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```Chapter III
Measure Theory
This chapter is devoted to the developement of measure theory. The goal is
to define a suitable tool which is analgous to the notion of area and volume in
Calculus. As the exposition suggests, this goal can be achieved in a great deal of
generality.
1. Set arithmetic: (σ-)rings, (σ-)algebras, and monontone classes
In this section we discuss various types of set collections used in Measure Theory.
Notation. Given a (non-empty) set X, we denote by P(X) the collection of
all subsets of X.
Definition. Let X be a non-empty set. For A ∈ P(X), we define the function
κ A : X → {0, 1} by
1 if x ∈ A
κ A (x) =
0 if x ∈ X r A
The function κ A is called the characteristic function of A.
The basic properties of characteristic functions are summarized in the following.
Exercise 1. Let X be a non-empty set. Prove:
(i) κ ∅ = 0 and κ X = 1.
(ii) For A, B ∈ P(X) one has
A ⊂ B ⇔ κA ≤ κB;
A = B ⇔ κA = κB.
(iii) κ A∩B = κ A &middot; κ B , ∀ A, B ∈ P(X).
(iv) κ ArB = κ A &middot; (1 − κ B ), ∀ A, B ∈ P(X).
(v) κ A∩B = κ A + κ B − κ A &middot; κ B , ∀ A, B ∈ P(X).
n
X
X
(vi) κ A1 ∪&middot;&middot;&middot;∪An =
(−1)k−1
κ Ai1 &middot; &middot; &middot; κ Ai , ∀, A1 , . . . , An ∈ P(X).
k
k=1
1≤i1 &lt;&middot;&middot;&middot;&lt;ik ≤n
(vii) κ A4B = |κ A −κ B | = κ A +κ B −2κ A &middot;κ B , ∀ A, B ∈ P(X). Here 4 stands
for the symmetric set difference, defined by A4B = (A r B) ∪ (B r A).
Property (vi) is called the Inclusion-Exclusion Formula.
Hint: (vi). Show that the right hand side is equal to 1 − (1 − κ A1 ) &middot; &middot; &middot; (1 − κ An ).
Remark 1.1. The Inclusion-Exclusion formula has an interesting application
in Combinatorics. If the ambient set X is finite, then the number of elements of
any subset A ⊂ X is given by
X
|A| =
κ A (x).
x∈X
201
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CHAPTER III: MEASURE THEORY
Using the Inclusion-Exclusion formula, we then get
n
X
X
|A1 ∪ &middot; &middot; &middot; ∪ An | =
(−1)k−1
k=1
|Ai1 ∩ &middot; &middot; &middot; Aik |.
1≤i1 &lt;&middot;&middot;&middot;&lt;ik ≤n
This is known as the Inclusion-Exclusion Principle.
Definition. Let X be a non-empty set, and let K be one of the fields1 Q, R
or C. An function φ : X → K is said to be elementary, if its range φ(X) is finite.
Remark that this gives
X
X
φ=
λ &middot; κ φ−1 ({λ}) =
λ &middot; κ φ−1 ({λ}) .
λ∈φ(X)
λ∈φ(X)r{0}
We define
ElemK (X) = {φ : X → K : φ elementary}.
Given a collection M ⊂ P(X), a function φ : X → K is said to be M-elementary,
if φ is elementary, and moreover,
φ−1 ({λ}) ∈ M, ∀ λ ∈ K r {0}.
We define
M-ElemK (X) = {φ : X → K : φ M-elementary}.
Exercise 2. With the above notations, prove that ElemK (X) is a unital Kalgebra.
Proposition 1.1. Given a non-empty set X, the collection P(X) is a unital
ring, with the operations
A + B = A4B and A &middot; B = A ∩ B, A, B ∈ P(X).
Proof. First of all, it is clear that 4 is commutative.
To prove the associativity of 4, we simply observe that
κ (A4B)4C = κ A4B + κ C − 2κ A4B κ C =
= κ A + κ B − 2κ A κ B + κ C − (κ A + κ B − 2κ A κ B ) &middot; κ C =
