Measure, Integration, and Differentiation...Measure, Integration, and Differentiation Emile...

PART TWO

Measure, Integration, andDifferentiation

Emile Felix-Edouard-Justin Borel(1871–1956)

Emile Borel was born at Saint-Affrique, France, onJanuary 7, 1871, the third child of Honore Borel, aProtestant minister, and Emilie Teissi-Solier. Emilehad two older sisters who were 14 and 16 years oldwhen he was born.

From an early age, Borel exhibited a strong pro-clivity for mathematics. At the age of 11, he wentto live with his eldest sister, Madame Lebeau. This

move allowed Borel to attend the Lycee at Montauben, where he showed ex-traordinary talents. Several years later, Borel went to Paris, where he tookcourses at the Lycee Louis-le-Grand in preparation for taking the examinationsto enter the Ecole Polytechnique and the Ecole Normale. He ranked first inboth of these examinations and could choose either of the two universities.In 1890, Borel entered the Ecole Polytechnique in Paris, where he graduatedin 1893. He received his doctorate from the Ecole Normale Superieure in 1894.

Borel’s most important research was done in the 1890s when he worked onprobability, infinitesimal calculus, divergent series, and measure theory. In 1896,he provided the proof of Picard’s theorem, a proof that mathematicians hadbeen seeking for nearly 20 years. Although John von Neumann is credited asthe founder of game theory, Borel completed a series of papers on the subjectbetween 1921 and 1927, thus being the first to define games of strategy.

After WW I, Borel developed an interest in politics, serving as Minister ofthe Navy from 1925–1940. He was arrested and briefly imprisoned by the Vichyregime in 1940, after which he worked in the Resistance. His honors includedthe Resistance medal in 1945, the Croix de Guerre in 1918, the Grand Cross ofthe Legion d’Honneur in 1950, and the first gold medal of the Centre Nationalde la Recherche Scientifique in 1955.

Borel was appointed to the faculty of the Ecole Normale Superieure in 1896,held the Chair in Function Theory at the Sorbonne from 1909 until 1940, andwas director and founding member of the Henri Poincare Institute from 1928until his death on February 3, 1956, in Paris.

80

3Lebesgue Measure on theReal Line

In Chapter 2, we discussed open sets, continuous functions, and the Riemannintegral. Those classical concepts have served mathematics and its applicationswell. However, for the purposes of modern mathematics, a more general andsophisticated framework is required. In this chapter and the next, we will takethe first steps toward obtaining that framework.

Here we will first expand the collection of continuous functions to the collec-tion of Borel measurable functions, the smallest algebra of functions that containsthe continuous functions and is closed under pointwise limits. In doing so, we willbe led to consider the collection of Borel sets, the smallest σ-algebra of subsetsof R that contains the open sets.

Next we will generalize the concept of length so that it applies to all Borelsets. In fact, that generalization will be to an even larger collection of sets thanthe Borel sets, namely, the Lebesgue measurable sets.

3.1 BOREL MEASURABLE FUNCTIONS AND BOREL SETS

In Chapter 2, we showed that the collection of continuous, real-valued functionsforms an algebra but is not closed under pointwise limits. As this latter propertyis essential in modern mathematical analysis, we will enlarge the collection ofcontinuous functions to a collection of functions that is closed under pointwiselimits.

Specifically, we will consider the smallest algebra of (real-valued) functionsthat contains the continuous functions and is closed under pointwise limits. Suchan algebra of functions exists — it is the intersection of all algebras of functionsthat contain the continuous functions and are closed under pointwise limits.

81A Course in Real Analysis. Second edition. Copyright © 1999, 2013, Elsevier Inc. All rights reserved.

82 Chapter 3 Lebesgue Measure on the Real Line

Note: The aforementioned intersection is not vacuous because the collection ofall real-valued functions is an algebra that contains the continuous functions andis closed under pointwise limits.

As we will see presently, the condition of being an algebra is superfluous. Thatis, the smallest collection of functions that contains the continuous functions andis closed under pointwise limits is necessarily an algebra of functions. Thus, wemake the following definition:

DEFINITION 3.1 Borel Measurable Functions

We denote by C the smallest collection of real-valued functions on R that con-tains the collection of continuous functions and is closed under pointwise limits.The members of C are called Borel measurable functions.

THEOREM 3.1

The collection C of Borel measurable functions forms an algebra. That is, if fand g are Borel measurable and α ∈ R, then

a) f + g is Borel measurable.b) αf is Borel measurable.c) f · g is Borel measurable.

PROOF We prove only part (a); parts (b) and (c) are left as exercises. First of all, letg ∈ C (the collection of continuous functions on R) and set

D = { f ∈ C : f + g ∈ C }.If f ∈ C, then f ∈ C and f + g ∈ C ⊂ C. Thus, D ⊃ C. Now suppose that{fn}∞n=1 ⊂ D and that fn → f pointwise. Then fn ∈ C and fn + g ∈ C for all

n ∈ N and fn + g → f + g pointwise. Since C is closed under pointwise limits,we conclude that f ∈ C and f + g ∈ C. Hence, f ∈ D. Therefore, we see thatD is closed under pointwise limits.

The previous paragraph shows that D contains the continuous functions andis closed under pointwise limits. Since, by definition, C is the smallest suchcollection of functions, it follows that D ⊃ C. But, by the definition of D, D ⊂ C.Thus, D = C; in other words, f + g ∈ C whenever f ∈ C and g ∈ C.

Next, let f ∈ C and set

E = { g ∈ C : f + g ∈ C }.It follows from what we just proved that E contains the continuous functions.Moreover, the same argument that we used to show that D is closed underpointwise limits shows that E is closed under pointwise limits. Hence, E = C;that is, f + g ∈ C whenever f ∈ C and g ∈ C.

Borel Sets

In Chapter 2, we discovered that there is a natural association between thecontinuous functions and the collection of open sets: A function is continuous

3.1 Borel Measurable Functions and Borel Sets 83

if and only if the inverse image of each open set is open. Now we ask whichcollection of sets corresponds naturally to the Borel measurable functions. Aswe will see, it is the collection of sets whose characteristic functions are Borelmeasurable functions.

DEFINITION 3.2 Borel Sets

A set B ⊂ R is called a Borel set if its characteristic function is Borel measur-able. The collection of all Borel sets is denoted B. So, B = {B ⊂ R : χB ∈ C }.

To begin, we will prove that the Borel sets form a σ-algebra of subsets of R.In order to accomplish this, we will need several lemmas. The proof of the firstlemma is left as an exercise for the reader (see Exercise 3.2).

LEMMA 3.1

Let h denote the absolute value function, that is h(x) = |x|, x ∈ R. Then thereis a sequence {pn}∞n=1 of polynomials such that pn → h pointwise.

Next we introduce the notation used for the maximum and minimum of twofunctions and prove that the Borel measurable functions are closed under thosetwo operations.

DEFINITION 3.3 Maximum and Minimum of Two Functions

Let f and g be real-valued functions on R. Then we define f ∨ g = max{f, g}and f ∧ g = min{f, g}. That is,

(f ∨ g)(x) = max{f(x), g(x)}and

(f ∧ g)(x) = min{f(x), g(x)}.

LEMMA 3.2

If f and g are Borel measurable functions, then so are f ∨ g and f ∧ g. More gen-erally, if f1, f2, . . . , fn are Borel measurable functions, then so are f1 ∨ · · · ∨ fnand f1 ∧ · · · ∧ fn.

PROOF We first note that the following identities hold:

f ∨ g =1

2

(f + g + |f − g| ) (3.1)

and

f ∧ g =1

2

(f + g − |f − g| ). (3.2)


Next we show that |F | ∈ C whenever F ∈ C. From Lemma 3.1, we can choosea sequence of polynomials {pn}∞n=1 such that pn(x) → |x| for all x ∈ R. Be-

cause C is an algebra of functions (Theorem 3.1), it follows that pn ◦ F ∈ C for

all n ∈ N . However, pn ◦ F → |F | pointwise and, consequently, as C is closed

under pointwise limits, |F | ∈ C.

Now suppose that f , g ∈ C. Then |f − g| ∈ C (why?). Using (3.1) and (3.2)

and the fact that C is an algebra, we deduce that f ∨ g ∈ C and f ∧ g ∈ C. Theremaining conclusions of the lemma follow by mathematical induction.

LEMMA 3.3

If {fn}∞n=1 is a sequence of Borel measurable functions with {fn(x)}∞n=1 boundedfor each x ∈ R, then supn fn and infn fn are Borel measurable.

PROOF By Lemma 3.2, if f1, f2, . . . , fn are Borel measurable, then so are f1 ∨ · · · ∨ fnand f1 ∧ · · · ∧ fn. But,

supn

fn = limn→∞ f1 ∨ · · · ∨ fn

and

infn

fn = limn→∞ f1 ∧ · · · ∧ fn.

The lemma now follows because C is closed under pointwise limits.

Now that we have established Lemmas 3.1–3.3, we can prove that the collec-tion B of Borel sets is a σ-algebra of subsets of R.

THEOREM 3.2

The collection of Borel sets B = {B ⊂ R : χB ∈ C } is a σ-algebra of subsetsof R.

