel sets Gn and Hn

such that Cn C Gn and D C Hn. Then {JnGn and f]nHn are disjoint Borel sets

with Un °n Q Un Gn and DCf]nHn. Hence assertion (a) is proved.

Next consider assertion (b). Assertion (a) implies that for each m the sets Em

and \Jn Fn can be separated by Borel sets. Again applying assertion (a) we have

that U m Em and \Jn Fn can be separated by Borel sets.

We turn to the proof of the theorem itself. Let X be a Polish space and let A\ and

A2 be disjoint analytic subsets of X. Without loss of generality assume that A\ and

67

A2 are non-empty. By Corollary 3.2.5 we can find continuous functions f,g: A f X

such that /(AO = Ai and g(N) = A2. Arguing by way of contradiction, suppose

that A\ and A2 cannot be separated by Borel sets. For each positive integer k and

ni, n2, •••, fik G N, let the set

N{ni, n2, nk) — {111 £ Af\mi = ni for i = 1,2,..., fc}.

Clearly A\ = Umi=i /(-^(mi)) anc^ = Um=i 9(N(n>i))-

By assertion (b) above, we can choose n i ,mi G N such that f(N(m\)) and

g(N(n1)) cannot be separated by Borel sets. Likewise since

OO

f(N(mi))= ( J /(iV(mi,m2)) 1712 = 1

and OO

g(N(ni)) = [ J g(N(n1,n2))

we can choose M2,^2 G N such that /(AT{TYI\, ^2)) g(iV(rii, ^2)) cannot be

separated by Borel sets. Similarly we can choose sequences m = {mt)t and n = (nt)t

such that for each k the sets /(A r(mi,m2,...,m) !)) and g(N(ni, n2,..., nk)) cannot

be separated by Borel sets. Suppose that / ( m ) ^ <7(11). Choose open balls U and

V such that / ( m ) G U, g(n) G V, and UC\V = 0. Since / is continuous there exists

a positive integer k] such that

i\r(mi,ra2,...,mA;1) C /_1(77)

and since g is continuous there exists a positive integer k2 such that

N(n\, n2,..., nfc2)

Without loss of generality suppose that k\ > k2. Then

N(rii,n2,...,njfcj) C iV(ni,n2, - ..,nk2) Q g 1(U).

68

This imples that f(N(m i , . . . ,m f c l ) ) and #(iV(ni, ...,nkl)) can be separated by the

Borel sets U and V. This is a contradiction. Therefore it must be that / ( m ) = fif(n).

However since / ( m ) £ f{N(mi)) C Ai and g(n) € fiWm)) C A2 we have that

Ai n A2 ^ 0. Again we have a contradiction and therefore we must conclude that

A\ and A2 can be separated by Borel sets. •

Corollary 4.1.2. Let X be a Polish space, and let be disjoint analytic

subsets of X. Then there are disjoint Borel subsets of X such that An C

Bn holds for each n.

Proof Let n G N. By Theorem 3.1.2, {jm^nAm is analytic. Since An and

Um^ nAm a f e disjoint analytic sets by Theorem 4.1.1 we can choose a Borel set

Cn such that An C Cn and [jm^nAm C Ccn. Now define the Borel sets BUB2,...

by letting Bn = C„ — Cm). Then each Bn is Borel and the sets Bi, B2,

are pairwise disjoint. If a; 6 An it is clear that x € Cn. Suppose m ^ n, then

Cm C O i^mAj. Since x $ Acn and n / m we have that x Cm- Therefore x G Bn

and An C Bn as desired. •

Corollary 4.1.3. Let X be a Polish space, and let A be a subset of X. If both A

and Ac are analytic} then A is Borel.

Proof. According to Theorem 4.1.1 there are disjoint Borel subsets B\ and B2 of X

such that A C Bi and Ac C B2. It follows immediately that A — B\ and Ac = B2*

Consequently A is Borel. •

The above corollary, along with Corollary 3.4.7 which gives us a non-Borel ana-

lytic set, implies that the complement of an analytic set is not necessarily analytic.

