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Chapter 7

Infinite product spaces

So far we have talked a lot about infinite sequences of independent random vari-ables but we have never shown how to construct such sequences within the frame-work of probability/measure theory. For that we will need to be able to constructinfinite product measure spaces and extend natural measures to them. Interest-ingly, the construction requires that the measure structure be tied to the naturaltopology on this space. A class of topologies which behave naturally under suchconstruction goes by the name Polish; the measure spaces arising thereby are calledstandard Borel.

7.1 Standard Borel spaces

Definition 7.1 Recall that a metric space (X, d) is a set with a symmetric functiond : X ⇥ X ! [0, •) such that d(x, y) = 0 is equivalent to x = y and the triangle in-equality d(x, y) d(x, z) + d(z, y) holds for all x, y, z 2 X.

(1) We say that a metric space (X, d) is Polish if it is separable (there exists a countabledense set) and complete (all Cauchy sequences have limits in X).

(2) A standard Borel space (X, B) is a measure space such that X is a Polish metricspace and B is the s-algebra of Borel sets induced by the Polish topology.

(3) A map f : X ! Y between topological spaces X, Y—of which metric spaces are examples—is called Borel isomorphism if f is bijective and both f and f�1 are Borel measurable.

The following are examples of Polish (and hence standard Borel) spaces:

(1) {1, . . . , n} with n 2 N or N itself; all equipped with discrete topology.

(2) [0, 1] with d(x, y) = |x� y|.

(3) Cantor space {0, 1}N with product (discrete) topology.

(4) Baire space NN with product (discrete) topology.

1

2 CHAPTER 7. INFINITE PRODUCT SPACES

(5) Hilbert cube [0, 1]N with product (Euclidean) topology.

(6) Rn with Euclidean metric.

(7) RN with product Euclidean topology.

(8) Lp(W, F , µ) with 1 p < • provided F is countably generated and µ iss-finite. (This includes the `p spaces with p < •.)

(9) `• fails to be Polish but its subspace of converging sequences

c0 =�

x = (xn) 2 `•(N) : xn ! 0

(7.1)

with supremum metric is Polish.

(10) C(X)—the set of all continuous functions on a compact metric space X is Polishin the topology generated by the supremum norm.

(11) L(X, Y)—the space of bounded linear operators T : X ! Y where X and Y areseparable Banach spaces is Polish in the strong topology (the norm topologygenerally fails this property). The metric is defined by

d(T1, T2) = Ân�1

2�n��T1( fn)� T2( fn)��

Y (7.2)

where ( fn) is a countable dense set in X.

(12) M1(W, F )—the set of all probability measures on a standard Borel space (W, F )in the topology of so called weak convergence. The metric is defined by

d(µ, n) = Ân�1

2�n��Eµ( fn)�En( fn)�� (7.3)

where ( fn) is a countable dense set in C(W)—which, as can be shown, is sepa-rable. Note that, in such cases, this leads to a hierarchy of spaces: measures on(W, F ), measures on measures on (W, F ), etc.

The reason why standard Borel spaces are good for us is that there are essentiallyonly three cases of them that we will ever have to consider:

Theorem 7.2 [Borel isomorphism theorem] Let (X, B(X)) be a standard Borel space.Then there is a Borel isomorphism f : (X, B(X)) ! (Y, B(Y)) onto a Polish space Ywhere

Y =

8><>:{1, . . . , n}, if |Y| = n,N, if Y is countable,[0, 1], otherwise,

(7.4)

and where B(Y) is the Borel s-algebra induced by the corresponding Polish topology.

7.2. PROOF OF BOREL ISOMORPHISM THEOREM 3

7.2 Proof of Borel isomorphism theorem

The proof invokes a lot of nice arguments that have their core in set theory. Webegin by developing a tool for proving equivalence of measure spaces:

Theorem 7.3 [Cantor-Schroder-Bernstein] Let A ⇢ X and B ⇢ Y be sets such thatthere exist bijections f : X ! B and g : Y ! A. Then there exists a bijection h : X ! Y.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 -measurable and f�1, g are G /F -measurable, then h can be taken F /G -measurable.

Proof. Let C = (g � f )(X) ⇢ A and note that the sets

{(g � f )n(X \ A)}n�0, {(g � f )n(A \ C)}n�0,\

n�0(g � f )n(X) (7.5)

form a disjoint partition of X. We define

h(x) =

8><>:f (x), if x 2 S

n�0(g � f )n(X \ A),g�1(x), if x 2 S

n�0(g � f )n(A \ C),g�1(x), if x 2 T

n�0(g � f )n(X).(7.6)

(Note that we could have also defined h(x) to be f (x) in the last case because g �f maps the intersection onto itself in one-to-one fashion.) Since g : Y ! A is abijection, we need to show that g � h : X ! A is a bijection. In the latter two cases g �

f

g

X

A

Y

B

Figure 7.1: The setting of Cantor-Schroder-Bernstein Theorem: f maps X injec-tively onto B ⇢ Y and g maps Y injectively onto A ⇢ X. The shaded oval inside Aindicates the result of infinite iteration of g � f on X, i.e., the set

Tn�1(g � f )n(X).

