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MATH 532 Outer Measure
Throughout, let X be a non-empty set, let R be a non-empty collection of subsets of X
that is closed under finite unions and set differences, and let be a finite-valued,
countably additive measure defined on R. We now shall see how to measure a larger
class of sets generated by R.
Definition 2.1. Let R be a non-empty collection of subsets of a non-empty set X. Then
R is called a Boolean ring if the following condition holds: If A, B ∈ R , then A ∪ B ∈ R
and B − A ∈ R. (That is, R is closed under finite unions and set differences.)
Lemma 2.1. A Boolean ring R is closed under symmetric differences and finite
intersections.
Proof. Let A, B ∈ R. Then A ∆ B = (A − B) ∪ (B − A) ∈ R because R is closed under
finite unions and set differences. Then A ∩ B = ( A ∪ B) − ( A ∆ B) ∈ R also. QED
(i) A ring is not necessarily closed under complements, and the set X may or may not
be in R.
(ii) If X ∈ R , then for any set A ∈ R we have A
c = X − A ∈ R. Thus, if X ∈ R , then
R is closed under complements and R is then called an algebra.
(iii) If R is closed under denumerable unions, then R is called a - ring.
(iv) If R is a -ring and X ∈ R , then R will be a -algebra.
(v) We have seen previously that a -algebra is a Boolean ring, because a -algebra is
closed under finite unions and intersections, as well as complements. Thus, a -algebra
is also closed under set differences using B − A = B ∩ A
c .
(vi) Because R is non-empty, there is some set A ∈ R. Thus, ∅ = A − A ∈ R.
(vii) The power set of X , P( X) , is a -ring that contains R. The intersection of all -
rings that contains R is a -ring that contains R and is denoted by ˜ (R ).
(viii) A ring R is not necessarily closed under denumerable unions. But a measure
on R is still assumed to be countably additive. That is, if A { (^) i}i= 1
∞ is a denumerable
sequence of disjoint events from R and ∪
i = 1
∞
A i
∈ R , then ( ∪
i = 1
∞
A i
) = ( A
i i = 1
∞
∑ ).^ Because
∅ ∈ R , also will be finitely additive.
We now give some results about that were previously proven for a countably
additive measure on a -algebra, but the proofs of which require only the weaker
assumptions about a ring R.
Lemma 2.2. Let R be Boolean ring and let : R → [0, ∞) be a measure.
(a) If A, B ∈ R with B ⊆ A , then (B) ≤ ( A).
(b) If A ∈ R and A ⊆ A i i = 1
n
U where^1 ≤^ n^ ≤ ∞^ with all^ Ai ∈^ R^ , then^ (A)^ ≤^ (Ai )
i = 1
n
Proof. (a) Because B ⊆ A , set A can be written as the disjoint union A = B ∪ ( A − B).
Thus, (A) = (B) + ( A − B) ≥ (B) because (A − B) ≥ 0.
(b) Let B 1
= A
1
, B
2
= A
2
− A
1 , and in general let B i
= A
i
− A
j j = 1
i− 1
U for^ i^ ≥^2.^ Because^ R^ is
a ring, B i ∈ R for all i. The sets {B i } are mutually disjoint and B i
⊆ A
i for all i ; thus,
(B
i
) ≤ (A
i ) for all i by Part (a). Moreover, B i i = 1
n
U =^ Ai
i = 1
n
U , so^ A^ ⊆^ Bi
i = 1
n
U.