Proposi t ion 4,1.4. Let X and Y be Polish spaces, let A be a Borel subset of X}

and let f be a function from A to Y. Then f is Borel measurable if and only if its

graph is a Borel subset of X x Y.

69

Proof. Suppose that / is Borel measurable. By Proposition 2.5.2, the graph of / is

a Borel subset of A x Y. By Lemma 3.2.2 there is a set B x C in B(X x 7 ) such

that

gr( / ) = (B x C) fl (A x F) .

Since by Proposition 2.5.1 A x Y is a Borel subset of X x Y, we have gr( / ) is a

Borel subset of X xY.

Now suppose that gr( /) is a Borel subset of X x Y , and that B is a Borel subset

of Y. Then gr(/) D {X x B) and gr( / ) 0 (X x Bc) are Borel and hence analytic

subsets of X x Y. Thus the projections of these sets on X are analytic. But these

projections are / _ 1 ( 5 ) and f~l(Bc) respectively. Furthermore the sets f~l{B) and

f~l(Bc) are disjoint, and so by Theorem 4.1.1 there are Borel sets and B2 such

that / _ 1 ( B ) C Bu f-^B0) C B2, and Bx n B2 = 0. Let x G Bx fl A. Then x G A

and f(x) G B. Consequently x G f l { B ) . This gives us that f~l(B) = Ar\Bx.

Since A and Bx are Borel, f~l{B) is Borel. Therefore / is Borel measurable. •

Proposi t ion 4.1.5. Let X and Y be Polish spaces, let A be a Borel subset of X,

let f : A Y be Borel measurable, and let B = f{A). If f is injective and if

B G B(Y), then f - 1 is Borel measurable.

Proof. Define h such that h(x, y) = (y, x). Clearly h is a homeomorphism from 1 x 7

onto Y x X. Let (y,x) G h(gv(f)). Then x G X and f(x) = y. Since / is injective

we have that f~1(y) — x. Consequently (y, x) G gr ( / 1 ) and h(gv(f)) C gr ( / ). A

similar argument shows the converse and consequently g r ( / 1 ) = h(gr(f)). Hence

g r ( / _ 1 ) is a Borel subset of Y x X if and only if gr( / ) is a Borel subset of 1 x 7 .

By Proposition 4.1.4 we have that f"1 is Borel measurable. •

We now show that the assumption that f(A) belongs to B(Y) can be removed

from Proposition 4.1.5. First we need the following lemma.

70

L e m m a 4.1 .6 . Let X and Y be Polish spaces, let A be a non-empty Borel subset

of X, and let f : A Y be Borel measurable and injective. Then there is a Borel

measurable function g \ Y —¥ X such that g{Y) C A and such that g{f{x)) %

holds at each x in A.

Proof. Let d be a metric for X , and let x be a fixed element of A. For each positive

integer n we define a function gn : Y —> X as follows. Choose a finite or countably

infinite partition {An,k}k of A into non-empty Borel subsets of diameter at most \

and in each An,k choose a point xri)k. We can construct this partition as follows.

Let n G N and let (xk)k be a sequence whose terms form a dense subset of A. Let

Ana = Bd{xu i ) n A. Let An,2 = (Bd(x2, i ) n A) - An,!. Similarly let

j J

An,j = ^ — U -^n,k-

Without loss of generality we can assume that each of these sets is non-empty,

because we could discard any set which was empty and renumber. The sets f(An,k),

k = 1,2,.. . , are disjoint and analytic by Proposition 3.2.3. Therefore by Corollary

4.1.2 we can choose disjoint Borel sets Bn,k, k = 1,2,... , such that f(An,k) C Bn,k

holds for each k. Now let gn{y) = xn,k if y G Bn,k and let gn(y) = x if y £

(Ufc Bn,k)c- Let B be a Borel set in X. Then ^ ( - B ) is the union of those Bn,k

such that xn,k G B and possibly (lj fe Bn,k)c if x G B. Therefore g'^iB) is Borel

and gn is Borel measurable. Define g : Y -4 A by letting g(y) = limn gn(y) if the

limit exists and belongs to A, and letting g(y) = x otherwise. Proposition 2.5.8 and

the statement which follows Proposition 2.5.8 imply that g is Borel measurable. If

x G A, then d(x,g„(f(x))) < ^ holds for each n. Consequently g(f(x)) = x. Thus

g is the required function. •

T h e o r e m 4.1 .7 . Let X and Y be Polish spaces, let A be a Borel subset of X, and

let f : A —>• Y be Borel measurable and injective. Then f(A) is a Borel subset ofY.