This set is invariant under g � f which, in fact, acts bijectively on it.


h is an identity, while in the first case we note that g � h is the bijection g � h : (g �f )n(A \ C) ! (g � f )n+1(A \ C). Hence g � h maps X bijectively onto⇣ [

n�1(g � f )n(X \ A)

⌘[✓⇣ [

n�0(g � f )n(A \ C)

⌘[

\n�0

(g � f )n(X)◆

= A. (7.7)

Since all of these maps are measurable, so is the above partition of A. It followsthat h is measurable too.

The crux of the above theorem is that we only need to demonstrate the existenceof Borel isomorphism(s) onto subsets of the relevant spaces. We will eventually useopen, continuous bijections, i.e., homeomorphisms. Here it is important to character-ize the image of a Polish space under a homeomorphism. Recall that a Gd is a classof sets in a topological space that are countable intersections of open sets.

Lemma 7.4 Let X and Y be Polish and let g : X ! B ⇢ Y be a homeomorphism—i.e., abijection that is open and continuous. Then B is a Gd subset of Y.

Proof. Let f = g�1 : B ! X. Define the oscilation of f by

osc f (y) = inf�

diamX( f (U)) : U 3 y open

. (7.8)

(This is defined at all y 2 Y.) Suppose y 2 ∂B is such that osc f (y) = 0. Thenfor any yn ! y, yn 2 B, xn = f (yn) converges to x = limn!• f (yn). Since X iscomplete, we have x 2 X and so y = limn!• yn = limn!• g(xn) = g(x). It followsthat y 2 B, i.e.,

B = B \�

y 2 Y : osc f (y) = 0

. (7.9)

But closed sets are Gd in metric spaces and {osc f = 0} is Gd because

{osc f = 0} =\

n�1{osc f < 1/n} (7.10)

and because osc f is upper semicontinuous, non-negative and so continuous on {osc f =0}. Hence B is also a Gd set.

The actual use of Theorem 7.3 is enabled by the following observation:

Lemma 7.5 The spaces [0, 1], C = [0, 1]N, N = NN and [0, 1]N, equipped with theirnatural Polish topologies, are Borel isomorphic.

Proof. This results from the string of open, continuous (and thus Borel) injections:

[0, 1] ,! C ,! N ,! [0, 1]N ,! [0, 1] (7.11)

The first map is the injection defined by writing each number in [0, 1] in dyadicform. The second map is a simple inclusion. The third map is induced by thecontinued fraction expansion: if x 2 [0, 1] \Q, there exists a unique sequence (an) 2{0, 1, . . . }N such that

x =1

a1 +1

a2 + 1a3+...

(7.12)

7.2. PROOF OF BOREL ISOMORPHISM THEOREM 5

and noting that, since an = b 1x�an�1

c, if x is irrational two distinct sequences willlead to two distinct numbers. The fourth injection is a diagonal map: Write anelement x = (x1, x2, . . . ) 2 [0, 1]N by means of a matrix am,n 2 {0, 1} such that

xm = Ân�1

am,n

2n , m � 1. (7.13)

Now consider the diagonal map

f(x) = Âk�1

k�1

Âm=1

am,m�k

21+···+k�2+m . (7.14)

It is easy to check that this map is continuous, one-to-one and the inverse is contin-uous. By Lemma 7.4, the images of these maps are Borel and so all maps are Borelmeasurable. By Theorem 7.3, all of these spaces are Borel isomorphic.

Recall the well-known fact—which enables the definition of dimension—that [0, 1]and [0, 1]2 are not homeomorphic. However, by the arguments in the above lemma,as standard Borel spaces they are indistinguishable.Having proved the Borel-equivalence of examples 2-4 in our list above, let us nowaddress the embedding of a general Polish space. Here it is convenient to workwith Hilbert cube:

Proposition 7.6 [Universality of Hilbert cube] Every Polish space is homeomorphicto a Gd-subset of [0, 1]N.

Proof. Let (X, d) be a Polish space and let us assume without loss of generalitythat d � 1 (otherwise maps (X, d) ! (X, d

1+d ) by identity map). Let (xn) be acountable dense set in X. Define the map f : X ! [0, 1]N by

x 2 X 7! f (x) =�d(x, xn)

�n�1 2 [0, 1]N. (7.15)

As is easy to check, f is continuous and one-to-one and so we have f�1 : f (X) ! X.We claim that f�1 is also continuous. Indeed, suppose that (zn) is a sequence suchthat f (zn) ! f (z) in f (X). Let e > 0 and find n � 1 such that d(xn, z) < e. Then

d(zm, z) d(zm, xn) + d(xn, z)= f (zm)n + d(z, xn) �!n!•

f (z)n + d(z, xn) < 2e. (7.16)

Hence zm ! z or, in explicit terms, f�1( f (zm)) ! f�1( f (z)), and so f�1 is continu-ous on f (X). Thus f : X ! f (X) is a homeomorphism and so, by Lemma 7.4, f (X)is a Gd subset of [0, 1]N.

