Chapter 2

Lebesgue Measure

This chapter is a stepping stone for the subject of measure and integration. Here, we define the concept of an outer measure of a subset of the real line as a generalization of the concept of length of an interval, and then define the concept of measure of certain subsets of the real line which must satisfy certain intuitively desirable properties. The properties of the class of subsets which can be measured is a motivation for the concept of a general measure in the context of an arbitrary set which we shall consider in the subsequent chapters.

2.1  Lebesgue Outer Measure

In Chapter 1 we defined the concept of a set of measure zero for subsets of R while stating a characterization of Riemann integrability of bounded real valued functions defined on closed and bounded intervals (Definition 1.2.1). We have also observed that every countable subset of R is of measure zero. In particular, the set of rationals in any interval is of measure zero. So, we may ask the following questions:

(1)If a subset E of R contains an open interval (of nonzero length), then can it be of measure zero?

(2)If a subset E of R contains no open interval, is it of measure zero?

(3)Can an uncountable subset E of R be of measure zero?

Let us wait for a while to answer these questions.

Given a set E⊆R, we shall denote by IE, the collection of all countable family {I_n} of open intervals which covers E, that is, E⊆∪nIn.

Definition 2.1.1 The Lebesgue outer measure of E⊆R is defined as

m∗(E):=infIE∑nℓ(In),

where the infimum is taken over IE.♢

Note that m*(E) ≥ 0, and m*(E) can take the value ∞ as well.

Definition 2.1.2 The map m∗:E↦m∗(E) from 2R, the power set of R, to [0, ∞] is called the Lebesgue outer measure on R.♢

In due course, we shall refer to m*, simply as outer measure.

One would expect that if A₁ and A₂ are any two disjoint subsets of R, then we must have

m∗(A1∪A2)=m∗(A1)+m∗(A2).

We shall see that, this is not necessarily true, and for the above to be true, we have to restrict the outer measure to a class of subsets of R, called Lebesgue measurable sets. Although, the class of Lebesgue measurable sets is going to be very large, it is important to know that there are non-Lebesgue measurable sets. Before, introducing this class of Lebesgue measurable sets, we observe some properties of the Lebesgue outer measure.

Theorem 2.1.3 The following results hold.

(i)If I is an open interval and A ⊆ I, then m*(A) ≤ ℓ(I).

(ii)m∗(∅)=0.

(iii)If E is a countable subset of R, then m*(E) = 0.

Proof. (i) Let A ⊆ I, where I is an open interval. Taking the singleton family {I_n} with I_n = I and n = 1, we obtain m*(A) ≤ ℓ(I).

(ii) For every ɛ > 0, we have ∅⊆(−ε,ε). Hence by (i), m∗(∅)≤2ε. This is true for every ɛ > 0. Hence, m∗(∅)=0.

(iii) First let E be a finite set, say E={a1,…,ak}⊆R. Then for every ɛ > 0, E⊆∪i=1kIi, where I_i = (a_i − ɛ, a_i + ɛ). Hence, m*(E) ≤ 2kɛ. Since this is true for every ɛ > 0, m*(E) = 0. Next suppose that E is a countably infinite set, say E={ai:i∈N}. Then taking

In:=(an−ε/2n+1,an+ε/2n+1),

we have E⊆∪n∈NIn so that

m∗(E)≤∑n∈Nℓ(In)=∑n∈N(ε/2n)≤ε.

Since this is true for every ɛ > 0, we obtain m*(E) = 0.▮

The following theorem is very useful in deriving many properties of the outer measure.

Theorem 2.1.4 Let E⊆R. For every ɛ > 0, there exists a countable family {I_n} of open intervals such that E⊆∪nIn and

∑nℓ(In)≤m∗(E)+ε.

Strict inequality occurs in the above if m*(E) < ∞.

Proof. If m*(E) = ∞, then the conclusion is true trivially. So, assume that m*(E) < ∞. Then the results follow from the definition of infimum of a subset of R.▮

By Theorem 2.1.4, for E⊆R, m*(E) = 0 if and only if for every ɛ > 0, there exists a countable family {I_n} of open intervals such that

E⊆∪nInand∑nℓ(In)≤ε.

Thus, we can conclude:

E is of measure zero if and only if its outer measure is zero.

Corollary 2.1.5 Let E⊆R. For every ɛ > 0, there exists an open set G in R such that

E⊆Gandm∗(G)≤m∗(E)+ε.

Proof. Let ɛ > 0 be given. By Theorem 2.1.4, there exists a countable family {I_n} of open intervals such that E⊆∪nIn and

∑nℓ(In)≤m∗(E)+ε.

Take G=∪nIn. Then, by the definition of m*, m∗(G)≤∑nℓ(In). Thus, we have m*(G) ≤ m*(E) + ɛ.▮

Remark 2.1.6 In view of Corollary 2.1.5, one may ask the following question:

If E⊆R, then for every ɛ > 0, does there exist an open set G⊃E such that m∗(G∖E)<ε?

We shall see, in Theorem 2.2.21, that the answer to the above question is in the affirmative only if E belongs to a particular class of subsets, called Lebesgue measurable sets.

