You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
<strong>Chapter</strong> 7<br />
<strong>Infinite</strong> <strong>product</strong> <strong>spaces</strong><br />
So far we have talked a lot about infinite sequences of independent random variables<br />
but we have never shown how to construct such sequences within the framework<br />
of probability/measure theory. For that we will need to be able to construct<br />
infinite <strong>product</strong> measure <strong>spaces</strong> and extend natural measures to them. Interestingly,<br />
the construction requires that the measure structure be tied to the natural<br />
topology on this space. A class of topologies which behave naturally under such<br />
construction goes by the name Polish; the measure <strong>spaces</strong> arising thereby are called<br />
standard Borel.<br />
7.1 Standard Borel <strong>spaces</strong><br />
Definition 7.1 Recall that a metric space (X, d) is a set with a symmetric function<br />
d : X ⇥ X ! [0, •) such that d(x, y) =0 is equivalent to x = y and the triangle inequality<br />
d(x, y) apple d(x, z)+d(z, y) holds for all x, y, z 2 X.<br />
(1) We say that a metric space (X, d) is Polish if it is separable (there exists a countable<br />
dense set) and complete (all Cauchy sequences have limits in X).<br />
(2) A standard Borel space (X, B) is a measure space such that X is a Polish metric<br />
space and B is the s-algebra of Borel sets induced by the Polish topology.<br />
(3) A map f : X ! Y between topological <strong>spaces</strong> X, Y—of which metric <strong>spaces</strong> are examples—<br />
is called Borel isomorphism if f is bijective and both f and f 1 are Borel measurable.<br />
The following are examples of Polish (and hence standard Borel) <strong>spaces</strong>:<br />
(1) {1, . . . , n} with n 2 N or N itself; all equipped with discrete topology.<br />
(2) [0, 1] with d(x, y) =|x y|.<br />
(3) Cantor space {0, 1} N with <strong>product</strong> (discrete) topology.<br />
(4) Baire space N N with <strong>product</strong> (discrete) topology.<br />
1
2 CHAPTER 7. INFINITE PRODUCT SPACES<br />
(5) Hilbert cube [0, 1] N with <strong>product</strong> (Euclidean) topology.<br />
(6) R n with Euclidean metric.<br />
(7) R N with <strong>product</strong> Euclidean topology.<br />
(8) L p (W, F , µ) with 1 apple p < • provided F is countably generated and µ is<br />
s-finite. (This includes the ` p <strong>spaces</strong> with p < •.)<br />
(9) ` • fails to be Polish but its subspace of converging sequences<br />
with supremum metric is Polish.<br />
c0 = x =(xn) 2 ` • (N) : xn ! 0<br />
(7.1)<br />
(10) C(X)—the set of all continuous functions on a compact metric space X is Polish<br />
in the topology generated by the supremum norm.<br />
(11) L(X, Y)—the space of bounded linear operators T : X ! Y where X and Y are<br />
separable Banach <strong>spaces</strong> is Polish in the strong topology (the norm topology<br />
generally fails this property). The metric is defined by<br />
d(T1, T2) =Â 2<br />
n 1<br />
n T1( fn) T2( fn) Y<br />
where ( fn) is a countable dense set in X.<br />
(7.2)<br />
(12) M1(W, F )—the set of all probability measures on a standard Borel space (W, F )<br />
in the topology of so called weak convergence. The metric is defined by<br />
d(µ, n) =Â 2<br />
n 1<br />
n Eµ( fn) En( fn) (7.3)<br />
where ( fn) is a countable dense set in C(W)—which, as can be shown, is separable.<br />
Note that, in such cases, this leads to a hierarchy of <strong>spaces</strong>: measures on<br />
(W, F ), measures on measures on (W, F ), etc.<br />
The reason why standard Borel <strong>spaces</strong> are good for us is that there are essentially<br />
only three cases of them that we will ever have to consider:<br />
Theorem 7.2 [Borel isomorphism theorem] Let (X, B(X)) be a standard Borel space.<br />
Then there is a Borel isomorphism f : (X, B(X)) ! (Y, B(Y)) onto a Polish space Y<br />
where<br />
8<br />
>< {1, . . . , n}, if |Y| = n,<br />
Y = N,<br />
>:<br />
[0, 1],<br />
if Y is countable,<br />
otherwise,<br />
(7.4)<br />
and where B(Y) is the Borel s-algebra induced by the corresponding Polish topology.