The previous proposition gives the desired embedding of X into the Hilbert cubeand, by Lemma 7.5, a Borel isomorphism into the Cantor space. It remains to con-struct an embedding of the Cantor space into a general (uncountable) Polish space.


Theorem 7.7 [Cantor-Bendixon] Let X be a second countable topological space (i.e.,with countable basis of the topology). Then there exists unique subsets P and O with P

perfect (i.e., having no isolated points) and O countable open such that

X = P[O and P\O = ∆. (7.17)

Proof. Let {Un} be a countable basis (of open sets) and let

O =[�

Un : Un countable

. (7.18)

Then O is open and countable while P = X \O is closed. Let x 2 P. If U 3 x is open,then U—being the union of a subcollection from {Un} is necessarily uncountable,because otherwise x would end up in O. This means that P\U \ {x} 6= ∆, i.e., x isnot isolated.If P0 and O0 are two other such sets, then U \P0 6= ∆ implies that U \P0 is uncount-able, for any U open. Hence P0 \O = ∆, i.e., P0 ⇢ P. On the other hand, O0 is count-able open and so it appears in the union defining O. Hence O0 ⇢ O. If P0 [O0 = X,then we must have P0 = P and O0 = O.

The relevant consequence of this general theorem is as follows:

Corollary 7.8 [Cantor space to Polish space] Every uncountable Polish space con-tains a homeomorphic copy of C (and so it has cardinality 2@0).

Proof. Let X be an uncountable Polish space. Every separable metric space is secondcountable, so let X = P\O be as in the previous theorem. Then P—being a closedsubset of X—is Polish too, so we may as well assume that X = P. We will use thestandard “tree-like” construction to identify a copy of C in X.Let (xn) be a countable dense set (of distinct points) in P. As is easily checked, thereexist two functions f0 and f1 from {x1, . . . } to itself such that

max�

d( f0(xn)), xn), d( f1(xn)), xn) 2�n (7.19)

andmin

�d( f0(xn)), xn), d( f1(xn)), xn)

� d

�f0(xn), f1(xn)

�> 0. (7.20)

We now define a collection of balls Bs, indexed by s{0, 1}n, centered at pointsof {x1, x2, . . . } as follows: Set B? to ball of radius 1/2 about x1. If Bs is definedfor all s{0, 1}n, and xs are their centers, let rn be the minimum of f0(xs) and f1(xs)for all s 2 {0, 1}n. Then let Bs0 be the ball centered at f0(xs) of radius rn

4 , andsimilarly let Bs1 be the corresponding ball centered at f1(xs). These balls are dis-joint with Bs0, Bs1 ⇢ Bs and, by induction, the same is true for {Bs : s{0, 1}n} forall n � 1.Now let s = (s1, s2, . . . ) 2 {0, 1}N. By construction, the intersection

Tn�1 B(s1,...,sn)

is non-empty and, by the completeness of X, it contains exactly one point ys.Moreover, if s and s agree in the first n digits and sn+1 = 0 and s0n+1 = 1,then ys 2 B(s1,...,sn,0) and ys 2 B(s1,...,sn,1) and so

rn

2 d(ys, ys) 2�n (7.21)

7.3. PRODUCT MEASURE SPACES 7

Thus the map s 7! yn is one-to-one and continuous. To show that the inversemap is continuous, let us note that if d(ys, ys) < e, then sn = s0n for all n forwhich rn/2 > e. So the smaller e, the larger part of s and s must be the same.Hence, the inverse map is continuous.

Now we are finally ready to put all pieces together:

Proof of Theorem 7.2. If X is finite or countable, then the desired Borel isomorphismis constructed directly. So let us assume that X is uncountable. Then Corollary 7.8says there is a homeomorphism C ,! X while Proposition 7.6 says there exists ahomeomorphism X ,! [0, 1]N. Both of these image into a Gd-set by Lemma 7.4and so the inverse is Borel measurable. Since [0, 1]N is Borel isomorphic to [0, 1]N,we thus we have Borel-embeddings C ,! X and X ,! C . From Theorem 7.3 itfollows that X is Borel isomorphic to C —and thus to any of the spaces listed inLemma 7.5.

