7.5 Measures And Regularity
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 7.5.1 A Polish space is a complete separable metric space. For a
Polish space E or more generally a metric space or even a general topological space,
denotes the Borel sets of E. This is defined to be the smallest σ algebra which
contains the open sets. Thus it contains all open 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
is a Borel set if the Ej 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 7.5.2 A measure, μ defined on ℬ
will be called inner regular if for all
A measure, μ defined on ℬ
will be called outer regular if for all F ∈ℬ
When a measure is both inner and outer regular, it is called regular. Actually, it is more useful
and likely more standard to refer to μ being inner regular as
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 7.5.3 Let μ be a finite measure defined on ℬ
where X is a metric
space. 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
Then clearly the union of the Kk equals V. Next, for K closed let
Clearly the intersection of the V k equals K. Therefore, letting V denote an open set and K a
Also since V
is open and K
In words, μ
is regular on open and closed sets. Let
Then ℱ contains the open sets and the closed sets.
Suppose F ∈ℱ. Then there exists V ⊇ F with μ
It follows V C ⊆ FC
Thus μ is inner regular on FC. Since F ∈ℱ, there exists K ⊆ F where K is closed and
. Then also KC ⊇ FC
Thus if F ∈ℱ so is FC.
Suppose now that
, the Fi
being disjoint. Is ∪Fi ∈ℱ
? There exists Ki ⊆ Fi
such that μ
2i > μ
is large enough. Thus it follows μ
is inner regular on ∪i=1∞Fi
. Why is it outer
regular? Let V i ⊇ Fi
such that μ
2i > μ
which shows μ is outer regular on ∪i=1∞Fi. 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
set of open sets. Hence ℱ
contains the Borel sets and is itself a subset of the Borel sets by
definition. Therefore, ℱ
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 7.5.4 Let μ be a finite measure on a σ algebra containing ℬ
, the Borel sets
of X, a separable complete metric space. Then if C is a closed set,
It follows that for a finite measure on ℬ
where X is a Polish space, μ is inner regular in
the sense that for all F ∈ℬ
be a countable dense subset of
there exists mn
Now let K = C ∩
is a subset of Cn
for each n
and so for each ε >
exists an ε
net for K
has a 1∕n
net, namely a1,
. Since K
is closed, it is
complete and so it is also compact since it is complete and totally bounded, Theorem 2.5.5
can be approximated by
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 7.5.5 A measurable function X :
→ Z a metric space is called a
random variable when μ
. For such a random variable, one can define a distribution
measure λX on the Borel sets of Z as follows.
This is a well defined measure on the Borel sets of Z because it makes sense for every G open
is a σ algebra which contains the open sets, hence the Borel
sets. Such a random variable is also called a random vector when Z is a vector
Corollary 7.5.6 Let X be a random variable with values in a separable complete
metric space, Z. Then λX is an inner and outer regular measure defined on ℬ
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 7.5.7 Let
be a measurable space with the measure μ.
Then μ is said to be σ finite if there is a sequence of disjoint Borel sets
= E and μ
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 7.5.8 Let Ω be a complete separable metric space (Polish space). Let μ
be a measure on ℬ
which has the property that μ
< ∞ for every ball B. Then μ
must be regular.
Proof: Let μK
. Then this is a finite measure if
is contained in a ball
and is therefore, regular.
x0 ∈ Ω and let
Thus the An are disjoint and have union equal to Ω, and the Bn are open sets having finite
measure which contain the respective An. (If x is a point, let n be the first such that
x ∈ B
. ) Also, for
E ⊆ An,
By Lemma 7.5.4, each μBn is regular. Let E be any Borel set with l < μ
Then for n
Choose r < 1 such that also
There exists a compact set Kk contained in E ∩ Ak such that
Then letting K be the union of these, K ⊆ E and
Thus this is inner regular.
To show outer regular, it suffices to assume μ
since otherwise there is nothing to
prove. There exists an open V n
containing E ∩ An
which is contained in Bn
Then let V be the union of all these V n.
It follows that