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
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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-
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.
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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\(BcU Yc)
= F n Bc.
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Since B E B(X), B° E B(X). Hence Y — A E B(X)y• Now suppose is a
sequence of elements from B(X)y. For each i choose Bi such that At = BtC\Y and
Bi EB{X). Then
U = U(5i n r ) = ( | J 5 0 n r .
i i i
Since (J iBi E B(X) we have (J^ Ai E B(X)y. Therefore B(X)y is a cr-algebra
containing all open subsets of Y. This implies that B(Y) C B(X)y. We now have
shown that B(Y) = B{X)y. •
Proposition 3.2.3. Let X and Y be Polish spaces, let A be an analytic subset
of X, and let f : A —>• Y be Borel measurable (that is measurable with respect to
B{A) and B(Y)). If A\ and A2 are analytic subsets of X and Y respectively, then
f ( A f ) A i ) and f~1(A2) are analytic subsets ofY and X respectively.
Proof. Suppose the hypotheses are satisfied. Proposition 2.5.2 implies that gr(/) E
B(A x Y) and Lemma 3.2.2 then implies that there is a Borel subset B of X xY such
that gr( / ) = B D {A x Y"). We have that Y is analytic by Proposition 3.1.4. Also
gr( / )n(A 1 x F ) is analytic by Proposition 3.1.2 and 3.2.1. Similarly, g r ( / )n(A 1 x Y)
is an analytic subset o f X x F and therefore is the image of a Polish space Z under
a continuous map, say h. Let Try be the projection o f l x F onto Y. Then tt y is
continuous and 7Ty o h is a continuous map from Z into Y.
We want to show that (7xy o h)(Z) = f(A Pi A\). 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
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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
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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 £,