Notation: Sets considered in this chapter are subsets of R. Also, for E⊆R, we shall use the notation IE to denote the class of all countable family of open intervals {I_n} such that E⊆∪nIn.

For subsets A and B of R, we define

A+B={x+y:x∈A,y∈B}.

Also for E⊆R and a∈R, we denote, E + a = E + {a} so that

E+a:={x+a:x∈E}.

Theorem 2.1.7 Let A and B be subsets of R. Then the following results hold.

(i)If A ⊆ B, then m*(A) ≤ m*(B).

(ii)m∗(A∪B)≤m∗(A)+m∗(B).

(iii)If E ⊆ A and m*(E) = 0, then m∗(A∖E)=m∗(A).

(iv)If E⊆R and x∈R, then m*(E + x) = m*(E).

Proof. (i) Let A ⊆ B and let {In}∈IB. Then {In}∈IA. Hence

m∗(A)≤∑nℓ(In).

Now, taking infimum over all {In}∈IB, we have m*(A) ≤ m*(B).

(ii) If either m*(A) or m*(B) is infinity, then the result holds. Next, assume that both m*(A) and m*(B) are finite. Hence, by Theorem 2.1.4, given ɛ > 0 there exist {I_n} and {J_n} in IA and IB, respectively, such that

∑nℓ(In)<m∗(A)+ε2,∑nℓ(Jn)<m∗(B)+ε2.

Then, the collection {In}n=1∞∪{Jk}k=1∞ of open intervals covers A∪B. Therefore,

m∗(A∪B)≤∑nℓ(In)+∑nℓ(Jn)≤(m∗(A)+ε2)+(m∗(B)+ε2)=m∗(A)+m∗(B)+ε.

This is true for all ɛ > 0, so that m∗(A∪B)≤m∗(A)+m∗(B).

(iii) If E ⊆ A and m*(E) = 0, then by (i) and (ii),

m∗(A)≤m∗(E)+m∗(A∖E)=m∗(A∖E)≤m∗(A).

Hence, m∗(A∖E)=m∗(A).

(iv) Suppose E⊆R and x∈R. Given ɛ > 0, let {I_n} in IE be such that ∑nℓ(In)≤m∗(E)+ε. Note that each I_n + x is an open interval with ℓ(I + x) = ℓ(I) and E+x⊆∪n(In+x). Hence,

m∗(E+x)≤∑nℓ(In+x)=∑nℓ(In)=m∗(E)+ε.

This is true for every ɛ > 0. Hence, m*(E + x) ≤ m*(E). Since E = (E + x) + (−x), it follows from the above that m*(E) ≤ m*(E + x). Thus the proof is complete.▮

The property (i) in Theorem 2.1.7 is called the monotonicity property of m*, and the property (iv) is called the translation invariance of m*.

Making use of the monotonicity of m*, we deduce the following.

Corollary 2.1.8 Let E⊆R. Then there exists a set G which is a countable intersection of open sets in R such that

E⊆Gandm∗(G)=m∗(E).

Proof. By Corollary 2.1.5, for each n∈N, there exists an open set G_n in R such that

E⊆Gnandm∗(Gn)≤m∗(E)+1n.

Take G=⋂nGn. Then, E ⊆ G and m*(E) ≤ m*(G) ≤ m*(G_n) for every n∈N, so that

m∗(G)≤m∗(E)+1n∀n∈N.

Letting n tend to infinity, we obtain, m*(G) ≤ m*(E). Thus, we have proved m*(G) = m*(E).▮

Definition 2.1.9 A subset of R is said to be a G_δ-set if it is a countable intersection of open sets, and a subset of R is said to be an F_σ-set if it is a countable union of closed sets.♢

Since a subset of R is closed if and only if its complement is open, using De Morgan’s law, it follows that a subset of R is a Gδ-set if and only if its complement is an Fσ-set.

Note that a Gδ-set need not be open and an Fσ-set need not be closed. For example, the set [0, 1) is neither open nor closed, but it is both a Gδ-set and an Fσ-set, as

[0,1)=∩n=1∞(−1n,1),[0,1)=∪n=1∞[0,nn+1].

In fact, every interval is both Gδ-set and Fσ-set.

In view of Definition 2.1.9, Corollary 2.1.8 can be stated as follows:

For every E⊆R, ∃ a Gδ-set G⊇E such that m*(G) = m*(E).

The following corollary is immediate from Theorem 2.1.7 (ii).

Corollary 2.1.10 For subsets A₁, …, A_n of R,

m∗(∪k=1nAk)≤∑k=1nm∗(Ak).

More generally, we have the following theorem.

Theorem 2.1.11 Suppose Ak⊆R for k∈N. Then

m∗(∪k=1∞Ak)≤∑k=1∞m∗(Ak).

Proof. If m*(A_k) = ∞ for some k∈N, then the result holds trivially. Hence, assume that m*(A_k) < ∞ for every k∈N. Then, by Theorem 2.1.4, for each k∈N and ɛ > 0, there exists {Ik,n}∈IAk such that

∑nℓ(Ik,n)<m∗(Ak)+ε/2k.