71

Proof. Assume without loss of generality that A ± 0 . According to Lemma 4.1.6

there is a Borel measurable function g :Y X such that g(Y) C A and such that

g(f(x)) = x holds at each x in A. We will now show that

/(A) = {ye Y\f(g(y)) = y}.

Let x e /(A). Then /_ 1(a;) G A and consequently

5(/(/_1 (®)) = 9(x) = /_ 1 (®).

This gives us that f{g(x)) = x and z G {y € Y\ f(g(y)) = y}- Now suppose

f(g(y)) = y. Then g(y) = f~l{y), giving f~~l{y) € A and y € /(A). Therefore

/(A) = {y € Y\ f(g(y)) = y}.

Thus Proposition 2.5.9 applied to the functions y y and f(g{y)) implies that

f(A) is a Borel subset of Y. •

4.2 Borel Isomorphisms

Let (X,A) and (Y,B) be measurable spaces. A bijection / : X Y is an

isomorphism if f is measurable with respect to *4 and B and / 1 is measurable

with respect to B and A* Equivalently, the bijection / is an isomorphism if the

subsets A of X that belong to A are exactly those for which f(A) belongs to B.

The spaces {X^A) and (Y,B) are isomorphic if there exists such an isomorphism.

We shall also call subsets X0 and Y0 of X and Y isomorphic if the spaces (X0, Ax0)

and (Y 0 lBy o) are isomorphic. In case (X,B(X)) and (Y,B(Y)) are Polish spaces

together with their Borel cr-algebras, we shall often use the term Borel isomorphism

instead of isomorphism.

Theo rem 4.2.1. Let A and B be Borel subsets of Polish spaces. Then A and B

are Borel isomorphic if and only if they have the same cardinality. Furthermore}

72

the cardinality of each uncountable Borel subset of a Polish space is that of the

continuum.

Proof. It is clear that if A and B are isomorphic then they have the same cardinality.

Suppose that A and B have the same cardinality. If these sets are finite or countably

infinite, then each of their subsets is a Borel set and each bijection between them

is an isomorphism; hence A and B are isomorphic.

Now suppose that A and B are uncountable. Then Proposition 3.4.2 implies that

there are continuous injective maps / : Af A and g : Af —»• B such that A — f(Af)

and B - g(Af) are Borel sets. Consequently f(Af) and g(Af) are also Borel sets.

Since / and g are continuous, / and g are Borel measurable. By Proposition 4.1.5, /

and g are Borel isomorphisms of Af onto f(Af) and g(.Af) respectively. Thus go f

is a Borel isomorphism of f(Af) onto g{Af). Now let I be a countably infinite subset

of / (Af) and let h be a bijection of the countably infinite set IU ( A - f (Af)) onto the

countably infinite set ( 5 ( / _ 1 ( ^ ) ) ) U (B - g{Af)). Let k be the function that agrees

with g o f-1 on f(Af) - I and with h on / U {A - f{Af)). Clearly k is a bijection

from A onto B. Let H be a Borel subset of A. Then

k(H) = k{H n (/(AO - I)) U k(H n (I U (A - /(A/"))))

= g o r 1 (H n (/(AO - 1 ) ) U h(H n (I u (A — /(Af))).

We have that f{Af), H, and I are Borel; therefore since g o f 1 is a Borel ismorphism

g o f - ^ H n i f i A T j - I ) )

is also Borel. Since h(Hn(IU(A-f{Af))) is countable, it is also Borel. Consequently

k(H) is Borel.