7.2. PROOF OF BOREL ISOMORPHISM THEOREM 3<br />
7.2 Proof of Borel isomorphism theorem<br />
The proof invokes a lot of nice arguments that have their core in set theory. We<br />
begin by developing a tool for proving equivalence of measure <strong>spaces</strong>:<br />
Theorem 7.3 [Cantor-Schröder-Bernstein] Let A ⇢ X and B ⇢ Y be sets such that<br />
there exist bijections f : X ! B and g : Y ! A. Then there exists a bijection h : X ! Y.<br />
Moreover, if F is a s-algebra on X and G is a s-algebra on Y, and if f , g 1 are F /G -<br />
measurable and f 1 , g are G /F -measurable, then h can be taken F /G -measurable.<br />
Proof. Let C =(g f )(X) ⇢ A and note that the sets<br />
{(g f ) n (X \ A)}n 0, {(g f ) n (A \ C)}n 0, \<br />
(g f ) n (X) (7.5)<br />
form a disjoint partition of X. We define<br />
n 0<br />
8<br />
>< f (x), if x 2<br />
h(x) =<br />
>:<br />
S<br />
n 0(g f ) n (X \ A),<br />
g 1 (x), if x 2 S<br />
n 0(g f ) n (A \ C),<br />
g 1 (x), if x 2 T<br />
n 0(g f ) n (X).<br />
(7.6)<br />
(Note that we could have also defined h(x) to be f (x) in the last case because g<br />
f maps the intersection onto itself in one-to-one fashion.) Since g : Y ! A is a<br />
bijection, we need to show that g h : X ! A is a bijection. In the latter two cases g<br />
X<br />
A<br />
Figure 7.1: The setting of Cantor-Schröder-Bernstein Theorem: f maps X injectively<br />
onto B ⇢ Y and g maps Y injectively onto A ⇢ X. The shaded oval inside A<br />
indicates the result of infinite iteration of g f on X, i.e., the set T<br />
n 1(g f ) n (X).<br />
This set is invariant under g f which, in fact, acts bijectively on it.<br />
f<br />
g<br />
B<br />
Y
4 CHAPTER 7. INFINITE PRODUCT SPACES<br />
h is an identity, while in the first case we note that g h is the bijection g h : (g<br />
f ) n (A \ C) ! (g f ) n+1 (A \ C). Hence g h maps X bijectively onto<br />
⇣ [<br />
(g f ) n ⌘ ✓ ⇣ [<br />
(X \ A) [ (g f ) n ⌘<br />
(A \ C) [ \<br />
(g f ) n ◆<br />
(X) = A. (7.7)<br />
n 1<br />
n 0<br />
Since all of these maps are measurable, so is the above partition of A. It follows<br />
that h is measurable too.<br />
The crux of the above theorem is that we only need to demonstrate the existence<br />
of Borel isomorphism(s) onto subsets of the relevant <strong>spaces</strong>. We will eventually use<br />
open, continuous bijections, i.e., homeomorphisms. Here it is important to characterize<br />
the image of a Polish space under a homeomorphism. Recall that a Gd is a class<br />
of sets in a topological space that are countable intersections of open sets.<br />
Lemma 7.4 Let X and Y be Polish and let g : X ! B ⇢ Y be a homeomorphism—i.e., a<br />
bijection that is open and continuous. Then B is a Gd subset of Y.<br />
Proof. Let f = g 1 : B ! X. Define the oscilation of f by<br />
n 0<br />
osc f (y) =inf diamX( f (U)): U 3 y open . (7.8)<br />
(This is defined at all y 2 Y.) Suppose y 2 ∂B is such that osc f (y) = 0. Then<br />
for any yn ! y, yn 2 B, xn = f (yn) converges to x = limn!• f (yn). Since X is<br />
complete, we have x 2 X and so y = limn!• yn = limn!• g(xn) =g(x). It follows<br />
that y 2 B, i.e.,<br />
B = B \ y 2 Y : osc f (y) =0 . (7.9)<br />
But closed sets are Gd in metric <strong>spaces</strong> and {osc f = 0} is Gd because<br />
{osc f = 0} = \<br />
n 1<br />
{osc f < 1/ n} (7.10)<br />
and because osc f is upper semicontinuous, non-negative and so continuous on {osc f =<br />
0}. Hence B is also a Gd set.<br />
The actual use of Theorem 7.3 is enabled by the following observation:<br />
Lemma 7.5 The <strong>spaces</strong> [0, 1], C =[0, 1] N , N = N N and [0, 1] N , equipped with their<br />
natural Polish topologies, are Borel isomorphic.<br />
Proof. This results from the string of open, continuous (and thus Borel) injections:<br />
[0, 1] ,! C ,! N ,! [0, 1] N ,! [0, 1] (7.11)<br />
The first map is the injection defined by writing each number in [0, 1] in dyadic<br />
form. The second map is a simple inclusion. The third map is induced by the<br />
continued fraction expansion: if x 2 [0, 1] \ Q, there exists a unique sequence (an) 2<br />
{0, 1, . . . } N such that<br />
1<br />
x =<br />
(7.12)<br />
1<br />
a1 +<br />
a2 + 1<br />
a3+...
7.2. PROOF OF BOREL ISOMORPHISM THEOREM 5<br />
1<br />
and noting that, since an = b x c, if x is irrational two distinct sequences will<br />
an 1<br />
lead to two distinct numbers. The fourth injection is a diagonal map: Write an<br />
element x =(x1, x2,...) 2 [0, 1] N by means of a matrix am,n 2{0, 1} such that<br />
Now consider the diagonal map<br />
xm = Â n 1<br />
k 1<br />
f(x) =Â Â<br />
k 1 m=1<br />
am,n<br />
, m 1. (7.13)<br />
2n a m,m k<br />
. (7.14)<br />
21+···+k 2+m<br />
It is easy to check that this map is continuous, one-to-one and the inverse is continuous.<br />
By Lemma 7.4, the images of these maps are Borel and so all maps are Borel<br />
measurable. By Theorem 7.3, all of these <strong>spaces</strong> are Borel isomorphic.<br />
Recall the well-known fact—which enables the definition of dimension—that [0, 1]<br />
and [0, 1] 2 are not homeomorphic. However, by the arguments in the above lemma,<br />
as standard Borel <strong>spaces</strong> they are indistinguishable.<br />
Having proved the Borel-equivalence of examples 2-4 in our list above, let us now<br />
address the embedding of a general Polish space. Here it is convenient to work<br />
with Hilbert cube:<br />
Proposition 7.6 [Universality of Hilbert cube] Every Polish space is homeomorphic<br />
to a Gd-subset of [0, 1] N .<br />
Proof. Let (X, d) be a Polish space and let us assume without loss of generality<br />
that d 1 (otherwise maps (X, d) ! (X, d<br />
1+d ) by identity map). Let (xn) be a<br />
countable dense set in X. Define the map f : X ! [0, 1] N by<br />
x 2 X 7! f (x) = d(x, xn) n 1 2 [0, 1] N . (7.15)<br />
As is easy to check, f is continuous and one-to-one and so we have f 1 : f (X) ! X.<br />
We claim that f 1 is also continuous. Indeed, suppose that (zn) is a sequence such<br />
that f (zn) ! f (z) in f (X). Let e > 0 and find n 1 such that d(xn, z) < e. Then<br />
d(zm, z) apple d(zm, xn)+d(xn, z)<br />
= f (zm)n + d(z, xn) !<br />
n!• f (z)n + d(z, xn) < 2e. (7.16)<br />
Hence zm ! z or, in explicit terms, f 1 ( f (zm)) ! f 1 ( f (z)), and so f 1 is continuous<br />
on f (X). Thus f : X ! f (X) is a homeomorphism and so, by Lemma 7.4, f (X)<br />
is a Gd subset of [0, 1] N .<br />
The previous proposition gives the desired embedding of X into the Hilbert cube<br />
and, by Lemma 7.5, a Borel isomorphism into the Cantor space. It remains to construct<br />
an embedding of the Cantor space into a general (uncountable) Polish space.
