Measure Theory and Integration - Mathematics &...

CHAPTER 1

Measure Theory and Integration

One fundamental result of measure theory which we will encounterin this Chapter is that not every set can be measured. More specifically,if we wish to define the measure, or size of subsets of the real line insuch a way that the measure of an interval is the length of the intervalthen, either we find that this measure does not extend all subsets of thereal line or some desirable property of the measure must be sacrificed.An example of a desirable property that might need to be sacrificedis this: the (extended) measure of the union of two disjoint subsetsmight not be the sum of the measures. The Banach-Tarski paradoxdramatically illustrates the difficulty of measuring general sets; it isstated in Section ???????? without proof. It is with this in mind thatour discussion of measure and integration begins, in this Section, bymaking precise what constitutes a measurable set and a measurablefunction and then an integrable function.

1. Algebras of Sets

Let Ω be a set and P(Ω) denote the power set of Ω which is theset of all subsets of Ω. The reader is referred to Chapter 0, Subsection1 if any notations or set theoretic notions are unfamiliar.

Notation: N0 is the set of nonnegative integers; Q is the set ofrational numbers; Z is the set of all integers.

Suppose Ω1 and Ω2 are two sets and f : Ω1 → Ω2 is a function withdomain Ω1 and range in Ω2 (but f need not be onto). For any B ⊆ Ω2,denote by

f−1(B) = {x ∈ Ω1 : f(x) ∈ B}.the inverse image of the set B under f . Of course f need not beone-to-one (injective) and so f−1 need not exist as a function.

Some of the elementary properties of the inverse image are worthyof note: If B,B′ ⊆ Ω2 then

f−1(B ∩ B′) = f−1(B) ∩ f−1(B′)f−1(B ∪ B′) = f−1(B) ∪ f−1(B′)

f−1(Bc) = f−1(B)c

25

26 1. MEASURE THEORY AND INTEGRATION

The verification is asked for in the Exercises at the end of the Section.Definition: If A ⊆ Ω then define the characteristic function χA on

Ω by χA(x) = 1 if x ∈ A and χA(x) = 0 if x ∈ Ω− A.Exercise: Determine χ−1A (B) for an arbitrary B ⊆ R. There are 4

cases, if ∅ 6= A 6= Ω.Exercise: (De Morgan’s Laws) If {Aα : α ∈ I} is a collection

of subsets of a set Ω then the operation C of complementation has theproperties.

C (∪α∈IAα) = ∩α∈IC(Aα)C (∩α∈IAα) = ∪α∈IC(Aα)

If I = N is countable then we can write ∪n∈NAn = ∪n∈NBn whereB1 = A1, B2 = A2 ∩ Ac1, . . .Bn = An ∩ Ac1 ∩ Ac2 ∩ . . . ∩ Acn−1. The Bnare then disjoint.

Definition: Let An, n ∈ N be a countable collection of sets in Ω.The limit supremum of {An} is

lim supn

An = ∩n∈N ∪k≥n Ak

whereas the limit infimum of {An} islim inf

nAn = ∪n∈N ∩k≥n Ak.

Intuitively lim supn An is all points that are in infinitely many ofthe An whereas lim infn An is all those points that are in all exceptpossibly finitely many of the An.

If An is a decreasing sequence of sets, A1 ⊇ A2 ⊇ A3 ⊇ . . . thenlim sup

nAn = ∩n∈NAn = lim inf

nAn

Definition: A collection F of subsets of Ω is an algebra (field) ifa. Ω ∈ F .b. If A ∈ F then Ac ∈ F .c. If {Aα : α ∈ I} ⊆ F and if I is finite then ∪α∈I ∈ F .

Condition c in the definition could equivalently be replaced by eitherof the following Conditions.

c′. If A,B ∈ F then A ∪B ∈ F .c′′. If A,B ∈ F then A ∩B ∈ F .

because Condition c′ is equivalent to c; and Condition c′′ in conjunctionwith Condition b is equivalent to Conditions c′ and b, as can be seenby observing that C(A ∩ B) = CA ∪ CB.

Definition: A collection F of subsets of Ω is a σ-algebra (or σ-field)if F is an algebra and

1. ALGEBRAS OF SETS 27

d. If {Aα : α ∈ I} ⊆ F and if I is countable, or finite, then∪α∈IAα ∈ F .

Of course condition d implies condition c in the definition of analgebra. Conditions b and d together are equivalent to b and d′ where

d′. If {Aα : α ∈ I} ⊆ F and if I is countable, or finite, then∩α∈IAα ∈ F .

Example 1.1. (i) The power set P(Ω) of a set Ω is a σ-algebra.

(ii) {∅,Ω} is a σ-algebra.(iii) {∅, A,Ac,Ω} is a σ-algebra, if A is any subset of Ω.Example (i) is the largest σ-algebra of subsets of Ω; Example (ii) is

the smallest and Example (iii) is the smallest σ-algebra that containsthe set A.

Example 1.2. Consider the set of all real intervals of the form[a, b) along with all intervals of the form [a,∞) and (−∞, b) where aand b are real numbers. Define F to be the set of all finite unions ofdisjoint intervals of this form along with the null set and R itself.

We claim that F is an algebra. Certainly ∅ ∈ F . If A ∈ F then wemust check that Ac is in F and that is obvious if A = ∅ or A = R and sowe may suppose A = ∪1≤j≤n[aj, bj) where −∞ ≤ a1 < b1 < a2 < b2 <a3 < . . . < bn ≤ ∞. Then Ac is of the same form: indeed if a1 > −∞and bn < ∞ then Ac = (−∞, a1) ∪ [b1, a2) ∪ . . . ∪ [bn−1, an) ∪ [bn,∞).Therefore Ac ∈ F ; the cases where a1 = −∞ or b1 = ∞ or both arehandled similarly. Therefore F is closed under taking complements.

