Recall Lemma 12.2.3 on Page 1087 which is stated here for convenience.
Lemma 58.1.1Let M be a metric space with the closed balls compact and supposeλ is a measure defined on the Borel sets of M which is finite on compact sets.Then there exists a unique Radon measure, λwhich equals λ on the Borel sets. Inparticular λ must be both inner and outer regular on all Borel sets.
Also important is the following fundamental result which is called the Borel Cantelli
lemma.
Lemma 58.1.2Let
(Ω,ℱ, λ)
be a measure space and let
{A }
i
be a sequence ofmeasurable sets satisfying
∞∑
λ(Ai) < ∞.
i=1
Then letting S denote the set of ω ∈ Ω which are in infinitely many A_{i}, it follows S is ameasurable set and λ
(S)
= 0.
Proof: S = ∩_{k=1}^{∞}∪_{m=k}^{∞}A_{m}. Therefore, S is measurable and also
∞∑
λ(S) ≤ λ (∪∞m=kAm ) ≤ λ (Ak )
m=k
and this converges to 0 as k →∞ because of the convergence of the series.
■
Here is another nice observation.
Proposition 58.1.3Suppose E_{i}is a separable Banach space. Then if B_{i}is a Borelset of E_{i}, it follows∏_{i=1}^{n}B_{i}is a Borel set in∏_{i=1}^{n}E_{i}.
Proof: An easy way to do this is to consider the projection maps.
πix ≡ xi
Then these projection maps are continuous. Hence for U an open set,
− 1 ∏n
πi (U ) ≡ Aj, Aj = Ej if j ⁄= i and Ai = U.
j=1
Thus π_{i}^{−1}
(open)
equals an open set. Let
S ≡{V ⊆ ℝ : π −i1(V) is Borel}
Then S contains all the open sets and is clearly a σ algebra. Therefore, S contains the
Borel sets. Let B_{i} be a Borel set in E_{i}. Then
n
∏ n −1
i=1Bi = ∩i=1πi (Bi),
a finite intersection of Borel sets. ■
Definition 58.1.4A probability spaceis a measure space,
(Ω, ℱ,P )
where P isa measure satisfying P
(Ω )
= 1. A random vector (variable)is a measurablefunction, X : Ω → Z where Z is some topological space. It is often the case that Zwill equal ℝ^{p}.Assume Z is a separable Banach space. Define the following σalgebra.
σ(X) ≡ {X− 1(E) : E is Borel in Z }
Thus σ
(X )
⊆ℱ. For E a Borel set in Z define
( )
λX (E) ≡ P X −1(E ) .
This is called the distributionof the random variable, X. If
∫
|X (ω )|dP < ∞
Ω
then define
∫
E (X) ≡ XdP
Ω
where the integral is defined as the Bochner integral.
Recall the following fundamental result which was proved earlier but which I will give
a short proof of now.
Proposition 58.1.5Let
(Ω,S,μ)
be a measure space and let X : Ω → Z whereZ is a separable Banach space. Then X is strongly measurable if and only ifX^{−1}
− 1 −1 ( −1(----------))
X (D (a,r)) = ∩iX fi B( (fi(a),r) )
= ∩ (f ∘X )− 1 B-(f-(a),r)
i i i
If X is strongly measurable, then it is weakly measurable and so each f_{i}∘ X is a real
(complex) valued measurable function. Hence the expression on the right in the
above is measurable. Now if U is any open set in Z, then it is the countable
union of such closed disks U = ∪_{i}D_{i}. Therefore, X^{−1}
(U )
= ∩_{i}X^{−1}
(Di)
∈S.
It follows that strongly measurable implies inverse images of open sets are in
S.
Conversely, suppose X^{−1}
(U )
∈S for every open U. Then for f ∈ Z^{′}, f ∘ X is real
valued and measurable. Therefore, X is weakly measurable. By the Pettis theorem, it
follows that f ∘ X is strongly measurable. ■
Proposition 58.1.6If X : Ω → Z is measurable, then σ
(X )
equals the smallest σalgebra such that X is measurable with respect to it. Also if X_{i}are random variableshaving values in separable Banach spaces Z_{i}, then σ
(X )
= σ
(X1, ⋅⋅⋅,Xn )
where Xis the vector mapping Ω to∏_{i=1}^{n}Z_{i}and σ
(X1,⋅⋅⋅,Xn)
is the smallest σ algebrasuch that each X_{i}is measurable with respect to it.