6 CHAPTER 7. INFINITE PRODUCT SPACES<br />
Theorem 7.7 [Cantor-Bendixon] Let X be a second countable topological space (i.e.,<br />
with countable basis of the topology). Then there exists unique subsets P and O with P<br />
perfect (i.e., having no isolated points) and O countable open such that<br />
X = P [ O and P \ O = ∆. (7.17)<br />
Proof. Let {Un} be a countable basis (of open sets) and let<br />
O = [ Un : Un countable . (7.18)<br />
Then O is open and countable while P = X \ O is closed. Let x 2 P. If U 3 x is open,<br />
then U—being the union of a subcollection from {Un} is necessarily uncountable,<br />
because otherwise x would end up in O. This means that P \ U \{x} 6= ∆, i.e., x is<br />
not isolated.<br />
If P 0 and O 0 are two other such sets, then U \ P 0 6= ∆ implies that U \ P 0 is uncountable,<br />
for any U open. Hence P 0 \ O = ∆, i.e., P 0 ⇢ P. On the other hand, O 0 is countable<br />
open and so it appears in the union defining O. Hence O 0 ⇢ O. If P 0 [ O 0 = X,<br />
then we must have P 0 = P and O 0 = O.<br />
The relevant consequence of this general theorem is as follows:<br />
Corollary 7.8 [Cantor space to Polish space] Every uncountable Polish space contains<br />
a homeomorphic copy of C (and so it has cardinality 2 @0).<br />
Proof. Let X be an uncountable Polish space. Every separable metric space is second<br />
countable, so let X = P \ O be as in the previous theorem. Then P—being a closed<br />
subset of X—is Polish too, so we may as well assume that X = P. We will use the<br />
standard “tree-like” construction to identify a copy of C in X.<br />
Let (xn) be a countable dense set (of distinct points) in P. As is easily checked, there<br />
exist two functions f0 and f1 from {x1,...} to itself such that<br />
and<br />
max d( f0(xn)), xn), d( f1(xn)), xn) apple 2 n<br />
min d( f0(xn)), xn), d( f1(xn)), xn)<br />
(7.19)<br />
d f0(xn), f1(xn) > 0. (7.20)<br />
We now define a collection of balls Bs, indexed by s{0, 1} n , centered at points<br />
of {x1, x2,...} as follows: Set B? to ball of radius 1/ 2 about x1. If Bs is defined<br />
for all s{0, 1} n , and ˜xs are their centers, let rn be the minimum of f0( ˜xs) and f1( ˜xs)<br />
for all s 2{0, 1} n . Then let Bs0 be the ball centered at f0( ˜xs) of radius rn<br />
4 , and<br />
similarly let Bs1 be the corresponding ball centered at f1( ˜xs). These balls are disjoint<br />
with Bs0, Bs1 ⇢ Bs and, by induction, the same is true for {Bs : s{0, 1} n } for<br />
all n 1.<br />
Now let s =(s1, s2,...) 2{0, 1} N . By construction, the intersection T<br />
n 1 B (s1,...,sn)<br />
is non-empty and, by the completeness of X, it contains exactly one point ys.<br />
Moreover, if s and ˜s agree in the first n digits and sn+1 = 0 and s0 then ys 2 B (s1,...,sn,0) and y ˜s 2 B (s1,...,sn,1) and so<br />
n+1 = 1,<br />
rn<br />
2 apple d(ys, y ˜s) apple 2 n<br />
(7.21)
7.3. PRODUCT MEASURE SPACES 7<br />
Thus the map s 7! yn is one-to-one and continuous. To show that the inverse<br />
map is continuous, let us note that if d(ys, y ˜s) < e, then sn = s 0 n for all n for<br />
which rn/ 2 > e. So the smaller e, the larger part of s and ˜s must be the same.<br />
Hence, the inverse map is continuous.<br />
Now we are finally ready to put all pieces together:<br />
Proof of Theorem 7.2. If X is finite or countable, then the desired Borel isomorphism<br />
is constructed directly. So let us assume that X is uncountable. Then Corollary 7.8<br />
says there is a homeomorphism C ,! X while Proposition 7.6 says there exists a<br />
homeomorphism X ,! [0, 1] N . Both of these image into a Gd-set by Lemma 7.4<br />
and so the inverse is Borel measurable. Since [0, 1] N is Borel isomorphic to [0, 1] N ,<br />
we thus we have Borel-embeddings C ,! X and X ,! C . From Theorem 7.3 it<br />
follows that X is Borel isomorphic to C —and thus to any of the <strong>spaces</strong> listed in<br />
Lemma 7.5.<br />
7.3 Product measure <strong>spaces</strong><br />
Definition 7.9 [General <strong>product</strong> <strong>spaces</strong>] Let (Wa, Fa)a2I be any collection of measurable<br />
<strong>spaces</strong>. Then W = ⇥a2I Wa can be equipped with the <strong>product</strong> s-algebra F =<br />
N<br />
