Bhattacharya R., Majumdar M. Random Dynamical Systems: Theory and Applications
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A2. Infinite Products of Metric Spaces 423
Theorem A.6 If {K
θ
} is a collection of compact subsets of a metric
space (S, d) such that the intersection of every finite subfamily of {K
θ
}
is nonempty, then
θ
K
θ
is nonempty.
Proof. For proofs of Theorems A.4–A.6, as well as the Bolzano–
Weirstrass theorem, refer to Dieudonn´e (1960, pp. 56 (3.16.1), 60
(3.17.6), 61 (3.17.9)) or Billingsley (1968, pp. 217, 218).
Corollary A.1 If {K
n
}is a sequence of nonempty compact sets such that
K
n
⊃ K
n+1
(n = 1, 2,...),
then
∞
n=1
K
n
is nonempty.
A2. INFINITE PRODUCTS OF METRIC SPACES AND
THE DIAGONALIZATION ARGUMENT
To begin with, let R
∞
be the space of sequences x = (x
1
, x
2
,...)of
real numbers. If d
0
(a, b) =
|
a − b
|
/(1 +
|
a − b
|
), then d
0
is a metric on
the line
R
1
equivalent to the ordinary metric
|
a − b
|
. The line (R
1
)is
complete under d
0
. It follows that, if d(x, y) =
∞
k=1
d
0
(x
k
, y
k
)2
−k
, then
d is a metric on
R
∞
.If
N
k,ε
(x) ={y :
|
y
i
− x
i
|
<ε, i = 1,...,k}, (A.2)
then N
k,ε
(x) is open in the sense of the metric d. Moreover, since
d(x, y) <ε2
−k
/(1 + ε) implies y ∈ N
k,ε
(x), which in turn implies
d(x, y) <ε+2
−k
, the sets (A.2) form a base for the topology given
by d.
This topology is the product topology, or the topology of coordinate-
wise convergence: for a sequence x(n)in
R
∞
, lim
n
x(n) = x if and only
if lim
n
x
k
(n) = x
k
, for each k.
The space
R
∞
is separable; one countable, dense subset consists of
those points with coordinates that are all rationals and that, with only
finitely many exceptions, vanish.
Suppose {x(n)} is a Cauchy sequence in
R
∞
. Since
d
0
(x
k
(m), x
k
(n)) ≤ 2
k
d(x(m), x(n)),
it follows easily that, for each k, {x
k
(1), x
k
(2),...} is a Cauchy sequence
on the line with the usual metric, so that the limit x
k
= lim
n
x
k
(n) exists.
If x = (x
1
, x
2
,...), then x(n) converges to x in the sense of R
∞
. Thus
R
∞
is complete.
424 Appendix
Theorem A.7 A subset A of R
∞
has compact closure if and only if the
set {x
k
: x ∈ A is, for each k, a bounded set on the line.
Proof. It is easy to show that the stated condition is necessary for com-
pactness. We prove sufficiency by the “diagonalization” argument. Given
a sequence {x(n)} in A, we may choose a sequence of subsequences
x(n
11
), x(n
12
), x(n
13
),...
x(n
21
), x(n
22
), x(n
23
),...
......................................
......................................
(A.3)
in the following way. The first row of (A.3) is a subsequence of {x(n)},so
chosen that x
1
= lim
i
x
1
(n
1i
) exists; there is such a subsequence because
{x
1
: x ∈ A} is a bounded set of real numbers. The second row of (A.3)
is a subsequence of the first row, so chosen that x
2
= lim
i
x
2
(n
2i
) exists;
there is such a subsequence because {x
2
: x ∈ A} is bounded.
We continue in this way; row k is a subsequence of row k − 1, and
x
k
= lim
i
x
k
(n
ki
) exists. Let x be the point of R
∞
with coordinates x
k
.If
n
i
= n
ii
, then {x(n
i
)} is a subsequence of {x(n)}. For each k, moreover,
x(n
k
), x(n
k+1
),...all lie in the kth row of (A.3), so that lim
i
x
k
(n
i
) = x
k
.