7.3 Product measure spaces

Definition 7.9 [General product spaces] Let (Wa, Fa)a2I be any collection of mea-surable spaces. Then W = ⇥a2I Wa can be equipped with the product s-algebra F =N

a2I Fa which is defined by

F = s

✓n⇥

a2IAa : Aa 2 Fa & #{a 2 I : Aa 6= Wa} < •

o◆. (7.22)

If Wa = W0 and Fa = F0 then we write W = WI0 and F = F⌦I

0 .

Next we note that standard Borel spaces behave well under taking countable prod-ucts:

Lemma 7.10 Let I be a finite or countably infinite set and let (Wa, Fa)a2I be a familyof standard Borel spaces. Let (W, F ) be the product measure space defined above. Then(W, F ) is also standard Borel. Moreover, if fa : Wa ! [0, 1] are Borel isomorphisms, thenf = ⇥a2I fa is a Borel isomorphism of W onto [0, 1]I .

Our goal is to show the existence of product measures. We begin with finite prod-ucts. The following is a restatement of Lemma 4.13:

Theorem 7.11 Let (W1, F1, µ1) and (W2, F2, µ2) be probability spaces. Then thereexists a unique probability measure µ1 ⇥ µ2 on F1 ⌦F2 such that for all A1 2 F1,A2 2 F2,

µ1 ⇥ µ2(A1 ⇥ A2) = µ1(A1)µ2(A2). (7.23)

The reason why this is not a trivial application of Caratheodory’s extension theo-rem is the fact that (7.26) defines µ1 ⇥ µ2 only on a structure called semialgebra andnot algebra.

Definition 7.12 A non-empty collection S ⇢ P(W) is called semialgebra if


(1) A, B 2 S implies A \ B 2 S .

(2) If A 2 S , then there exist Ai 2 S disjoint such that Ac =Sn

i=1 Ai.

It is easy to verify that semialgebras on W automatically contain W and ∆.

Lemma 7.13 The following are semialgebras:

(1) S = {(a, b] : �• a b •}

(2) S = {A1 ⇥ · · ·⇥ An : Ai 2 Ai} where Ai are algebras.

(3) S =�⇥

a2IAa : Aa 2 Aa & #{a : Aa 6= Wa} < •

provided Aa’s are algebras.

Proof. Let us focus on (2): Since all Ai are p-systems, S is closed under inter-sections. To show the complementation property in the definition of semialgebra,let D be a collection of all sets of the form B1 ⇥ · · ·⇥ Bn, where Bi is either Ai or Ac

i ,but such that A1 ⇥ · · ·⇥ An is not included in D . Then

(A1 ⇥ · · ·⇥ An)c =[

B2BB.

From here the complementation property follows by noting that D is finite andD ⇢ S .

Next we will show that via finite disjoint unions semialgebras give rise to algebras:

Lemma 7.14 Let S be a semialgebra and let S be the set of finite unions of sets Ai 2 S .Then S is an algebra.

Proof. First we show that S is closed under intersections. Indeed, A is the (finite)union of some Ai 2 S and B is the (finite) union of some Bj 2 S . Therefore,

A \ B =✓[

iAi

◆\✓[

jBj

◆=

[i,j

Ai \ Bj.

Since Ai, Bj 2 S , then Ai \ Bj 2 S and A \ B is the finite union of sets from S .Consequently, A \ B 2 S .Next we will show that S is closed under taking the complement. Let A be asabove. Then

Ac =✓[

iAi

◆c

=\

iAc

i =\

i

[j

Bi,j,

where Bi,j 2 S are such that Aci is the union of Bi,j—see the definition of semial-

gebra. The union on the extreme right is finite and, therefore, it belongs to S . ButS is closed under finite intersections, so Ac 2 S as well.

Now we will learn how (and under what conditions) a set function on a semialge-bra can be extended to a measure on the associated algebra:

7.3. PRODUCT MEASURE SPACES 9

Theorem 7.15 Let S be a semialgebra and let µ : S ! [0, 1] be a set function. Supposethe properties hold:

(1) Finite additivity: A, Ai 2 S , A =Sn

i=1 Ai disjoint ) µ(A) = Âni=1 µ(Ai),

(2) Countable subadditivity: A, Ai 2 S , A =S•

i=1 Ai disjoint ) µ(A) Â•i=1 µ(Ai).

Then there exists a unique extension µ : S ! [0, 1] of µ, which is a measure on S .

Proof. Note that (1) immediately implies that µ(∆) = 0. The proof comes in foursteps:

Definition of µ: For A 2 S , then A =Sn

i=1 Si for some Si 2 S . Then we letµ(A) = Ân

i=1 µ(Si). This definition does not depend on the representation of A;if A =

Smj=1 Ti for some Tj 2 S , then the fact that S is a p-system and finite

additivity of µ ensure that

n

Âi=1

µ(Si) =n

Âi=1

µ⇣ m[

j=1Si \ Tj

⌘=

n

Âi=1

m

Âj=1

µ(Si \ Tj) =m

Âj=1

µ(Tj).