a2I Fa which is defined by<br />
✓n o<br />
F = s ⇥ Aa : Aa 2 Fa &#{a 2I: Aa 6= Wa} < •<br />
a2I<br />
◆<br />
. (7.22)<br />
If Wa = W0 and Fa = F0 then we write W = W I 0<br />
and F = F ⌦I<br />
0 .<br />
Next we note that standard Borel <strong>spaces</strong> behave well under taking countable <strong>product</strong>s:<br />
Lemma 7.10 Let I be a finite or countably infinite set and let (Wa, Fa)a2I be a family<br />
of standard Borel <strong>spaces</strong>. Let (W, F ) be the <strong>product</strong> measure space defined above. Then<br />
(W, F ) is also standard Borel. Moreover, if fa : Wa ! [0, 1] are Borel isomorphisms, then<br />
f = ⇥a2I fa is a Borel isomorphism of W onto [0, 1] I .<br />
Our goal is to show the existence of <strong>product</strong> measures. We begin with finite <strong>product</strong>s.<br />
The following is a restatement of Lemma 4.13:<br />
Theorem 7.11 Let (W1, F1, µ1) and (W2, F2, µ2) be probability <strong>spaces</strong>. Then there<br />
exists a unique probability measure µ1 ⇥ µ2 on F1 ⌦ F2 such that for all A1 2 F1,<br />
A2 2 F2,<br />
µ1 ⇥ µ2(A1 ⇥ A2) =µ1(A1)µ2(A2). (7.23)<br />
The reason why this is not a trivial application of Carathéodory’s extension theorem<br />
is the fact that (7.26) defines µ1 ⇥ µ2 only on a structure called semialgebra and<br />
not algebra.<br />
Definition 7.12 A non-empty collection S ⇢ P(W) is called semialgebra if
8 CHAPTER 7. INFINITE PRODUCT SPACES<br />
(1) A, B 2 S implies A \ B 2 S .<br />
(2) If A 2 S , then there exist Ai 2 S disjoint such that A c = S n i=1 Ai.<br />
It is easy to verify that semialgebras on W automatically contain W and ∆.<br />
Lemma 7.13 The following are semialgebras:<br />
(1) S = {(a, b] : • apple a apple b apple •}<br />
(2) S = {A1 ⇥···⇥An : Ai 2 Ai} where Ai are algebras.<br />
(3) S = ⇥ Aa : Aa 2 Aa &#{a : Aa 6= Wa} < • provided Aa’s are algebras.<br />
a2I<br />
Proof. Let us focus on (2): Since all Ai are p-systems, S is closed under intersections.<br />
To show the complementation property in the definition of semialgebra,<br />
let D be a collection of all sets of the form B1 ⇥···⇥Bn, where Bi is either Ai or A c i ,<br />
but such that A1 ⇥···⇥An is not included in D. Then<br />
(A1 ⇥···⇥An) c = [<br />
From here the complementation property follows by noting that D is finite and<br />
D ⇢ S .<br />
Next we will show that via finite disjoint unions semialgebras give rise to algebras:<br />
Lemma 7.14 Let S be a semialgebra and let S be the set of finite unions of sets Ai 2 S .<br />
Then S is an algebra.<br />
Proof. First we show that S is closed under intersections. Indeed, A is the (finite)<br />
union of some Ai 2 S and B is the (finite) union of some Bj 2 S . Therefore,<br />
✓<br />
[<br />
A \ B =<br />
i<br />
Ai<br />
◆<br />
\<br />
B2B<br />
B.<br />
✓ ◆<br />
[<br />
Bj =<br />
j<br />
[<br />
Ai \ Bj.<br />
i,j<br />
Since Ai, Bj 2 S , then Ai \ Bj 2 S and A \ B is the finite union of sets from S .<br />
Consequently, A \ B 2 S .<br />
Next we will show that S is closed under taking the complement. Let A be as<br />
above. Then<br />
A c ✓<br />
[<br />
=<br />
◆c = \ \ [<br />
= Bi,j,<br />
i<br />
Ai<br />
where Bi,j 2 S are such that A c i is the union of Bi,j—see the definition of semialgebra.<br />
The union on the extreme right is finite and, therefore, it belongs to S . But<br />
S is closed under finite intersections, so A c 2 S as well.<br />
Now we will learn how (and under what conditions) a set function on a semialgebra<br />
can be extended to a measure on the associated algebra:<br />
i<br />
A c i<br />
i<br />
j
7.3. PRODUCT MEASURE SPACES 9<br />
Theorem 7.15 Let S be a semialgebra and let µ : S ! [0, 1] be a set function. Suppose<br />
the properties hold:<br />
(1) Finite additivity: A, Ai 2 S ,A= S n i=1 Ai disjoint ) µ(A) =Â n i=1 µ(Ai),<br />
(2) Countable subadditivity: A, Ai 2 S ,A= S • i=1 Ai disjoint ) µ(A) apple  • i=1 µ(Ai).<br />
Then there exists a unique extension ¯µ : S ! [0, 1] of µ, which is a measure on S .<br />
Proof. Note that (1) immediately implies that µ(∆) =0. The proof comes in four<br />
steps:<br />
Definition of ¯µ: For A 2 S , then A = S n i=1 Si for some Si 2 S . Then we let<br />