PROOF We first show that the collection of Borel sets is closed under complementation.Assume B ∈ B. Then, by definition, χB ∈ C. Since 1 ∈ C and C is an algebra,we conclude that 1 − χB ∈ C. But 1 − χB = χBc and, consequently, Bc ∈ B.

Next we show that the collection of Borel sets is closed under countable unions.Suppose that Bn ∈ B, for n ∈ N . Then χBn ∈ C for n ∈ N and, therefore, by

Lemma 3.3, supn χBn ∈ C. But supn χBn = χ⋃∞n=1

Bn. Hence,

⋃∞n=1 Bn ∈ B.

Further Properties of Borel Sets and Borel Measurable Functions

It is left as an exercise for the reader to show that if O is an open set, then χO isa Borel measurable function. In other words, every open set is a Borel set. Wewill prove shortly that, in fact, B is the smallest σ-algebra that contains all theopen sets.

But first, we will justify our contention that it is natural to associate theBorel sets with the Borel measurable functions, as we do the open sets withthe continuous functions. Specifically, we will show that a function is Borel


measurable if and only if the inverse image of each open set is a Borel set. Inorder to accomplish this, we will introduce another collection of functions which,as we will see, turns out to be identical to the collection of Borel measurablefunctions.

LEMMA 3.4

Let F = { f : f−1(O) ∈ B for all open sets O }. Then F contains the continuousfunctions.

PROOF Suppose that f is a continuous function on R. Then, by Theorem 2.5 on page 55,f−1(O) is open whenever O is open. But every open set is a Borel set (Ex-ercise 3.3). Therefore, f−1(O) ∈ B whenever O is open. This shows that Fcontains the continuous functions.

LEMMA 3.5

f ∈ F if either of the following conditions hold:a) For each a ∈ R, f−1

((−∞, a)

) ∈ B.

b) For each a ∈ R, f−1((a,∞)

) ∈ B.

PROOF We will prove part (a). The proof of part (b) is similar and is left as an exercise.So, suppose that f satisfies the condition in part (a). We claim that f ∈ F ; thatis, f−1(O) ∈ B for all open sets O. Set

A = {A ⊂ R : f−1(A) ∈ B }.Because f−1(Ac) = [f−1(A)]c, f−1(

⋃n An) =

⋃n f

−1(An), and B is a σ-algebra,it follows that A is a σ-algebra.

Now, by assumption, A contains all sets of the form (−∞, α), where α ∈ R. Ifa ∈ R, then we can write (−∞, a] =

⋂∞n=1(−∞, a + 1/n); therefore, (−∞, a] ∈ A

because A is a σ-algebra. This in turn implies that (a, b) ∈ A for each a, b ∈ R,since (a, b) = (−∞, b) ∩ (−∞, a]c. It now follows easily that A contains all openintervals. But, by Proposition 2.13 on page 48, every open set is a countableunion of open intervals. Consequently, A contains all open sets. This meansthat f−1(O) ∈ B for all open sets O; that is, f ∈ F .

LEMMA 3.6

F is closed under pointwise limits.

PROOF Suppose that {gn}∞n=1 ⊂ F and let g = supn gn. Then, for each a ∈ R, wehave g−1

((a,∞)

)=⋃∞

n=1 g−1n

((a,∞)

) ∈ B. Therefore, by the preceding lemma,supn gn ∈ F . Similarly, infn gn ∈ F .

Now suppose that {fn}∞n=1 ⊂ F and that fn → f pointwise. Then, for eachx ∈ R, limn→∞ fn(x) = f(x) and, so, lim supn→∞ fn(x) = f(x). Consequently,f = infn{supk≥n fk}. Let gn = supk≥n fk. Then the previous paragraph showsin turn that gn ∈ F for all n ∈ N and infn gn ∈ F . Hence, f ∈ F .

COROLLARY 3.1

F contains the Borel measurable functions.


PROOF By Lemmas 3.4 and 3.6, F contains the continuous functions and is closed underpointwise limits. Since the collection of Borel measurable functions C is thesmallest collection of functions that contains the continuous functions and isclosed under pointwise limits, it must be that F ⊃ C.

LEMMA 3.7

Let f ∈ F . Then there is a sequence {fn}∞n=1 of Borel measurable functions suchthat fn → f pointwise.

PROOF First of all, note that if a, b ∈ R, then f−1([a, b)) ∈ B (why?). For n ∈ N , let

Enk =

{x :

k

n≤ f(x) <

k + 1

n

}= f−1

([k

n,k + 1

n

))for k = 0, ±1, ±2, . . . . Then Enk ∈ B and so χEnk

∈ C. Since C is an algebraof functions and is closed under pointwise limits, the function

fn =

∞∑k=−∞

k

nχEnk

= limk→∞

k∑j=−k

j

nχEnj

is in C. It is easy to see that |f(x) − fn(x)| < 1/n for all x ∈ R. Hence fn → fpointwise (in fact, the convergence is uniform).

Using the preceding results, it is now evident that F and the collection of Borelmeasurable functions are identical. Specifically, we have the following theorem.

THEOREM 3.3

A function f is Borel measurable if and only if the inverse image of each openset under f is a Borel set; that is, if and only if f−1(O) ∈ B for all open sets O.

PROOF By Corollary 3.1, F ⊃ C. Conversely, suppose that f ∈ F . Then, by Lemma 3.7,f is the pointwise limit of functions in C. Since C is closed under pointwise limits,this implies f ∈ C. Thus, C ⊃ F .

We mentioned previously that the collection of Borel sets B is the smallestσ-algebra of sets that contains the open sets. Here is a proof of that result.

THEOREM 3.4

The collection of Borel sets B is the smallest σ-algebra of subsets of R thatcontains all the open sets.

PROOF We already know that B is a σ-algebra that contains all the open sets. Let Abe any σ-algebra that contains all the open sets. We claim that B ⊂ A. LetG = { f : f−1(O) ∈ A for all open sets O }. The arguments used for Lemmas 3.4and 3.6 depend only on the fact that B is a σ-algebra containing the open sets.It follows that G contains the continuous functions and is closed under pointwiselimits; thus, G ⊃ C. This last fact implies that χB ∈ G for all B ∈ B. But then,for each B ∈ B, B = χ−1

B

((1/2, 3/2)

) ∈ A. Thus B ⊂ A.


Remarks: In many texts, the collection of Borel sets B is defined to be thesmallest σ-algebra of sets that contains the open sets; and a function is definedto be Borel measurable if the inverse image of each open set is a Borel set. Aswe see from Theorems 3.3 and 3.4, the definitions presented here (Definitions 3.1and 3.2) are equivalent to those. It seems more natural, though, to introducethe Borel measurable functions in a way that is motivated by a defect in thecollection of continuous functions; namely, the defect of not being closed underpointwise limits. Once this introduction is accomplished, however, it may indeedbe easier to think of Borel measurable functions via Theorem 3.3 and Borel setsvia Theorem 3.4. Moreover, those characterizations will be used as a means todefine Borel sets and Borel measurable functions in more general contexts.

Here now are several examples that illustrate Borel measurable functions andBorel sets. We have left some of the justifications as exercises for the reader.

EXAMPLE 3.1 Illustrates Borel Measurable Functions and Borel Sets

a) Because B is a σ-algebra containing the open sets, it follows that all opensets, closed sets, and intervals are Borel sets.

b) If B is a countable set, then B ∈ B; in particular, Q ∈ B. From this, it alsofollows that the set of irrational numbers R \Q is in B.

c) By definition, any continuous function is Borel measurable.d) χQ is Borel measurable because Q ∈ B. Indeed, if B ∈ B, then χB is Borel

measurable by the definition of B.e) If B1, . . . , Bn ∈ B and α1, . . . , αn ∈ R, then f =

∑nk=1 αkχBk

is Borel mea-

surable. This fact follows immediately from part (d) and the fact that C is analgebra of functions. In particular, all step functions are Borel measurable.

f) Every monotone function is Borel measurable, as the reader can easily verifyby applying Lemma 3.5.

g) A real-valued function f on R that is 0 except on a countable set is Borel mea-surable. Indeed, suppose K is countable and f(x) = 0 for x /∈ K. Let {xn}nbe an enumeration of K. Then f =

∑n f(xn)χ{xn}. If K is finite, then f is

Borel measurable by parts (a) and (e). If K is infinite, then f is the pointwiselimit of the Borel measurable functions

∑nk=1 f(xk)χ{xk}, n ∈ N , and, hence,

is itself Borel measurable.

Borel Measurable Functions and Borel Sets on Subsets of RWe conclude this section with a brief discussion of Borel measurable functionsand Borel sets when the underlying space is some subset D ⊂ R. The pertinentdefinitions and theorems are obvious modifications of those discussed earlier.

DEFINITION 3.4 Borel Measurable Functions on D

We denote by C(D) the smallest collection of real-valued functions on D thatcontains the continuous functions on D and is closed under pointwise limits. Themembers of C(D) are called Borel measurable functions on D.


DEFINITION 3.5 Borel Sets of D

A set B ⊂ D is called a Borel set of D if its characteristic function is a Borelmeasurable function on D. The collection of all Borel sets of D is denoted B(D).