Since Ak⊆∪nIk,n, we have ∪k=1∞Ak⊆∪k=1∞∪nIk,n. Therefore,

m∗(∪k=1∞Ak)≤∑k=1∞∑nℓ(Ik,n)≤∑k=1∞(m∗(Ak)+ε/2k)=∑k=1∞m∗(Ak)+ε.

Thus, m∗(∪k=1∞Ak)≤∑k=1∞m∗(Ak)+ε. This is true for every ɛ > 0. Hence the result follows.

The property of m* in Theorem 2.1.11 is called the countable subadditivity of m*.

Now we prove a result that we are waiting for.

Theorem 2.1.12 The Lebesgue outer measure of any interval is its length.

Proof. Let I be an interval.

Case 1. Suppose I = [a, b] with −∞ < a < b < ∞:

Let ɛ > 0 and let I_ɛ := (a − ɛ, b + ɛ). Then,

m∗(I)≤m∗(Iε)≤b−a+2ε.

This is true for all ɛ > 0. Hence, m*(I) ≤ b − a. Thus, it remains to show that m*(I) ≥ b − a. For this, it is enough to show that

b−a≤∑nℓ(In)∀{In}∈II,	(*)

because, in that case we can take infimum over all such {In}∈II and obtain b − a ≤ m*(I). So, let {In}∈II. If ℓ(I_n) = ∞ for some n∈N, then (*) holds trivially. So, we may assume that each I_n is of finite length. By the compactness of I, there exists a finite sub-collection {In1,,…,Ink} of {I_n} such that I⊆∪i=1kIni. Let Ini:=(ai,bi) for i ∈ {1, …, k}. We may assume, without loss of generality, that

(ai,bi)∩[a,b]≠∅andai+1<bi

for i ∈ {1, …, k − 1}. Then

∑i=1kℓ(Ini)=∑i=1k(bi−ai)=bk−a1+∑i=1k−1(bi−ai+1)≥bk−a1≥b−a,

so that

b−a≤∑i=1kℓ(Ini)≤∑nℓ(In).

Thus, we have proved (*).

Case 2. Suppose I is an interval of finite length whose end points a and b with a < b, which is not necessarily a closed interval:

Then for sufficiently small ɛ > 0, we have [a + ɛ, b − ɛ] ⊆ I ⊆ [a − ɛ, b + ɛ]. Hence,

m∗([a+ε,b−ε])≤m∗(I)≤m∗([a−ε,b+ε])

so that by Case 1,

b−a−2ε≤m∗(I)≤b−a+2ε.

Since this is true for every ɛ > 0, it follows that m*(I) = b − a.

Case 3. Suppose I is of infinite length:

In this case, for every M > 0 there exists a closed interval I_M of length M such that I_M ⊆ I. By Case 1, m*(I_M) = M. Thus,

M=m∗(IM)≤m∗(I).

Since this is true for every M > 0, m*(I) = ∞.▮

By Theorem 2.1.3 (iii), outer measure of every countable subset of R is zero, and by Theorem 2.1.12, every interval is of positive measure. Hence, we can conclude:

Every interval is an uncountable set.

Can an uncountable set be of measure 0?

The answer is in the affirmative as the following example of Cantor’s ternary set or simply Cantor set shows.

Example 2.1.13 (Cantor ternary set) Let us first recall how the Cantor ternary set is constructed. Consider the unit interval [0, 1]. Let C₁ be the set obtained after removing its “middle third” J1:=(13,23) from C₀ := [0, 1]. That is

C1=[0,13]∪[23,1].

Next, let C₂ be the set obtained from C₁ after removing the “middle thirds” from each of the two subintervals in C₁. Let the removed set be J₂. Thus,

C2=[0,19]∪[29,13]∪[23,79]∪[89,1].

Continue this procedure to obtain C₃, C₄, and so on. At the n^th stage, we obtain Cn=Cn−1∖Jn, where J_n is the union of the middle thirds of each of the subintervals in C_n−1. Now the Cantor set C is defined by

C=∩n=1∞Cn.

Note that

C1⊃C2⊃C3⊃⋯.

and m*(C₁) = 2/3, m*(C₂) = (2/3)²,     m*(C₃) = (2/3)³, etc., and more generally,

m∗(Cn)=(23)n,n∈N.

Hence,

m∗(C)≤m∗(∩n=1kCn)=m∗(Ck)=(23)k∀k∈N.

Thus, m*(C) = 0. To see that C is an uncountable set, first we recall that every number a ∈ [0, 1] can be written as a series ∑n=1∞an3n, where a_n ∈ {0, 1, 2} for n∈N. It is possible that a number a ∈ [0, 1] can have two different representations; one with only a finite number of terms and the other with an infinite number of terms. For example, the number 13 can be written as

13=13+032+033+034+⋯and13=232+233+234+⋯.

Similarly, the number 23 is represented as

23=03+23+032+033+⋯and23=13+232+233+234+⋯.