It remains to check that F is closed under finite unions and itsuffices to show that if A,B ∈ F then A ∪ B ∈ F . If A = ∅ orA = R then this is obvious and so we suppose A = ∪1≤j≤n[aj, bj) where−∞ ≤ a1 < b1 < a2 < b2 < a3 < . . . < bn ≤ ∞ and further it sufficesto consider the cases that B = [c, d) where c < d or B = (−∞, d) orB = [c,∞) or B = R. If A ∩ B = ∅ then there is nothing to showbut if B meets any subinterval [aj, bj) of A or shares an endpoint thenB ∪ [aj, bj) is again an interval of the same type, that is half open andhalf closed. (Alternatively one argues that there are four cases: c ∈ Aor not and d ∈ A or not.) Therefore A ∪B is in F .

Example 1.3. Sometimes it is more convenient to work with Fconsisting of all finite disjoint intervals of the form (a, b] (as opposedto [a, b) above) and (−∞, b] and (a,∞). Again in this case F is analgebra but not a σ-algebra.


Proposition 1.4. Let {Fα : α ∈ I} be a non empty collectionof σ-algebras (resp. algebras) on a given set Ω indexed by a set I. Thenthe intersection ∩α∈IFα is also a σ-algebra (resp. algebra).

Proof. We need only check that ∩α∈IFα satisfies the three prop-erties of σ-algebras The set Ω is certainly in all σ-algebras and hencein ∩α∈IFα. If A ∈ ∩α∈IFα then Ac is in every Fα. Finally if (An)n∈Nis a sequence of sets in ∩α∈IFα then ∪n∈NAn is in every Fα and thiscompletes the proof in the case of σ-algebras and the proof for algebrasis analogous. �

Suppose now that S ⊆ P(Ω) is any collection of subsets of a setΩ. Then there is a smallest σ-algebra which contains S which we shalldenote this σ(S). To see this observe that the set of all σ-algebras thatcontain S is non empty because it contains P(Ω) and so the intersectionof all σ-algebras that contain S is a σ-algebra, by the Proposition andit certainly contains S and so it must be the minimal such σ-algebra.

Definition: If F is a σ-algebra then a subset S ⊆ F with theproperty that F = σ(S) is said to be a system of generators of F .

Consider now the case that Ω = Rn, where n ≥ 1. Similar consider-ations apply if Ω is any topological space. On Rn ∋ x = (x1, . . . , xn), wedefine the norm |x| = (∑1≤j≤n |xj|2)1/2 and the corresponding (usual)topology where a base for the neighborhoods of a point x ∈ Rn is{Vr(x) : r > 0} where Vr(x) = {y ∈ Rn : |x − y| < r} is an openball of radius r centered at x. Then a set U is said to be open in Rn,if, for every x ∈ U there is r > 0 so that Vr(x) ⊆ U . The set of allopen sets is denoted O and we define the Borel subsets of Rn to beB = σ(O), that is the smallest σ-algebra that contains O. We shallsometimes write B(Rn) for B when wishing to emphasize that it is theBorel σ-algebra on Rn. In general, if Ω is a topological space and Ois the set of all open subsets of Ω then the Borel σ-algebra on Ω isσ(O).

Remark: If C denotes the set of all closed subsets of Rn thenσ(C) = B. For recall that a set is closed if and only if its complement isopen. Since B contains the open sets and is closed under the operationof taking complements, it follows that B ⊇ C and hence B ⊇ σ(C).Conversely the open sets are contained in σ(C): O ⊆ σ(C). ThereforeB = σ(O) ⊆ σ(C) and so B = σ(C). Therefore O is a system ofgenerators of B by the definition of B and C is a second system ofgenerators.

Example: The set S of all intervals of the form S = {(−∞, a) : a ∈R} is a system of generators of the Borel σ-algebra B of R: B = σ(S)Indeed, since S consists solely of open sets it is obvious that B ⊇ σ(S).


It therefore suffices to show that any open set is contained in σ(S).Certainly if a < b then [a, b[=] − ∞, a[c∩] − ∞, b[ belongs to σ(S).Therefore it suffices to show that any open set can be written as a unionof at most countably many intervals of the form [a, b). We can do this asfollows: For each n, consider all intervals of the form [k2−n, (k+1)2−n).where k ∈ Z. Given an open set U ⊆ R, we let An be the union of allsuch intervals that are entirely contained in U . Then An is an increasingsequence of sets and U = ∪n≥0An and this expresses U as a countableunion of the desired type of intervals. It follows that B = σ(S).

Other generating sets for the Borel subsets of R are

(1) S1 = {]−∞, a] : a ∈ R}(2) S2 = {]a,∞[: a ∈ R}(3) S3 = {[a,∞[: a ∈ R}(4) S4 = {]a, b] : a, b ∈ R, a < b}(5) S5 = {[a, b[: a, b ∈ R, a < b}Example: For each a = (a1, a2, . . . , an) ∈ Rn let

I(a) = {x ∈ Rn : x1 < a1, x2 < a2, . . . , xn < an}.Here x ∈ Rn has components x = (x1, x2, . . . , xn) is used. Further letI0(a, b) = {x ∈ Rn : a1 ≤ x1 < b1, a2 ≤ x2 < b2, . . . , an ≤ xn < bn}.

whenever a, b ∈ Rn and, presumably a1 < b1, a2 < b2, . . . an < bnbecause I0(a, b) is void otherwise. Define S = {I(a) : a ∈ Rn} andS0 = {I0(a, b) : a, b ∈ Rn}. Therefore S and S0 are two collections ofsubsets of Rn and we claim that σ(S) = σ(S0) = B. The rectangu-lar sets I0(a, b) can be expressed in terms of unions, intersections andcomplements of the sets I(a) so that σ(S) ⊇ σ(S0). Conversely any ofthe sets I(a) is the limit of an increasing sequence of sets I0(bn, a) sothat σ(S) = σ(S0). Also σ(S) ⊆ B since each I(a) is open. Thereforeit remains only to show that B ⊆ σ(S) and for this it suffices to showthat an arbitrary open set is in σ(S0). Proceeding as in the previousexample, it can be shown that any σ-algebra open set is a countableunion of the rectangular sets I0(a, b) and so this completes the proof.