Thus, B(D) = {B ⊂ D : χB ∈ C(D) }.

Using arguments similar to those used earlier, we can obtain the followingtheorems:

THEOREM 3.5

A function f is Borel measurable on D if and only if the inverse image of eachopen set under f is a Borel set in D, that is, f−1(O) ∈ B(D) for all open sets O.

THEOREM 3.6

The collection of Borel sets of D, B(D), is the smallest σ-algebra of subsets of Dthat contains all open sets in D.

An interesting and useful characterization of B(D) is given by the followingtheorem. Note the analogy with open sets in D (see Theorem 2.3 on page 51).

THEOREM 3.7

B ∈ B(D) if and only if there is an A ∈ B such that B = D ∩A. That is, theBorel sets of D are precisely the Borel sets (of R) intersected with D.

PROOF Let A = {D ∩A : A ∈ B }. We claim that A = B(D). It is easy to see that A isa σ-algebra of subsets of D and, since B contains all open sets of R, A containsall open sets of D. Thus, by Theorem 3.6, A ⊃ B(D).

Now, let C be any σ-algebra of subsets of D that contains the open sets of Dand set

D = {A ∈ B : D ∩A ∈ C }.Then D contains the open sets of R because C contains the open sets of D.Also, D is a σ-algebra because B and C are. Consequently, D ⊃ B. But, bydefinition, D ⊂ B. Hence, D = B. It now follows that A ⊂ C and, since C was anarbitrary σ-algebra of subsets of D that contains the open sets, we conclude thatA ⊂ B(D). This last result and the previous paragraph show that A = B(D).

Exercises for Section 3.1Note: A ★ denotes an exercise that will be subsequently referenced.

3.1 Prove parts (b) and (c) of Theorem 3.1 on page 82.

★3.2 Let h(x) = |x|. Prove Lemma 3.1 on page 83 by proceeding as follows:a) Show that there exists a sequence of polynomials that converges uniformly to h

on [−1, 1]. Hint: Consider the Taylor series expansion for (1 − t)1/2 on [0, 1].b) Use part (a) to conclude that for each compact subset K of R, there exists a

sequence of polynomials that converges uniformly to h on K. Hint: If b > 0, we canwrite |x| = b · |x/b|.

3.2 Lebesgue Outer Measure 89

c) Use part (b) to conclude that there exists a sequence of polynomials that convergesto h uniformly on each compact subset of R.

d) Deduce Lemma 3.1 from part (c).

3.3 Prove that every open set is a Borel set by showing that for each open set O, χO isa Borel measurable function. Hint: Begin by showing that χI is Borel measurable foreach open interval I.

3.4 Verify part (b) of Lemma 3.5 on page 85.

3.5 Show that f is Borel measurable if and only if f−1(B) ∈ B for all Borel sets B.

3.6 Let D be a dense subset of R. Show that f is Borel measurable if either of the followingconditions holds:a) For each d ∈ D, f−1

((−∞, d)

)∈ B.

b) For each d ∈ D, f−1((d,∞)

)∈ B.

3.7 Show that all closed sets and all intervals are Borel sets.

3.8 Prove that every monotone function is Borel measurable.

3.9 Prove Theorems 3.5–3.7.

3.10 Verify (3.1) and (3.2) on page 83.

3.11 Show that any countable subset of R is a Borel set.

3.12 For subsets A and B of R, define

A+B = { a+ b : a ∈ A and b ∈ B }.

Suppose that B is a Borel set. Prove that A+B is a Borel set if A isa) countable. b) open.

3.13 Most functions encountered in a calculus course can be obtained from the identity func-tion i(x) = x by using the standard operations of algebra (sums, products, quotients,and the extraction of roots) together with the operation of passing to the limit in asequence of functions. For example,

ex1/2

= limn→∞

n∑k=0

(x1/2)k

k!.

Explain why any function obtained using the forementioned operations is a Borel mea-surable function.

3.2 LEBESGUE OUTER MEASURE

In the previous section, we enlarged our basic collection of functions from thecontinuous functions to the Borel measurable functions. Although both of thosecollections of functions are algebras, the latter collection has the advantage ofbeing closed under pointwise limits.

Our next goal is to extend the Riemann integral to an integral that applies toall Borel measurable functions. The extension is not trivial since there are Borelmeasurable functions that are not Riemann integrable. Indeed, as we learned inTheorem 2.7 on page 72, a bounded function is Riemann integrable if and onlyif it is continuous except on a set of measure zero. There are certainly Borelmeasurable functions that do not satisfy this last condition (e.g., χQ).


Referring to Section 2.6, beginning on page 67, we see that the Riemann inte-gral is developed by first defining the integral of a step function h =

∑nk=1 akχIk

on [a, b] to be ∫ b

a

h(x) dx =

n∑k=1

ak�(Ik),

where �(I) denotes the length of an interval I. Therefore, the definition of theRiemann integral ultimately depends on the concept of length, which appliesonly to intervals of real numbers.

To obtain an integral that applies to all Borel measurable functions, weproceed by analogy with the development of the Riemann integral. Specifi-cally, we must first define the integral of a Borel measurable function of theform s =

∑akχBk

, where the Bks are Borel sets. If the Bks are intervals, then sis a step function and we simply define the integral to equal the Riemann inte-gral

∑ak�(Bk). If the Bks are not intervals, then what? It seems that we need

to generalize the concept of length so that it applies to arbitrary Borel sets.

The Definition of Lebesgue Outer Measure

The concept of length will be extended and replaced by that of measure. Aswe will see, this is by no means a simple procedure. Let us denote the requiredmeasure by the Greek letter μ, and the collection of subsets of R to which itapplies by the letter A. Subsets of R that belong to A are called measurablesets. We will now list some properties that μ and A should satisfy.

Since measure is to be a generalization of length, we require that the measureof an interval be its length; that is, μ(I) = �(I) for all intervals I. Also, forpurely mathematical reasons, we require that A be a σ-algebra; and as we wantall Borel sets to be measurable, we require that A ⊃ B.

Now, clearly, the measure of the union of two disjoint intervals should bethe sum of their lengths (measures). More generally, then, we require that themeasure of the union of two disjoint measurable sets be the sum of their measures.That is, if A, B ∈ A and A ∩B = ∅, then

μ(A ∪B) = μ(A) + μ(B). (3.3)

Using mathematical induction, we can show that the previous condition impliesthat if A1, A2, . . . , An ∈ A and Ai ∩Aj = ∅ for i �= j, then

μ

( n⋃k=1

Ak

)=

n∑k=1

μ(Ak). (3.4)

This condition on μ is called finite additivity.For purposes of modern mathematical analysis, we need to impose a somewhat

stronger condition on our measure than finite additivity; namely, that (3.4) holdnot only for finite collections of pairwise disjoint measurable sets but also forcountably infinite collections of pairwise disjoint measurable sets. This conditionis called countable additivity.


In summary, if μ is the required generalization of length and A is the collectionof subsets of R that have a length in this extended sense, then the followingconditions should be satisfied:

(M1) A is a σ-algebra and A ⊃ B.

(M2) μ(I) = �(I), for all intervals I.

(M3) If A1, A2, . . . are in A, with Ai ∩Aj = ∅ for i �= j, then

μ

(⋃n

An

)=∑n

μ(An).

Conditions (M1)–(M3) provide us with the means for extending the notion oflength to all open sets. First, since every open set is a Borel set, Condition (M1)implies that every open set should be measurable. Now, let O be an open set.Then O is a countable union of disjoint open intervals, say O =

⋃n In. Now

applying, in turn, Conditions (M3) and (M2), we infer that

μ(O) = μ

(⋃n

In

)=∑n

μ(In) =∑n

�(In).

So, we now see how to extend the notion of length to all open sets.For sets that are more complicated than open sets, however, it is not at all

obvious what to do. In fact, defining a suitable measure for subsets of R consti-tuted a major problem for mathematicians until the beginning of the twentiethcentury, when Henri Lebesgue found the key. His idea was as follows: For asubset A ⊂ R, consider all open sets that contain A as a subset. Then define themeasure of A to be the greatest lower bound of the measures of all those opensets:

inf{μ(O) : O open, O ⊃ A }. (3.5)

With this definition, we “close down on A” or “come at A from the outside,”so we call this measure of A its outer measure. Outer measure is definedfor all subsets of R. But, as we will see, it is countably additive (i.e., satisfiesCondition (M3)) only when restricted to a proper subcollection of subsets of R.Consequently, we will denote outer measure not by μ, but instead by λ∗.

Below we give a formal definition of outer measure. The definition that wepresent does not use (3.5) but is equivalent to it.

DEFINITION 3.6 Lebesgue Outer Measure

For each subset A ⊂ R, the Lebesgue outer measure of A, denoted by λ∗(A),is defined by

λ∗(A) = inf

{∑n

�(In) : {In}n open intervals,⋃n

In ⊃ A

}.

Note: A sequence of open intervals {In}n appearing in Definition 3.6 can beeither a finite or infinite sequence.


Basic Properties of Lebesgue Outer Measure

Some basic properties of Lebesgue outer measure are proved in the next twopropositions.

