In this section we consider theorems about the Borel sets.
Definition 5.4.1The Borel sets on ℝ^{p}, denoted by ℬ
(ℝp)
consist of the smallest σ algebracontaining 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
like
( ∞ ∞ )C
∩i=1 ∪j=i Ej
is a Borel set if the E_{j} are. This is useful.
Definition 5.4.2A measure, μ defined on ℬ
p
(ℝ )
will be called inner regularif for all F ∈ℬ
p
(ℝ )
,
μ (F ) = sup{μ (K ) : K ⊆ F and K is closed}
A measure, μ defined on ℬ
p
(ℝ )
will be called outer regularif for all F ∈ℬ
p
(ℝ )
,
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”.A measure space
(Ω, ℱ,μ)
is called complete ifwhenever A ⊆ B for B ∈ℱ and μ
(B)
= 0, it follows that A ∈ℱ also.A measure which is defined on theBorel 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 5.4.3Let μ be a finite measure defined on ℬ
p
(ℝ )
where ℝ^{p}.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
Vk ≡ {x ∈ ℝp : dist(x,K) < 1∕k}.
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,
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
p
ℱ ≡ {F ∈ ℬ (ℝ ) 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) ( C )
μ F ∖V = 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 5.3.2, 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
(ℝp)
.■
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 5.4.4Let μ be a finite measure on a σ algebra containing ℬ
(ℝp )
, the Borel sets of ℝ^{p}. Then if Cis a closed set,
μ(C ) = sup{μ(K ) : K ⊆ C and K is compact.}
It follows that for a finite measure on ℬ
(ℝp)
, μ is inner regularin the sense that for all F ∈ℬ
(ℝp )
,
μ (F ) = sup{μ (K ) : K ⊆ F and K is compact}
Proof: This follows from noting that a closed set C is of the form
C = ∪∞ C ∩ D (0,k)
k=1
where D
(0,k)
is the closed disk
{x : ∥x∥ ≤ k}■
The above theorem can be considerably extended although I will not do so in this book. One can
replace ℝ^{p} with a complete separable metric space for example.
An important example of the above is the case of a random vector and its distribution
measure.
Definition 5.4.5A measurable function X :
(Ω,ℱ, μ)
→ ℝ^{p}is called a randomvariable when μ
(Ω )
= 1.For such a random variable, one can define a distribution measure λ_{X}on the Borel sets of ℝ^{p}asfollows.
( )
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 open andG≡
{ }
G ⊆ ℝp : 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.
Corollary 5.4.6Let X be a random variable with values in ℝ^{p}. Then λ_{X}is an inner and outerregular measure defined on ℬ
(ℝp )
.
The following is useful.
Corollary 5.4.7Let μ
(B )
< ∞ for every ball B ⊆ ℝ^{p}. 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
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 5.4.4, 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
There is an interesting application of regularity to approximation of a measurable function with one
that is continuous.
Lemma 5.4.8Suppose f : ℝ^{p}→ [0,∞) is measurable where μ is a regular measure as in the abovetheorem having the measure of any ball finite and the closures of balls compact. Then there is a setof measure zero N and a sequence of functions
{hn}
,h_{n} : ℝ^{p}→ [0,∞) each in C_{c}
(ℝp )
such thatfor all x ∈ ℝ^{p}∖ N, h_{n}
(x)
→ f
(x)
. Also, for x
∕∈
N,h_{n}
(x)
≤ f
(x)
for all n large enough.
Proof:Consider f_{n}
(x)
≡X_{Bn}
(x)
min
(f (x),n)
where B_{n} is a ball centered at x_{0} which has radius n.
Thus f_{n}
(x)
is an increasing sequence and converges to f
(x)
for each x. Also by Corollary 5.1.9, there
exists a simple function s_{n} such that
1
sn(x) ≤ fn (x ), sup |fn(x)− sn(x)| <-n
x∈ℝp 2
Let
∑mn
sn(x) = cnkXEnk (x), cnk > 0
k=1
Then it must be the case that μ
(Enk )
< ∞ because ∫f_{n}dμ < ∞.
By regularity, there exists a compact set K_{k}^{n} and an open set V _{k}^{n} such that
m∑n
Knk ⊆ Enk ⊆ Vkn, μ (Vnk ∖Knk) <-1n
k=1 2
Now let K_{k}^{n}≺ ψ_{k}^{n}≺ V _{k}^{n} and let
mn
h (x) ≡ ∑ cnψn(x)
n k=1 k k
Thus for N_{n} = ∪_{k=1}^{mn}V _{k}^{n}∖ K_{k}^{n}, it follows μ
(N )
n
< 1∕2^{n} and
sup |f (x)− h (x)| <-1
x∕∈Nn n n 2n
If h_{n}
(x )
fails to converge to f
(x )
, then x must be in infinitely many of the N_{n}. That is,
= 0. If x is not in N, then eventually x fails to be in N_{n} and also x ∈ B_{n} so
h_{n}
(x )
= s_{n}
(x)
for all n large enough. Now f_{n}
(x)
→ f
(x )
and
|sn (x)− fn(x)|
< 1∕2^{n} so also
s_{n}
(x)
= h_{n}
(x)
→ f
(x)
.■
Note that each N_{k} is an open set and so, N is a Borel set. Thus the above lemma leads to the following
corollary.
Corollary 5.4.9Let f be measurable. Then there exists a Borel measurable function g and a Borelset of measure zero N such that f
(x)
= g
(x)
for all x
∕∈
N. In fact, off N,f
(x)
= lim_{n→∞}h_{n}
(x)
where h_{n}is continuous and
|hn(x)|
≤
|f (x)|
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 X_{NC} is the limit
of a sequence h_{n}X_{NC} where h_{n} is continuous and for large enough n,
|hn (x )|
≤
|f (x)|
for
x
∕∈
N. Thus h_{n}X_{NC} is Borel and it follows that fX_{NC} is Borel measurable. Let g = fX_{NC}. ■