3-01-sets

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 The-
ory.
Notation. Given a (non-empty) set X, we denote by P(X) the collection of
all subsets of X.
Definition. Let Xbe a non-empty set. For A∈P(X), we define the function
κA:X→ {0,1}by
κA(x) = 1 if x∈A
0 if x∈XrA
The function κAis called the characteristic function of A.
The basic properties of characteristic functions are summarized in the following.
Exercise 1.Let Xbe a non-empty set. Prove:
(i) κ∅= 0 and κX= 1.
(ii) For A, B ∈P(X) one has
A⊂B⇔κA≤κB;
A=B⇔κA=κB.
(iii) κA∩B=κA·κB,∀A, B ∈P(X).
(iv) κArB=κA·(1 −κB), ∀A, B ∈P(X).
(v) κA∩B=κA+κB−κA·κB,∀A, B ∈P(X).
(vi) κA1∪···∪An=
n
X
k=1
(−1)k−1X
1≤i1<···<ik≤n
κAi1· · · κAik,∀, A1, . . . , An∈P(X).
(vii) κA4B=|κA−κB|=κA+κB−2κA·κB,∀A, B ∈P(X). Here 4stands
for the symmetric set difference, defined by A4B= (ArB)∪(BrA).
Property (vi) is called the Inclusion-Exclusion Formula.
Hint: (vi). Show that the right hand side is equal to 1 −(1 −κA1)···(1 −κAn).
Remark 1.1. The Inclusion-Exclusion formula has an interesting application
in Combinatorics. If the ambient set Xis finite, then the number of elements of
any subset A⊂Xis given by
|A|=X
x∈X
κA(x).
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202 CHAPTER III: MEASURE THEORY
Using the Inclusion-Exclusion formula, we then get
|A1∪ · · · ∪ An|=
n
X
k=1
(−1)k−1X
1≤i1<···<ik≤n
|Ai1∩ · · · Aik|.
This is known as the Inclusion-Exclusion Principle.
Definition. Let Xbe a non-empty set, and let Kbe one of the fields1Q,R
or C. An function φ:X→Kis said to be elementary, if its range φ(X) is finite.
Remark that this gives
φ=X
λ∈φ(X)
λ·κφ−1({λ})=X
λ∈φ(X)r{0}
λ·κφ−1({λ}).
We define
ElemK(X) = {φ:X→K:φelementary}.
Given a collection M⊂P(X), a function φ:X→Kis said to be M-elementary,
if φis elementary, and moreover,
φ−1({λ})∈M,∀λ∈Kr{0}.
We define
M-ElemK(X) = {φ:X→K:φM-elementary}.
Exercise 2.With the above notations, prove that ElemK(X) is a unital K-
algebra.
Proposition 1.1. Given a non-empty set X, the collection P(X)is a unital
ring, with the operations
A+B=A4Band A·B=A∩B, A, B ∈P(X).
Proof. First of all, it is clear that 4is commutative.
To prove the associativity of 4, we simply observe that
κ(A4B)4C=κA4B+κC−2κA4BκC=
=κA+κB−2κAκB+κC−(κA+κB−2κAκB)·κC=
=κA+κB+κC−2(κAκB+κAκC+κBκC)+2κAκBκC.
Since the final result is symmetric in A, B, C, we see that we get
κA4(B4C)=κ(A4B)4C,
so we indeed get
(A4B)4C=A4(B4C).
The neutral element for 4is 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)=κA∩C+κB∩C−2κA∩CκB∩C=
=κAκC+κBκC−2κAκBκC= (κA+κB−2κAκB)κC=
=κA4BκC=κ(A4B)∩C,
1Kcan 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 Xbe a non-empty set. A ring on Xis a non-empty sub-ring
R⊂P(X). We do not require the unit Xto belong to R, but we do require ∅∈R.
An algebra on Xis a ring Awhich contains the unit X.
Rings and algebras of sets are characterized as follows.
Proposition 1.2. Let Xbe a non-empty set.
A. For a non-empty collection R⊂P(X), the following are equivalent:
(i) Ris a ring on X;
(ii) For any A, B ∈R, we have ArB∈Rand A∪B∈R.
B. For a non-empty collection A⊂P(X), the following are equivalent:
(i) Ais an algebra on X;
(ii) For any A∈A, we have XrA∈A, and for any A, B ∈A, we have
A∪B∈A.
Proof. A. (i)⇒(ii). Assume Ris a ring on X, and let A, B ∈R. Then A∩B
belongs to R, so
ArB=A4(A∩B)
also belongs to R. It the follows that
A∪B= (A4B)4(A∩B)
again belongs to R.
(ii)⇒(i). Assume Rsatisfies property (ii). Start with A, B ∈R. Then ArB
belongs to R, and
A∩B=Ar(ArB)
again belongs to R. Since A∪Balso 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 Asatisfies property (ii). Start with two sets A, B ∈A.
Then the complements XrAand XrBboth belong to A, hence their union
(XrA)∪(XrB) = Xr(A∩B)
belongs to A, and the complement of this union
XrXr(A∩B)=A∩B
will also belong to A.
If A, B ∈A, then since XrBbelongs to A, by the above considerations, it
follows that the intersection
A∩(XrB) = ArB
also belongs to A. Likewise, the difference BrAalso belongs to A, hence the union
(ArB)∪(BrA) = A4B
also belongs to A. By part A, it follows that Ais a ring.

204 CHAPTER III: MEASURE THEORY
Finally, since Ais non-empty, if we choose some A∈A, then A4A=∅belongs
to A, so its complement Xr ∅ =Xalso belongs to A.
It will be useful to introduce the following terminology.
Definition. A system of sets (Ai)i∈Iis said to be pair-wise disjoint, if Ai∩
Aj=∅, for all i, j ∈Iwith i6=j.
Lemma 1.1. Let Xbe a non-empty set, let Kbe one of fields Q,Ror C, and
let Rbe a ring on X. For a function φ:X→K, the following are equivalent:
(i) φis R-elementary;
(ii) there exist an integer n≥1and sets A1, . . . , An∈R, and numbers
λ1, . . . , λn∈K, such that
φ=λ1κA1+· · · +λnκAn.
(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=µ1κB1+· · · +µmκBm.
Proof. (i)⇒(ii). Assume φis R-elementary. If φ= 0, there is nothing to
prove, because we have φ=κ∅. If φis not identically zero, then we can obviously
write
φ=X
λ∈φ(X)r{0}
λκφ−1({λ}),
with all sets φ−1({λ}) in R.
(ii)⇒(iii). Define
E={ψ:X→K:ψsatisfies property (iii)}.
Assume φsatisfies (ii), i.e.
φ=λ1κA1+· · · +λnκAn,
with A1, . . . , An∈Rand 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=λκA
for some A∈Rand λ6= 0, in which case we also have φ∈E).
Assume
α1κD1+· · · +αkκDk∈E,
for all D1, . . . , Dk∈R,α1, . . . , αk∈K. Start with a function
φ=λ1κA1+· · · +λkκAk+λk+1κAk+1 ,
with A1, . . . , Ak+1 ∈Rand λ1, . . . , λk+1 ∈K, and based on the above inductive
hypothesis, let us show that φ∈E. Using the inductive hypothesis, the function
ψ=λ2κA2+· · · +λkκAk+λk+1κAk+1
belongs to E, so there exist scalars η1, . . . , ηp∈K, an integer p≥1, and a pair-wise
disjoint system (Cj)p
j=1 ⊂R, such that
ψ=η1κC1+· · · +ηpκCp.
With this notation, we have
φ=λ1κA1+η1κC1+· · · +ηpκCp.