It is often the case that Ω has more going on than to simply be a set. In particular, it is often the case that
Ω is some sort of topological space, often a metric space. In this case, it is usually if not always the
case that the open sets will be in the σ algebra of measurable sets. This leads to the following
definition.
Definition 19.4.1A Polish spaceis a complete separable metric space. For a Polish space E ormore generally a metric space or even a general topological space, ℬ
(E )
denotes the Borel sets ofE. This isdefined to be the smallest σ algebra which contains the open sets. Thus it contains allopen sets and closed sets and compact sets and many others.
Don’t ever try to describe a generic Borel set. Always work with the definition that it is
the smallest σ algebra containing the open sets. Attempts to give an explicit description of
a “typical” Borel set tend to lead nowhere because there are so many things which can be
done.You can take countable unions and complements and then countable intersections of what
you get and then another countable union followed by complements and on and on. You just
can’t get a good useable description in this way. However, it is easy to see that something
like
(∩∞ ∪∞ E )C
i=1 j=i j
is a Borel set if the E_{j} are. This is useful.
For example, ℝ is a Polish space as is any separable Banach space. Amazing things can be said about
finite measures on the Borel sets of a Polish space. First the case of a finite measure on a metric space will
be considered.
Definition 19.4.2A measure, μ defined on ℬ
(E)
will be called inner regularif for all F ∈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 outerregular, it is calledregular. Actually, it is more useful and likelymore standard to refer to μ being inner regular as
μ (F ) = sup{μ (K ) : K ⊆ F and K is compact}
Thus the word “closed” is replaced with “compact”.
For finite measures, defined on the Borel sets of a metric space, the first definition of regularity is
automatic. These are always outer and inner regular provided inner regularity refers to closed
sets.
Lemma 19.4.3Let μ be a finite measure defined on ℬ
(X)
where X is a metric space.Then μ isregular.
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
K ≡ {x ∈ V : dist(x,V C) ≥ 1∕k} .
k
Then clearly the union of the K_{k} equals V. Next, for K closed let
Vk ≡ {x ∈ X : dist(x,K ) < 1∕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 (X) 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 FC ∖ 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
{F }
i
⊆ℱ, the F_{i} being disjoint. Is ∪F_{i}∈ℱ? There exists K_{i}⊆ F_{i} such that
μ
which shows μ is outer regular on ∪_{i=1}^{∞}F_{i}. It follows ℱ contains the π system consisting of open sets and
so by the Lemma on π systems, Lemma 19.3.2, 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
(X )
.■
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 where the closed sets are replaced with compact
sets.
Lemma 19.4.4Let μ be a finite measure on a σ algebra containing ℬ
(X)
, the Borel sets of X, aseparable complete metric space. Then if C is a closed set,
μ(C ) = sup{μ(K ) : K ⊆ C and K is compact.}
It follows that for a finite measure on ℬ
(X )
where X is a Polish space, μ is inner regularin the sense thatfor all F ∈ℬ
(X )
,
μ (F ) = sup{μ (K ) : K ⊆ F 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, Theorem 11.1.38. 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. The last claim follows from
Lemma 19.4.3. ■
An important example is the case of a random vector and its distribution measure.
Definition 19.4.5A measurable function X :
(Ω,ℱ, μ)
→ Z a metric space is called a randomvariablewhen μ
(Ω )
= 1. For such a random variable, one can define a distribution measure λ_{X}on the Borel sets ofZ as follows.
( −1 )
λX (G ) ≡ μ X (G )
This is a well defined measure on the Borel sets of Z because it makes sense for every G open andG≡
{ }
G ⊆ Z : X−1 (G ) ∈ ℱ
is a σ algebra which contains the open sets, hence the Borel sets. Such arandom variable is also called a random vector when Z is a vector space.
Corollary 19.4.6Let X be a random variable with values in a separable complete metric space,Z. Then λ_{X}is an inner and outer regular measure defined on ℬ
(Z)
.
What if the measure μ is defined on a Polish space but is not finite. Sometimes one can still get the
assertion that μ is regular. In every case of interest in this book, the measure will also be σ
finite.
Definition 19.4.7Let
(E,ℬ (E),μ)
be a measurablespace with the measure μ. Then μ is saidto be σ finite if there is a sequence of disjoint Borel sets
{Bi}
_{i=1}^{∞}such that ∪_{i=1}^{∞}B_{i} = E andμ
(Bi )
< ∞.
One such example of a complete metric space and a measure which is finite on compact sets is the
following where the closures of balls are compact. Thus, this involves finite dimensional situations
essentially.
Corollary 19.4.8Let Ω be a complete separable metric space (Polish space). Let μ be a measureon ℬ
(Ω )
which has the property that μ
(B)
< ∞ for every ball B. Then μ must be regular.
Proof: Let μ_{K}
(E)
≡ μ
(K ∩ E)
. Then this is a finite measure if K is contained in a ball and is
therefore, regular.
Let
A ≡ B (x ,n)∖B (x ,n− 1),
n 0 0
x_{0}∈ Ω and let
-----------
Bn = B (x0,n+ 1)∖B (x0,n− 2)
Thus the A_{n} are disjoint and have union equal to Ω, and the B_{n} are open sets having finite measure which
contain the respective A_{n}. (If x is a point, let n be the first such that x ∈ B
(x0,n)
. ) Also, for
E ⊆ A_{n},
μ (E ) = μBn (E )
By Lemma 19.4.4, each μ_{Bn} is regular. Let E be any Borel set with l < μ
(E )
. Then for n large
enough,
∑n ∑n
l < μ (E ∩ Ak) = μBk (E ∩ Ak)
k=1 k=1
Choose r < 1 such that also
∑n
l < r μBk (E ∩ Ak)
k=1
There exists a compact set K_{k} contained in E ∩ A_{k} such that
μ (K ) > rμ (E ∩ A ).
Bk k Bk k
Then letting K be the union of these, K ⊆ E and
∑n ∑n ∑n
μ (K ) = μ (Kk) = μBk (Kk) > r μBk (E ∩ Ak) > l
k=1 k=1 k=1
Thus this is inner regular.
To show outer regular, it suffices to assume μ
(E)
< ∞ since otherwise there is nothing to prove. There
exists an open V_{n} containing E ∩ A_{n} which is contained in B_{n} such that