Proof: Let G denote the smallest σ algebra such that X is measurable with
respect to this σ algebra. By definition X^{−1}
(open)
∈G. Furthermore, the set of
all E such that X^{−1}
(E )
∈G is a σ algebra. Hence it includes all the Borel
sets. Hence X^{−1}
(Borel)
∈G and so G⊇ σ
(X)
. However, σ
(X )
defined above
is a σ algebra such that X is measurable with respect to σ
(X )
. Therefore,
G = σ
(X )
.
Letting B_{i} be a Borel set in Z_{i},∏_{i=1}^{n}B_{i} is a Borel set by Proposition 58.1.3 and
so
( n )
X− 1 ∏ Bi = ∩n X −1(Bi) ∈ σ (X1, ⋅⋅⋅,Xn )
i=1 i=1 i
If G denotes the Borel sets F ⊆∏_{i=1}^{n}Z_{i} such that X^{−1}
(F )
∈ σ
(X1,⋅⋅⋅,Xn)
, then G is
clearly a σ algebra which contains the open sets. Hence G = ℬ the Borel sets of ∏_{i=1}^{n}Z_{i}.
This shows that σ
(X )
⊆ σ
(X1, ⋅⋅⋅,Xn )
. Next we observe that σ
(X )
is a σ algebra with
the property that each X_{i} is measurable with respect to σ
(X )
. This follows
from X_{i}^{−1}
(Bi)
= X^{−1}
(∏n )
j=1 Aj
∈ σ
(X )
, where each A_{j} = Z_{j} except for
A_{i} = B_{i}.Since σ
(X1,⋅⋅⋅,Xn )
is defined as the smallest such σ algebra, it follows that
σ
(X )
⊇ σ
(X1,⋅⋅⋅,Xn )
.■
For random variables having values in a separable Banach space or even more
generally for a separable metric space, much can be said about regularity of
λ_{X}.
Definition 58.1.7A measure, μ defined on ℬ
(E)
will be called inner regularif for allF ∈ℬ
(E )
,
μ (F ) = sup{μ (K ) : K ⊆ F and K is closed}
A measure, μ defined on ℬ
(E)
will be called outer regularif for all F ∈ℬ
(E)
,
μ (F) = inf{μ(V ) : V ⊇ F and V is open}
When a measure is both inner and outer regular, it is called regular.
For probability measures, the above definition of regularity tends to come free. Note it
is a little weaker than the usual definition of regularity because K is only assumed to be
closed, not compact.
Lemma 58.1.8Let μ be a finite measure defined on ℬ
(E )
where E is a metricspace. Then μ is regular.
Proof: First note every open set is the countable union of closed sets and every
closed set is the countable intersection of open sets. Here is why. Let V be an open set
and let
{ ( ) }
Kk ≡ x ∈ V : dist x,V C ≥ 1∕k .
Then clearly the union of the K_{k} equals V. Next, for K closed let
V ≡ {x ∈ E : dist(x,K ) < 1∕k} .
k
Clearly the intersection of the V_{k} equals K. Therefore, letting V denote an open set and
K a closed set,
μ(V ) = sup {μ(K ) : K ⊆ V and K is closed}
μ(K ) = inf{μ (V ) : V ⊇ K and V is open} .
Also since V is open and K is closed,
μ(V) = inf{μ(U ) : U ⊇ V and V is open}
μ (K) = sup {μ (L) : L ⊆ K and L is closed}
In words, μ is regular on open and closed sets. Let
ℱ ≡ {F ∈ ℬ (E) such that μ is regular on F }.
Then ℱ contains the open sets and the closed sets.
Suppose F ∈ℱ. Then there exists V ⊇ F with μ
(V ∖F )
< ε. It follows V^{C}⊆ F^{C}
and
μ(F C ∖ VC) = μ(V ∖F ) < ε.
Thus F^{C} is inner regular. Since F ∈ℱ, there exists K ⊆ F where K is closed and
μ
(F ∖ K)
< ε. Then also K^{C}⊇ F^{C} and
μ(KC ∖ FC) = μ(F ∖K ) < ε.
Thus if F ∈ℱ so is F^{C}.