There will also be occasion to work in the extended real numbersR = [−∞,∞], regarded as a two point compactification of the real line.The open sets in this case are all those sets that are subsets of R andare open in R as well as the sets ]a,∞] and ]−∞, b] for arbitrary real aand b or is a union of the three types of sets. Again we denote the opensets by O and we define the Borel σ-algebra of R to be B(R) = σ(O).We have A ∈ B(R) if and only if A−{−∞,∞}, regarded as a subset ofR is in B(R), for it is straightforward to check that the set of such setsis closed under complementation and countable union. The generating


sets listed above for B(R) are also generating sets for B(R) if one adjoinsthe singleton set {∞} (resp. {−∞}).

Monotone Classes We say that a sequence (Ap)p∈N of subsets of aset Ω is increasing if A1 ⊆ A2 ⊆ A3 ⊆ . . . and decreasing if A1 ⊇ A2 ⊇A3 ⊇ . . . and monotone if it is either increasing or decreasing. Everymonotone sequence has a limit: A = ∪p∈NAp if (Ap)p∈N is increasingand A = ∩p∈NAp in the decreasing case. A set M of subsets of Ω issaid to be a monotone class if, for every monotone sequence in M , thelimit is also in M .

Every σ-algebra is a monotone class. For recall that an increasingsequence of sets can be written as the union of disjoint subsets. Thecomplement of a decreasing sequence forms an increasing sequence andso a σ-algebra is a monotone class. On the other hand if F0 is both analgebra and a monotone class then it must also be a σ-algebra.

If {Mα : α ∈ I} is a nonempty collection of monotone classesthen ∩α∈IMα is also a monotone class. Since the power set P(Ω) is amonotone class it follows that every subset S ⊆ P(Ω) is contained ina smallest monotone class which we denote M(S)

Theorem 1.5. (Monotone Class Lemma) If F0 is an alge-bra of subsets of a set Ω then σ(F0) = M(F0).

Proof: It has already been remarked that a σ-algebra is a monotoneclass and so it follows that a M(F0) ⊆ σ(F0).

To prove the converse it suffices to show that M(F0) is an algebra.Introduce the notation

MA = {B ∈ M(F0) : A ∩B,A ∩ Bc and Ac ∩ B are in M(F0)}It is not difficult to check that MA is a monotone class. Now if wesuppose that A ∈ F0 then MA ⊇ F0. By minimality MA = M(F0)if A ∈ F0. Next we observe that, by symmetry that B ∈ MA impliesA ∈ MB. Therefore if A ∈ F0 and B ∈ MA then A ∈ MB. Thisshows that, whenever B ∈ M(F0) = MA, then we have MB ⊇ F0.But then, by minimality again we must have MB = M(F0) even forB ∈ M(F0). Therefore for every A,B ∈ M(F0)

A ∩B; Ac ∩ B, and A ∩ Bc are in M(F0).We check now that M(F0) is an algebra. Certainly ∅ ∈ F0 ⊆ M(F0)and if A ∈ M(F0) then Ac ∈ M(F0) for take B = Ω. Finally ifA,B ∈ M(F0) then Ac, B ∈ M(F0) so that Ac ∩Bc ∈ M(F0) and wemay take the complement to get A∪B ∈ M(F0). This proves M(F0)is an algebra and as a monotone class it is therefore a σ-algebra. 2


We shall have occasion to use the Monotone Class Lemma whenverifying uniqueness of measures.

Exercises:

(1) Show that, if B,B′ ⊆ Ω2 thenf−1(B ∩ B′) = f−1(B) ∩ f−1(B′)f−1(B ∪ B′) = f−1(B) ∪ f−1(B′)

f−1(Bc) = f−1(B)c

(2) Show that, in general, lim infn An ⊆ lim supnAn(3) Recall the definition of the limit supremum and limit infimum

of a sequence of real numbers an.

lim supn∈N

an = inf{sup{ak : k ≥ n} : n ≥ 1} and

lim infn∈N

an = sup{inf{ak : k ≥ n} : n ≥ 1}

where the “value” ∞ (resp. −∞) is possible if the sequenceis unbounded above (resp. below). (See page two of Ash’sbook Measure, Integration, and Functional Analysis.) Howis lim supχAn related to χB where B = lim supAn? Whathappens if we replace lim supχAn by lim inf χAn?

(4) Suppose that An = {x ∈ R3 : |x− (0, 0, (−1)n/n)| < 1}. Findlim supn An and lim infn An. Compare this with Exercise 3,page 3 of Ash’s book.

(5) Show that F of Example 1.2 is not a σ-algebra.(6) Suppose that E ⊆ Ω is an arbitrary subset of a set Ω and S is

a collection of subsets of Ω. Define E ∩ S = {E ∩A : A ∈ S}.Show thata. If F0 is an algebra of subsets of Ω then E ∩ F0 is also an

algebra of subsets of E.b. σ(E ∩ S) = E ∩ σ(S). That is the smallest σ-algebra in

P(E) containing E ∩ S is the intersection of E with thesmallest σ-algebra in P(Ω) containing S.

c. Conclude further that, if F is a σ-algebra of subsets of Ω,then E ∩ F is a σ-algebra of subsets of E.

(Reference: Halmos’s Measure Theory)


2. Measurability

It is an objective of this Chapter to define the integral of a function.If f is a nonnegative function defined on an interval in the real linethen the integral should be able to tell us the area under the graph off just as the Riemann integral does in calculus. It will tell us muchmore but even in this limited context we shall discover that f mustbe restricted: the notion of area under the graph of f will not makesense for every f and we will be forced to restrict the class of functionsconsidered. We introduce in this Section the concept of a “measurable”function. We shall see that f must be measurable if we are to makesense of the notion of area under the graph. Of course this difficultyarises already in the case of Riemann integration: the Riemann integralis not defined for arbitrary functions f . We shall clarify this remarkand discuss the Riemann integral and its relation to the “Lebesgueintegral,” introduced here, at the end of this Chapter.

Definition: Let Ω be a set and F be a σ-algebra of subsets of Ω.The we shall refer to (Ω,F) as a measurable space. The sets in F willbe referred to as measurable sets.