PROPOSITION 3.1

Lebesgue outer measure λ∗ has the following properties:a) λ∗(A) ≥ 0, for all A ⊂ R. (nonnegativity)b) λ∗(∅) = 0.c) A ⊂ B ⇒ λ∗(A) ≤ λ∗(B). (monotonicity)d) λ∗(x + A) = λ∗(A) for x ∈ R, A ⊂ R, where x + A = {x + y : y ∈ A }.

(translation invariance)e) If {An}n is a sequence of subsets of R, then

λ∗(⋃

n

An

)≤∑n

λ∗(An). (3.6)

In particular if A, B ⊂ R, then λ∗(A ∪B) ≤ λ∗(A) + λ∗(B). The relationin (3.6) is called countable subadditivity.

PROOF For each E ⊂ R, let

SE =

{∑n

�(In) : {In}n open intervals,⋃n

In ⊃ E

}.

Then, by definition, λ∗(E) = inf{x : x ∈ SE }.a) If A ⊂ R, then SA ⊂ [0,∞] so that λ∗(A) = inf{x : x ∈ SA } ≥ 0.b) For ε > 0, the interval Iε = (−ε/2, ε/2) contains ∅; so, ε = �(Iε) ∈ S∅. Hence,

λ∗(∅) = inf{x : x ∈ S∅ } ≤ ε for all ε > 0. This implies that λ∗(∅) = 0.c) Let u ∈ SB . Then there is a sequence {In} of open intervals such that

B ⊂ ⋃ In and u =∑

�(In). But B ⊂ ⋃ In ⇒ A ⊂ ⋃ In ⇒ u ∈ SA. There-fore, SB ⊂ SA and, consequently,

λ∗(A) = inf{x : x ∈ SA } ≤ inf{x : x ∈ SB } = λ∗(B).

d) The proof of this part is left to the reader as an exercise.e) If λ∗(An) = ∞ for some n, then, by part (c), λ∗(

⋃An) = ∞; consequently,

(3.6) holds. So, assume that λ∗(An) < ∞ for all n. Let ε > 0 be given. Foreach n, choose a sequence {Ink}k of open intervals such that

⋃k Ink ⊃ An and∑

k �(Ink) < λ∗(An) + ε/2n. The collection of intervals {Ink}n,k is countable(because N ×N is countable) and

⋃n,k Ink =

⋃n(⋃

k Ink) ⊃⋃

n An. Hence,

λ∗(⋃

n

An

)≤∑n,k

�(Ink) =∑n

∑k

�(Ink)

≤∑n

(λ∗(An) +

ε

2n

)≤∑n

λ∗(An) + ε.

As ε > 0 was arbitrary, this proves that λ∗(⋃

n An) ≤∑n λ∗(An).


As we have noted, the domain of λ∗ is P(R); that is, every subset of R has anouter measure. Our question now is whether λ∗ is the desired extension of length.That is, do Conditions (M1)–(M3) hold with μ = λ∗ and A = P(R)? Certainly,Condition (M1) holds; and the next proposition shows that Condition (M2)holds also.

PROPOSITION 3.2

The outer measure of an interval is its length. That is, λ∗(I) = �(I) for everyinterval I.

PROOF First assume I = [a, b], that is, that I is a bounded and closed interval. If ε > 0,then (a− ε/2, b + ε/2) ⊃ [a, b] and so

λ∗([a, b]) ≤ �

((a− ε

2, b +

ε

2

))= b− a + ε.

Thus, for any ε > 0, λ∗([a, b]) ≤ b− a + ε and, hence, λ∗([a, b]) ≤ b− a.Consequently, it remains to establish that λ∗([a, b]) ≥ b− a. Let {In} be a se-

quence of open intervals such that⋃In ⊃ [a, b]. We claim that

∑�(In) > b− a.

Since {In} is an open cover for [a, b], the Heine-Borel theorem implies that there

is a finite subcover, say {Ik}Nk=1. Now, clearly,∑N

k=1 �(Ik) ≤∑

n �(In). So, we

need only show that∑N

k=1 �(Ik) > b− a.Because a ∈ [a, b], there must be an interval, say J1 = (a1, b1), in the collec-

tion {Ik}Nk=1 with a1 < a < b1. If b < b1, then

N∑k=1

�(Ik) ≥ �(J1) = b1 − a1 > b− a.

Otherwise, b1 ∈ [a, b], so there must be an interval, say J2 = (a2, b2), in thecollection {Ik}Nk=1 with a2 < b1 < b2. Note that, necessarily, J2 �= J1. If b < b2,then

N∑k=1

�(Ik) ≥ �(J1) + �(J2) = (b1 − a1) + (b2 − a2)

= (b2 − a1) + (b1 − a2) > b2 − a1 > b− a.

Otherwise, b2 ∈ [a, b], so there must be an interval, say J3 = (a3, b3), in thecollection {Ik}Nk=1 such that a3 < b2 < b3 and, necessarily, J3 �= J2 and J3 �= J1.

This process can continue at most N times. Consequently, there is an m ∈ Nwith m ≤ N such that Ji = (ai, bi) ∈ {Ik}Nk=1 for 1 ≤ i ≤ m and

a1 < a, a2 < b1 < b2, . . . , am < bm−1 < bm, b < bm.

Therefore,

N∑k=1

�(Ik) ≥m∑i=1

�(Ji) = (b1 − a1) + (b2 − a2) + · · · + (bm − am)

= (bm − a1) + (b1 − a2) + (b2 − a3) + · · · + (bm−1 − am)

> bm − a1 > b− a.


So, if {In} is a sequence of open intervals with⋃In ⊃ [a, b], then

∑�(In) > b− a.

But then, by definition, λ∗([a, b]) ≥ b− a. This last fact and the previouslyestablished reverse inequality show λ∗([a, b]) = b− a.

Now, let I be any finite (i.e., bounded) interval. Then for each ε > 0, there isa closed interval J with J ⊂ I and �(I) < �(J) + ε (why?). Thus,

�(I) − ε < �(J) = λ∗(J) ≤ λ∗(I).

Since ε > 0 was arbitrary, it follows that λ∗(I) ≥ �(I). But, on the other hand,λ∗(I) ≤ λ∗ (I) = �

(I)

= �(I), so that λ∗(I) ≤ �(I).Finally, if I is an infinite (i.e., unbounded) interval, then for each real num-

ber M , there is a closed interval J of length M with J ⊂ I. It follows thatλ∗(I) ≥ λ∗(J) = �(J) = M . Hence, λ∗(I) = ∞.

We have observed that Conditions (M1) and (M2) are satisfied with μ = λ∗

and A = P(R). Therefore, our question now is: Does Condition (M3) holdwith μ = λ∗ and A = P(R)? If the answer to this question were yes, then λ∗

would be the desired extension of length and every subset of R would be mea-surable. The answer, however, is no! In fact, as we will discover in the nextsection, λ∗ is not even finitely additive.

Exercises for Section 3.2

3.14 Let I be any finite interval. Show that for each ε > 0, there is a closed interval Jwith J ⊂ I and �(I) < �(J) + ε.

3.15 Prove part (d) of Proposition 3.1. That is, show that λ∗(x+A) = λ∗(A) for x ∈ R,A ⊂ R, where x+A = {x+ y : y ∈ A }.

3.16 Suppose that A is a set with λ∗(A) <∞. Show that the function g defined byg(x) = λ∗(A ∩ (−∞, x]

)is uniformly continuous on R.

3.17 Show that the Cantor set P has Lebesgue outer measure zero.

3.18 Show that, for each E ⊂ R, there is a sequence of open sets {On}∞n=1 such thatO1 ⊃ O2 ⊃ · · · ⊃ E and

λ∗(E) = λ∗( ∞⋂

n=1

On

)= lim

n→∞λ∗(On).

3.19 For A ⊂ R and b ∈ R, define bA = { ba : a ∈ A }. Show that λ∗(bA) = |b|λ∗(A).

3.20 Suppose that f :R → R is differentiable at each point of R.a) If |f ′(x)| ≤ 1 for each x ∈ R, prove that for each A ⊂ R, λ∗(f(A)

)≤ λ∗(A). Hint:

Use the mean-value theorem.b) Provide an example to show that the preceding inequality may fail to hold if

|f ′(x)| > 1 for some x ∈ R.

3.3 FURTHER PROPERTIES OF LEBESGUE OUTER MEASURE

Recall that we are trying to extend the notion of length so that it applies toall Borel sets. Specifically, we are searching for a function μ defined on some

3.3 Further Properties of Lebesgue Outer Measure 95

collection A of subsets of R such that




μ

(⋃n

An

)=∑n

μ(An).

In Section 3.2, we discovered that Conditions (M1) and (M2) are satisfiedwith μ = λ∗ (Lebesgue outer measure) and A = P(R). We will prove in thissection that Condition (M3) does not hold with μ = λ∗ and A = P(R).

In fact, we will show that even finite additivity does not obtain. That is, it ispossible to find disjoint subsets A and B of R such that the equation

λ∗(A ∪B) = λ∗(A) + λ∗(B) (3.7)

fails to hold. The idea is to choose A and B so that they are disjoint but“sufficiently intermingled.”