¯µ(A) = n i=1 µ(Si). This definition does not depend on the representation of A;<br />
if A = S m j=1 Ti for some Tj 2 S , then the fact that S is a p-system and finite<br />
additivity of µ ensure that<br />
n<br />
 µ(Si) =<br />
i=1<br />
n ⇣ [ m ⌘<br />
 µ Si \ Tj =<br />
i=1 j=1<br />
n m<br />
 Â<br />
i=1 j=1<br />
µ(Si \ Tj) =<br />
m<br />
 µ(Tj).<br />
j=1<br />
Finite additivity of ¯µ: By the very definition, ¯µ is also finitely additive, because if<br />
An is given as the disjoint union S<br />
i Si,n, then S<br />
n An = S<br />
i,n Si,n and thus ¯µ( S<br />
n An) =<br />
Âi,n µ(Si,n) =Ân ¯µ(An). (All indices are finite.)<br />
Countable subadditivity of ¯µ: We claim that ¯µ is countably subadditive, whenever<br />
the disjoint union lies in S . Let Ai 2 S and A = S<br />
k 1 Ak 2 S . Writing Ak =<br />
S<br />
n Sj—note the efficient way to index the Sj’s contributing to A kapplejn Ak is also in S<br />
because S is an algebra. Since we already showed that ¯µ is finitely additive, we<br />
have<br />
n<br />
¯µ(A) = ¯µ(Bn)+  ¯µ(A k)<br />
n<br />
 ¯µ(A k).<br />
k=1<br />
k=1<br />
Letting n ! •, the opposite inequality follows.
10 CHAPTER 7. INFINITE PRODUCT SPACES<br />
Thus we have a well defined function ¯µ : S ! [0, 1] which is finitely additive<br />
on S and countably additive on disjoint unions that belong to S . These are the<br />
properties required to be a measure on an algebra.<br />
Note that the combination of Lemma 7.13, Lemma 7.14 and the construction of<br />
the measure ¯µ from Theorem 7.15 provide an alternative proof of the fact that<br />
Lebesgue-Stieltjes measures on R are defined by the corresponding “distribution”<br />
function.<br />
Proof of Theorem 7.11. The set S = {A1 ⇥ A2 : Ai 2 Fi} is a semialgebra by<br />
Lemma 7.13 and s(S ) is the <strong>product</strong> s-algebra F1 ⌦ F2. We will show that the set<br />
function µ : S ! [0, 1] defined by<br />
µ(A1 ⇥ A2) =µ1(A1)µ2(A2), Ai 2 Fi,<br />
is countably additive on S —this implies the desired properties (1-2) in Theorem 7.15.<br />
Let A 2 F1 and B 2 F2 and suppose that A ⇥ B is the disjoint union S<br />
j Aj ⇥ Bj,<br />
where Aj 2 F1 and Bj 2 F2, and where j is either finite or countable index. The<br />
proof uses integration: For each x 2 A, let I(x) ={j : Aj 3 x}. Now since {x}⇥<br />
Bj ⇢ Aj ⇥ Bj whenever j 2I(x), the sets {x}⇥Bj with j 2I(x) are disjoint and<br />
{x}⇥ S<br />
j2I(x) Bj = {x}⇥B. It follows that B is the disjoint union<br />
and thus<br />
B = [<br />
1A(x)µ2(B) =Â j<br />
j2I(x)<br />
Bj<br />
1A j (x)µ2(Bj)<br />
for any x 2 W1. Both sides are F1-measurable, and integrating with respect to µ1<br />
we get<br />
µ1(A)µ2(B) =Â µ1(Aj)µ2(Bj).<br />
j<br />
This shows that µ(A ⇥ B) equals the sum of µ(Aj ⇥ Bj) and so µ is countably additive.<br />
Lemma 7.14, Theorem 7.15 and Carathéodory’s extension theorem then guarantee<br />
that µ has a unique extension to F1 ⌦ F2.<br />
General finite <strong>product</strong> <strong>spaces</strong> are handled by induction using the fact that<br />
nO ⇣nO 1 ⌘<br />
Fi = Fi ⌦ Fn<br />
i=1 i=1<br />
Note that this proves the associativity of the <strong>product</strong> of measures:<br />
We leave the details of these calculations implicit.<br />
(7.24)<br />
(µ1 ⇥ µ2) ⇥ µ3 = µ1 ⇥ (µ2 ⇥ µ3). (7.25)
7.4. KOMOGOROV’S EXTENSION THEOREM 11<br />
7.4 Komogorov’s extension theorem<br />
We now address the corresponding extension of measures to infinite <strong>product</strong> <strong>spaces</strong>:<br />
Theorem 7.16 [Kolmogorov’s extension theorem] Let M be a compact metric space,<br />
B = B(M) the Borel s-algebra on M and, for each n 1, let µn be a probability measure<br />
on (M n , B ⌦n ). We suppose these measures are consistent in the following sense:<br />
8n 1, 8A1,...,An 2 B : µn+1(A1 ⇥···⇥An ⇥ M) =µn(A1 ⇥···⇥An).<br />
Then there exists a unique probability measure µ on (M N , B ⌦N ) such that<br />
µ(A1 ⇥···⇥An ⇥ R ⇥ R ...)=µn(A1 ⇥···⇥An) (7.26)<br />
holds for any n 1 and any Borel sets Ai 2R.<br />
As many extension arguments from measure theory, the proof boils down to proving<br />
that if a sequence of sets decreases to an empty set, then the measures decrease<br />
to zero. In our case this will be ensure through the following lemma:<br />