Definition: Suppose (Ω1,F1) and (Ω2,F2) are two measurablespaces. Then a mapping f : Ω1 → Ω2 is measurable, with respectto F1 and F2 if f−1(B) ∈ F1 whenever B ∈ F2. We shall sometimessay f is measurable function from (Ω1,F1) to (Ω2,F2) to clarify thechoice of σ-algebras F1 and F2. Unless otherwise specified a functionf : Ω1 → Rn is said to be measurable or Borel measurable if f is mea-surable from (Ω1,F1) to (Rn,B).

An elementary example of a measurable function is given by thecharacteristic function χA of a set A . Then χA is measurable as amapping from (Ω,F) to (R,B) if and only if A ∈ F , that is if and onlyif A is a measurable set. Further examples of measurable functions willbe apparent once some elementary properties have been established.

Lemma 2.1. Suppose that (Ω1,F1), (Ω2,F2) and (Ω3,F3) arethree measurable spaces and that f : Ω1 → Ω2 and g : Ω2 → Ω3 are twomappings. If f and g are both measurable then the composed functiong ◦ f is measurable from (Ω1,F1) to (Ω3,F3).

Proof: We must show that, for an arbitrary set C ∈ F3, (g ◦f)−1(C) ∈ F1. It suffices to show that

(2.1) (g ◦ f)−1(C) = f−1(g−1(C))

2. MEASURABILITY 33

because g−1(C) ∈ F2 because g is measurable and so f−1(g−1(C)) ∈ F1because f is measurable. The verification of (2.1) is a straightforwardexercise in checking the equality of sets. 2

Proposition 2.2. Suppose that f : Ω1 → Ω2 is a function andF1 and F2 are σ-algebras on Ω1 and Ω2 respectively. Suppose furtherthat S is a system of generators of F2, which is to say F2 = σ(S). Thenf is measurable from (Ω1,F1) to (Ω2,F2) if and only if f−1(B) ∈ F1for every B ∈ S.

Proof : It is obvious that, if f is measurable then f−1(B) ∈ F1 forevery B ∈ S simply because S ⊆ F2. Conversely, suppose f−1(B) ∈ F1for every B ∈ S. Let T = {B ∈ F2 : f−1(B) ∈ F1} so that T ⊇ S. Weshall show that T is a σ-algebra which implies that T = F2 and so, bythe definition of T , f is measurable. This will complete the proof.

We check therefore that T is a σ-algebra. Certainly ∅ ∈ T , becausef−1(∅) = ∅ ∈ F1. We check next that if B ∈ T then f−1(B) ∈ F1and so f−1(Bc) = f−1(B)c ∈ F1 because F1 is closed under takingcomplements. (Recall f−1(Bc) = f−1(B)c by an Exercise.) We checkfinally that if (Bn)n∈N is a sequence of sets in T so that f−1(Bn) ∈ Tfor each n ∈ N then

f−1(∪n∈NBn) = ∪n∈Nf−1(Bn) ∈ F1because F1 is a σ-algebra. This shows that T is closed under countableunions and is therefore a σ-algebra and the proof is complete. 2

Remark: This set T in the proof constitute “good” sets and theargument that there are many good sets is an instance of what Ashcalls the good sets principle in his text page 5.

Remark: The above proof uses a special case of the followingobservation: If f is a mapping f : Ω1 → Ω2 and if (Bα)α∈I is a collectionof subsets of Ω2 indexed by a set I then

f−1(∪α∈IBα) = ∪α∈If−1(Bα)f−1(∩α∈IBα) = ∩α∈If−1(Bα).

On the other hand if (Aα)α∈I is a collection of subsets of Ω1 then

f(∪α∈IAα) = ∪α∈If(Aα)but it may happen that f(∩α∈IAα) 6= ∩α∈If(Aα).

Corollary 2.3. A continuous function f : Rm → Rn is Borelmeasurable.

Of course the understood σ-algebras here are the Borel σ-algebraB(Rm) and B(Rn).


Proof: Since the B(Rn) is generated by the open sets, it sufficesto show that f−1(U) ∈ B(Rm) for an arbitrary open set U . The resultwill then follow from the Proposition. However f is continuous meansthat f−1(U) is open whenever U is and so f−1(U) ∈ B(Rm). 2

The same reasoning applies in a more general setting.

Corollary 2.4. If f is a continuous function from one topolog-ical space (X,Σ) to another (Y, T ) then f is measurable from (X,B(X))to (Y,B(Y )) where B(X) is the Borel σ-algebra which is generated bythe open sets Σ of X and similarly B(Y ) is generated by the open setsT .

Proposition 2.5. Suppose that (Ω,F) is a measurable spaceand f1, f2, . . . fm are m Borel measurable real valued functions, fj :(Ω,F) → (R,B), for 1 ≤ j ≤ m. Suppose that g : (Rm,B(Rm)) →(R,B(R)) is measurable. Then

φ(x) = g(f1(x), f2(x), . . . , fm(x))

defines a measurable function φ : (Ω,F) → (R,B(R)).We shall set aside the proof of the Proposition until later and con-

sider its consequences.

Corollary 2.6. If fj : (Ω,F) → (R,B), j = 1, 2 are Borelmeasurable functions, then f1 + f2 is also Borel measurable.

Proof. This is an application of the Proposition with g : R2 → Rdefined by g(x1, x2) = x1 + x2. Because g is continuous it is Borelmeasurable. �

Corollary 2.7. If fj : (Ω,F) → (R,B), j = 1, 2 are Borelmeasurable functions, then the product f1f2 is also Borel measurable.In particular, if k is a real constant then kf2 is Borel measurable.

Proof. In this case g(x, y) = xy. The special case follows bydefining f1(x) = k which makes f1 measurable because it is continuous.

�

Corollary 2.8. If f : (Ω,F) → (R,B), is a Borel measurablefunction then |f | is also Borel measurable.

Proof. In this case g(x) = |x|. �Corollary 2.9. If fj : (Ω,F) → (R,B), j = 1, 2 are Borel

measurable functions, and if f2(x) 6= 0 for all x ∈ Ω then the ratiof1/f2 is also Borel measurable.

