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 7.5.1A Polish spaceis a complete separable metric space. For aPolish space E or more generally a metric space or even a general topological space,ℬ
(E)
denotes the Borel sets of E. This isdefined to be the smallest σ algebra whichcontains the open sets. Thus it contains all open sets and closed sets and compact setsand 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
( ∞ ∞ )C
∩i=1 ∪j=i Ej
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 7.5.2A 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 outerregular, it is calledregular. Actually, it is more usefuland likely more 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 7.5.3Let μ be a finite measure defined on ℬ
(X)
where X 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
K ≡ {x ∈ V : dist(x,VC ) ≥ 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
( C C)
μ F ∖V = μ(V ∖F ) < ε.
Thus μ is inner regular on F^{C}. Since F ∈ℱ, there exists K ⊆ F where K is closed and
μ
(F ∖ K)
< ε. Then also K^{C}⊇ F^{C} and
( C C)
μ K ∖F = 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
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 7.4.2, 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 7.5.4Let μ be a finite measure on a σ algebra containing ℬ
(X)
, the Borel setsof X, a separable 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 regularinthe sense that for 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
( )
ak, 1n
⊇ C. Therefore,
there exists m_{n} such that
( --(-----)-)
μ C ∖∪mn B ak, 1 ≡ μ(C ∖Cn) < -ε.
k=1 n 2n
Now let K = C ∩
(∩∞ C )
n=1 n
. 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 2.5.5.
Now
for K a compact subset of C. The last claim
follows from Lemma 7.5.3. ■
An important example is the case of a random vector and its distribution measure.
Definition 7.5.5A measurable function X :
(Ω,ℱ, μ)
→ Z a metric space is called arandomvariable when μ
(Ω )
= 1. For such a random variable, one can define a distributionmeasure 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 G openand G≡
{G ⊆ Z : X −1(G) ∈ ℱ}
is a σ algebra which contains the open sets, hence the Borelsets. Such a random variable is also called a random vector when Z is a vectorspace.
Corollary 7.5.6Let X be a random variable with values in a separable completemetric 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 7.5.7Let
(E,ℬ (E),μ)
be a measurablespace with the measure μ.Then μ is said to be σ finite if there is a sequence of disjoint Borel sets
{Bi}
_{i=1}^{∞}suchthat ∪_{i=1}^{∞}B_{i} = E and μ
(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 7.5.8Let Ω be a complete separable metric space (Polish space). Let μbe a measure on ℬ
(Ω)
which has the property that μ
(B)
< ∞ for every ball B. Then μmust 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
An ≡ B (x0,n )∖B (x0,n − 1),
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 7.5.4, 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∑ μ (E ∩ A )
k=1 Bk k
There exists a compact set K_{k} contained in E ∩ A_{k} such that
μBk (Kk) > rμBk (E ∩Ak ).
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