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 · κ B , ∀ A, B ∈ P(X).
(iv) κ ArB = κ A · (1 − κ B ), ∀ A, B ∈ P(X).
(v) κ A∩B = κ A + κ B − κ A · κ B , ∀ A, B ∈ P(X).
n
X
X
(vi) κ A1 ∪···∪An =
(−1)k−1
κ Ai1 · · · κ Ai , ∀, A1 , . . . , An ∈ P(X).
k
k=1
1≤i1 <···<ik ≤n
(vii) κ A4B = |κ A −κ B | = κ A +κ B −2κ A ·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 ) · · · (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
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CHAPTER III: MEASURE THEORY
Using the Inclusion-Exclusion formula, we then get
n
X
X
|A1 ∪ · · · ∪ An | =
(−1)k−1
k=1
|Ai1 ∩ · · · Aik |.
1≤i1 <···<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 · κ φ−1 ({λ}) =
λ · 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 · 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 ) · 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.
§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 + · · · + 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 µ1 , . . . , µm ∈ K, such that
φ = µ1 κ B1 + · · · + µm 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 + · · · + 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 + · · · + αk κ Dk ∈ E,
for all D1 , . . . , Dk ∈ R, α1 , . . . , αk ∈ K. Start with a function
φ = λ1 κ A1 + · · · + 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 + · · · + 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 + · · · + ηp κ Cp .
With this notation, we have
φ = λ1 κ A1 + η1 κ C1 + · · · + ηp κ Cp .
§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 ∪ · · · ∪ 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 ∪ · · · ∪ 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 µ1 , . . . , µm ∈ K, such that
φ = µ1 κ B1 + · · · + µm κ Bm ,
and let us prove that φ is R-elemntary.
If all the µ’s are zero, there is noting to prove, since φ = 0.
Assume the µ’s are not all equal to zero. Since the µ’s that are equal to zero
do not have any contribution, we can in fact assume that all the µ’s are non-zero.
Notice that
φ(X) r {0} = {µj : 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 µj = 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 · κ B ∈ R-ElemK (X).
But this implication is trivial, since κ A · 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 · 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.
§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 · 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
·
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
property (σ). Start with an
∞
arbitrary sequence (An )n≥1 in R, and let S
us prove that n=1 An again belongs to R.
n
For every integer n ≥ 1, we define Bn = k=1 An . Since R is a ring, it follows that
Bn ∈ R,S∀ n ≥ 1. Moreover, the sequence (Bn )n≥1 is increasing, so by assumption,
∞
the set n=1 An = limn→∞ Bn indeed belongs to R.