Finite additivity of µ: By the very definition, µ is also finitely additive, because ifAn is given as the disjoint union

Si Si,n, then

Sn An =

Si,n Si,n and thus µ(

Sn An) =

Âi,n µ(Si,n) = Ân µ(An). (All indices are finite.)

Countable subadditivity of µ: We claim that µ is countably subadditive, wheneverthe disjoint union lies in S . Let Ai 2 S and A =

Sk�1 Ak 2 S . Writing Ak =S

nkj<nk+1Sj—note the efficient way to index the Sj’s contributing to Ak—we have

A =S

j�1 Sj where now Sj 2 S . But A 2 S implies that A =Sn

i=1 Ti and thusTi =

Sj�1 Sj \ Ti—all sets again disjoint. Now countable subadditivity of µ gives

us thatµ(Ti) Â

j�1µ(Sj \ Ti)

and thus

µ(A) n

Âi=1

Âj�1

µ(Sj \ Ti) = Âj�1

n

Âi=1

µ(Sj \ Ti) = Âj�1

µ(Sj) = Âk�1

µ(Ak),

so µ is countably subadditive as well.

Countable superadditivity of µ: To prove the opposite inequality we note that Acan be written as the disjoint union Bn [

Snk=1 Ak, where Bn =

Sk>n Ak is also in S

because S is an algebra. Since we already showed that µ is finitely additive, wehave

µ(A) = µ(Bn) +n

Âk=1

µ(Ak) �n

Âk=1

µ(Ak).

Letting n ! •, the opposite inequality follows.


Thus we have a well defined function µ : S ! [0, 1] which is finitely additiveon S and countably additive on disjoint unions that belong to S . These are theproperties required to be a measure on an algebra.

Note that the combination of Lemma 7.13, Lemma 7.14 and the construction ofthe measure µ from Theorem 7.15 provide an alternative proof of the fact thatLebesgue-Stieltjes measures on R are defined by the corresponding “distribution”function.

Proof of Theorem 7.11. The set S = {A1 ⇥ A2 : Ai 2 Fi} is a semialgebra byLemma 7.13 and s(S ) is the product s-algebra F1 ⌦F2. We will show that the setfunction µ : S ! [0, 1] defined by

µ(A1 ⇥ A2) = µ1(A1)µ2(A2), Ai 2 Fi,

is countably additive on S —this implies the desired properties (1-2) in Theorem 7.15.Let A 2 F1 and B 2 F2 and suppose that A⇥ B is the disjoint union

Sj Aj ⇥ Bj,

where Aj 2 F1 and Bj 2 F2, and where j is either finite or countable index. Theproof uses integration: For each x 2 A, let I(x) = {j : Aj 3 x}. Now since {x}⇥Bj ⇢ Aj ⇥ Bj whenever j 2 I(x), the sets {x}⇥ Bj with j 2 I(x) are disjoint and{x}⇥S

j2I(x) Bj = {x}⇥ B. It follows that B is the disjoint union

B =[

j2I(x)Bj

and thus1A(x)µ2(B) = Â

j1Aj(x)µ2(Bj)

for any x 2 W1. Both sides are F1-measurable, and integrating with respect to µ1we get

µ1(A)µ2(B) = Âj

µ1(Aj)µ2(Bj).

This shows that µ(A⇥ B) equals the sum of µ(Aj ⇥ Bj) and so µ is countably addi-tive. Lemma 7.14, Theorem 7.15 and Caratheodory’s extension theorem then guar-antee that µ has a unique extension to F1 ⌦F2.

General finite product spaces are handled by induction using the fact that

nOi=1

Fi =⇣n�1O

i=1Fi

⌘⌦Fn (7.24)

Note that this proves the associativity of the product of measures:

(µ1 ⇥ µ2)⇥ µ3 = µ1 ⇥ (µ2 ⇥ µ3). (7.25)

We leave the details of these calculations implicit.

7.4. KOMOGOROV’S EXTENSION THEOREM 11

7.4 Komogorov’s extension theorem

We now address the corresponding extension of measures to infinite product spaces:

Theorem 7.16 [Kolmogorov’s extension theorem] Let M be a compact metric space,B = B(M) the Borel s-algebra on M and, for each n � 1, let µn be a probability measureon (Mn, B⌦n). We suppose these measures are consistent in the following sense:

8n � 1, 8A1, . . . , An 2 B : µn+1(A1 ⇥ · · ·⇥ An ⇥M) = µn(A1 ⇥ · · ·⇥ An).

Then there exists a unique probability measure µ on (MN, B⌦N) such that

µ(A1 ⇥ · · ·⇥ An ⇥R⇥R . . . ) = µn(A1 ⇥ · · ·⇥ An) (7.26)

holds for any n � 1 and any Borel sets Ai 2 R.