Now consider the disjoint sequence of sets {A ∩ B i
i = 1
n , all of which are in R
because R is closed under finite intersections, and for which we have
(A ∩ B
i
i = 1
n
U =^ A^ ∩^ Bi
i = 1
n
U =^ A^ ∈^ R^. By countable additivity, we have
(A) = ( A ∩ B
i i = 1
n
∑ )^ ≤^ (^ Bi
i = 1
n
∑ )^ ≤^ (^ Ai
i = 1
n
∑ ).^ QED
Part (c) also implies that A i i = 1
n
U
≤^ (^ Ai )
i = 1
n
∑ whenever^ A 1 , .. .,^ An ∈^ R^.^ But when
n = ∞ , then A i i = 1
∞
U may not be in^ R^ so we may not be able to compute its measure.^ So
we can only say that for A ∈ R , A ⊆ A i i = 1
∞
U with all^ Ai ∈^ R^ , then^ (A)^ ≤^ (Ai )
i = 1
∞
The Hereditary -ring
Definition 2.2. Let R be a Boolean ring of a non-empty set X and let A ⊆ X. A
collection of sets { A } is a covering of A if A ⊆ U A. The hereditary -ring of R is
C(R ) = A ⊆ X : A ⊆ A
i i = 1
n
U , for some^ n, where 1^ ≤^ n^ ≤ ∞, and all^ Ai ∈R
In other words, C(R ) is the collection of subsets of X that have a countable covering
of sets from R. The covering A i
i = 1
n could be finite (when n < ∞ ) or it could be
denumerable (when n = ∞ ).
Proof. Let A ∈ R. Using n = 1 and A 1 = A in the definition of * , we have A ⊆ A i i = 1
1
U ;
thus, * ( A) ≤ ( A i
i = 1
1
∑ =^ (^ A 1 )^ =^ (A)^ because^ * (^ A)^ is a lower bound of all such
sums (A i
i = 1
n
∑.^ On the other hand, for any collection^ {Ai}i = 1
n from R with 1 ≤ n ≤ ∞
and A ⊆ A i i = 1
n
U , we have^ (A)^ ≤^ (Ai )
i = 1
n
∑ by Lemma 2.2 (b).^ Because^ * (^ A)^ is the
greatest of the lower bounds of such sums (A i
i = 1
n
∑ , we must have that^ (A)^ ≤^ * (^ A).
Thus, * ( A) = ( A). QED
An outer measure is not necessarily countably additive nor even finitely additive.
But there are various properties that hold that we shall need.
Lemma 2.5. Let * be the outer measure induced by a measure on a ring R.
(a) If A, B ∈ C(R ) with A ⊆ B , then * ( A) ≤ * (B).
(b) For any sequence of subsets {B j
j = 1
∞ from C(R ), * B j j = 1
∞
U
j = 1
∞
∑ * (Bj )^.
Proof. (a) For A ⊆ B , any covering {A i
i = 1
n from R of set B is also a covering of set A ;
thus, * ( A) ≤ (A i
i = 1
n
∑.^ But because^ * (B)^ is the greatest lower bound of the set of
sums (A i
i = 1
n
∑ taken over all coverings from^ R^ of^ B^ , we must have^ * (^ A)^ ≤^ * (B)^.
(b) If
j = 1
∞
∑ * (Bj )^ =^ +∞, then we are done.^ So we may assume that^
j = 1
∞
∑ * (Bj )^ <^ ∞
and therefore * (B j ) < ∞ for all j. Now let > 0 be given. For all j , the value
* (B
j
j cannot be an upper bound of the set of sums (A i
i = 1
n
∑ taken over all
coverings from R of B j
. So for all j , B j has a covering {A ji
i = 1
nj of sets from R such
that (A ji
i = 1
n j
∑ <^ * (Bj )^ +^ / 2^
j .
Then B j j = 1
∞
U ⊆^ Aji
i = 1
n j
U
j = 1
∞
U which is a denumerable covering of^ Bj
j = 1
∞
U by sets from^ R^.
Therefore,
* B
j j = 1
∞
U
i = 1
n j
∑ (^ Aji )
j = 1
∞
∑ ≤^
j = 1
∞
∑ (^ * (Bj )^ +^ / 2^
j ) =
j = 1
∞
∑ * (Bj )^ +^.