2. MEASURABILITY 35

Proof. In this case we define

g(x, y) =

{

x/y if y 6= 00 if y = 0

so that g is defined on all of R2 but of course it is not continuous.We shall check however that g is measurable, as a mapping from(R2,B(R2)) to (R,B). It suffices to show that g−1(−∞, a) ∈ B(R2)for any real a by Proposition 2.2. If a < 0 then g−1(−∞, a) = {(x, y) ∈R2 : x/y < a} is easily seen to be open in R2 and hence in B(R2). Ifa ≥ 0 then g−1(−∞, a) = {(x, y) ∈ R2 : x/y < a} ∪ {(x, y) : y = 0}which is the union of an open and a closed set and is therefore in B(R2).This proves that g is measurable and so the above Proposition applieswhich completes the proof. �

Exercise: Show that if fj : (Ω,F) → (R,B), j = 1, 2 are Borelmeasurable functions then max{f1, f2} and min{f1, f2} are also mea-surable. Remark: max{a, b} = (|a − b| + a + b)/2 for any reals a andb.

It remains to establish the Proposition.Proof of the Proposition: It suffices to show that the map-

ping, f say, defined by f(x) = (f1(x), f2(x), . . . fm(x)) is measurableas a mapping from (Ω,F) to (Rm,B(Rm) because the compositionof measurable functions is measurable. To verify f is measurable,it suffices to show that, for an arbitrary a = (a1, a2, . . . , am) ∈ Rm,f−1({x1 < a1, x2 < a2, . . . , xm < am}) ∈ F by Proposition 2.2. How-ever

f−1({x1 < a1, x2 < a2, . . . , xm < am}) = ∩1≤j≤mf−1j (−∞, aj)and as the intersection of m measurable sets, this set is, itself measur-able. 2

Definition 2.10. A function f : (Ω,F) → R is said to besimple if there exist finitely many sets Aj ∈ F , 1 ≤ j ≤ m and scalarsλj so that

f(x) =∑

1≤j≤mλjχAj(x)

The set of all simple functions is denotes S(Ω,F) or S when the contextis clear.

Observe that a simple function is measurable because it is the sumof measurable functions. Also a simple function takes on only finitelymany values, λj, 1 ≤ j ≤ m and possibly 0. Conversely a measurablefunction that takes on finitely many values is simple. To see this, sim-ply define, for each λj in the image of f , Aj = f

−1({λj}). The set S


is closed under addition, scalar multiplication and multiplication andtherefore forms a linear algebra over R of functions. (Recall that a lin-ear algebra is a vector space with a multiplication operation (f, g) 7→ fgthat is associative and distributes over addition from the left and rightand, for any scalar α, α(fg) = (αf)g = f(αg). Reference: Naimark’sNormed Algebras §7.) It is further worth noting that the representationf(x) =

∑

1≤j≤m λjχAj(x) in the definition of f ∈ S is not unique.Proposition 2.11. Let fn, n ∈ N be a sequence of Borel mea-

surable real valued functions defined on a measurable space (Ω,F). Sup-pose that

limn∈N

fn(x) = f(x) exists in R for every x ∈ Ω.

Then the function f : (Ω,F) → R, so defined, is measurable.This result says, briefly, that the pointwise limit of measurable func-

tions is measurable.Proof: We will show that

(2.2) f−1((−∞, a]) = ∩p∈N ∪n∈N ∩j≥nf−1j ((−∞, a+ 1/p])for every a ∈ R The set of all sets of the form (−∞, a] generate theBorel σ-algebra B because any open set is generated. Therefore, wewill have shown f is measurable by Proposition 2.2.

To verify (2.2), let x ∈ f−1((−∞, a]) so that f(x) ≤ a. Then,for any p ∈ N there is n so that fj(x) ≤ a + 1/p for all j ≥ n:x ∈ f−1j ((−∞, a + 1/p]) for all j ≥ n or x ∈ ∩j≥nf−1j ((−∞, a + 1/p]).But p was arbitrary and so x belongs to the right side of (2.2) and thisshows that f−1((−∞, a]) is a subset or equal to the right side of (2.2).

Conversely suppose that x belongs to the right hand side of (2.2).Then, for every p ∈ N, there exists n ∈ N so that fj(x) ≤ a + 1/p forall j ≥ n. Taking limits in this last expression as j → ∞ we see thatf(x) ≤ a+ 1/p. Since p is arbitrary f(x) ≤ a and this shows the rightside of (2.2) is in f−1((−∞, a]) which verifies (2.2). This proves that fis measurable. 2.

The Extended Reals R: We introduce the extended real line R =[−∞,∞], sometimes referred to as the two point compactification ofthe real line, and this is just R with two points ±∞ adjoined: R = R∪{∞}∪{−∞}. This will be a convenience when discussing convergence.We introduce the following topology on R. The neighborhoods of ∞(resp. −∞) are those sets which contain an interval of the form (a,∞]for some a ∈ R (resp. [∞, a)) and the neighborhoods of x ∈ R are theusual: those sets that contain an interval of the form {y : |y − x| < δ}

2. MEASURABILITY 37

for some δ > 0. Of course the topology that R inherits as a subset of Ris its usual topology. As a consequence of these definitions we see thatR is homeomorphic to the compact interval [−1, 1] with the (usual)topology it inherits as a subset of R. Indeed

(2.3) Φ(x) =

x√x2+1

if x ∈ R1 if x = ∞−1 if x = −∞

is continuous from R to [−1, 1] with inverse

Φ−1(x) =

x√1−x2 if −1 < x < 1

∞ if x = 1−∞ if x = −1

which is also continuous.

Corollary 2.12. Let fn, n ∈ N be a sequence of Borel mea-surable extended real valued functions defined on a measurable space(Ω,F). Suppose that

limn∈N

fn(x) = f(x) exists in R for every x ∈ Ω.

Then the function f : (Ω,F) → R, so defined, is measurable.Proof. An extended real valued function g is Borel measurable

if and only if Φ ◦ g is Borel measurable for Φ as in (2.3) because Φand Φ−1 are continuous and hence measurable. Therefore the previousProposition 2.11 applied to Φ◦fn implies implies the present result �

Measurable functions can be written as the pointwise limit of simplefunctions. We begin by considering nonnegative bounded functions andlater we will extend to all measurable functions.