Suppose now that
{Fi}
⊆ℱ, the F_{i} being disjoint. Is ∪F_{i}∈ℱ? There exists K_{i}⊆ F_{i}
such that μ
provided N is large enough. Thus it follows ∪_{i=1}^{∞}F_{i} is inner regular. Why is it outer
regular? Let V_{i}⊇ F_{i} such that μ
(Fi)
+ ε∕2^{i}> μ
(Vi)
and
∞ ∞
μ(∪∞ F ) = ∑ μ (F ) > − ε+ ∑ μ (V) ≥ − ε+ μ (∪ ∞ V )
i=1 i i=1 i i=1 i i=1 i
which shows ∪_{i=1}^{∞}F_{i} is outer regular. It follows ℱ contains the π system consisting of
open sets and so by the Lemma on π systems, Lemma 10.12.3, ℱ contains σ
(τ)
where τ
is the set of open sets. Hence ℱ contains the Borel sets and is itself a subset of the Borel
sets by definition. Therefore, ℱ = ℬ
(E )
. ■
One can say more if the metric space is complete and separable. In fact in this
case the above definition of inner regularity can be shown to imply the usual
one.
Lemma 58.1.9Let μ be a finite measure on a σ algebra containing ℬ
(E )
,the Borel sets of E, a separable complete metric space. Then if C is a closedset,
μ(C ) = sup{μ (K ) : K ⊆ C and K is compact.}
Proof: Let
{ak}
be a countable dense subset of C. Thus ∪_{k=1}^{∞}B
. Then K is a subset of C_{n} for each n and so for each ε > 0
there exists an ε net for K since C_{n} has a 1∕n net, namely a_{1},
⋅⋅⋅
,a_{mn}. Since K is closed,
it is complete and so it is also compact since it is complete and totally bounded.
Now
∑∞
μ(C ∖K ) = μ (∪ ∞n=1 (C ∖ Cn)) < ε-= ε.
n=1 2n
Thus μ
(C )
can be approximated by μ
(K )
for K a compact subset of C. This proves the
lemma.
Definition 58.1.10A measurable function X :
(Ω,ℱ, μ)
→ Z a topological space iscalled a random variable when μ
(Ω )
= 1. For such a random variable, one can define adistribution measure λ_{X}on the Borel sets of Z as follows.
λX (G ) ≡ μ (X −1 (G ))
This is a well defined measure on the Borel sets of Z because it makes sense for every Gopen and G≡
{G ⊆ Z : X −1(G) ∈ ℱ }
is a σ algebra which contains the opensets, hence the Borel sets. Such a measurable function is also called a randomvector.
Corollary 58.1.11Let X be a random variable (random vector) with values in acomplete metric space, Z. Then λ_{X}is an inner and outer regular measure definedon ℬ
(Z )
.
Proposition 58.1.12For X a random vector defined above, X having values ina complete separable metric space Z, then λ_{X}is inner and outer regular andBorel.
(Ω,P) X→ (Z,λ ) h→ E
X
If h is Borel measurable and h ∈ L^{1}
(Z,λX; E)
for E a Banach space, then
∫ ∫
h(X (ω ))dP = h (x)dλ . (58.1.1)
Ω Z X
(58.1.1)
In the case where Z = E, a separable Banach space, if X is measurable thenX ∈ L^{1}
(Ω; E)
if and only if the identity map on E is in L^{1}
(E;λX )
and
∫ ∫
X (ω)dP = xdλX (x) (58.1.2)
Ω E
(58.1.2)
Proof: The regularity claims are established above. It remains to verify
58.1.1.
Since h ∈ L^{1}
(Z,E)
, it follows there exists a sequence of simple functions
{hn}
such
that
∫
hn (x) → h (x), ||hm − hn||dλX → 0 as m,n → ∞.
Z
The first convergence above implies
hn ∘X → h∘ X pointwise on Ω (58.1.3)
(58.1.3)
Then letting h_{n}
(x )
= ∑_{k=1}^{m}x_{k}X_{Ek}
(x )
, where the E_{k} are disjoint and Borel, it follows
easily that h_{n}∘X is also a simple function of the form h_{n}∘X
(ω )
= ∑_{k=1}^{m}x_{k}X_{X−1(Ek)
}
(ω)
and by assumption X^{−1}
(Ek )
∈ℱ. From the definition of the integral, it is easily
seen
∫ ∫ ∫ ∫
hn ∘XdP = hnd λX, ||hn||∘ XdP = ||hn||dλX
Also, h_{n}∘ X − h_{m}∘ X is a simple function and so