Now suppose J is a Borel subset of B. Then

k - \ j ) = k~\j n k{f(Af) -1)) u k-\J n k{i u (A- f(Af)))

73

= (9 o r ' T V n k{f{M) - I)) U h-\J n (g{f ^J)) u {B - g{N))).

Since gof-1 is a Borel isomorphism, from above we have that k maps Borel sets to

Borel sets, and the sets f ( A f ) — I a n ( l J are Borel, (g o / ) (J H k[f(M) I))

Borel. Since h~\Jf\{g{f~l (I)P(B - g{Af))) is countable it is also Borel. Therefore

krl(,J) is Borel. Therefore k is a Borel isomorphism from A onto B. In particular

each uncountable Borel subset of a Polish space is Borel isomorphic to R, and so

has the cardinality of the continuum. •

It follows from Theorem 4.2.1 that a Borel subset of a Polish space is Borel

isomorphic to R, to the set N of all positive integers, to the set {1,2,..., n} for some

positive integer n, or to the empty set. In Corollary 3.4.7 we showed that there is

an analytic subset of Af which is not Borel. The following proposition gives us that

each uncountable Polish space has an analytic subset which is not Borel.

Proposition 4.2.2. If X is an uncountable Polish space, then there is an analytic

subset of X that is not Borel.

Proof. By Theorem 4.2.1 each of M and X has the cardinality of the continuum and

therefore Af and X are Borel isomorphic. By Corollary 3.4.7 there is an analytic

subset S of J\f that is not a Borel subset. Let / be an isomorphism from M onto X.

Then f(S) is not a Borel subset of X. There exists a continuous function g and a

Polish space Z such that g : Z -»> M and g{Z) = S. Then ( / o g)(Z) = f(S) giving

that f(S) is analytic. Consequently there is an analytic subset of X that is not a

Borel set. •

CHAPTER V

THE MEASURABILITY OF ANALYTIC SETS

Let (X, A) be a measurable space, and let ^ be a measure on A. The measure ft

(or the measure space (X,A,fJ.)) is complete if the relations A £ A, (J-{A) = 0, and

B C A together imply that B G A. The completion of A under is the collection

A^ of subsets A of X for which there are sets E and F in A such that

1) E C A C F and

2) (jl(F-E) = 0.

A set that belongs to Ap is said to be /i-measurable.

Suppose that A C I and there are sets E and F in A that satisfy (1) and (2).

Then fx(E) = f/,(F). Furthermore if B C A and B G A then fJ-(B) < fi(F) = fJ>(E).

Hence

fi(E) = sup{/u(j5)| B G A and B C A}.

Therefore the common value of fi(E) and fJ-(F) depends only on the set A and the

measure jj, and not on the choice of sets E and F satisfying (1) and (2). Thus we

can define a function Ji : Afl —>• [0, +oo] by letting ji{A) be the common value of

jj,(E) and /J,(F) where E and F belong to A and satisfy (1) and (2). This function

/J is called the completion of fx.

We also define the outer measure fx*(A) and the inner measure of an

arbitrary subset A of X by

3) /u*(A) = mi{/j,(B)\ AC B and B G A},

and

4) n*{A) = sup{ju(5)| B C A and B G *4}.

75

on Propos i t ion 5.0.1. Let (X,A) be a measurable space, and let // be a measure

A. Then A^ is a a-algebra on X that includes A and ~p xs a measure on A^ that

is complete and whose restriction to A is fi. Furthermore /J is the only measure on

A^ that agrees with jj, on A.

Proof. It is clear that A? includes A and hence that X € A ft- Note that the relations

E C A C F and jx(F - E) = 0 imply that Fc C Ac C Ec and fj,(Ec - Fc) = 0;

thus Ap is closed under complementation. Now suppose that (An)n is a sequence

of sets in AFor each n choose sets En and Fn in A such that En C An C Fn and

fi(Fn - En) = 0. Then U n E n and U n F n belong to A and satisfy

UnEn C UnAn C Urai^n-

Also,

/ i (U nF n - U n E n ) = fi{\JnFn n ( U n E n ) c )

= /u ( (u n F n ) n ( n „ ^ ) )

= /u(un{Fn n (n„££)))

< /u(Un{Fn n Ecn))

— )J,(Un(Fn - En))

< - En) n

= o.