= κ A + κ B + κ C − 2(κ A κ B + κ A κ C + κ B κ C ) + 2κ A κ B κ C .
Since the final result is symmetric in A, B, C, we see that we get
κ A4(B4C) = κ (A4B)4C ,
so we indeed get
(A4B)4C = A4(B4C).
The neutral element for 4 is the empty set ∅. Since we obviously have A4A = ∅,
it follows that P(X), 4 is indeed an abelian group.
The operation ∩ is clearly commutative, associative, and has the total set X
as the unit.
To check distributivity, we again use characteristic functions:
κ (A∩C)4(B∩C) = κ A∩C + κ B∩C − 2κ A∩C κ B∩C =
= κ A κ C + κ B κ C − 2κ A κ B κ C = (κ A + κ B − 2κ A κ B )κ C =
= κ A4B κ C = κ (A4B)∩C ,
1 K can be any field.
&sect;1. Set arithmetic: (σ-)rings, (σ-)algebras, and monontone classes
203
so we indeed have the equality
(A ∩ C)4(B ∩ C) = (A4B) ∩ C.
Definitions. Let X be a non-empty set. A ring on X is a non-empty sub-ring
R ⊂ P(X). We do not require the unit X to belong to R, but we do require ∅ ∈ R.
An algebra on X is a ring A which contains the unit X.
Rings and algebras of sets are characterized as follows.
Proposition 1.2. Let X be a non-empty set.
A. For a non-empty collection R ⊂ P(X), the following are equivalent:
(i) R is a ring on X;
(ii) For any A, B ∈ R, we have A r B ∈ R and A ∪ B ∈ R.
B. For a non-empty collection A ⊂ P(X), the following are equivalent:
(i) A is an algebra on X;
(ii) For any A ∈ A, we have X r A ∈ A, and for any A, B ∈ A, we have
A ∪ B ∈ A.
Proof. A. (i) ⇒ (ii). Assume R is a ring on X, and let A, B ∈ R. Then A ∩ B
belongs to R, so
A r B = A4(A ∩ B)
also belongs to R. It the follows that
A ∪ B = (A4B)4(A ∩ B)
again belongs to R.
(ii) ⇒ (i). Assume R satisfies property (ii). Start with A, B ∈ R. Then A r B
belongs to R, and
A ∩ B = A r (A r B)
again belongs to R. Since A ∪ B also belongs to R, it follows that the set
A4B = (A ∪ B) r (A ∩ B)
again belongs to R.
B. (i) ⇒ (ii). This is clear from the implication A.(i) ⇒ (ii).
(ii) ⇒ (i). Assume A satisfies property (ii). Start with two sets A, B ∈ A.
Then the complements X r A and X r B both belong to A, hence their union
(X r A) ∪ (X r B) = X r (A ∩ B)
belongs to A, and the complement of this union
X r X r (A ∩ B) = A ∩ B
will also belong to A.
If A, B ∈ A, then since X r B belongs to A, by the above considerations, it
follows that the intersection
A ∩ (X r B) = A r B
also belongs to A. Likewise, the difference B rA also belongs to A, hence the union
(A r B) ∪ (B r A) = A4B
also belongs to A. By part A, it follows that A is a ring.
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CHAPTER III: MEASURE THEORY
Finally, since A is non-empty, if we choose some A ∈ A, then A4A = ∅ belongs
to A, so its complement X r ∅ = X also belongs to A.
It will be useful to introduce the following terminology.
Definition. A system of sets (Ai )i∈I is said to be pair-wise disjoint, if Ai ∩
Aj = ∅, for all i, j ∈ I with i 6= j.
Lemma 1.1. Let X be a non-empty set, let K be one of fields Q, R or C, and
let R be a ring on X. For a function φ : X → K, the following are equivalent:
(i) φ is R-elementary;
(ii) there exist an integer n ≥ 1 and sets A1 , . . . , An ∈ R, and numbers
