5.4 Measures And Regularity
In this section we consider theorems about the Borel sets.
Definition 5.4.1 The Borel sets on ℝp, denoted by ℬ
consist of the smallest σ algebra
containing the open sets.
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.
Definition 5.4.2 A measure, μ defined on ℬ
will be called inner regular if for all F ∈ℬ
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”. A measure space
is called complete if
whenever A ⊆ B for B ∈ℱ and μ
, it follows that A ∈ℱ also. A measure which is defined on the
Borel sets which is also complete, finite on compact sets, and regular is called a Radon measure.
For finite measures, defined on the Borel sets, the first definition of regularity is automatic. These are
always outer and inner regular provided inner regularity refers to closed sets.
Lemma 5.4.3 Let μ be a finite measure defined on ℬ
where ℝp. 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
Then clearly the union of the Kk equals V. Next, for K closed let
Clearly the intersection of the V k equals K and they are open sets by the discussion by Proposition 2.2.13.
Therefore, letting V denote an open set and K a closed set,
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 μ
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
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 5.3.2, ℱ contains σ
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
One can say more due to the fact that every closed set is the union of compact sets. 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 5.4.4 Let μ be a finite measure on a σ algebra containing ℬ
, the Borel sets of ℝp. Then if C
is a closed set,
It follows that for a finite measure on ℬ
, μ is inner regular in the sense that for all F ∈ℬ
Proof: This follows from noting that a closed set C is of the form
is the closed disk
The above theorem can be considerably extended although I will not do so in this book. One can
replace ℝp with a complete separable metric space for example.
An important example of the above is the case of a random vector and its distribution
Definition 5.4.5 A measurable function X :
→ ℝp is called a random variable when μ
For such a random variable, one can define a distribution measure λX on the Borel sets of ℝp as
This is a well defined measure on the Borel sets of Z because it makes sense for every G open and
is a σ algebra which contains the open sets, hence the Borel sets. Such a
random variable is also called a random vector.
Corollary 5.4.6 Let X be a random variable with values in ℝp. Then λX is an inner and outer
regular measure defined on ℬ
The following is useful.
Corollary 5.4.7 Let μ
< ∞ for every ball B ⊆ ℝp. Then μ must be regular.
Proof: Let μK
. Then this is a finite measure if
is contained in a ball and is
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 5.4.4, 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
There is an interesting application of regularity to approximation of a measurable function with one
that is continuous.
Lemma 5.4.8 Suppose f : ℝp → [0,∞) is measurable where μ is a regular measure as in the above
theorem having the measure of any ball finite and the closures of balls compact. Then there is a set
of measure zero N and a sequence of functions
: ℝp →
) each in Cc
for all x ∈ ℝp ∖ N, hn
. Also, for x
for all n large enough.
Proof: Consider fn
is a ball centered at x0
which has radius n
is an increasing sequence and converges to
. Also by Corollary 5.1.9
exists a simple function sn
Then it must be the case that μ
fndμ < ∞
By regularity, there exists a compact set Kkn and an open set V kn such that
Now let Kkn ≺ ψkn ≺ V kn and let
Thus for Nn = ∪k=1mnV kn ∖ Kkn, it follows μ
fails to converge to
must be in infinitely many of the Nn
. That is,
However, this set N is contained in
and so μ
= 0. If
is not in N
, then eventually x
fails to be in Nn
and also x ∈ Bn
large enough. Now fn
Note that each Nk is an open set and so, N is a Borel set. Thus the above lemma leads to the following
Corollary 5.4.9 Let f be measurable. Then there exists a Borel measurable function g and a Borel
set of measure zero N such that f
for all x
N. In fact, off N,f
where hn is continuous and
for all n large enough.
Proof: Apply the above lemma to the positive and negative parts of the real and imaginary parts of
f. Let N be the union of the exceptional Borel sets which result. Thus, f XNC is the limit
of a sequence hnXNC where hn is continuous and for large enough n,
. Thus hnXNC
is Borel and it follows that fXNC
is Borel measurable. Let g