As many extension arguments from measure theory, the proof boils down to prov-ing that if a sequence of sets decreases to an empty set, then the measures decreaseto zero. In our case this will be ensure through the following lemma:

Lemma 7.17 Let W be a metric space and let A be an algebra of subsets of W (which isnot necessarily a s-algebra). Suppose that µ : A ! [0, 1] is a finitely-additive set functionwhich is regular in the following sense: For each set B 2 A and each e > 0, there exists acompact set C 2 A such that

C ⇢ B (7.27)

andµ(B \ C) e. (7.28)

Then for each decreasing sequence Bn 2 A with Bn # ∆ we have µ(Bn) # 0.

Proof. Let Bn be a decreasing sequence of sets with limn!• µ(Bn) = d > 0. We willshow that

Tn Bn 6= ∆. By the assumptions, we can find compact sets Cn 2 A such

that Cn ⇢ Bn and µ(Bn \ Cn) < d2�n�1. Now since Bn � Bn+1, we have

Bn \n[

k=1(Bk \ Ck) ⇢

n\k=1

Ck.

Indeed, if x 2 Bn then x 2 Bk for all k n and then x can only belong to the set onthe left if x 2 Ck for all k n. Using finite additivity of µ, it follows that

µ⇣ n\

k=1Ck

⌘� µ(Bn)�

n

Âk=1

µ(Bk \ Ck) � d� Âk�1

µ(Bk \ Ck) �d

2.

Consequently, since finite additivity implies that µ(∆) = 0, the intersectionTn

k=1 Ckis non-empty for any finite n. But the sets Cn are compact and thus

T•k=1 Ck 6= ∆.

The inclusions Cn ⇢ Bn imply that alsoT

n Bn 6= ∆.

Lemma 7.17 links measure with topology (which is the reason why we often con-sider standard Borel spaces). Next we observe:


Lemma 7.18 Let (M, r) be a compact metric space. Equip MN with the metric

d(x, y) = Ân�1

r(xn, yn)2�n (7.29)

whenever x = (xn) 2 MN and y = (yn) 2 MN. Then (MN, d) is a compact metricspace. Moreover, if C1, . . . , Cn are compact sets in (M, r), then C1 ⇥ · · · ⇥ Cn ⇥ M ⇥M⇥ · · · is compact in (MN, d).

Proof. The first part is a reformulation of Tychonoff’s theorem for product spaces.The second part is checked directly.

Now we are finally ready to prove Kolmogorov’s theorem:

Proof of Theorem 7.16. Consider the set

S =�

A1 ⇥ · · · An ⇥M⇥ · · · : Ai compact in M 8i

(7.30)

Then, as is easy to check, S is a semialgebra. Let µ : S ! [0, 1] be defined by

µ(A1 ⇥ · · ·⇥ An ⇥M⇥ · · · ) = µn(A1 ⇥ · · ·⇥ An) (7.31)

The consistency condition for µn shows that this gives the same number regardlessof the representation of the set A1 ⇥ · · · ⇥ An ⇥M ⇥ · · · . Let S be the set of allfinite disjoint unions of elements from S . By Lemma 7.14, S is an algebra; weextend µ to a set function µ : S ! [0, 1] by additivity. A similar argument to thatused in the proof of Theorem 7.15 now shows that µ is well-defined and finitelyadditive.It remains to show that µ is countably additive. To that end, let A1, A2, · · · 2 S bedisjoint with A =

Sn�1 An 2 S . Define DN =

Sn>N An and note that DN 2 S as

well. Then A1, . . . , An, Bn are disjoint and so finite additivity of µ implies

µ(A) = µ(A1) + · · ·+ µ(An) + µ(Dn) (7.32)

We thus need to show that µ(Dn) ! 0 as n ! •. Here we use the topology: ByLemma 7.17 it suffices to show

8e > 0 8B 2 S 9C 2 S : C ⇢ B compact & µ(B \ C) < e. (7.33)

But B 2 S is a finite disjoint union of Bi 2 S and so it clearly suffices to provethis for B 2 S . However, every B 2 S is compact by Lemma 7.18 and so wecan take C = B above and the regularity property holds trivially. By Lemma 7.17,we have µ(Dn) ! 0 and µ is thus countably additive on S . The Caratheodoryextension theorem now implies the existence of a unique extension of µ to s(S ) =B⌦N.

The proof in the textbook bundles Lemma 7.17 with the proof of Tychonoff’s theo-rem. We prefer to separate these to highlight the role of Borel regularity (Lemma 7.17)which is a property of separate interest. Note also that the consistency conditionholds only for probability; extensions of other measures are not meaningful. Infact, it is known that one cannot construct an infinite product of Lebesgue mea-sures on R.

7.5. TWO ZERO-ONE LAWS 13

7.5 Two zero-one laws

We will demonstrate the usefulness of the infinite product spaces by proving twoZero-One Laws. We demonstrate these on an example of percolation.