λ1 , . . . , λn ∈ K, such that
φ = λ 1 κ A1 + &middot; &middot; &middot; + λ n κ A n .
(ii) there exist an integer m ≥ 1, and a finite pair-wise disjoint system (Bj )m
j=1 ⊂
R, and numbers &micro;1 , . . . , &micro;m ∈ K, such that
φ = &micro;1 κ B1 + &middot; &middot; &middot; + &micro;m κ Bm .
Proof. (i) ⇒ (ii). Assume φ is R-elementary. If φ = 0, there is nothing to
prove, because we have φ = κ ∅ . If φ is not identically zero, then we can obviously
write
X
φ=
λκ φ−1 ({λ}) ,
λ∈φ(X)r{0}
with all sets φ ({λ}) in R.
(ii) ⇒ (iii). Define
−1
E = {ψ : X → K : ψ satisfies property (iii)}.
Assume φ satisfies (ii), i.e.
φ = λ 1 κ A1 + &middot; &middot; &middot; + λ n κ A n ,
with A1 , . . . , An ∈ R and λ1 , . . . , λn ∈ K. We are going to prove that φ ∈ E, by
induction on n. The case n = 1 is trivial (either φ = 0, so φ = κ ∅ ∈ E, or φ = λκ A
for some A ∈ R and λ 6= 0, in which case we also have φ ∈ E).
Assume
α1 κ D1 + &middot; &middot; &middot; + αk κ Dk ∈ E,
for all D1 , . . . , Dk ∈ R, α1 , . . . , αk ∈ K. Start with a function
φ = λ1 κ A1 + &middot; &middot; &middot; + λk κ Ak + λk+1 κ Ak+1 ,
with A1 , . . . , Ak+1 ∈ R and λ1 , . . . , λk+1 ∈ K, and based on the above inductive
hypothesis, let us show that φ ∈ E. Using the inductive hypothesis, the function
ψ = λ2 κ A2 + &middot; &middot; &middot; + λk κ Ak + λk+1 κ Ak+1
belongs to E, so there exist scalars η1 , . . . , ηp ∈ K, an integer p ≥ 1, and a pair-wise
disjoint system (Cj )pj=1 ⊂ R, such that
ψ = η1 κ C1 + &middot; &middot; &middot; + ηp κ Cp .
With this notation, we have
φ = λ1 κ A1 + η1 κ C1 + &middot; &middot; &middot; + ηp κ Cp .
&sect;1. Set arithmetic: (σ-)rings, (σ-)algebras, and monontone classes
205
Put then
B2j = A1 ∩ Cj and B2j−1 = Cj r A1 , for all j ∈ {1, . . . , p};
B2p+1 = A1 r (C1 ∪ &middot; &middot; &middot; ∪ CP ).
It is clear that (Bk )2p+1
k=1 ⊂ R is pair-wise disjoint. Notice now that the equalities
Cj = B2j−1 ∪ B2j , ∀ j ∈ {1, . . . , p},
A1 = B1 ∪ B3 ∪ &middot; &middot; &middot; ∪ B2p+1 ,
combined with the fact that the B’s are pairwise disjoint, give
κ Cj = κ B2j−1 + κ B2j , ∀ j ∈ {1, . . . , p},
κ A1 = κ B1 + κ B3 + . . . κ B2p+1 ,
which give
φ=
p
X
j=1
ηj κ B2j +
p
X
(ηj + λ1 )κ B2j−1 + λ1 κ B2p+1 ,
j=1
which proves that φ indeed belongs to E.
(iii) ⇒ (i). Assume there exists a finite pair-wise disjoint system (Bj )m
j=1 ⊂ R,
and numbers &micro;1 , . . . , &micro;m ∈ K, such that
φ = &micro;1 κ B1 + &middot; &middot; &middot; + &micro;m κ Bm ,
and let us prove that φ is R-elemntary.
If all the &micro;’s are zero, there is noting to prove, since φ = 0.
Assume the &micro;’s are not all equal to zero. Since the &micro;’s that are equal to zero
do not have any contribution, we can in fact assume that all the &micro;’s are non-zero.
Notice that
φ(X) r {0} = {&micro;j : 1 ≤ j ≤ m}.
In particular φ is elementary.
If we start with an arbitrary λ ∈ K r {0}, then either λ 6∈ φ(X), or λ ∈
φ(X) r {0}. In the first case we clearly have φ−1 ({λ}) = ∅ ∈ R. In the second
case, we have the equality
[
φ−1 ({λ}) =
Bj ,
j∈Mλ
where
Mλ = {j : 1 ≤ j ≤ m and &micro;j = λ}.