Thus, lim
i
x(n
i
) = x, and it follows that
¯
A is compact.
Theorem A.8 Every separable metric space S is homeomorphic to a
subset of
R
∞
.
Proof. An adaptation of the proof of Lemma C11.1, Chapter 2, is left
as an exercise.
Example A.1 (The space C[0, 1]) Let C = C[0, 1] be the space of
continuous, real-valued functions on the unit interval [0, 1] with the
uniform metric. The distance between two elements x = x(t) and y =
y(t)ofC is
d(x, y) = sup
t∈[0,1]
|
x(t) − y(t)
|
;
it is easy to check that d is a metric. Convergence in the topology is
uniform pointwise convergence of (continuous) functions.
The space C is separable; one countable, dense set consists of the
(polygonal) functions that are linear on each subinterval [(i − 1)/k,
A3. Measurability 425
i/ k], i = 1,...,k, for some integer k, and assume rational values at
the points i/k, i = 0, 1,...,k.
If {x
n
} is a Cauchy sequence in C, then, for each value of t, {x
n
(t)} is
a Cauchy sequence on the line and hence has a limit x(t). It is easy to
show that the convergence x
n
(t) → x(t ) is uniform in t, so that x lies in
C and is the limit in C of {x
n
}. Thus C is complete.
Example A.2 Let (S, d) be a metric space and C(S) the set of all real-
valued bounded continuous functions on S. The distance between two
elements x = x(t) and y = y(t)ofC is defined as
d(x, y) = sup
t∈S
|x(t) − y(t)|.
The separability of C[0, 1] (Example A.1) is a special case of the
following:
Theorem A.9 C(S) is separable if and only if S is compact.
Proof. See Dieudonn´e (1960, p. 134), Taylor (1985, p. 175, Problem 4
with hints), Kuller (1969, p. 120, Theorem 4.6.5), and Kelly (1955, p. 245,
Problem S(d)).
A3. MEASURABILITY
A sigmafield F of subsets of a nonempty set is a family of subsets of
which contains φ (the empty set) and and is closed under the operations
of complementation, countable union, and countable intersection. The
pair (,F) consisting of a set and a sigmafield of subsets of is
called a measurable space. Subsets of that belong to F are called
F-measurable.
Given a class C of subsets , the smallest sigmafield containing C (i.e.,
the intersection of all sigmafields containing C) is called the sigmafield
generated by C.
Let (, F) and (
, F
) be measurable spaces; F[F
] is a sigmafield
of subsets of [
]. A mapping h : →
from into
is said to be
measurable (F, F
) if the inverse image h
−1
M
belongs to F for each M
in F
.Ifh
−1
F
denotes the family {h
−1
M
: M
∈ F
}, this condition is
formally stated as h
−1
F
⊂F. Since {M
: h
−1
M
∈ F} is a sigmafield,
if F
0
is contained in F
and generates it, then h
−1
F
0
⊂ F implies
426 Appendix
h
−1
F
⊂ F. To simplify notation, measurable stands for measurable
(F, F
).
Let (
, F
) be a third measurable space, let j :
→
map
into
, and denote by jh the composition of h and j :(jh)(ω) =
j(h(ω)). It is easy to show that, if h
−1
F
⊂F and j
−1
F
⊂ F
, then
( jh)
−1
F
⊂ F. Hence, the composition of two (or more) measurable
maps is measurable.
Let S be a metric space. The Borel sigmafield of S, denoted by B(S)
(with abandon, by S, to simplify notation), is the sigmafield generated
by the open sets of S. Since every closed set is the complement of an
open set, S is also the smallest sigmafield of subsets of S which contains
all closed subsets of S.
If = S and
= S
are metric spaces, h is continuous when h
−1
G
is open in S for each open G
in S
. Let S and S
be the sigmafield of
Borel sets in S and S
.Ifh is continuous, then, since h
−1
G
∈ S for G
open in S
, and since the open sets in S
generate S
, h
−1
S
⊂ S, so that
h is measurable.