Example 7.19 Percolation : Let p 2 [0, 1] and let B denote the set of edges of Zd.(We regard the latter as a graph with vertices at points of Rd with integer coordi-nates and edges between any two points with Euclidean distance one.) Let (hb)b2B

be i.i.d. Bernoulli(p). We may view each h = (hb) as a subgraph of (Zd, B)—withedge set {b 2 B : hb = 1}. The question is: Does this graph have an infinite con-nected component? We define

E• = {h : h has an infinite connected component}. (7.34)

We refer to edges with hb = 1 as occupied and those with hb = 0 as vacant. To setthe notation, we will denote by Pp the law of h with parameter p.

First we have to check that E• is an event:

Lemma 7.20 E• is measurable (w.r.t. the product s-algebra on {0, 1}B).

Proof. Let F denote the product s-algebra on {0, 1}B. We have to show E• 2 F .For each n � 1 and each x 2 Zd, consider the event

En(x) =�

x connected to (x + [�n, n]d)c in h

(7.35)

that x is connected by a path of occupied edges to the boundary of a box of side 2ncentered at x. Since En(x) depends only on a finite number of hb’s, it is measurable.But

E• =\

n�1

[x2Zd

En(x) (7.36)

and so E• 2 F as well.

Next we observe that E• does not depend on the status on any given finite numberof edges. Indeed, if an edge is occupied, making it vacant may increase the numberof infinite component but it won’t destroy them while. Similarly, making a vacantedge occupied may connect two infinite components together but if there is none,it will not create one. In a more technical language, E• is a tail event according tothe following definition:

Definition 7.21 Let X1, X2, . . . be random variables. Then T =T

n�1 s(Xn, Xn+1, . . . )is the tail s-algebra and events from T are called tail events.

Here is our first zero-one law:

Theorem 7.22 [Komogorov’s Zero-One Law] Let X1, X2, . . . be independent and letT =

Tn�1 s(Xn, Xn+1, . . . ) be the tail s-algebra. Then

8A 2 T : P(A) 2 {0, 1}. (7.37)


Figure 7.2: The infinite connected component in 40⇥40 box centered at the originfor percolation with parameter p = 0.55. The finite components are not depicted.

Proof. We will show that T is independent of F . First pick n � 1 and note

A 2 s(X1, . . . , Xn) & B 2 s(Xn+1, Xn+2, . . . ) ) A, B independent (7.38)

But T ⇢ s(Xn+1, Xn+2, . . . ) and so

A 2 s(X1, . . . , Xn) & B 2 T ) A, B independent (7.39)

This now holds for all n � 1 and so

A 2[

n�1s(X1, . . . , Xn) & B 2 T ) A, B independent (7.40)

ButS

n�1 s(X1, . . . , Xn) is a p-system and F = s(S

n�1 s(X1, . . . , Xn)) and so The-orem 4.6 implies

A 2 F & B 2 T ) A, B independent (7.41)

Let now A 2 T . Then A is independent of itself which yields

P(A) = P(A \ A) = P(A)2. (7.42)

This permits only P(A) = 0 or P(A) = 1.

Here is a consequence of this law for percolation:


Corollary 7.23 For each d � 1 there exists pc 2 [0, 1] such that

Pp(E•) =

(0, if p < pc,1, if p > pc,

(7.43)

At p = pc we have Ppc(E•) 2 {0, 1}.

Proof. We have already shown that E• 2 T and so Pp(E•) 2 {0, 1} for all p. Itis also clear that the larger p the more edges there are and so Pp(E•) should beincreasing. However, to prove this we have to work a bit.The key idea is that we can actually realize percolation for all p’s on the same prob-ability space. Consider i.i.d. random variables U = (Ub)b2B whose law is uniformon [0, 1] and define

h(p)b = 1{Ubp}. (7.44)

Clearly, h(p) = (h(p)b ) are i.i.d. Bernoulli(p) so they realize a sample of percolation at

parameter p. Now let E•(p) = {U : h(p) contains infinite connected component}.Since p 7! h

(p)b increases, if E•(p) occurs, then so does E•(p0) for p0 > p. In other

wordsp0 > p ) E•(p) ⇢ E•(p0) (7.45)

This impliesPp(E•) = P

�E•(p)

� P

�E•(p0)

�= Pp0(E•), (7.46)

i.e., p 7! Pp(E•) is non-decreasing. Since it takes values zero or one, there exists aunique point where Pp(E•) jumps from zero to one. This defines pc.

Let us remark what really happens: It is known that

pc

(= 1, d = 1,2 (0, 1), d � 2.