Because > 0 was arbitrary, we must have * B j j = 1
∞
U
j = 1
∞
∑ * (Bj )^.^ QED
Note : Because * (∅) = 0 , by taking B j = ∅ for j ≥ 3 , we also have for all B 1
, B
2 in
C(R ) that * B 1
∪ B
2
( ) ≤ * (B
1
) + * (B
2
Although we can evaluate * ( A) for every set A in C(R ), we want to restrict our
measurements to subsets that behave in a certain way. We will show that on these sets
that * is actually countably additive.
Definition 2.4. Let * be the outer measure induced by a measure on a ring R. A
set E ∈ C(R ) is called * – measurable provided * ( A) = * ( A ∩ E) + * ( A ∩ E
c ) for
all A ∈ C(R ). The set of all * – measurable sets is denoted by M
(C(R )).
Note : For all sets A , we have A = (A ∩ E ) ∪ ( A ∩ E
c ). Thus, we already have that
* ( A) ≤ * ( A ∩ E) + * ( A ∩ E
c ) for all A, E in C(R ). Thus, we need only show that
* ( A) ≥ * ( A ∩ E) + * ( A ∩ E
c ) for all A ∈ C(R ) in order to show that E is * –
measurable.
Theorem 2.1. M
(C(R )) contains all sets with outer measure 0.
Proof. Suppose * (E ) = 0. Then for all A ∈ C(R ) we have by Lemma 2.5 (a) that
* ( A ∩ E) + * ( A ∩ E
c ) ≤ * (E ) + * ( A) = * ( A).
Thus, E is * – measurable by the preceding note.
Theorem 2.3. For 1 ≤ n ≤ ∞ , let {B i
i = 1
n be a sequence of disjoint sets in M
(C(R )) with
B = B
i i = 1
n
U. Then^ * (^ A^ ∩^ B)^ =^ * (^ A^ ∩^ Bi )
i = 1
n
∑ for all^ A^ ∈^ C(R^ ).
Proof. For n = 1 , the result is obvious. In line (v) of the previous proof, we had
* ( A ∩ (E ∪ F )) = * ( A ∩ E ∩ F) + * ( A ∩ E ∩ F
c ) + * ( A ∩ E
c ∩ F),
for all A ∈ C(R ) if E , F ∈ M
(C(R )). So if E , F are disjoint, then E ⊆ F
c and
F ⊆ E
c and we obtain
* ( A ∩ (E ∪ F )) = * ( A ∩ E) + * ( A ∩ F).
Thus the result holds for n = 2.
Assume now that the result holds for some n ≥ 2 , and let B = B i i = 1
n+ 1
U.^ Then by the
result for n = 2 and the induction hypothesis, we have
* ( A ∩ B) = * A ∩ B
i
i = 1
n
U ∪^ Bn+ 1
= * A ∩ B
i
i = 1
n
U
+ * A ∩ B
( n+ 1 )
= *( A ∩ B
i
i = 1
n
∑ +^ *^ (^ A^ ∩^ Bn+ 1 ) =^ *(^ A^ ∩^ Bi )
i = 1
n+ 1
Thus, by induction, the result holds when 1 ≤ n < ∞.
Now for B = B i i = 1
∞
U , we have by the result for finite^ n^ and Lemma 2.5 (a) that
* ( A ∩ B
i
i = 1
n
∑ =^ *^ A^ ∩^ Bi
i = 1
n
U
≤ * A ∩ B
i i = 1
∞
U
By taking the limit as n → ∞ and applying Lemma 2.5 (b), we obtain
* ( A ∩ B
i
i = 1
∞
∑ ≤^ *^ A^ ∩^ Bi
i = 1
∞
U
= * (A ∩ B
i
i = 1
∞
U
≤ * ( A ∩ B
i
i = 1
∞
QED
Corollary 2.1. M
(C(R )) is closed under denumerable unions of disjoint sets.