More generally, if x ∈ (0, 1] has a finite sum representation, say x=∑j=1kaj3j for some k∈N with ak≠0, then a_k ∈ {1, 2} and hence

ak3k=ak−13k+∑j=1∞23k+j

so that x has an infinite expansion as well.

Now, let [0, 1]. Then a ∈ C if and only if a ∈ C_n for every n. Recall that each C_n consists of 2ⁿ disjoint closed intervals, and the 2ⁿ⁺¹ closed intervals of C_n+1 is obtained by removing the middle one-third from each of the 2ⁿ intervals of C_n. Hence, it is seen that a_n is either 0 or 2. Hence we obtain

C={∑n=1∞bn3n:bn∈{0,2},n∈N}.

Thus, C is in one-one correspondence with the family of all sequences (b_n) with b_n ∈ {0, 2}, and hence, C is an uncountable set.

2.2  Lebesgue Measurable Sets

Now investigate the question whether the equality

m∗(A1∪A2)=m∗(A1)+m∗(A2).	(1)

holds for any two disjoint subsets A₁ and A₂ of R.

Suppose for a moment that (1) is true for any two disjoint sets A₁ and A₂ of R. Then we also have

m∗(∪i=1nAi)=∑i=1nm∗(Ai)	(2)

for any pairwise disjoint sets A₁, …, A_n. Now, consider a denumerable disjoint family {An}n=1∞, that is, An∩Am=∅ whenever n≠m. Then using the equality (2) above, monotonicity (Theorem 2.1.7 (i)), and the subadditivity (Theorem 2.1.11) of m*, we obtain

∑i=1nm∗(Ai)=m∗(∪i=1nAi)≤m∗(∪i=1∞Ai)≤∑i=1∞m∗(Ai)

for every n∈N, so that

∑i=1nm∗(Ai)≤m∗(∪i=1∞Ai)≤∑i=1∞m∗(Ai)

for every n∈N. Taking limit as n → ∞, we obtain

m∗(∪i=1∞Ai)=∑i=1∞m∗(Ai).	(3)

We now show that (3) does not hold for certain denumerable disjoint family {An}n=1∞ of subsets of R, and consequently, there are disjoint subsets A₁ and A₂ for which (1) does not hold.

For this, first we prove the following.

Theorem 2.2.1 There exists a set E ⊂ [0, 1] such that if {r₁, r₂, …} is an enumeration of the rational numbers in [−1, 1] and E_n := E + r_n for n∈N, then {E_n} is a disjoint family satisfying

1≤m∗(∪n=1∞En)≤3.

Proof. Consider the relation ∼ on R by defining

x∼y⟺x−y∈Q.

It can be easily seen that ∼ is an equivalence relation on R. Hence, R is the disjoint union of equivalence classes. Let E be the subset of [0, 1] such that its intersection with each equivalence class is a singleton set. Such a set E exists by using the axiom of choice on the collection

E:={[x]∩[0,1]:x∈[0,1]},

where [x] is the equivalence class of x. We note that if x ∼ y, then the rational number r := x − y satisfies −1 ≤ r ≤ 1. Let {r₁, r₂, …} be the set of all rational numbers in [ − 1, 1]. Let E_n: = E + r_n for n∈N. Then {E_n} is a disjoint family. Indeed, if En∩Em≠∅ for some n≠m, then there exist x, y ∈ E such that x + r_n = y + r_m so that x ∼ y; consequently, x = y so that r_n = r_m, which is not possible. Since E ⊆ [0, 1] and r_n ∈ [ − 1, 1], we have

En=E+rn⊆[−1,2]∀n∈N.

Hence, ∪n=1∞En⊆[−1,2]. Also, for each x ∈ [0, 1], there exists y ∈ E such that x ∼ y so that x ∈ E_n for some n∈N. Hence, [0,1]⊆∪n=1∞En. Thus,

[0,1]⊆∪n=1∞En⊆[−1,2]

so that, using the monotonicity of m*, the required inequality follows.▮

From the above theorem, we deduce the following.

Theorem 2.2.2 Let E and {E_n} be as in Theorem 2.2.1. Then m*(E) > 0 and

m∗(∪n=1∞En)<∑n=1∞m∗(En).

Proof. By the sub-additivity of m* and by Theorem 2.2.1, we have

1≤m∗(∪n=1∞En)≤∑n=1∞m∗(En).

From this, using the fact that m*(E_n) = m*(E) for all n∈N, it follows that m*(E) > 0. Therefore, the relation 1≤m∗(∪n=1∞En)≤3 in Theorem 2.2.1 shows that m∗(∪n=1∞En)=∑n=1∞m∗(En) is not possible.▮

By the above theorem, the relation (3), and hence the relation (1) does not hold for some disjoint family of sets involved. Thus, we have also proved the following result.

Theorem 2.2.3 There exist disjoint subsets A₁ and A₂ of R such that

m∗(A1∪A2)<m∗(A1)+m∗(A2).

The above discussion motivates us to look for a family of sets in which the relation (1), and hence (2) and (3) hold for all possible disjoint family of sets involved.