Since all B’s belong to R, it follows that φ−1 ({λ}) again belongs to R. Having
shown that φ is elementary, and φ−1 ({λ}) ∈ R, for all λ ∈ K r {0}, it follows that
φ is indeed R-elementary.
Proposition 1.3. Let X be a non-empty set, and let K be one of the fields Q,
R, or C.
A. For a non-empty collection R ⊂ P(X), the following are equivalent:
(i) R is a ring on X;
(ii) R-ElemK (X) is a K-subalgebra of ElemK (X).
B. For a non-empty collection A ⊂ P(X), the following are equivalent:
(i) A is an algebra on X;
(ii) A-ElemK (X) is a K-subalgebra of ElemK (X), which contains the constant
function 1.
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CHAPTER III: MEASURE THEORY
Proof. A. (i) ⇒ (ii). Assume R is a ring on X. Using Lemma 1.1 we see that
we have the equality:
R-ElemK (X) = Span{κ A : A ∈ R}.
In particular, this shows that R-ElemK (X) is a K-linear subspace of ElemK (X).
Moreover, in order to prove that R-ElemK (X) is a K-subalgebra, it suffices to prove
the implication
A, B ∈ R =⇒ κ A &middot; κ B ∈ R-ElemK (X).
But this implication is trivial, since κ A &middot; κ B = κ A∩B , and A ∩ B belongs to R.
(ii) ⇒ (i). Assume R-ElemK (X) is a K-subalgebra of ElemK (X). First of all,
since κ ∅ = 0 ∈ R-ElemK (X), it follows that ∅ ∈ R.
Start now with two sets A, B ∈ R. Then κ A and κ B belong to R-ElemK (X).
Since R-ElemK (X) is an algebra, the function
κ A∩B = κ A &middot; κ B
belongs to R-ElemK (X), so we immediately see that A ∩ B ∈ R.
Likewise, the function
κ A4B = κ A + κ B − 2κ A κ B
belongs to R-ElemK (X), so we also get A4B ∈ R.
B. This equivalence is clear from part A, plus the identity κ X = 1.
Algebras of elementary functions give in fact a complete description for rings
or algebras of sets, as indicated in the result below.
Proposition 1.4. Let X be a non-empty set, and let K be one of the fields Q,
R, or C.
A. The map
R 7−→ R-ElemK (X)
is a bijective correspondence from the collection of all rings on X, and the collection
of all K-subalgebras of ElemK (X).
B. The map
A 7−→ A-ElemK (X)
is a bijective correspondence from the collection of all algebras on X, and the collection of all K-subalgebras of ElemK (X) that contain 1.
Proof. A. We start by proving surjectivity. Let E ⊂ ElemK (X) be an arbitrary
K-subalgebra. Define the collection
R = {A ⊂ X : κ A ∈ E}.
If A, B ∈ R, then the equalities
κ A∩B = κ A κ B and κ A4B = κ A + κ B − 2κ A κ B ,
combined with the fact that E is a subalgebra, prove that κ A∩B and κ A4B both
belong to E, hence A ∩ B and A4B both belong to R. This shows that R is a ring.
It is pretty clear (see Lemma 1.1) that R-ElemK (X) ⊂ E. To prove the other
inclusion, start with some arbitrary function φ ∈ E, and let us prove that φ ∈
R-ElemK (X). If φ = 0, there is nothing to prove. Assume φ is not identically zero.
&sect;1. Set arithmetic: (σ-)rings, (σ-)algebras, and monontone classes
207
We write φ(X) r {0} as {λ1 , . . . , λn }, with λi 6= λj for all i, j ∈ {1, . . . , n} with
i 6= j. For each i ∈ {1, . . . , n}, we set Ai = φ−1 ({λi }), so that
n
X
φ=
λ i &middot; κ Ai .
i=1
Since all λ’s are different, the matrix