Proposition 2.13. Suppose f is be a nonnegative, bounded,measurable function defined on a measurable space (Ω,F). Then thereis a sequence (fn)n∈N of simple functions such that

(1) (fn)n∈N is increasing.(2) 0 ≤ fn ≤ f , for every n ∈ N(3) (fn)n∈N converges to f pointwise and even uniformly.

Proof. Let ǫ > 0 be given. We shall construct a simple functiongǫ so that gǫ ≤ f and |f(x) − gǫ(x)| < ǫ for all x ∈ Ω. Since f isbounded we have 0 ≤ f ≤ M for some positive constant M . Wechoose a0 = 0 < a1 < a2 < . . . < am < am+1] so that aj+1 − aj < ǫ, for0 ≤ j ≤ m+ 1 and M < am+1 < M + ǫ We define

Aj = {x ∈ Ω : aj ≤ f(x) < aj+1} = f−1([aj, aj+1))


for 0 ≤ j ≤ m. Then Aj is measurable and Aj ∩ Ak = ∅ if j 6= k and∪0≤j≤mAj = Ω. We define

gǫ(x) =∑

0≤j≤majχAj

Then gǫ ≤ f < gǫ + ǫ, in fact, for any x ∈ Ω, x ∈ Aj for some j and sogǫ(x) = aj and aj ≤ f(x) < aj+1.

Define f1 = g1, f2 = max{f1, g1/2}, . . . fn = max{fn−1, g1/n} Itfollows that fn ≤ f < fn + 1/n Also that fn is measurable as themaximum of two simple functions. And (fn)n∈N is increasing and so fnhas the required properties. �

Next consider the case that f is not necessarily nonnegative but isstill bounded.

Proposition 2.14. Suppose f is a real valued, bounded, mea-surable function defined on a measurable space (Ω,F). Then there is asequence (fn)n∈N of simple functions such that

(1) |fn| ≤ |f |, for every n ∈ N(2) (fn)n∈N converges to f pointwise and even uniformly.

Remark: The proof is based on the observation that any realvalued, measurable function f = f+ − f− where f+ = max{f, 0} =(|f | + f)/2 is nonnegative and measurable and f− = −min{f, 0} =max{−f, 0} is also nonnegative and measurable.

Proof of the Proposition 2.14 : Let gn be a sequence of non-negative simple functions convergent to f+ = max{f, 0} as guaranteedby the preceding Proposition. Similarly let hn be simple functions con-vergent to f− and define fn = gn − hn. Then |fn| = |gn − hn| ≤gn + hn ≤ f+ + f− = |f |. Moreover |f − fn| = |f+ − gn − (f− − hn)| ≤|f+−gn|+ |f−−hn| and since f+−gn and f−−hn go to zero uniformlyon Ω, fn converges uniformly to f . 2

Finally we consider the general case when f need not be bounded.

Theorem 2.15. Let f be a nonnegative extended value measur-able function, f : (Ω,F) → R+. Then there is an increasing sequencefn of simple functions (measurable on (Ω,F)) so that

(1) 0 ≤ fn ≤ f , for every n ∈ N(2) (fn)n∈N converges to f pointwise.

Theorem 2.16. Let f be a measurable function, f : (Ω,F) →R. Then there is a sequence fn of simple functions so that

(1) |fn| ≤ |f |, for every n ∈ N(2) (fn)n∈N converges to f pointwise.

2. MEASURABILITY 39

Remark: It is not true in general that the convergence is uniformin this more general setting. Indeed, if f is unbounded then it cannotbe approximated uniformly by bounded functions. For suppose fn be asequence of functions, uniformly convergent to a function f and supposeeach fn is bounded (with bound depending on n). If |f − fn| < 1 then||f | − |fn|| < 1 which says that |f | is bounded. In the setting of theTheorem above, simple functions are bounded and so we cannot expectuniform convergence.

Proof of Theorem 2.15: For each p ∈ N, define hp = min{f, p}so that hp is a bounded, nonnegative measurable function and so itis possible to choose a simple function gp ≥ 0 so that hp − 1p ≤ gp ≤hp by Proposition 2.13. Then we define f1 = g1, f2 = max{f1, g2},. . . fp = max{fp−1, gp} so that fp is simple, nonegative and increasing.We also see by induction on p, fp ≤ hp ≤ f . To check the pointwiseconvergence, suppose x0 ∈ Ω. If f(x0) < ∞ then we may choose p0 ∈ Nso that p0 ≥ f(x0). Then for p ≥ p0, f(x0) = hp(x0) so that f(x0)− 1p =hp(x0)− 1p ≤ gp ≤ fp(x0) ≤ f(x0) and so f(x0) = limp∈N fp(x0). On theother hand, if f(x0) = ∞ then hp(x0) = p and gp(x0) ≥ p − (1/p) sothat fp(x0) ≥ gp(x0) ≥ p− (1/p). Therefore limp fp(x0) = ∞ = f(x0).2

Proof of Theorem 2.16 The proof of Proposition 2.14 applieshere except the uniform convergence there must be replaced by point-wise convergence. Let gn be a sequence of nonnegative simple functionsconvergent to f+ = max{f, 0} as guaranteed by the preceding Theo-rem. Similarly let hn be simple functions convergent to f

− and definefn = gn − hn. Then |fn| = |gn − hn| ≤ gn + hn ≤ f+ + f− = |f |.Moreover, if f(x) ≥ 0 then hn(x) = 0 and

limn∈N

fn(x) = limn∈N

gn(x) = f+(x) = f(x)

Similarly f(x) < 0, then gn(x) = 0 and

limn∈N

fn(x) = limn∈N

−hn(x) = −f−(x) = f(x)

Note that limn gn(x) = ∞ and limn hn(y) = ∞ are both possible, butonly if x 6= y. 2

The set of simple functions is an algebra as we have seen. Moreoverif φ : R → R is Borel measurable then φ ◦ f is simple whenever f is.In particular |f | is simple if f is and max{f, g} = 1

2(f + g + |f − g|) is

simple whenever f and g are. Similarly min{f, g} is simple.