Finite Additivity Properties of λ∗

It is best to begin by determining conditions on disjoint sets A and B that implythat (3.7) holds. Our first result is that if A and B are not only disjoint but area positive distance apart, then (3.7) is true. Before proving that fact, we needsome preliminary definitions and lemmas.

DEFINITION 3.7 Distance Between a Point and a Set or Two Sets

If x ∈ R and E ⊂ R, then the distance from x to E, denoted by d(x,E), isdefined to be

d(x,E) = inf{ |y − x| : y ∈ E }.If E and F are subsets of R, then the distance from E to F , denoted byd(E, F ), is defined to be

d(E,F ) = inf{ |y − x| : y ∈ E, x ∈ F }.

It is left as an exercise for the reader to show that (1) for fixed E ⊂ R, thefunction d(x,E) is continuous, (2) d(E,F ) = inf{ d(y, F ) : y ∈ E }, and (3) ifA ⊂ E and B ⊂ F , then d(E,F ) ≤ d(A,B). The proof of the following lemmais also left to the reader as an exercise.

LEMMA 3.8

Suppose that I is a finite open interval and let ε, δ > 0 be given. Then there area finite number of open intervals, say J1, . . . , Jn, such that �(Jk) < δ, 1 ≤ k ≤ n,I ⊂ ⋃n

k=1 Jk, and∑n

k=1 �(Jk) < �(I) + ε.


LEMMA 3.9

Suppose that A is a subset of R with λ∗(A) < ∞. Then for each ε, δ > 0, thereis a sequence {In} of open intervals such that �(In) < δ for all n,

⋃In ⊃ A,

and∑

�(In) < λ∗(A) + ε.

PROOF Given ε > 0, there is a sequence {Jn} of open intervals such that⋃Jn ⊃ A

and∑

�(Jn) < λ∗(A) + ε/2. By Lemma 3.8, for each Jn, there are a finitenumber of open intervals, say Jn1, Jn2, . . . , Jnkn , with �(Jnj) < δ, 1 ≤ j ≤ kn,⋃kn

j=1 Jnj ⊃ Jn, and∑kn

j=1 �(Jnj) < �(Jn) + ε/2n+1.Now, the collection⋃

n

{Jnj}knj=1 = {J11, J12, . . . , J1k1

, J21, J22, . . . , J2k2, . . . }

is countable, being a countable union of finite collections. We have �(Jnj) < δ,for each n and j, and

∑n,j

�(Jnj) =∑n

kn∑j=1

�(Jnj) ≤∑n

(�(Jn) +

ε

2n+1

)< λ∗(A) +

ε

2+∑n

ε

2n+1≤ λ∗(A) + ε.

Also,⋃

n,j Jnj =⋃

n

(⋃kn

j=1 Jnj

)⊃ ⋃n Jn ⊃ A. If we now reindex the collection

{J11, J12, . . . , J1k1 , J21, J22, . . . , J2k2 , . . . } by using a single subscript andobtain {In}n, then this sequence satisfies the conclusions of the lemma.

THEOREM 3.8

Suppose that A and B are subsets of R that are a positive distance apart; thatis, d(A,B) > 0. Then

λ∗(A ∪B) = λ∗(A) + λ∗(B).

PROOF Let δ = d(A,B). If λ∗(A ∪B) = ∞, then it follows from Proposition 3.1(e) thatthe conclusion of the theorem holds. So, assume that λ∗(A ∪B) < ∞. Let ε > 0be given. By Lemma 3.9, there is a sequence {In} of open intervals such that�(In) < δ for all n,

⋃In ⊃ A ∪B, and

∑�(In) < λ∗(A ∪B) + ε.

Now, let {Jn} denote the members of {In} that contain a point of A andlet {Kn} denote the ones that do not contain a point of A. As A ⊂ A ∪B ⊂ ⋃ In,it follows that A ⊂ ⋃ Jn. Also, because d(A,B) = δ and �(In) < δ for all n, therecan be no points of B in any Jn. Therefore, because B ⊂ A ∪B ⊂ ⋃ In, it mustbe that B ⊂ ⋃Kn.

Using the definition of outer measure, we conclude that

λ∗(A) + λ∗(B) ≤∑

�(Jn) +∑

�(Kn) =∑

�(In) < λ∗(A ∪B) + ε.

Because ε > 0 was arbitrary, λ∗(A) + λ∗(B) ≤ λ∗(A ∪B). The reverse inequalityis true by Proposition 3.1(e).


We can, in fact, improve on Theorem 3.8. Indeed, suppose that A and Bare two subsets of R with the property that there is an open set O with A ⊂ Oand B ⊂ Oc. Then the conclusion of Theorem 3.8 obtains.

Roughly speaking, the reason is as follows. Since O is open, it can be writtenas a countable union of disjoint open intervals. Because the points of A mustlie within these open intervals and the points of B must lie outside of them, thesets A and B cannot be too intermingled. Before we can provide a rigorous proofof the improvement of Theorem 3.8, we need two more lemmas.

LEMMA 3.10

Let O be a proper open subset of R (i.e., O is open, nonempty, and not equalto R). For each n ∈ N , let

On =

{x : d(x,Oc) >

1

n

}.

Then,a) On is open and On ⊂ O for all n ∈ N .b) O1 ⊂ O2 ⊂ · · · and

⋃n On = O.

c) If On �= ∅, then d(On, Oc) = 1/n.

d) If On �= ∅, then d(On, Ocn+1) = 1/n(n + 1).

PROOF The proofs of parts (a) and (b) are left to the reader.c) Since d(On, O

c) = inf{ d(x,Oc) : x ∈ On }, we see that d(On, Oc) ≥ 1/n. To

prove the reverse inequality, we first note that because O is open, it can be ex-pressed as a countable union of disjoint open intervals, say the intervals {Ij}j .

Now, by assumption, On �= ∅. This means that there is a y ∈ O such thatd(y,Oc) > 1/n. Since y ∈ O, there is a k such that y ∈ Ik. Clearly, the dis-tance from y to Oc is the same as the distance from y to the nearest endpointof Ik. Therefore, if we write Ik = (ak, bk), then y ∈ (ak + 1/n, bk − 1/n). Itfollows that ∅ �= (ak + 1/n, bk − 1/n) ⊂ On. Note that at least one of the twonumbers ak and bk must be finite. We will assume that ak is finite. (If ak isinfinite, a similar argument holds.)

Since (ak + 1/n, bk − 1/n) ⊂ On and ak ∈ Oc, we conclude by applyingExercise 3.21(c) that

d(On, Oc) ≤ d

((ak +

1

n, bk − 1

n

), {ak}

)=

1

n.

This completes the proof of part (c).d) We first show that d(On, O

cn+1) ≥ 1/n(n + 1). Suppose y ∈ On and z ∈ Oc

n+1.By definition, d(y,Oc) > 1/n and d(z,Oc) ≤ 1/(n + 1). Let ε > 0 be given.Then there is a w ∈ Oc with |w − z| < 1/(n + 1) + ε. Also, w ∈ Oc impliesthat |w − y| > 1/n. Thus,

|z − y| ≥ |w − y| − |w − z| > 1

n− 1

n + 1− ε =

1

n(n + 1)− ε.

As z and y do not depend on ε, we conclude that |z − y| ≥ 1/n(n + 1). Conse-quently, because y ∈ On, z ∈ Oc

n+1 were arbitrary, d(On, Ocn+1) ≥ 1/n(n + 1).


To prove the reverse inequality, let Ik = (ak, bk) be as in the proof ofpart (c), and assume as before that ak is finite. Because ak ∈ Oc, we haveak + 1/(n + 1) ∈ Oc

n+1. Therefore, by Exercise 3.21(c),

d(On, Ocn+1) ≤ d

((ak +

1

n, bk − 1

n

),{ak +

1

n + 1

})=

1

n(n + 1).

This completes the proof of part (d).

LEMMA 3.11

Suppose that A ⊂ R and λ∗(A) < ∞. Assume there is a proper open subset Oof R with A ⊂ O. Let On = {x : d(x,Oc) > 1/n }. Then

λ∗(A) = limn→∞λ∗(A ∩On).

PROOF Let An = A ∩On. Then, by Lemma 3.10(b), A1 ⊂ A2 ⊂ · · · and, consequently,λ∗(A1) ≤ λ∗(A2) ≤ · · · . Also, since An ⊂ A for all n, λ∗(An) ≤ λ∗(A) for all n.By assumption, λ∗(A) < ∞. Thus, {λ∗(An)}n is a monotone nondecreasing,bounded sequence of real numbers; and, hence, converges to a real number,say α. Clearly, α ≤ λ∗(A).

Now, let Bn = A \An and Cn = An+1 \An. Then we have A = An ∪Bn

and Bn = Cn ∪ Cn+1 ∪ · · · . Thus,

λ∗(A) ≤ λ∗(An) + λ∗(Bn) (3.8)

andλ∗(Bn) ≤ λ∗(Cn) + λ∗(Cn+1) + · · · . (3.9)

Now, for n ≥ 2, An+1 = An ∪ Cn ⊃ An−1 ∪ Cn, so that

λ∗(An−1 ∪ Cn) ≤ λ∗(An+1). (3.10)

Also, An−1 ⊂ On−1 and Cn ⊂ Ocn. So, by Lemma 3.10(d),

d(An−1, Cn) ≥ d(On−1, Ocn) = 1/n(n− 1) > 0.