Lemma 7.17 Let W be a metric space and let A be an algebra of subsets of W (which is<br />
not necessarily a s-algebra). Suppose that µ : A ! [0, 1] is a finitely-additive set function<br />
which is regular in the following sense: For each set B 2 A and each e > 0, there exists a<br />
compact set C 2 A such that<br />
C ⇢ B (7.27)<br />
and<br />
µ(B \ C) apple e. (7.28)<br />
Then for each decreasing sequence Bn 2 A with Bn # ∆ we have µ(Bn) # 0.<br />
Proof. Let Bn be a decreasing sequence of sets with limn!• µ(Bn) =d > 0. We will<br />
show that T<br />
n Bn 6= ∆. By the assumptions, we can find compact sets Cn 2 A such<br />
that Cn ⇢ Bn and µ(Bn \ Cn) < d2 n 1 . Now since Bn Bn+1, we have<br />
Bn \<br />
n[<br />
(Bk \ Ck) ⇢<br />
k=1<br />
n\<br />
Ck. Indeed, if x 2 Bn then x 2 B k for all k apple n and then x can only belong to the set on<br />
the left if x 2 C k for all k apple n. Using finite additivity of µ, it follows that<br />
⇣ \ n<br />
µ<br />
k=1<br />
C k<br />
⌘<br />
µ(Bn)<br />
k=1<br />
n<br />
 µ(Bk \ Ck) d  µ(Bk \ Ck) k=1<br />
k 1<br />
Consequently, since finite additivity implies that µ(∆) =0, the intersection T n k=1 C k<br />
is non-empty for any finite n. But the sets Cn are compact and thus T • k=1 C k 6= ∆.<br />
The inclusions Cn ⇢ Bn imply that also T<br />
n Bn 6= ∆.<br />
Lemma 7.17 links measure with topology (which is the reason why we often consider<br />
standard Borel <strong>spaces</strong>). Next we observe:<br />
d<br />
2 .
12 CHAPTER 7. INFINITE PRODUCT SPACES<br />
Lemma 7.18 Let (M, r) be a compact metric space. Equip M N with the metric<br />
d(x, y) =Â r(xn, yn)2<br />
n 1<br />
n<br />
(7.29)<br />
whenever x =(xn) 2 M N and y =(yn) 2 M N . Then (M N , d) is a compact metric<br />
space. Moreover, if C1,...,Cn are compact sets in (M, r), then C1 ⇥···⇥Cn ⇥ M ⇥<br />
M ⇥··· is compact in (M N , d).<br />
Proof. The first part is a reformulation of Tychonoff’s theorem for <strong>product</strong> <strong>spaces</strong>.<br />
The second part is checked directly.<br />
Now we are finally ready to prove Kolmogorov’s theorem:<br />
Proof of Theorem 7.16. Consider the set<br />
S = A1 ⇥···An ⇥ M ⇥···: Ai compact in M 8i<br />
Then, as is easy to check, S is a semialgebra. Let µ : S ! [0, 1] be defined by<br />
(7.30)<br />
µ(A1 ⇥···⇥An ⇥ M ⇥···)=µn(A1 ⇥···⇥An) (7.31)<br />
The consistency condition for µn shows that this gives the same number regardless<br />
of the representation of the set A1 ⇥···⇥An ⇥ M ⇥···. Let S be the set of all<br />
finite disjoint unions of elements from S . By Lemma 7.14, S is an algebra; we<br />
extend µ to a set function ¯µ : S ! [0, 1] by additivity. A similar argument to that<br />
used in the proof of Theorem 7.15 now shows that ¯µ is well-defined and finitely<br />
additive.<br />
It remains to show that ¯µ is countably additive. To that end, let A1, A2, ···2S be<br />
disjoint with A = S<br />
n 1 An 2 S . Define DN = S<br />
n>N An and note that DN 2 S as<br />
well. Then A1,...,An, Bn are disjoint and so finite additivity of ¯µ implies<br />
¯µ(A) = ¯µ(A1)+···+ ¯µ(An)+ ¯µ(Dn) (7.32)<br />
We thus need to show that ¯µ(Dn) ! 0 as n ! •. Here we use the topology: By<br />
Lemma 7.17 it suffices to show<br />
8e > 0 8B 2 S 9C 2 S : C ⇢ B compact & ¯µ(B \ C) < e. (7.33)<br />
But B 2 S is a finite disjoint union of Bi 2 S and so it clearly suffices to prove<br />
this for B 2 S . However, every B 2 S is compact by Lemma 7.18 and so we<br />
can take C = B above and the regularity property holds trivially. By Lemma 7.17,<br />
we have ¯µ(Dn) ! 0 and ¯µ is thus countably additive on S . The Carathéodory<br />
extension theorem now implies the existence of a unique extension of ¯µ to s(S )=<br />
B ⌦N .<br />
The proof in the textbook bundles Lemma 7.17 with the proof of Tychonoff’s theorem.<br />
We prefer to separate these to highlight the role of Borel regularity (Lemma 7.17)<br />
which is a property of separate interest. Note also that the consistency condition<br />
holds only for probability; extensions of other measures are not meaningful. In<br />
fact, it is known that one cannot construct an infinite <strong>product</strong> of Lebesgue measures<br />
on R.