Definition 2.2.4 A set E⊆R is said to be Lebesgue measurable if

m∗(A)=m∗(A∩E)+m∗(A∩Ec)

for every A⊆R.

Notation: We denote the set of all Lebesgue measurable sets by M.♢

The proof of the following theorem is easy and hence it is left as an exercise (Problem 10).

Theorem 2.2.5 The following results hold.

(i)E∈M⟺m∗(A)≥m∗(A∩E)+m∗(A∩Ec)∀A⊆R.

(ii)E∈M⇒Ec∈M.

(iii)∅∈M,R∈M.

(iv)m∗(E)=0⇒E∈M.

Since countable sets are of zero outer measure, Theorem 2.2.5 implies:

M contains all countable sets. In particular, Q∈M and Qc∈M.

Theorem 2.2.6 Let A₁, A₂ be subsets of R such that A1∩A2=∅. If either A₁ or A₂ belongs to M, then

m∗(A1∪A2)=m∗(A1)+m∗(A2).

Proof. Suppose A1∩A2=∅. Assume A1∈M. Then

m∗(A1∪A2)=m∗((A1∪A2)∩A1)+m∗((A1∪A2)∩A1c).

Since (A1∪A2)∩A1=A1 and (A1∪A2)∩A1c=A2, we obtain the result.▮

The following corollary is immediate from the above theorem.

Corollary 2.2.7 If {A₁, …, A_n} is a disjoint family in M, then

m∗(∪i=1nAi)=∑i=1nm∗(Ai).

The proof of the following theorem is along the same lines as we have deduced (3) from (2). However, we give its detailed proof.

Theorem 2.2.8 Let {An:n∈N} be a disjoint family in M. Then

m∗(∪n=1∞An)=∑n=1∞m∗(An).

Proof. If {A_n} is a finite family, then by Corollary 2.2.7,

m∗(∪i=1nAi)=∑i=1nm∗(Ai)

for every n∈N. Hence, by using the monotonicity of m*, we have

∑n=1∞m∗(An)≥m∗(∪n=1∞An)≥m∗(∪i=1nAi)=∑i=1nm∗(Ai)

for all n∈N. Letting n tend to infinity, the result follows.▮

In view of Theorem 2.2.8 and Theorem 2.2.2, we have the following corollary.

Corollary 2.2.9 The set E in Theorem 2.2.1 does not belong to M.

Definition 2.2.10 Those subsets of R which do not belong to M are called non-measurable sets.

Corollary 2.2.9 shows that there are non-measurable sets. The following theorem shows that non-measurable sets are, in fact, plenty.

Theorem 2.2.11 Let A⊆R be such that m*(A) > 0. Then there exists a subset E ⊆ A such that E∉M.

Proof. First let us assume that A is a bounded set. We follow the arguments used in the proof of Theorem 2.2.2.

Consider the equivalence relation ∼ on R as in the proof of Theorem 2.2.2, that is,

x∼y⟺x−y∈Q.

Let E be the subset of A such that its intersection with each equivalence class [x] with x ∈ A is a singleton set. Such a set E exists by using the axiom of choice on the collection

E:={[x]∩A:x∈A},

where [x] is the equivalence class of x. Let a = inf A and b = sup A. Then A ⊆ [a, b]. We note that if x ∼ y, then the rational number r := x − y satisfies a − b ≤ r ≤ b − a. Let {r₁, r₂, …} be the set of all rational numbers in [a − b, b − a]. Let E_n: = E + r_n for n∈N. Then {E_n} is a disjoint family and, for every x ∈ E and n∈N, 2a − b ≤ x + r_n ≤ 2b − a so that

A⊆∪n=1∞En⊆[2a−b,2b−a].

Hence,

0<m(A)≤m∗(∪n=1∞En)≤3(b−a).

Since m*(E_n) = m*(E + r_n) = m*(E) for all n∈N, as in the proof of Theorem 2.2.2,

∑n=1∞m∗(En)=limk→∞∑n=1km∗(En)={0ifm∗(E)=0,∞ifm∗(E)>0.

Hence, m∗(∪n=1∞En)≠∑n=1∞m∗(En). In particular E∉M.

Next, suppose that A is not bounded. Then for every α > 0, Aα:=A∩[−α,α] is a bounded set. Since A=∪α>0Aα and m*(A) > 0, there exists β > 0 such that m*(A_β) > 0. By the arguments in the above paragraph, there exists Eβ⊆Aβ⊆A such that Eβ∉M.

This completes the proof.▮

We have seen that M contains all countable subsets of R. Since the Cantor ternary set is of zero outer measure, it also belongs to M. Now we show that M contains a lot more subsets of R.

Theorem 2.2.12 Let A1,A2∈M. Then A1∪A2∈M. More generally, if {A1,…,An}⊆M, then ∪i=1nAi∈M.

Proof. Let A⊆R. We have to show that

m∗(A)≥m∗(A∩(A1∪A2))+m∗(A∩A1c∩A2c).	(1)

Note that A∩(A1∪A2)=(A∩A1)∪(A∩A2∩A1c), so that

m∗(A∩(A1∪A2))≤m∗(A∩A1)+m∗(A∩A2∩A1c).