Therefore, Theorem 3.8 implies that

λ∗(An−1 ∪ Cn) = λ∗(An−1) + λ∗(Cn).

Using (3.10) and this last equation, we conclude that, for n ≥ 2,

λ∗(Cn) ≤ λ∗(An+1) − λ∗(An−1).

Then (3.9) implies

λ∗(Bn) ≤∞∑k=1

[λ∗(An+k) − λ∗(An+k−2)]


so that by (3.8)

λ∗(A) ≤ λ∗(An) + λ∗(Bn) ≤ λ∗(An) +

∞∑k=1

[λ∗(An+k) − λ∗(An+k−2)].

But,

λ∗(An) +

∞∑k=1

[λ∗(An+k) − λ∗(An+k−2)]

= limm→∞

{λ∗(An) +

m∑k=1

[λ∗(An+k) − λ∗(An+k−2)]

}= lim

m→∞{−λ∗(An−1) + λ∗(An+m−1) + λ∗(An+m)}= −λ∗(An−1) + 2α.

Consequently, we have shown that λ∗(A) ≤ −λ∗(An−1) + 2α for all n. Lettingn → ∞, reveals that λ∗(A) ≤ α. As we know, however, α ≤ λ∗(A). Thus, wehave λ∗(A) = α = limn→∞ λ∗(An), as required.

THEOREM 3.9

Suppose that A and B are subsets of R with the property that there is an openset O with A ⊂ O and B ⊂ Oc. Then

λ∗(A ∪B) = λ∗(A) + λ∗(B).

PROOF If either λ∗(A) or λ∗(B) is infinite, then the result is trivial. So, assume thatboth are finite. If O = ∅, then A = ∅; and if O = R, then B = ∅. In either ofthose cases, the result is also trivial.

Consequently, we can assume that O is a proper open subset of R. Asbefore, let On = {x : d(x,Oc) > 1/n } and An = A ∩On. Because An ⊂ On

and B ⊂ Oc, Lemma 3.10(c) implies that d(An, B) ≥ d(On, Oc) = 1/n and, thus,

by Theorem 3.8, λ∗(An ∪B) = λ∗(An) + λ∗(B). Since An ∪B ⊂ A ∪B, we haveλ∗(An ∪B) ≤ λ∗(A ∪B). Therefore, for each n ∈ N ,

λ∗(A ∪B) ≥ λ∗(An ∪B) = λ∗(An) + λ∗(B).

Letting n → ∞ and applying Lemma 3.11, we get that

λ∗(A ∪B) ≥ λ∗(A) + λ∗(B).

Proposition 3.1(e) shows that the reverse inequality holds.

Lebesgue Outer Measure Is Not Finitely Additive

We have now seen that, under certain conditions,

λ∗(A ∪B) = λ∗(A) + λ∗(B) (3.11)

for disjoint subsets A and B of R. Our next theorem, which we will state andprove shortly, shows that (3.11) does not hold for every pair of disjoint subsets Aand B of R, that is, that λ∗ is not finitely additive.

In view of Theorem 3.9, it is clear that if (3.11) fails to hold for disjointsubsets A and B, then those sets must be considerably intermingled. To obtainthis intermingling, we proceed as follows.


LEMMA 3.12

For x, y ∈ R, define x ∼ y if and only if y − x ∈ Q. Then ∼ is an equivalence re-lation and, hence, partitions R into disjoint equivalence classes. Moreover, thereis a set S ⊂ [0, 1) containing exactly one element from each equivalence class.

PROOF That ∼ is an equivalence relation is left as an exercise for the reader. By theaxiom of choice (see page 14), we can select a set T ⊂ R that contains ex-actly one element from each equivalence class. Let us set S = {x− [x] : x ∈ T }where [x] denotes the greatest integer in x. Because for each x, x− [x] ∈ [0, 1)and x− [x] ∼ x, the proof is complete.

LEMMA 3.13

Let S be the set defined in Lemma 3.12 and W = (−1, 1) ∩Q. Thena) {S + r}r∈Q forms a collection of pairwise disjoint sets.b) (0, 1) ⊂ ⋃r∈W (S + r) ⊂ (−1, 2).

PROOF

a) Suppose q, r ∈ Q and (S + q) ∩ (S + r) �= ∅. Let y ∈ (S + q) ∩ (S + r). Thenthere exist u, v ∈ S such that y = u + q = v + r. Hence, u ∼ v. Since Scontains only one element from each equivalence class, we must have u = v,which, in turn, implies q = r.

b) Let x ∈ (0, 1). Then there is a u ∈ S such that x ∼ u. Put r = x− u. Thenr ∈ Q and x ∈ S + r. Moreover, since x ∈ (0, 1) and S ⊂ [0, 1), −1 < r < 1.Thus, (0, 1) ⊂ ⋃r∈W (S + r). That

⋃r∈W (S + r) ⊂ (−1, 2) follows immedi-

ately from the fact that S ⊂ [0, 1).

Note: Lemma 3.13(a) shows that the sets {S + r}r∈Q are pairwise disjoint. Theyare also considerably intermingled as is shown in Exercise 3.27.

THEOREM 3.10

Lebesgue outer measure λ∗ is not finitely additive.

PROOF Suppose to the contrary that λ∗ is finitely additive. Let {qn}∞n=1 be an enumer-ation of the rationals in (−1, 1) and set En = S + qn. Using the assumed finiteadditivity of λ∗, Proposition 3.2, Lemma 3.13, and Proposition 3.1(c) and (e),we conclude that

1 = λ∗((0, 1)) ≤ λ∗

( ∞⋃n=1

En

)≤

∞∑n=1

λ∗(En)

= limn→∞

n∑k=1

λ∗(Ek) = limn→∞λ∗

( n⋃k=1

Ek

)≤ λ∗((−1, 2)

)= 3.

(3.12)

This shows 1 ≤ limn→∞∑n

k=1 λ∗(Ek) ≤ 3. But, by Proposition 3.1(d),

n∑k=1

λ∗(Ek) =

n∑k=1

λ∗(S + qk) =

n∑k=1

λ∗(S) = nλ∗(S).

Consequently, 1 ≤ limn→∞ nλ∗(S) ≤ 3, which is impossible (why?). Hence, λ∗ isnot finitely additive.

3.4 Lebesgue Measure 101

COROLLARY 3.2

Lebesgue outer measure λ∗ is not countably additive. That is, Condition (M3)does not hold with μ = λ∗ and A = P(R).


3.21 Prove the following facts:a) For fixed E ⊂ R, the function d(x,E) is continuous.b) If E and F are subsets of R, then d(E,F ) = inf{ d(y, F ) : y ∈ E }.c) If A ⊂ E and B ⊂ F , then d(E,F ) ≤ d(A,B).d) d(E,F ) = d(E,F ).

3.22 Prove the following facts:a) Suppose that F is a closed subset of R, K is a compact subset of R, and F ∩K = ∅.

Then d(F,K) > 0.b) Show that part (a) is not true if it is assumed only that K is closed.

3.23 Prove Lemma 3.8 on page 95.

3.24 Verify parts (a) and (b) of Lemma 3.10 on page 97.

3.25 Suppose that O is open. Prove that

λ∗(W ) = λ∗(W ∩O) + λ∗(W ∩Oc)

for all subsets W of R.

3.26 Define x ∼ y if and only if x− y ∈ Q. Show that ∼ is an equivalence relation.

3.27 Let N be a positive integer, {rn}∞n=1 an enumeration of Q, and S as in Lemma 3.12. Foreach n ∈ N , define Sn = S + rn. Prove that there is no open set O with the propertythat

⋃N

n=1Sn ⊂ O and

⋃∞n=N+1

Sn ⊂ Oc.

3.28 Suppose that 0 ≤ a < b ≤ 1. Prove that it is possible to select the elements of the set Sin Lemma 3.12 so that S ⊂ (a, b).

3.29 Provide a detailed justification for each step in (3.12).

3.4 LEBESGUE MEASURE

For ease in reference, we repeat once more that we are searching for a function μdefined on some collection A of subsets of R such that




μ

(⋃n

An

)=∑n

μ(An).

As we have seen, Conditions (M1) and (M2) hold with μ = λ∗ and A = P(R),but Condition (M3) does not (Corollary 3.2). Note, however, that we do not needto have our measure μ defined for all subsets of R; Condition (M1) requires onlythat it be defined on a σ-algebra A of subsets of R that contains the Borel sets.


Thus, one way to get Condition (M3) to hold might be to restrict λ∗ to someproper subcollection of subsets of R; that is, select A to be a proper subsetof P(R). And, to do that, we need to identify a criterion for deciding whethera subset of R is measurable, that is, is a member of A. By Condition (M1), wemust have B ⊂ A; so, in particular, A must contain all open sets. Hence, thecriterion we select must be satisfied by all open sets.

The Caratheodory Criterion

Theorem 3.9 states that if A and B are subsets of R with the property that thereis an open set O with A ⊂ O and B ⊂ Oc, then

λ∗(A ∪B) = λ∗(A) + λ∗(B).

As a consequence of Theorem 3.9, we obtain the following proposition.