7.5. TWO ZERO-ONE LAWS 13<br />
7.5 Two zero-one laws<br />
We will demonstrate the usefulness of the infinite <strong>product</strong> <strong>spaces</strong> by proving two<br />
Zero-One Laws. We demonstrate these on an example of percolation.<br />
Example 7.19 Percolation : Let p 2 [0, 1] and let B denote the set of edges of Z d .<br />
(We regard the latter as a graph with vertices at points of R d with integer coordinates<br />
and edges between any two points with Euclidean distance one.) Let (h b) b2B<br />
be i.i.d. Bernoulli(p). We may view each h =(h b) as a subgraph of (Z d , B)—with<br />
edge set {b 2 B : h b = 1}. The question is: Does this graph have an infinite connected<br />
component? We define<br />
E• = {h : h has an infinite connected component}. (7.34)<br />
We refer to edges with h b = 1 as occupied and those with h b = 0 as vacant. To set<br />
the notation, we will denote by Pp the law of h with parameter p.<br />
First we have to check that E• is an event:<br />
Lemma 7.20 E• is measurable (w.r.t. the <strong>product</strong> s-algebra on {0, 1} B ).<br />
Proof. Let F denote the <strong>product</strong> s-algebra on {0, 1} B . We have to show E• 2 F .<br />
For each n 1 and each x 2 Z d , consider the event<br />
En(x) = x connected to (x +[ n, n] d ) c in h<br />
(7.35)<br />
that x is connected by a path of occupied edges to the boundary of a box of side 2n<br />
centered at x. Since En(x) depends only on a finite number of hb’s, it is measurable.<br />
But<br />
E• = \ [<br />
En(x) (7.36)<br />
and so E• 2 F as well.<br />
n 1 x2Zd Next we observe that E• does not depend on the status on any given finite number<br />
of edges. Indeed, if an edge is occupied, making it vacant may increase the number<br />
of infinite component but it won’t destroy them while. Similarly, making a vacant<br />
edge occupied may connect two infinite components together but if there is none,<br />
it will not create one. In a more technical language, E• is a tail event according to<br />
the following definition:<br />
Definition 7.21 Let X1, X2,... be random variables. Then T = T<br />
n 1 s(Xn, Xn+1,...)<br />
is the tail s-algebra and events from T are called tail events.<br />
Here is our first zero-one law:<br />
Theorem 7.22 [Komogorov’s Zero-One Law] Let X1, X2,... be independent and let<br />
T = T<br />
n 1 s(Xn, Xn+1,...) be the tail s-algebra. Then<br />
8A 2 T : P(A) 2{0, 1}. (7.37)
14 CHAPTER 7. INFINITE PRODUCT SPACES<br />
Figure 7.2: The infinite connected component in 40⇥40 box centered at the origin<br />
for percolation with parameter p = 0.55. The finite components are not depicted.<br />
Proof. We will show that T is independent of F . First pick n 1 and note<br />
A 2 s(X1,...,Xn) & B 2 s(Xn+1, Xn+2,...) ) A, B independent (7.38)<br />
But T ⇢ s(Xn+1, Xn+2,...) and so<br />
A 2 s(X1,...,Xn) & B 2 T ) A, B independent (7.39)<br />
This now holds for all n 1 and so<br />
A 2 [<br />
s(X1,...,Xn) & B 2 T ) A, B independent (7.40)<br />
n 1<br />
But S<br />
n 1 s(X1,...,Xn) is a p-system and F = s( S<br />
n 1 s(X1,...,Xn)) and so Theorem<br />
4.6 implies<br />
A 2 F & B 2 T ) A, B independent (7.41)<br />
Let now A 2 T . Then A is independent of itself which yields<br />
This permits only P(A) =0 or P(A) =1.<br />
Here is a consequence of this law for percolation:<br />
P(A) =P(A \ A) =P(A) 2 . (7.42)
7.5. TWO ZERO-ONE LAWS 15<br />
Corollary 7.23 For each d 1 there exists pc 2 [0, 1] such that<br />
(<br />
0, if p < pc,<br />
Pp(E•) =<br />
1, if p > pc,<br />
At p = pc we have Ppc(E•) 2{0, 1}.<br />
(7.43)<br />
Proof. We have already shown that E• 2 T and so Pp(E•) 2{0, 1} for all p. It<br />
is also clear that the larger p the more edges there are and so Pp(E•) should be<br />
increasing. However, to prove this we have to work a bit.<br />
The key idea is that we can actually realize percolation for all p’s on the same probability<br />
space. Consider i.i.d. random variables U =(U b) b2B whose law is uniform<br />
on [0, 1] and define<br />
h (p)<br />
b = 1 {U bapplep}. (7.44)<br />
Clearly, h (p) =(h (p)<br />
b ) are i.i.d. Bernoulli(p) so they realize a sample of percolation at<br />
parameter p. Now let E•(p) ={U : h (p) contains infinite connected component}.<br />
Since p 7! h (p)<br />
b increases, if E•(p) occurs, then so does E•(p 0 ) for p0 > p. In other<br />
words<br />
p 0 > p ) E•(p) ⇢ E•(p 0 ) (7.45)<br />
This implies<br />
Pp(E•) =P E•(p) apple P E•(p 0 ) = P p 0(E•), (7.46)<br />
i.e., p 7! Pp(E•) is non-decreasing. Since it takes values zero or one, there exists a<br />
unique point where Pp(E•) jumps from zero to one. This defines pc.<br />
Let us remark what really happens: It is known that<br />
(<br />
= 1, d = 1,<br />
pc<br />
2 (0, 1), d 2.<br />
(7.47)<br />
At d = 2 it is known that pc = 2 (Kesten’s Theorem) but the values for d 3<br />
are not known explicitly (and presumably are not of any special form). Concerning<br />
Ppc(E•), it is widely believed that this probability is zero—there is no infinite<br />
cluster at the critical point—but the proof exists only in d = 2 and d 19.<br />
Our next object of interest is the random variable:<br />
N = number of infinite connected components in h (7.48)<br />
The values N can take are {0, 1, . . . }[{•}. The random variable N definitely<br />
depends on any finite number of edges and so N is not T -measurable. (The reason<br />
why we care is if it were tail measurable then it would have to be constant a.s.)<br />
However, N is clearly translation invariant in the following sense:<br />
Definition 7.24 Let tx : {0, 1} B !{0, 1} B be the map defined by<br />
(txh) b = h b+x<br />
(7.49)<br />
We call tx the translation by x. An event A is translation invariant if t 1<br />
x (A) =A for<br />
all x 2 Z d . A random variable N is translation invariant if N tx = N.