A3.1. Subspaces
Let S be a metric space, and let S be its sigmafield of Borel sets. A subset
S
0
(not necessarily in S) is a metric space in the relative topology. Let
S
0
be the sigmafield of Borel sets in S
0
. We shall prove
S
0
={S
0
∩ A: A ∈ S}. (A.4)
Proof. If h(x) = x for x ∈ S
0
, then h is a continuous mapping from S
0
to S and hence h
−1
S ⊂ S
0
, so that S
0
∩ A ∈ S
0
if A ∈ S. Since {S
0
∩ A:
A ∈ S} is a sigmafield in S
0
and contains all sets S
0
∩ G with G open in
S, that is, all open sets in S
0
, (A.4) follows.
If S
0
lies in S, (A.4) becomes
S
0
={A : A ⊂ S
0
, A ∈ S}. (A.5)
A3.2. Product Spaces: Separability Once Again
Let S
and S
be metric spaces with metrics d
and d
and sigmafields
S
and S
of Borel sets. The rectangles
A
× A
(A.6)
A3. Measurability 427
with A
open in S
and A
open in S
are a basis for the product topology
in S = S
× S
. This topology may also be described as the one under
which (x
n
, x
n
) → (x
, x
) if and only if x
n
→ x
and x
n
→ x
. Finally,
the topology may be specified by various metrics, for example,
d((x
, x
), (y
, y
)) =
/
[d
(x
, y
)]
2
+ [d
(x
, y
)]
2
(A.7)
and
d((x
, x
), (y
, y
)) = max{d
(x
, y
), d
(x
, y
)}. (A.8)
Let the product sigmafield S
⊗ S
be the sigmafield generated by the
measurable rectangles (sets (A
× A
) with A
∈ S
and A
∈ S
), and
let S be the sigmafield of Borel sets in S for the product topology. One
can show (see Billingsley 1968, pp. 224, 225)
S
⊗ S
⊂ S. (A.9)
Now S is separable if and only if S
and S
are both separable. Let us
show that, if S is separable, then
S
⊗ S
= S. (A.10)
Proof. In view of (A.9) it remains to prove that if G is open in S, then
G ∈ S
⊗ S
. But G is a union of rectangles of the form A
× A
with
A
open in S
. A
open in S
(so that A
× A
∈ S
⊗ S
), and if S is
separable G is a countable such union.
It should be stressed that without separability (A.10) may fail.
Let (S, d) be a separable metric space, and S its Borel sigmafield.
Consider the (Kolmogorov) product sigmafield S
⊗∞
. Endow S
∞
with
the product topology, and the metric
d
∞
(x, y) =
∞
n=0
1
2
n
(d(x
m
, y
n
)1) ∀(x, y) ∈ S
∞
.
For the next two exercises, see Parthasarathy (1967, pp. 5–6).
Exercise A.1
(a) Show that d
∞
is a metric on S
∞
.
(b) Show that (S
∞
, d
∞
) is separable if (S, d) is separable.
428 Appendix
Exercise A.2 Prove that the Borel sigmafield of (S
∞
, d
∞
) is the same
as the product (Kolmogorov) sigmafield S
⊗∞
, where S is the Borel
sigmafield of S). [Hint: One needs to prove that the class O
1
of open sets
of S
∞
under the product topology is the same as the class O
2
of open
sets of S
∞
under the metric d
∞
.]
A3.3. The Support of a Measure
Let S be a separable metric space and P a probability measure on B(S).
Theorem A.10 There is a unique closed set supp(P) satisfying
(i) P[supp(P)] = 1.
(ii) If D is any closed set such that P(D) = 1, then supp(P) ⊂ D.
(iii) supp(P) is the set of all points s ∈ S having the property that
P(U) > 0 for each open set U containing s.
Proof. Let U ={U : U open, P(U ) = 0}. Since S is separable, there are
countably many open sets U
1
, U
2
,...such that
>
n
U
n
=
>
{U: U ∈ U}.