PROPOSITION 3.3

Let O be an open set. Then

λ∗(W ) = λ∗(W ∩O) + λ∗(W ∩Oc) (3.13)

for every subset W of R.

PROOF For every subset W of R, we have W = (W ∩O) ∪ (W ∩Oc). Since W ∩O ⊂ Oand W ∩Oc ⊂ Oc, we see that (3.13) is a simple consequence of Theorem 3.9.

Equation (3.13) provides an additivity relation for Lebesgue outer measurethat is satisfied by all open sets. That relation shows the way to the requiredcriterion for deciding whether a subset of R is measurable.

DEFINITION 3.8 Caratheodory Criterion

A set E ⊂ R is said to satisfy the Caratheodory criterion if

λ∗(W ) = λ∗(W ∩ E) + λ∗(W ∩ Ec) (3.14)

for all subsets W of R. We denote by M the collection of all subsets of R thatsatisfy the Caratheodory criterion.

Note: By Proposition 3.1(e), the inequality

λ∗(W ) ≤ λ∗(W ∩ E) + λ∗(W ∩ Ec)

always holds. Consequently, to prove that a subset E of R is a member of M,it suffices to establish the inequality

λ∗(W ) ≥ λ∗(W ∩ E) + λ∗(W ∩ Ec) (3.15)

for all subsets W of R.The next theorem demonstrates that Condition (M1) holds for the collec-

tion M of subsets of R that satisfy the Caratheodory criterion.


THEOREM 3.11

M is a σ-algebra and M ⊃ B.

PROOF That M is closed under complementation is clear. First we prove that M isclosed under finite unions. So, assume A,B ∈ M. We claim that A ∪B ∈ M.Let W ⊂ R. Then, we must show that

λ∗(W ) ≥ λ∗(W ∩ (A ∪B))

+ λ∗(W ∩ (A ∪B)c). (3.16)

(See the note following Definition 3.8.)Now, we can write W ∩ (A ∪B) = (W ∩A) ∪ (W ∩Ac ∩B) and, hence, by

the subadditivity of λ∗,

λ∗(W ∩ (A ∪B)) ≤ λ∗(W ∩A) + λ∗(W ∩Ac ∩B).

Consequently,

λ∗(W ∩ (A ∪B))

+ λ∗(W ∩ (A ∪B)c)

≤ λ∗(W ∩A) + λ∗(W ∩Ac ∩B) + λ∗(W ∩ (A ∪B)c)

= λ∗(W ∩A) +[λ∗((W ∩Ac) ∩B

)+ λ∗((W ∩Ac) ∩Bc

)].

Because B ∈ M, the quantity between the square brackets in the previous ex-pression equals λ∗(W ∩Ac). Thus,

λ∗(W ∩ (A ∪B))

+ λ∗(W ∩ (A ∪B)c) ≤ λ∗(W ∩A) + λ∗(W ∩Ac).

This last sum equals λ∗(W ) because A ∈ M. Hence, (3.16) holds. We have nowestablished that M is an algebra of sets.

Next, we show that M is closed under countable unions. To that end,let {En}∞n=1 ⊂ M. We must prove that

⋃∞n=1 En ∈ M. To begin, we disjointize

the sets En, n = 1, 2, . . . . Let A1 = E1, A2 = E2 \ E1, A3 = E3 \ (E1 ∪ E2),

and, in general, An = En \ (⋃n−1k=1 Ek

). Then, see Exercise 3.30, Ai ∩Aj = ∅,

for i �= j, and⋃∞

n=1 An =⋃∞

n=1 En. Moreover, because M is an algebra of setsand En ∈ M for n ∈ N , it follows that An ∈ M for n ∈ N .

Now, let W be any subset of R and set E =⋃∞

n=1 En =⋃∞

n=1 An. We mustshow that λ∗(W ) ≥ λ∗(W ∩ E) + λ∗(W ∩ Ec). By the subadditivity of λ∗,

λ∗(W ∩ E) = λ∗(W ∩

( ∞⋃n=1

An

))

= λ∗( ∞⋃

n=1

(W ∩An)

)≤

∞∑n=1

λ∗(W ∩An).

(3.17)

For each n ∈ N , set Bn =⋃n

k=1 Ak. Then, because M is an algebra, Bn ∈ Mfor all n ∈ N . Consequently, for all n,

λ∗(W ) = λ∗(W ∩Bn) + λ∗(W ∩Bcn). (3.18)


Because Bn ⊂ ⋃∞m=1 Am = E, it follows that Ec ⊂ Bc

n. This last fact and (3.18)imply that

λ∗(W ) ≥ λ∗(W ∩Bn) + λ∗(W ∩ Ec). (3.19)

We will now prove by induction that for all n ∈ N ,

λ∗(W ∩Bn) =

n∑k=1

λ∗(W ∩Ak). (3.20)

The equation holds trivially when n = 1. So, assume that it holds for n. SinceAn+1 ∈ M, we have

λ∗(W ∩Bn+1) = λ∗((W ∩Bn+1) ∩An+1

)+ λ∗((W ∩Bn+1) ∩Ac

n+1

).

(3.21)

Because the Aks are pairwise disjoint, we have W ∩Bn+1 ∩An+1 = W ∩An+1

and W ∩Bn+1 ∩Acn+1 = W ∩Bn. Thus, by (3.21) and the induction hypothesis,

λ∗(W ∩Bn+1) = λ∗(W ∩An+1) + λ∗(W ∩Bn)

= λ∗(W ∩An+1) +

n∑k=1

λ∗(W ∩Ak) =

n+1∑k=1

λ∗(W ∩Ak),

as required.Employing (3.19) and (3.20), we conclude that

λ∗(W ) ≥n∑

k=1

λ∗(W ∩Ak) + λ∗(W ∩ Ec)

for all n ∈ N and, consequently,

λ∗(W ) ≥∞∑

n=1

λ∗(W ∩An) + λ∗(W ∩ Ec).

Applying (3.17) to the previous inequality, we deduce that

λ∗(W ) ≥ λ∗(W ∩ E) + λ∗(W ∩ Ec).

This shows E ∈ M. We have now established that M is a σ-algebra.It remains to prove that M ⊃ B. By Proposition 3.3, M contains all open

sets and, as we have just seen, M is a σ-algebra. Consequently, since B is thesmallest σ-algebra that contains all open sets, it must be that M ⊃ B.

Our next theorem shows that Condition (M3) is satisfied when Lebesgue outermeasure λ∗ is restricted to M. We denote by λ the restriction of Lebesgue outermeasure to M; that is, λ:M → R is defined by λ(E) = λ∗(E).


THEOREM 3.12

If A1, A2, . . . are in M, with Ai ∩Aj = ∅ for i �= j, then

λ

(⋃n

An

)=∑n

λ(An).

PROOF We first prove that λ is finitely additive on M. So, let A, B ∈ M with A ∩B = ∅.Set W = A ∪B. Then W ∩A = A and W ∩Ac = B. Consequently, as A ∈ M,we have by (3.14) that

λ(A ∪B) = λ(W ) = λ∗(W ) = λ∗(W ∩A) + λ∗(W ∩Ac)

= λ∗(A) + λ∗(B) = λ(A) + λ(B).

This shows that λ is finitely additive.Suppose now that {An}∞n=1 ⊂ M with Ai ∩Aj = ∅ for i �= j. Using the fact

that λ is finitely additive on M and the monotonicity of Lebesgue outer measure,we conclude that

m∑k=1

λ(Ak) = λ

( m⋃k=1

Ak

)≤ λ

( ∞⋃n=1

An

)for all m ∈ N . Letting m → ∞ gives

∑∞n=1 λ(An) ≤ λ (

⋃∞n=1 An). The reverse

inequality obtains because of the countable subadditivity of Lebesgue outer mea-sure.

Lebesgue Measurable Sets and Lebesgue Measure

From Proposition 3.2 on page 93 and Theorems 3.11 and 3.12, we see thatConditions (M1)–(M3) are satisfied with μ = λ and A = M; that is,

(L1) M is a σ-algebra and M ⊃ B.

(L2) λ(I) = �(I), for all intervals I.

(L3) If A1, A2, . . . are in M, with Ai ∩Aj = ∅ for i �= j, then

λ

(⋃n

An

)=∑n

λ(An).

Consequently, the set function λ:M → R is the required extension of length.We will employ the following terminology:

DEFINITION 3.9 Lebesgue Measurable Sets and Lebesgue Measure

The members of M are called Lebesgue measurable sets. That is, E is aLebesgue measurable set if and only if for every subset W of R,

λ∗(W ) = λ∗(W ∩ E) + λ∗(W ∩ Ec).

The restriction of Lebesgue outer measure to M is denoted by λ and is calledLebesgue measure.

In the next three propositions, we establish some additional properties ofLebesgue measure and Lebesgue measurable sets.


PROPOSITION 3.4

A subset of R with Lebesgue outer measure zero is a Lebesgue measurable set;that is, λ∗(E) = 0 ⇒ E ∈ M.