Thus U n A n belongs to A? and Ap is a cr-algebra on X that includes A.

Clearly Ji is an extension of /', and ]J takes on non-negative values with ^(0) = 0.

Let (An)n be a disjoint sequence of sets in A^ and for each n again choose sets En

and Fn in A that satisfy En C An C Fn and fi(Fn — En) = 0. Then the disjointness

of the sets An implies the disjointness of the sets En. Therefore

^(Unj4.ji) = /i(Un-E'n) ^ y jJ-^En) ^ ^ fiAn.

76

Thus /I is a measure. Now suppose B is a subset of X such that there exists A £ A

with B C A and fi(A) = 0. Then 0 C B C A with fi(A - 0) = 0. Therefore B is

~p-measurable and Ji is complete.

Suppose v is a measure on A^ that agrees with fj, on A. Let B G A^. If B € A

then

-p(B) = v{B) =

Now suppose that B A. Choose sets E and F in A such that E C B C F and

— E) = 0. Then u(F — E) — 0 which implies that v{F) — v{E). Since v must

be monotone it must be true that

u(E) = n(E) = u{B).

Consequently,

u(B) = n(E) = -p(B).

Now we have that JI is a unique extention of /j, to A d

Propos i t ion 5.0.2. Let (X,A) be a measurable space, let fi be a measure on A

and let A be a subset of X such that /u*(A) < oo. Then A belongs to A^ if and only

if H*(A) — fJ>*{A).

Proof. If A belongs to AM then there are sets E and F that belong to A and satisfy

E C A C F and (i{F - E) = 0. Then

Since fjt(E) = /j,(F), the relation yU*(A) = ft* (A) follows.

Now suppose A is a subset of X with /U*(A) < oo and fJ,*(A) = ji*{A). for each

positive integer n choose a set En and a set Fn from A such that En C 4 C Fn and

77

f**(A) > fi(Fn) - J and /i*(A) < fx(En) + J . Let E = U n £ „ and F = O n F n . Then

E and F belong to A, and E C. A C F. Then

2 H(F -E)< n(Fn - En) < - .

Since n was arbitrary, ji(F — E) = 0. Therefore A E A •

L e m m a 5.0.3. Let (X, A) be a measurable space, let n be a finite measure on

(X,A) and let fi* be defined by equation (3). If (An)n is an increasing sequence of

subsets of X, then

fj,*(UnAn) = lim ft* (An). n

Proof. The monotonicity of pi* implies that lim„ fi*(An) exists and that

l i m n * ( A n ) < fx*(UnAn). n

We now only need to show that

Y\mjj*(An) > jj,*((JnAn). n

Let e > 0, and for each positive integer n, use (3) to choose a set Bn that belongs

to A such that An C Bn and

fJ'(Bn) < ( A n ) + e.

By replacing Bn with (~)<j°.nBj we can assume that the sequence {Bn\ is increasing.

By Lemma 2.5.10 we have that /j,(UnBn) = l i m n f j , ( B n ) . Therefore we have

/i*(U„An) < jj,(UnBn) = l i m n ( B n ) < limfj,*(An) + e. n n

Since e is arbitrary the proof is complete. •

78

T h e o r e m 5.0.4. Let X be a Polish space, and let fj, be a finite Borel measure on

X. Then every analytic subset of X is fx-measurable.

Proof. Let A be an analytic subset of X. We shall show that A is ^-measurable by

showing that /i*(A) = ji*(A), and we shall do this by producing, for an arbitrary

e > 0, a compact subset K of A such that /x(A') > /i*(A) — e.