Write U
p
≡
>
n
U
n
, and supp(P) = S\U
p
. Since P(U
p
) = P(
>
n
U
n
) ≤
n
P(U
n
) = 0, P[supp(P)] = 1.
If D is any closed set with P(D) = 1, then S\D ∈ U; hence, S\D ⊂
U
p
, i.e., supp(P) ⊂ D.
To prove (iii), let s ∈ U
p
= S\supp(P). U
p
is an open set containing
s and P(U
p
) = 0. On the other hand, if s ∈ supp(P), and U is an open
set containing s, it must be that P(U) > 0: otherwise U ⊂ U
p
,bythe
definition of U
p
. The uniqueness of supp(P) is obvious.
We refer to the set supp(P)asthesupport of P.
A3.4. Change of Variable
If P is a probability measure on (, F), and if h
−1
F
⊂ F, Ph
−1
denotes
the probability measure on (
, F
) defined by (Ph
−1
)(M
) = P(h
−1
M
)
for M
∈ F
.If f is a real function on
, measurable F
, then the real
function fhon is measurable F.
A3. Measurability 429
Under these circumstances, f is integrable with respect to Ph
−1
if
and only if f h is intergrable with respect to P, in which case we have
"
h
−1
M
f (h(ω))P(dω) =
"
M
f (ω
)Ph
−1
(dω
) (A.11)
for each M
in F
.
Note that (A.11) holds for indicator functions F = 1
A
(A
∈ F
) and,
therefore, for simple functions f =
n
1
c
i
1
A
i
(A
i
∈F
are pairwise dis-
joint). The general equality (A.11) now follows on approximating f by
simple functions.
If X takes into a metric space S and X
−1
S
⊂ S, so that X is a ran-
dom element of S, we generally write E{ f (X)}to denote the expectation
of f (X ) in place of
!
f (X(ω))P(dω). If P = PX
−1
is the distribution
of X in S, (A.11) implies
E{ f (X )}=
"
S
f (x)P(dx). (A.12)
Let X be a nonnegative and integrable random variable on a probability
space (, F, P).
Then, as proved under Exercise 3.1, Chapter 5,
E{X }=
"
∞
0
P{X ≥ t}dt. (A.13)
(The equation also holds in the extended sense that if one side of (A.13)
is infinite, so is the other.)
Let X be a real-valued random variable and g a nonnegative Borel
function such that E{g(X )} is finite. Assume that g(−x) = g(x), and g
is nondecreasing on [0, ∞). Then for every ε>0
P[|X|≥ε] ≤
E{g(X)}
g(ε)
. (A.14)
Proof. Let A ={|X |≥ε}. Then
E{g(X)}=
"
A
g(X ) dP +
"
A
C
g(X ) dP
≥ g(ε)P(A).
430 Appendix
In particular, with g(x) =|x|
λ
(λ>0), one has the Chebyshev
inequality; if E|X |
λ
is finite for some λ>0, then for every ε>0,
P[|X|≥ε] ≤
E|X |
λ
ε
λ
. (A.14
)
A4. BOREL-CANTELLI LEMMA
Let (, F, P) be a probability space. Here (,F) is a measurable space.
We refer to the elements of F as events, for example, as the “sure
event,” φ as the “impossible event.”
Two immediate consequences of the countable additivity property of
probability measures are the so-called continuity properties:
If A
1
⊂ A
2
⊂··· is a nondecreasing sequence of events in F, then
P
∞
?
i=1
A
i
= lim
n→∞
P(A
n
). (A.15)
To prove this, define B
n
≡ A
n
\A
n−1
for n ≥ 1, A
0
≡ φ and apply
countable additivity to
>
∞
n=1
B
n
=
>
∞
n=1
A
n
.
By considering complements we can also conclude the following:
Let A
1
⊃ A
2
⊃···be a nonincreasing sequence of events in F, then
P
∞
i=1
A
i
= lim
n→∞
P(A
n
). (A.16)
We now come to the powerful Borel-Cantelli lemma that has been
repeatedly invoked.