Therefore, the right-hand side of (1) is less than or equal to

m∗(A∩A1)+m∗(A∩A2∩A1c)+m∗(A∩A1c∩A2c).	(2)

Now, since A2∈M, we get

m∗(A∩A2∩A1c)+m∗(A∩A1c∩A2c)=m∗(A∩A1c).

Thus, the expression in (2) is less than or equal to

m∗(A∩A1)+m∗(A∩A1c)

which is equal to m*(A), since A1∈M. Thus,

m∗(A∩(A1∪A2))+m∗(A∩A1c∩A2c)≤m∗(A)

which completes the proof of (1). The last part follows by repeated application of the first part.▮

Corollary 2.2.13 If A1,A2∈M, then A1∖A2∈M.

Proof. Let A1,A2∈M. Since

A1∖A2=A1∩A2c=(A1c∪A2)c,

the result follows from Theorem 2.2.12, a complement of any element in M is an element in M.▮

By Theorem 2.2.12 and Corollary 2.2.13, M is closed under finite unions and complementations. Next we show that M is closed under countable unions as well. For this purpose we shall make use of the following lemma which is more general than Corollary 2.2.7.

Lemma 2.2.14 Let {A₁, …, A_n} be a disjoint family in M. Then for any A⊆R,

m∗(A∩∪i=1nAi)=∑i=1nm∗(A∩Ai).

Proof. Let n = 2. Since A1∈M,

m∗(A∩(A1∪A2))=m∗(A∩(A1∪A2)∩A1)+m∗(A∩(A1∪A2)∩A1c).

But,

A∩(A1∪A2)∩A1=A∩A1,A∩(A1∪A2)∩A1c=A∩A2.

Hence, we have

m∗(A∩(A1∪A2))=m∗(A∩A1)+m∗(A∩A2).

Thus, the result is proved for n = 2. The result for general n follows by induction.▮

Theorem 2.2.15 If En∈M for n∈N, then ∪n=1∞En∈M.

Proof. Let E=∪nEn, where En∈M for n∈N, and let A⊆R. We have to show that

m∗(A)≥m∗(A∩E)+m∗(A∩Ec).	(1)

We write E as a disjoint union E=∪nAn where A₁ = E₁ and for n ≥ 2,

An=En∖∪i=1n−1Ei.

Let Fn=∪i=1nAi. Note that Fn∈M, by Theorem 2.2.12 and Corollary 2.2.13. Hence,

m∗(A)≥m∗(A∩Fn)+m∗(A∩Fnc).	(2)

Now, Lemma 2.2.14 implies

m∗(A∩Fn)=∑i=1nm∗(A∩Ai)

and the relation Fnc⊇Ec implies m∗(A∩Fnc)≥m∗(A∩Ec). Thus, from (2) we obtain

m∗(A)≥∑i=1nm∗(A∩Ai)+m∗(A∩Ec).

This is true for all n∈N. Letting n tend to infinity,

m∗(A)≥∑i=1∞m∗(A∩Ai)+m∗(A∩Ec).

Therefore, using the subadditivity of m*,

m∗(A)≥∑i=1∞m∗(A∩Ai)+m∗(A∩Ec)≥m∗(∪i=1∞(A∩Ai))+m∗(A∩Ec)=m∗(A∩∪i=1∞Ai)+m∗(A∩Ec)=m∗(A∩E)+m∗(A∩Ec).

Thus, (1) is proved.▮

M is closed under countable unions and complementations.

In view of Theorem 2.2.15, the property of m* stated in Theorem 2.2.8 is called the countable additivity of m* on M.

Corollary 2.2.16 If En∈M for n∈N, then ⋂n=1∞En∈M.

Proof. By De Morgan’s law, we have ⋂n=1∞En=(∪n=1∞Enc)c. Hence, the result follows by Theorem 2.2.15 and the fact that E∈M implies Ec∈M.▮

Definition 2.2.17 The function m* restricted to M is called the Lebesgue measure on R, and it is denoted by m. For E∈M, m(E): = m*(E) is called the Lebesgue measure of E.

Let us list some of the important properties of the family M that we have proved:

• ∅∈M;
• E∈M⇒Ec∈M;
• {Ei}disjoint family inM⇒∪iEi∈M&m∗(∪iEi)=∑im∗(Ei).

Remark 2.2.18 We may observe that, for a set A∈M, the family

MA:={E⊆A:E∈M}

also has the properties listed in the above box by replacing M by MA. It is to be observed that, in this case, E^c appearing in the box corresponds to the complement of E ⊆ A in A, that is, the set A∖E.

The following theorem shows that the class M is very large.

Theorem 2.2.19 The following results hold.

(i)For any a∈R, the intervals (a, ∞), [a, ∞), (−∞, a) and (−∞, a] belong to M.

(ii)For any a,b∈R with a < b, the intervals (a, b), [a, b), (a, b], and [a, b] belong to M.

(iii)Open subsets of R belong to M.