(7.47)

At d = 2 it is known that pc = 2 (Kesten’s Theorem) but the values for d � 3are not known explicitly (and presumably are not of any special form). Concern-ing Ppc(E•), it is widely believed that this probability is zero—there is no infinitecluster at the critical point—but the proof exists only in d = 2 and d � 19.Our next object of interest is the random variable:

N = number of infinite connected components in h (7.48)

The values N can take are {0, 1, . . . } [ {•}. The random variable N definitelydepends on any finite number of edges and so N is not T -measurable. (The reasonwhy we care is if it were tail measurable then it would have to be constant a.s.)However, N is clearly translation invariant in the following sense:

Definition 7.24 Let tx : {0, 1}B ! {0, 1}B be the map defined by

(txh)b = hb+x (7.49)

We call tx the translation by x. An event A is translation invariant if t�1x (A) = A for

all x 2 Zd. A random variable N is translation invariant if N � tx = N.


First we note that the measure Pp is translation invariant:

Lemma 7.25 Let Pp be the law of Bernoulli(p) on B. Then Pp is translation invariant,i.e., P � t�1

x = Pp for all x 2 Zd.

Proof. This is verified similarly as translation invariance of the Lebesgue measure.First, a direct calculation shows that Pp(t�1

x (A)) = Pp(A) for all cylinder events.But the class of sets in F satisfying this identity is a l-system, and since cylin-der sets form a p-system, we have invariance of Pp on the s-algebra generated bycylinder events. This is all of F .

Next we will prove that every translation invariant event in F is a zero-one event:

Theorem 7.26 [Ergodicity of i.i.d. measures] Let (Xn)n2Z be i.i.d. and let t be theleft-shift acting on the sequence (Xn) via

t(· · · , X�1, X0, X1, · · · ) = (· · · , X0, X1, X2, · · · ). (7.50)

Then8A 2 s(Xn : n 2 Z) : t�1(A) = A ) P(A) 2 {0, 1}. (7.51)

Proof. We proceed in a way similar to Komogorov’s law; the goal is to show that atranslation invariant set is independent of itself. However, since not all translationinvariant events are tail events, we have use approximation to derive the result.Let F = s(Xn : n 2 Z), Fn = s(X�n, · · · , Xn) and let A 2 F be of positivemeasure. The construction of Caratheodory extension of the measure to F showsthat there exists a sequence An 2 Fn such that An approximates A in the sense

P(A4An) �!n!•0, (7.52)

where A4An denotes symmetric difference. Suppose now that t�1(A) = A andrecall that P is t-invariant. This means that t�1(An) approximates t�1(A) as wellas An,

P�(A4t�1(An)

�= P(A4An). (7.53)

Iterating this 2n + 1 times we get a set t�(2n+1)(An) which is in s(X�3n�1, . . . , X�n�1)and is thus independent of An. Thus, we may approximate A equally well by twodisjoint sets An and t�(2n+1)(An).Now we will derive the independence of A of itself: The key will be the identity

P(A \ B) P(C \ D) + P(A4C) + P(B4D) (7.54)

which we will apply to B = A, C = An and D = t�(2n+1)(An). This yields

P(A) = P(A \ A) P�

An \ t�(2n+1)(An)�+ P(A4An) + P

�A4t�(2n+1)(An)

�.

(7.55)Since An and t�(2n+1)(An) are independent have equal probability, (7.53) gives us

P(A) P(An)2 + 2P(A4An). (7.56)


However P(An) P(A) + P(A4An) and so (7.52) gives

P(A) P(A)2. (7.57)

This is only possible if P(A) = 0 or P(A) = 1.

Going back to percolation, the above zero-one law tells us:

Corollary 7.27 One of the events {N = 0}, {N = 1}, and {N = •} has probabilityone.

Proof. All of these events are translation invariant (or tail) and so they have proba-bility zero or one. We have to show that P(N = k) = 0 for all k 2 {2, 3, . . . }. Again,if this is not the case then P(N = k) = 1 because {N = k} is translation invariant;for simplicity let us focus on k = 2. If P(N = 2) = 1 then there exists n � 1 suchthat the event

Cn =

(box [�n, n]d is connected to infinity by two disjoint paths that are

not connected to each other in the complement of the box [�n, n]d

)(7.58)

occurs with probability at least 1/2. Indeed, {N = 2} ⇢ Sn�1 Cn and Cn is increas-

ing. However, the event Cn is independent of the edges inside the box [�n, n]d andso with positive probability, the two disjoint paths get connected inside the box.This means that N = 1 with positive probability on the event {N = 2} \ Cn, i.e.,P(N = 1) > 0. This contradicts P(N = 2) = 1.

Let us conclude by stating what really happens: we have P(N = •) = 0 so weeither have no infinite cluster or one infinite cluster a.s. The most beautiful versionof this argument comes in a theorem of Burton and Keane which combines rathersoft arguments not dissimilar to what we have seen throughout this section.

inﬁnite product spaces - ucla department of · pdf fileinﬁnite product spaces ......

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