16 CHAPTER 7. INFINITE PRODUCT SPACES<br />
First we note that the measure Pp is translation invariant:<br />
Lemma 7.25 Let Pp be the law of Bernoulli(p) on B. Then Pp is translation invariant,<br />
i.e., P t 1<br />
x = Pp for all x 2 Z d .<br />
Proof. This is verified similarly as translation invariance of the Lebesgue measure.<br />
First, a direct calculation shows that Pp(t 1<br />
x (A)) = Pp(A) for all cylinder events.<br />
But the class of sets in F satisfying this identity is a l-system, and since cylinder<br />
sets form a p-system, we have invariance of Pp on the s-algebra generated by<br />
cylinder events. This is all of F .<br />
Next we will prove that every translation invariant event in F is a zero-one event:<br />
Theorem 7.26 [Ergodicity of i.i.d. measures] Let (Xn)n2Z be i.i.d. and let t be the<br />
left-shift acting on the sequence (Xn) via<br />
Then<br />
t(··· , X 1, X0, X1, ···)=(··· , X0, X1, X2, ···). (7.50)<br />
8A 2 s(Xn : n 2 Z) : t 1 (A) =A ) P(A) 2{0, 1}. (7.51)<br />
Proof. We proceed in a way similar to Komogorov’s law; the goal is to show that a<br />
translation invariant set is independent of itself. However, since not all translation<br />
invariant events are tail events, we have use approximation to derive the result.<br />
Let F = s(Xn : n 2 Z), Fn = s(X n, ··· , Xn) and let A 2 F be of positive<br />
measure. The construction of Carathéodory extension of the measure to F shows<br />
that there exists a sequence An 2 Fn such that An approximates A in the sense<br />
P(A4An) !<br />
n!• 0, (7.52)<br />
where A4An denotes symmetric difference. Suppose now that t 1 (A) =A and<br />
recall that P is t-invariant. This means that t 1 (An) approximates t 1 (A) as well<br />
as An,<br />
P (A4t 1 (An) = P(A4An). (7.53)<br />
Iterating this 2n + 1 times we get a set t (2n+1) (An) which is in s(X 3n 1,...,X n 1)<br />
and is thus independent of An. Thus, we may approximate A equally well by two<br />
disjoint sets An and t (2n+1) (An).<br />
Now we will derive the independence of A of itself: The key will be the identity<br />
P(A \ B) apple P(C \ D)+P(A4C)+P(B4D) (7.54)<br />
which we will apply to B = A, C = An and D = t (2n+1) (An). This yields<br />
P(A) =P(A \ A) apple P An \ t (2n+1) (An) + P(A4An)+P A4t (2n+1) (An) .<br />
(7.55)<br />
Since An and t (2n+1) (An) are independent have equal probability, (7.53) gives us<br />
P(A) apple P(An) 2 + 2P(A4An). (7.56)
7.5. TWO ZERO-ONE LAWS 17<br />
However P(An) apple P(A)+P(A4An) and so (7.52) gives<br />
This is only possible if P(A) =0 or P(A) =1.<br />
P(A) apple P(A) 2 . (7.57)<br />
Going back to percolation, the above zero-one law tells us:<br />
Corollary 7.27 One of the events {N = 0}, {N = 1}, and {N = •} has probability<br />
one.<br />
Proof. All of these events are translation invariant (or tail) and so they have probability<br />
zero or one. We have to show that P(N = k) =0 for all k 2{2, 3, . . . }. Again,<br />
if this is not the case then P(N = k) =1 because {N = k} is translation invariant;<br />
for simplicity let us focus on k = 2. If P(N = 2) =1 then there exists n 1 such<br />
that the event<br />
( )<br />
d<br />
box [ n, n] is connected to infinity by two disjoint paths that are<br />
Cn =<br />
not connected to each other in the complement of the box [ n, n] d<br />
(7.58)<br />
occurs with probability at least 1/ 2. Indeed, {N = 2} ⇢ S<br />
n 1 Cn and Cn is increasing.<br />
However, the event Cn is independent of the edges inside the box [ n, n] d and<br />
so with positive probability, the two disjoint paths get connected inside the box.<br />
This means that N = 1 with positive probability on the event {N = 2}\Cn, i.e.,<br />
P(N = 1) > 0. This contradicts P(N = 2) =1.<br />
Let us conclude by stating what really happens: we have P(N = •) =0 so we<br />
either have no infinite cluster or one infinite cluster a.s. The most beautiful version<br />
of this argument comes in a theorem of Burton and Keane which combines rather<br />
soft arguments not dissimilar to what we have seen throughout this section.