Proof. Let {B i
i = 1
∞ be a sequence of disjoint sets in M
(C(R )) with B = B i i = 1
∞
U.^ We must
show that B is * – measurable. For n ≥ 1 , let F n
= B
i i = 1
n
U which is in^ M
(C(R )) because
M
(C(R )) is a ring. Because F n ⊆ B, we have B
c ⊆ F n
c
. So for all A ∈ C(R ), we have
by Lemma 2.5 (a) and Theorem 2.3 that
* ( A) = *( A ∩ F
n
) + * ( A ∩ F
n
c )
≥ *( A ∩ F
n
) + *( A ∩ B
c )
= * A ∩ B
i i = 1
n
U
+ * ( A ∩ B
c )
= *( A ∩ B
i
i = 1
n
∑ +^ *(^ A^ ∩^ B
c ).
By taking the limit as n → ∞ and applying Theorem 2.3 again, we obtain
* ( A) ≥ *( A ∩ B
i
i = 1
∞
∑ +^ * (^ A^ ∩^ B
c ) = * ( A ∩ B) + * ( A ∩ B
c ).
It follows that B is * - measurable by the Note after Definition 2.4. QED
Corollary 2.2. M
(C(R )) is a -ring.
Proof. Let {E i
i = 1
∞ be a sequence of sets in M
(C(R )). We must show that their union is
in M
(C(R )). But we simply disjointify the sets by letting B 1
= E
1
, B
2
= E
2
− E
1 , and
in general let B n
= E
n
− E
i i = 1
n− 1
U for^ n^ ≥^2.^ Because^ M
(C(R )) is a ring, B i
∈ M
(C(R ))
for all i. The sets {B i } are mutually disjoint and E i i = 1
∞
U =^ Bi
i = 1
∞
U ∈^ M
(C(R )) by
Corollary 2.1. QED
We conclude with the result that shows that * is actually a measure that extends
from being defined only on a ring R to being defined on the -ring M
(C(R )).
Synopsis
- X is a non-empty set. The power set P( X) is the set of all subsets of X.
- R^ is a ring of subsets of^ X^ ; i.e.,^ R^ is a non-empty collection that is closed under
finite unions and set differences.
- : R → [0, ∞) is a finite-valued, countably additive measure.
- C(R ) is the -ring of all subsets of X that have countable coverings of sets from R ,
and C(R ) contains R.
- : C(R ) → [0, ∞] is the outer measure and * = on R.
6. M
(C(R )) is a smaller class of subsets within C(R ) that consists of * –measurable
sets. M
(C(R )) is also a -ring that contains R.
7. * : M
(C(R )) → [0, ∞ ] is a countably additive measure, and still * = on R.
- ˜ (R ) is the smallest -ring containing R : R ⊆ ˜ (R ) ⊆ M
(C(R )) ⊆ C(R ) ⊆ P( X)
Exercises
Exercise 2.1. Let S be a -ring of sets and let : S → [0, ∞] be a measure. Let
R = {A ∈ S : (A) < ∞}. Prove that R is a ring.
Exercise 2.2. Let R be a Boolean ring of sets and let : R → [0, ∞) be a measure.
Suppose E ∈ R , E i ∈ R for all i with all E i mutually disjoint, and E i i = 1
∞
U ⊆^ E^.^ Prove
that (E i
i = 1
∞
∑ ≤^ (E^ ).
Exercise 2.3. Let R be a Boolean ring of sets and let : R → [0, ∞) be a measure. For
A, B ∈ R , say that A ~ B if and only if (A ∆ B) = 0.
(a) Prove that ~ is an equivalence relation (i.e., that it is reflexive, symmetric, and
transitive.)
(b) Prove: If A ~ B , then (A ) = (B ) = ( A ∩ B).
(c) Define the distance between equivalence classes of sets in R by
d ([A], [ B]) = ( A ∆ B). Prove that d satisfies the axioms of a metric.
(d) Suppose A 1
~ A
2 and B 1
~ B
2
. Prove that d ([A 1
], [B
1 ]) = d([A 2
], [B
2
]).