λ1
 λ21

T = .
 ..
λ2
λ22
..
.
...
...
..
.
λn
λ2n
..
.





λn1 λn2 . . . λnn
n
is invertible. Take αij i,j=1 to be the inverse of T . The obvious equalities
φk =
n
X
λkj κ Aj , ∀ k = 1, . . . , n
j=1
can be written in matrix form as

φ
 φ2

 ..
 .
φn
so multiplying by T
−1



=T




&middot;

κ A1
κ A2
..
.



,

κ An
yields
κ Aj =
n
X
αjk φk , ∀ j = 1, . . . , n,
k=1
which proves that κ A1 , . . . , κ An ∈ E, so A1 , . . . , An ∈ R. This then shows that
φ ∈ R-ElemK (X).
We now prove injectivity. Suppose first that R and S are rings such that
R-ElemK (X) = S-ElemK (X), and let us prove that R = S. For every A ∈ R, the
function κ A ∈ R-ElemK (X) is also S-elementary, which means that A ∈ S. This
proves the inclusion R ⊂ S. By symmetry we also have the inclusion S ⊂ R, so
indeed R = S.
B. This part is obvious from A.
Definitions. Let X be a (non-empty) set. A collection U ⊂ P(X) is called a
σ-ring, if it is a ring, and it has the property:
S∞
(σ) Whenever (An )∞
n=1 is a sequence in U, it follows that
n=1 An also belongs
to U.
A collection S ⊂ P(X) is called a σ-algebra, if it is an algebra, and it has property
(σ).
Clearly, every σ-algebra is a σ-ring.
Remarks 1.2. A. For σ-rings and σ-algebras, one of the properties in the
definition of rings and algebras is redundant. More explicitly:
(i) A collection U ⊂ P(X) is a σ-ring, if and only if it has the property (σ)
and the property: A, B ∈ U =⇒ A r B ∈ U.
(ii) A collection S ⊂ P(X) is a σ-algebra, if and only if it has the property
(σ) and the property: A ∈ S =⇒ X r A ∈ S.
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CHAPTER III: MEASURE THEORY
B. If U is a σ-ring, then it also has the property
T∞
(δ) (An )∞
n=1 ⊂ U =⇒
n=1 ∈ U.
Since σ-algebras are σ-rings, they will also have property (δ).
Definitions. Let X be a non-empty set. A sequence (An )n≥1 of subsets of X
is said to be monotone, if it satisfies one of the following conditions:
(↑) An ⊂ An+1 , ∀ n ≥ 1,
(↓) An ⊃ An+1 , ∀ n ≥ 1.
In the case (↑) the sequence is said to be increasing, and we define
∞
[
lim An =
An .
n→∞
n=1
In the case (↓) the sequence is said to be decreasing, and we define
∞
\
lim An =
An .
n→∞
n=1
A collection M ⊂ P(X) is said to be a monotone class on X, if it satisfies the
condition:
(m) whenever (An )n≥1 is a monotone sequence in M, it follows that its limit
limn→∞ An also belongs to M.
Proposition 1.5. Let R be a ring on X. Then the following are equivalent:
(i) R is a σ-ring;
(ii) R is a monotone class.
Proof. (i) ⇒ (ii). This is immediate from the definition and Remark 1.2.B.
(ii) ⇒ (i). Assume R is a monotone class, an let us prove that it is a σ-ring.
By Remark 1.2.A, we only need to prove that R has S