(iv)Closed subsets of R belong to M.

(v)Every Gδ-set belongs to M.

(vi)Every Fσ-set belongs to M.

Proof. We first prove that (a,∞)∈M for any a∈R, and then deduce other results by using some of the properties of m*. So, let a∈R and E = (a, ∞). Let A⊆R and ɛ > 0. By Theorem 2.1.4, there exists a countable family {I_n} of open intervals such that

A⊆∪nIn,∑nℓ(In)≤m∗(A)+ε.

Note that

A∩E⊆∪n(In∩(a,∞)),A∩Ec⊆∪n(In∩(−∞,a]).

Hence, taking In′:=In∩(a,∞) and In″:=In∩(−∞,a], we have

m∗(A∩E)+m∗(A∩Ec)≤∑nm∗(In′)+∑nm∗(In″)=∑n[m∗(In′)+m∗(In″)].

Note that I_n′ and I_n″ are intervals such that In′∩In″=∅ and In′∪In″=In so that

m∗(In′)+m∗(In″)=ℓ(In′)+ℓ(In″)=ℓ(In).

Thus, we have proved that

m∗(A∩E)+m∗(A∩Ec)≤∑nℓ(In)≤m∗(A)+ε.

This is true for any ɛ > 0, so that we get

m∗(A∩E)+m∗(A∩Ec)≤∑nℓ(In)≤m∗(A).

Hence, (a,∞)=E∈M. Next we observe that for any a∈R,

[a,∞)=∩n=1∞(a−1n,∞),

(−∞,a)=R∖[a,∞),(−∞,a]=R∖(a,∞),

and for any a,b∈R with a < b,

(a,b)=(a,∞)∩(−∞,b),[a,b]=[a,∞)∩(−∞,b].

Thus, every interval listed in (i) and (ii) belongs to M. Also, using the facts that M is closed under countable unions, countable intersections and complementation, and the fact that every open subset of R is a countable union of open intervals, the results listed in (iii)-(vi) follow.▮

Now, we prove a companion result to Corollary 2.1.5 and Corollary 2.1.8, which also answers the question raised in Remark 2.1.6.

First, let us observe the following result.

Proposition 2.2.20 Let E∈M. Then for every ɛ > 0, there exists an open set G in R such that E ⊆ G and m∗(G∖E)≤ε.

Proof. Let ɛ > 0 be given. By Corollary 2.1.5, there exists an open set G in R such that E ⊆ G and m*(G) ≤ m*(E) + ɛ. Since E and G are in M and G∖E=G∩Ec is also in M. Therefore,

m∗(E)+m∗(G∖E)=m∗(G)≤m∗(E)+ε.

Since m*(E) < ∞, we obtain m∗(G∖E)≤ε.▮

Theorem 2.2.21 Let E⊆R. Then the following are equivalent:

(i)E∈M.

(ii)For every ɛ > 0, there exists an open set G in R containing E such that E ⊆ G and m∗(G∖E)≤ε.

(iii)There exists a Gδ-set G in R such that E ⊆ G and m∗(G∖E)=0.

Proof. (i) ⇒ (ii): Suppose E∈M and ɛ > 0 be given. If m*(E) < ∞, then by Proposition 2.2.20, there exists an open set G in R such that E ⊆ G and m∗(G∖E)≤ε. For the case when m*(E) is not necessarily finite, we write E=∪n=1∞En, where m*(E_n) < ∞ for every n∈N. For example, we can take En:=E∩[−n,n] for n∈N. Since m*(E_n) < ∞ for each n∈N, again by Proposition 2.2.20, there exists open set Gn⊇En such that m∗(Gn∖En)<ε/2n. Taking G=∪n=1∞Gn, we have G is open, G⊇E and

G∖E=[∪n=1∞Gn]∖[∪n=1∞En]⊆∪n=1∞(Gn∖En).

Therefore,

m∗(G∖E)≤∑n=1∞m∗(Gn∖En)≤∑n=1∞ε2n=ε.

Thus, (ii) holds.

(ii) ⇒ (iii): Assume (ii). Then, for every n∈N, there exists an open set V_n in R such that

E⊆Vnandm∗(Vn∖E)≤1n.

Let G=⋂nVn. Then E ⊆ G ⊆ V_n and G∖E⊆Vn∖E for all n∈N so that

m∗(G∖E)≤1n∀n∈N.

Letting n tend to infinity, we obtain m∗(G∖E)=0. Thus, (iii) holds.

(iii) ⇒ (i): Assume (iii). Then there exists a Gδ-set G in R such that E ⊆ G and m∗(G∖E)=0. Therefore, G∖E∈M. We know that G∈M so that by Corollary 2.2.13,

E=G∖(G∖E)∈M.

This completes the proof.▮

Corollary 2.2.22 Let E⊆R. Then E∈M if and only if there exist a Gδ-set G and an Fσ-set F such that F ⊆ E ⊆ G and m∗(G∖F)=0.