3. Measures and Premeasures

A measure, or more generally a set function is a mapping from aσ-algebra into R = [−∞,∞]. Arithmetic on R is defined as follows.Significantly ∞−∞ is not defined but

∞+ a = ∞ if a 6= −∞−∞+ a = −∞ if a 6= ∞

a · ∞ = ∞ if a > 0

Definition: A measure µ on a measurable space (Ω,F) is a map-ping of F to R+ = [0,∞] such that

(1) µ(∅) = 0(2) µ(∪α∈IAα) =

∑

α∈I µ(Aα)

whenever {Aα ∈ F : α ∈ I} is a countable collection of sets in F whichare pairwise disjoint which means that Aα ∩ Aβ = ∅ whenever α 6= β.

A triple (Ω,F , µ) where µ is a measure on the measurable space(Ω,F) is ameasure space or sometimes measured space to distinguishit from a measurable space. In the special case that µ(Ω) < ∞ thenµ is said to be a finite measure and if µ(Ω) = 1 then µ is said to be aprobability measure and (Ω,F , µ) as a probability space.

Remark: If {xα : α ∈ I} is a set of nonnegative numbers then∑

α∈I xα = supF∑

α∈F xα where the supremum is taken over all finitesubsets F of I and the sum could be ∞. It is not necessary for thisdefinition that I be countable but of course if {xα > 0 : α ∈ I}is uncountable then the sum is necessarily ∞. (reference: GeneralTopology by Bourbaki, Chapter IV, §4.3 and §7.1.)

A somewhat more primitive concept than a measure is a premeasureDefinition: Let F0 be an algebra of subsets of a set Ω. A premea-

sure µ0 is a mapping of to R+ = [0,∞] such that(1) µ(∅) = 0(2) µ(∪α∈IAα) =

∑

α∈I µ(Aα)

whenever {Aα ∈ F0 : α ∈ I} is a countable collection of sets in F0which are pairwise disjoint and provided ∪α∈IAα is in F0.

Of course a measure is a premeasure and a premeasure is a measureif its domain is a σ-algebra.

Property 2 for measures and premeasures is referred to as countableadditivity. If a set function µ0 is countably additive on an algebra F0then

µ0(A) = µ0(A) +∑

α∈Nµ(∅) for any A ∈ F0

3. MEASURES AND PREMEASURES 41

Therefore if µ0(A) < ∞ we must have µ0(∅) = 0 that is, property 2of premeasures implies property 1. Therefore property 1 could equallywell be replaced by µ0(A) < ∞ for some A ∈ F .

A measure or premeasure must be finitely additive (that is I couldbe taken to be finite in property 2) because, by property 2

µ0(∪α∈IAα) = µ0([∪α∈IAα] ∪ [∪n∈N∅])=

∑

α∈Iµ0(Aα) +

∑

n∈Nµ0(∅) =

∑

α∈Iµ0(Aα)

An elementary consequence of this is that, whenever A ⊆ B and A,B ∈F then µ0(A) ≤ µ(B), because µ0(B) = µ0(A) + µ0(B − A) ≥ µ0(A)(Of course, R is ordered so that ∞ ≥ a if a ∈ R.) As a consequence,when we define a measure or premeasure µ0 to be finite if µ0(Ω) < ∞,we assure that indeed µ0(A) < ∞ for every A.

Example 3.1. Discrete Measures: Let Ω be any set andF = P(Ω) (the power set of Ω). For every x ∈ Ω assign a nonnegativeweight 0 ≤ px ≤ ∞ and define

µ(A) =∑

x∈Apx.

Property 1 of measures is easily checked and property 2 is left as anexercise to the reader. In the special case that px = 1 for every x thenµ is called the counting measure on Ω.

Example 3.2. Let Ω be any nonvoid set and F = P(Ω) be thepower set and let µ(∅) = 0 and µ(A) = ∞ if Ω ⊇ A 6= ∅. Then µ is ameasure.

More interesting examples will be constructed by starting with adistribution function which we now define.

Definition: A real valued function F defined on R is a distributionfunction if

(1) F is increasing which means, whenever x < y, F (x) ≤ F (y)(2) F is right continuous, which means that limx→a+ F (x) = F (a),

for every real a .

The archtypal example of a distribution function is F (x) = x butgiven any distribution function there is a corresponding “LebesgueStieltjes” premeasure which we now introduce.

Example 3.3. Lebesgue Stieltjes Premeasures Considerthe algebra F0 of all finite unions of half open, half closed intervals ofthe form (a, b] or (−∞, b] or (a,∞) for any reals a and b. We saw in§1.1 that indeed these sets do form an algebra. If F is a distribution


function then we define µ0((a, b]) = F (b)−F (a) and extend µ0 by finiteadditivity to F0. Later we shall see that µ0 is indeed a premeasure, thatis we will check the countable additivity property 2. It will further beshown, in §1.4 that µ0 has an extension defined on the Borel subsetsof the real line which is a measure. This extension is a “LebesgueStieltjes” measure and its construction and properties is an objective ofthis Chapter.

Some properties of measures and premeasures will now be recorded.

Proposition 3.4. If µ and ν are measures on a measurablespace (Ω,F) and a ≥ 0 and b ≥ 0 the aµ + bν is also a measure.Similarly if µ0 and ν0 are premeasures on an algebra of sets F0 thenaµ0 + bν0 is also a measure.

Proof: The proof is left as an exercise.

Theorem 3.5. Suppose that µ0 is a premeasure on an algebraof subsets F0 of a set Ω. Then

(1) If B ⊆ A and A,B ∈ F then µ0(B) ≤ µ0(A).(2) If {Aα : α ∈ I} is a countable collection of sets in F0 and if

∪α∈IAα ∈ F0 thenµ0(∪α∈IAα) ≤

∑

α∈Iµ0(Aα)

(3) If A1 ⊆ A2 ⊆ A3 ⊆ . . . is an increasing sequence of sets in F0and if ∪n∈NAα ∈ F0 then

limn∈N

µ0(An) = µ0(∪n∈NAn)

(4) If B1 ⊇ B2 ⊇ B3 ⊇ . . . is a decreasing sequence of sets in F0such that ∩n∈NBn is also in F0 and if µ0(Bk) < ∞ for somek ∈ N then

limn∈N

µ0(Bn) = µ0(∩n∈NBn)

Proof. Part 1 was discussed early in this section. For Part 2 werecall from §1.1 that ∪α∈IAα can be written as the union of disjointssets Bn ∈ F . Indeed if α : N → I is an enumeration of I and if withBn = Aα(n) ∩ Acα(n−1) ∩ . . . ∩ Acα(1).