We can assume that A is non-empty and let e > 0. Choose a continuous function

/ : Af —>• X such that f(Af) = A. For positive integers k and n i , n 2 , n * let

£ ( n i , n-2, ..njt) be the set of those elements m of Af that satisfy m; < n{ for i =

1 , 2 , k . We shall construct an element p = (pk)k of Af such that

/**(/(£(Pi>P2,-,Pfc))) > f**(A) ~ e

holds for each k. Let us begin by choosing the first term p\ of the sequence p. Note

that { £ ( n ) } ^ | is an increasing sequence of sets whose union is Af. Consequently

{ / ( £ ( n ) ) } ^ = 1 is an increasing sequence of sets whose union is A. By Lemma 5.0.3

/ i*(A) = lim n*(f(C(n))) n

and therefore we can pick a positive integer p\ such that

^*(/(£0i))) > - e.

Since C(pi) = U r e(£(pi ,n)) , a similar argument provides a positive integer p2 such

that

V*{f(£{Pi,P2))) > P*(A) - e.

Continuing in this way we obtain a sequence p = {pk)k of positive integers such

that

^*(/(£(Pl,P2, -,Pfc))) > V*(A) - €

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holds for each k. Now let L = Difc£(pi,p2? a n ( i ^ ^ — / ( ^ ) - Then

L = {m £ < p*, for each z} = € N|rrii < pi}. i

Since each set {m; G N|m; < p;} is finite, it is also compact. Tychonoff's Theorem

implies that L is also compact. Since a continuous image of a compact space is

compact, we have that K is a compact subset of A. We need to show that p ( i i ) >

p*(A) — e.

Let us begin by showing that K = fl £./(£&), where for each k we have abbreviated

C(pi,p2,...,pk) by Ck- It is clear that K C flfc/(£fc). Let d be a metric for the

topology on X. If x G n&/(£fc) then for each k we can choose an mfc = of

Ck such that d(f(mk), x) < | . Note that for each k the ith components of the terms

of {mfc}fc form a bounded subset of N; hence the terms of {m^} form a relatively

compact subset of Af. Consequently we can choose a convergent subsequence of

{mfc}. Let 1 be the limit of this subsequence. Then /(1) = x because we can

choose a k such that the distance from / ( m ^ ) and / ( l ) are arbitrarily small because

of the continuity of / and also we can choose k such that d(f(m.k ),ar) < \ is

arbitrarily small. Now fix k. Suppose 1 ^ Ck, then for some i < k, li > pi. But

for each j, rriji < pt. This contradicts the fact that some subsequence of {m?); }JL,

converges to Consequently m G Ck and m G C\k£>k- Hence x G K. Consequently

K = Hjkf{Ck) as desired. For each k the set f(Ck) is closed and includes /(£&).

Therefore

H(K) - n(r\kf{Ck)) - l i m n f { C k ) > H*(A) - e. k

Since fj,*A > fi*(K) > /i * (A) — e and e is arbitrary Clearly

and therefore we have equality. Consequently A is ^-measurable. •

Let (X, *4) be a measurable space. Recall from the introduction that a subset of

X is universally measurable with respect to (X, A) if it is //-measurable for every

80

finite measure (j, on (X,A). Let A* be the family of all universally measurable

subsets of X. Then A* = Hp Ap where fi ranges over the family of finite measures

on (X, A); hence A* is a cr-algebra. Also for every finite measure /J, on (X, .4) there

is a unique measure on (X,A*) that agrees with /x on A.

Now suppose that X is a Polish space. The universally measurable subsets of X

are those that are universally measurable with respect to (X,B(X)). Theorem 5.0.4

can be reformulated as follows.

Corollary 5.0.5. Every analytic subset of a Polish space is universally measurable.

Proof. The corollary is simply a restatement of Theorem 5.0.4. •

Using the above corollary and Proposition 4.2.2 we can now conclude that each

uncountable Polish space contains a non-Borel subset which is universally measur-

able.

REFERENCES

Cohn, Donald L. Measure Theory. Stuttgart : Birkhauser, 1980.

Kuratowski, K. Topology, vol. 1. New York: Academic Press, 1966.

Munkres, James. Topology: A First Course. Englewood Cliffs: Prentice-Hall, 1975.

Royden, H. L. Real Analysis. 3rd ed. New York: Macmillan, 1963.