Lemma A.1 (Borel–Cantelli) Let {A
n
}be any sequence of events in F.
Part 1. If
∞
n=1
P(A
n
) < ∞ then
P(A
n
i.o.):= P
∞
n=1
∞
?
k=n
A
k
= 0.
Part 2. If A
1
, A
2
,... are independent events and if
∞
n=1
P(A
n
)
diverges, then P(A
n
i.o.) = 1.
A5. Convergence 431
Proof. For Part 1 observe that the sequence of events B
n
=
>
∞
k=n
A
k
,
n = 1, 2,..., is a decreasing sequence. Therefore, we have by the
continuity property of P
P
∞
n=1
∞
?
k=n
A
k
= lim
n→∞
P
∞
?
k=n
A
k
lim
n→∞
∞
k=n
P(A
k
) = 0.
For Part 2 note that
P({A
n
i.o.}
c
)=P
∞
?
n=1
∞
k=n
A
c
k
= lim
n→∞
P
∞
k=n
A
c
k
= lim
n→∞
∞
8
k=n
P(A
c
k
).
But
∞
8
k=n
P(A
c
k
) = lim
m→∞
m
8
k=n
(1 − P(A
k
))
lim
m→∞
exp
−
m
k=n
P(A
k
)
= 0.
A5. CONVERGENCE
In what follows, (,F, P) is a probability space and {X
n
, n ≥ 1}, X are
all real-valued random variables on (, F, P).
A sequence {X
n
}is said to converge almost surely to a random variable
X (written X
n
a.s.
−→
X) if and only if there is a set E ∈ F with P(E) = 0
such that for every ω ∈ E
c
, |X
n
(ω) − X(ω)|→0asn →∞.
A weaker notion of convergence is that of convergence in probability.
A sequence {X
n
} is said to converge in probability to a random variable
X (written X
n
P
−→
X) if for every ε>0
P[ω ∈ : |X
n
(ω) − X(ω)|≥ε] → 0
as n →∞.
One can show that
“X
n
a.s
−→
X” implies “X
n
P
−→
X.”
Write S
n
≡
n
k=1
X
k
. We state the weak law of large numbers.
432 Appendix
Theorem A.11 Let X
n
be a sequence of independent, identically dis-
tributed random variables with mean µ and a finite variance σ
2
. Then
S
n
n
P
−→
µ.
Proof. Use Chebyshev’s inequality.
As an application of the Borel-Cantelli lemma, we first have the fol-
lowing version of the strong law of large numbers:
Theorem A.12 Let X
1
, X
2
,...be independent random variables having
a common distribution with mean µ and finite fourth moment. Then
S
n
n
a.s.
−→
µ. (A.17)
Proof. (Sketch). By expanding the fourth power, it is not difficult to
calculate that
E
n
i=1
(X
i
− µ)
4
= nE([X
1
− µ]
4
) + 6
n
2
σ
4
≤ Cn
2
. (A.18)
(Here σ
2
= E((X
i
− µ)
2
) can be shown to be finite.) But applying the
Chebyshev inequality we get
P
)
)
)
)
)
n
i=1
(X
i
− µ)
)
)
)
)
)
>εn
≤
Cn
2
(εn)
2
, (A.19)
and the sum on n of the right-hand side is finite. From the Borel-Cantelli
lemma we therefore can conclude that with probability 1, only finitely
many of the events
A
(ε)
n
=
ω:
)
)
)
)
S
n
n
− µ
)
)
)
)
>ε
occur; that is, P(B
ε
) = 0, where B
ε
= lim sup A
(ε)
n
. The sets B
ε
increase
as ε $ 0 to the ω set on which S
n
/n µ. Thus, letting ε $ 0 through
a countable set of values such as k
−1
,wehave
P
ω:
X
1
+···+X
n
n
− µ
0
= P
?
k
B
k
−1
= 0,
which proves Theorem A.11.