PROOF Suppose that λ∗(E) = 0. Let W be an arbitrary subset of R. As W ∩ E ⊂ E, themonotonicity of Lebesgue outer measure implies that λ∗(W ∩ E) ≤ λ∗(E) = 0.Using the fact that W ∩ Ec ⊂ W , we now conclude that

λ∗(W ) ≥ λ∗(W ∩ Ec) = λ∗(W ∩ E) + λ∗(W ∩ Ec).

This last inequality shows that E ∈ M.

PROPOSITION 3.5

Every countable subset of R has Lebesgue measure zero.

PROOF Let E ⊂ R be countable, say E = {xn}∞n=1. Then we can write E =⋃∞

n=1{xn}.Note that if a ∈ R, then, by (L2), λ({a}) = λ([a, a]) = a− a = 0. Therefore,applying (L3), we conclude that

λ(E) = λ

( ∞⋃n=1

{xn})

=

∞∑n=1

λ({xn}) = 0,

as required.

The next proposition shows that the converse of Proposition 3.5 does not hold.

PROPOSITION 3.6

The Cantor set P has Lebesgue measure zero.

PROOF Let G = [0, 1] \ P . From Chapter 2 (page 62), we know that G can be writ-ten as a countable union of disjoint open intervals {In}∞n=1 with the propertythat

∑∞n=1 �(In) = 1. Hence, by (L3) and (L2),

λ(G) =∞∑

n=1

λ(In) =

∞∑n=1

�(In) = 1.

Clearly, P and G are disjoint and P ∪G = [0, 1]. Therefore,

1 = λ(P ) + λ(G) = λ(P ) + 1,

which shows that λ(P ) = 0.

Another useful result is the following.

THEOREM 3.13

If {En}∞n=1 is a sequence of Lebesgue measurable sets with E1 ⊂ E2 ⊂ · · · , then

λ

( ∞⋃n=1

En

)= lim

n→∞λ(En).


PROOF If λ(En) = ∞ for some n, then both sides of the previous equation equal ∞. So,assume λ(En) < ∞ for all n.

We begin by disjointizing the Ens as follows. Let A1 = E1 and, for n ≥ 2,let An = En \ En−1. Then it is easy to see that {An}∞n=1 ⊂ M, Ai ∩Aj = ∅for i �= j, and

⋃∞n=1 An =

⋃∞n=1 En. Therefore, by countable additivity,

λ

( ∞⋃n=1

En

)= λ

( ∞⋃n=1

An

)=

∞∑n=1

λ(An).

As En−1 ⊂ En, we have λ(An) = λ(En \ En−1) = λ(En) − λ(En−1) for n ≥ 2.Consequently,

λ

( ∞⋃n=1

En

)=

∞∑n=1

λ(An) = limn→∞

n∑k=1

λ(Ak)

= limn→∞

(λ(E1) +

n∑k=2

[λ(Ek) − λ(Ek−1)]

)= lim

n→∞λ(En),

as required.

The Relation Between B and MWe close this section by discussing the relationship between the collection ofBorel sets B and the collection of Lebesgue measurable sets M. By Theorem 3.11,B ⊂ M. The question now is: Does B = M? In other words, is every Lebesguemeasurable set a Borel set or are there Lebesgue measurable sets that are notBorel sets?

It is not easy to answer that question. In fact, Lebesgue and Borel arguedthe question without finding the answer. It turns out that the answer to thequestion is no — there are Lebesgue measurable sets that are not Borel sets. Inother words, we have the following theorem:

THEOREM 3.14

The σ-algebra of Borel sets B is a proper subcollection of the σ-algebra ofLebesgue measurable sets M.

PROOF See Exercise 3.50.


3.30 Let {En}∞n=1 be a sequence of subsets of R. Define A1 = E1 and An = En \(⋃n−1

k=1Ek

)for n ≥ 2. Prove that Ai ∩Aj = ∅, for i = j, and that

⋃∞n=1

An =⋃∞

n=1En.

3.31 In Chapter 2, we introduced the concept of measure zero. Prove that this concept isequivalent to that of Lebesgue measure zero. In other words, show that a subset E ⊂ Rhas measure zero in the sense of Definition 2.19 on page 70 if and only if λ(E) = 0.

★3.32 Verify that if A ∈ M, λ(A) = 0, and B ⊂ A, then B ∈ M and λ(B) = 0.

3.33 Let A, B ∈ M with A ⊂ B and λ(A) <∞. Show that λ(B \A) = λ(B) − λ(A).


3.34 Use properties of Lebesgue measure to supply a simple proof that any (nondegenerate)interval of R is uncountable.

3.35 Suppose that {En}∞n=1 ⊂ M and that E1 ⊃ E2 ⊃ · · · . Also suppose that λ(E1) <∞.Prove that

λ

( ∞⋂n=1

En

)= lim

n→∞λ(En).

Can the assumption that λ(E1) <∞ be dropped? Why?

3.36 Show that if A,B ∈ M and λ(A ∩B) <∞, then

λ(A ∪B) = λ(A) + λ(B) − λ(A ∩B).

3.37 Suppose λ∗(A) = 0.a) Show that for any set B, λ∗(A ∪B) = λ∗(B).b) Show that if A ∪B ∈ M, then B ∈ M.

3.38 Find a sequence of pairwise disjoint sets {An}∞n=1 such that strict inequality holds inthe relation

λ∗( ∞⋃

n=1

An

)≤

∞∑n=1

λ∗(An).

Hint: Is {An}∞n=1 ⊂ M possible?

★3.39 If 0 < α < 1, construct a set Pα in a manner similar to that in which the Cantor set isconstructed, except that at the nth step remove open intervals of length α/3n insteadof 1/3n. Show that Pα is closed and that λ(Pα) = 1 − α > 0.

3.40 Prove that there is a sequence of continuous functions {fn}∞n=1 on [0, 1] that convergepointwise to a function f ∈ R([0, 1]). Hint: Use Exercise 3.39.

3.41 Prove that there is a Riemann integrable function on [0, 1] that is not a Borel measurablefunction. Hint: The proof of Theorem 3.14, which is carried out in Exercise 3.50, showsthere is a subset of the Cantor set that is not a Borel set.

3.42 Suppose that E ∈ M. Show that for each ε > 0, there is an open set O with O ⊃ Eand λ(O \ E) < ε. Hint: First consider the case where λ(E) <∞ and use the definitionof Lebesgue outer measure.

★3.43 Suppose that E ∈ M. Show that for each ε > 0, there is a closed set F with F ⊂ Eand λ(E \ F ) < ε.

★3.44 A set is called a Gδ-set if it is the intersection of a countable number of open sets; anda set is called an Fσ-set if it is the union of a countable number of closed sets. Notethat Gδ-sets and Fσ-sets are Borel sets. Now suppose that E ∈ M.a) Establish that there is a Gδ-set G and an Fσ-set F such that F ⊂ E ⊂ G and

λ(E \ F ) = λ(G \ E) = 0.b) Referring to part (a), deduce that λ(F ) = λ(E) = λ(G).

3.45 Let E ∈ M. Prove that λ(E) = inf{λ(O) : O ⊃ E, O open }.3.46 Let E ∈ M. Prove that λ(E) = sup{λ(K) : K ⊂ E, K compact }.3.47 Let E ⊂ R.

a) Suppose that there is a Borel set B such that B ⊂ E and λ∗(E \B) = 0. Showthat E ∈ M.

b) Suppose that λ∗(E) <∞ and that

λ∗(E) = sup{λ(F ) : F ⊂ E, F closed } = inf{λ(O) : O ⊃ E, O open }.

Show that E ∈ M.


3.48 Suppose that {En}∞n=1 is a sequence of pairwise disjoint Lebesgue measurable sets.Prove that

λ∗(A ∩

( ∞⋃n=1

En

))=

∞∑n=1

λ∗(A ∩ En)

for all subsets A of R.

3.49 Suppose that E ∈ M and that λ(E) <∞. Show that for each ε > 0, there are a finitenumber of pairwise disjoint intervals I1, I2, . . . , In such that

λ

(E �

( n⋃k=1

Ik

))< ε.

★3.50 Prove Theorem 3.14 on page 107. Proceed by establishing each of the following facts:a) If C ∈ M and x ∈ R, then C + x ∈ M and λ(C + x) = λ(C).b) Let S be the set defined in Lemma 3.12 on page 100. If C ∈ M and C ⊂ S,

then λ(C) = 0. Hint: Consider {C + r : r ∈ (−1, 1) ∩Q }.c) If D ⊂ R and λ∗(D) > 0, then there is a nonmeasurable subset of D. Hint: For

each r ∈ Q, let Dr = D ∩ (S + r). Use parts (a) and (b) to show that, if Dr ∈ M,then λ(Dr) = 0.

d) Define f : [0, 1] → R by f(x) = x+ ψ(x), where ψ denotes the Cantor function (seepage 64). Then f is a strictly increasing function and maps [0, 1] onto [0, 2].

e) The function g = f−1 is continuous and, hence, Borel measurable.f) f maps the Cantor set onto a set A with λ(A) = 1.g) Let E ⊂ A with E ∈ M. [Such an E exists by parts (f) and (c).] Then f−1(E) ∈ M

but f−1(E) ∈ B.

3.51 Prove that the set S defined in Lemma 3.12 on page 100 is not a Lebesgue measur-able set.

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