Proof. Suppose E∈M. Then by Theorem 2.2.21, there exists a Gδ-set G such that E ⊆ G and m∗(G∖E)=0. Also, by taking E^c in place of E, there exists a Gδ-set H such that E^c⊆ H and m∗(H∖Ec)=0. Then F := H^c is an Fσ-set satisfying

F=Hc⊆EandE∖F=E∖Hc=E∩H=H∖Ec

so that m∗(E∖F)=m∗(H∖Ec)=0. Since G∖F=(G∖E)∪(E∖F) we obtain m∗(G∖F)=0.

Conversely, suppose that there exists a Gδ-set G and an Fσ-set F such that F ⊆ E ⊆ G and m∗(G∖F)=0. In particular, G∖E⊆G∖F so that m∗(G∖E)=0 and hence G∖E∈M. Therefore, by Corollary 2.2.13, E=G∖(G∖E)∈M.▮

Using Theorem 2.2.21, we obtain the following theorem. The details of its proof are left as an exercise.

Theorem 2.2.23 Let E⊆R. Then the following are equivalent:

(i)E∈M.

(ii)For every ɛ > 0, there exists a closed set F in R such that

F⊆Eandm∗(E∖F)≤ε.

(iii)There exists an Fσ-set F in R such that

F⊆Eandm∗(E∖F)=0.

2.3  Problems

1. Let X be a set and A be a family of subsets of X. Show that A is an algebra on X if and only if X∈A, and A,B∈A implies A∖B∈A and A∪B∈A.
2. Prove that, in Definition 2.1.1, m*(E) remains the same if we take IE to be the collection of all countable family {I_n} of intervals of finite length, not necessary open intervals.
   [Hint: Given any open interval I and ɛ > 0, there exist intervals I₁ and I₂ containing one or both end points such that I₁ ⊆ I ⊆ I₂ and |ℓ(I) − ℓ(I_i)| < ɛ for i = 1, 2.]
3. Prove that, for any E⊆R, m∗(E)=inf{m∗(G):G⊇E,Gopen}.
   [Hint: Use Corollary 2.1.5.]
4. Show that, if E ⊆ A and m*(E) = 0, then m∗(A∪E)=m∗(A).
5. Show that, every interval is a Gδ-set and an Fσ-set.
6. Prove that if E is a countable subset of R, then R∖E is a Gδ set, and in particular, R∖Q is a Gδ set.
7. From Theorem 2.1.11, deduce that outer measure of every countable set is 0.
8. Deduce Corollary 2.1.8 from Theorem 2.1.11.
9. If E is a subset of an interval I such that m*(E) = 0, then prove that I∖E is dense in I.
   [Hint: Note that, if J ⊆ I is an interval which does not contain any point from I∖E, then J ⊆ E and hence m*(J) = 0.]
10. Prove Theorem 2.2.5.
11. Suppose {A₁, …, A_n} is a disjoint family of subsets of R such that A₁, …, A_n−1 belong to M. Then show that m∗(∪i=1nAi)=∑i=1nm∗(Ai).
    [Hint: Theorem 2.2.6.]
12. Let M~=M∪{E0}, where E0∉M. Prove that if {A_n} is a countable disjoint family in M~, then m∗(∪nAn)=∑nm∗(An).
    [Hint: Use Problem 11.]
13. Justify the statement: There exists a countably infinite disjoint family N of subsets of R such that N∩M=∅.
    [Hint: Theorem 2.2.11.]
14. Deduce Corollary 2.2.7 from Theorem 2.2.8.
15. Let E⊆R be such that m*(E) < ∞. Prove that the following are equivalent:
    

    (a)E∈M.

    (b)There exists a Gδ set G⊇E and E0⊆R with m*(E₀) = 0 such that E=G∖E0.

    (c)There exists an Fσ set F ⊆ E and F0⊆R with m*(F₀) = 0 such that E=F∪F0.

16. If A,B∈M with A ⊆ B and if m(A) < ∞, then show that
    m(B∖A)=m(B)−m(A).
17. If A,B∈M, prove that
    m(A∪B)+m(A∩B)=m(A)+m(B).
18. Let A,B∈M such that A ⊂ B and m(B) = m(A) < ∞. If C⊆R is such that A ⊆ C ⊆ B, then prove that C∈M and m(C) = m(A).
19. If E∈M, then prove that m(E)=sup{m(K):K⊆E,Kcompact}.
    [Hint: You may use Problem 3. Recall that a subset of R is compact if and only if it is closed and bounded.]
20. Find an open dense subset A of R with m(A) < ∞.
    [Hint: Use density of the set of rational numbers.]
21. Show that for every ɛ > 0, there exists an open dense subset A of R such that m(A) < ɛ.
    [Hint: Use arguments similar to those used in solving Problem 20.]
22. Find a closed subset B of R with empty interior and with m(B) = ∞.
    [Hint: Use Problem 21.]
23. Find a subset E of [0, 1] which is a countable union of closed nowhere dense sets such that m(E) = 1.
    [Hint: Use Problem 20. Recall that a set A is said to be nowhere dense if its closure has an empty interior.]
24. Verify the statement in Remark 2.2.18.
25. Supply details of the proof of Theorem 2.2.23.