µ0(∪α∈IAα) = µ0(∪n∈NBn) =∑

n∈Nµ0(Bn) ≤

∑

n∈Nµ0(Aα(n))

where the last inequality follows because Bn ⊆ Aα(n). Of course theorder of summation is irrelevant since the terms are nonnegative.


Consider Part 3. We observe that, if µ0(Ak) = ∞ for some k, thenµ0(∪n∈NAn) ≥ µ0(Ak) = ∞ and so there is nothing to check and so wecan suppose that µ0(Ak) is finite for every k. We have, by the additivityproperty for measures,

µ0(∪n∈NAn) = µ0(A1) + µ0(A2 − A1) + . . .+ µ0(Ap − Ap−1) + . . .= lim

p∈Nµ0(A1) + µ0(A2 − A1) + . . .+ µ0(Ap − Ap−1)

= limp∈N

µ0(Ap)

We shall suppose that µ0(B1) < ∞; the general case is very similar.Let B = ∩n∈NBn. Then B1−Bn is an increasing sequence of sets whoseunion is B1 −B and so by Part 3

µ0(B1)− µ0(B) = µ0(B1 −B) = µ0(∪n∈N(B1 − Bn)= lim

n∈Nµ0(B1 −Bn)

= µ0(B1)− limn∈N

µ0(Bn)

This implies Part 4 because µ0(B1) < ∞. �

Remark: In Part 4 of the preceding Theorem, the hypothesisµ0(Bk) < ∞ for some k cannot be dispensed with. For example if µ isthe counting measure on the rational numbers Q then µ(−1/n, 1/n) =∞ for every n but µ(∩n∈N(−1/n, 1/n)) = µ({0}) = 1.

The Theorem is of course true for measures in place of premea-sures in which case F0 would be replaced by a σ-algebra F say and sothe ∪α∈IAα and ∩n∈NBn are automatically in F and so the statementsimplifies slightly in this case.

There is a partial converse to parts 3 and 4 of the Theorem that ishelpful for checking the properties of premeasures.

Proposition 3.6. Suppose that F0 is an algebra of subsets ofa set Ω and µ0 : F0 → R+ is a mapping so that

(1) µ0(∅) = 0(2) µ(∪α∈IAα) =

∑

α∈I µ(Aα)

whenever {Aα ∈ F0 : α ∈ I} is a finite collection of sets in F0 which arepairwise disjoint. Suppose in addition that either one of the followingtwo conditions is valid.

a. Whenever A1 ⊆ A2 ⊆ A3 ⊆ . . . is an increasing sequence ofsets in F0 such that the limit A = ∪α∈IAα is also in F0 thenlimn∈N µ0(An) = µ0(A)

b. Whenever B1 ⊇ B2 ⊇ B3 ⊇ . . . is a decreasing sequence of setsin F0 such that the limit ∩n∈NBn = ∅ then limn∈N µ0(Bn) = 0


Then, in either case, µ0 is a premeasure.

Proof. Let C1, C2, C3 . . . be a sequence of pairwise disjoint setsin F0 and such that C = ∪p∈NCp is also in F0. Define An = ∪1≤p≤nCp.

Assume Condition a. It implies limn∈N µ0(An) = µ0(C) becauseAn ր C and C ∈ F0. By finite additivity, µ0(An) =

∑

1≤p≤n µ0(Cp) sothat

∑

p∈Nµ0(Cp) = lim

n∈N

∑

1≤p≤nµ0(Cp) = µ0(C)

which verifies µ0 is countably additive and hence a premeasure underCondition a.

Assume Condition b. We define Bn = C −An so that Bn ց ∅ andtherefore Condition b implies limn µ0(Bn) = 0 We have

µ0(C) = µ0(An) + µ0(Bn) =∑

1≤p≤nµ0(Cp) + µ0(Bn)

Now if we take the limit in n ∈ N and we have µ0(C) =∑

n µ0(Cp)which says µ0 is countably additive. �

We recall from Example 3.3 that each distribution function F de-fines a set function µ0 so that µ0((a, b]) = F (b)−F (a) which is a finitelyadditive set function on the set F0 of all finite disjoint unions of inter-vals of the form (a, b] and (−∞, b]) and (a,∞). Of course µ0((−∞, b]) =limn∈N = µ0((−n, b]) = limn∈N F (b)−F (−n), with the value of ∞ pos-sible. In general, if A ∈ F0 then µ0(A) = limn∈N µ0(A ∩ (−n, n]). Weare now ready to show that µ0 is a premeasure.

Lemma 3.7. Let F be any distribution function and let µ0 bethe corresponding finitely additive set function defined on the algebraF0 as in Example 3.3 (so that, in particular µ0((a, b]) = F (b)− F (a)).Then µ0 is a premeasure.

Proof: Consider first the finite case where F (x) is constant out-side (−N,N ] for some N so that µ0 is finite. We shall use Part b ofthe preceding proposition and so we assume that Bp ∈ F0 forms a de-creasing sequence of sets and Bp ց ∅. Given ǫ > 0, we will show thatµ0Bp) < ǫ for all suffiiciently large p and this will imply the countableadditivity of µ0. We claim that there is another sequence Cp ∈ F0 withcompact closures Cp ⊆ Bp and such that µ0(Bp − Cp) < ǫ/2p This ispossible because for each subinterval (a, b] of Bp we have µ0((a, b]) =F (b) − F (a) = lima′→a+ F (b) − F (a′) = lima′→a+ µ((a′b]) by the rightcontinuity of F . We may also suppose that Cp ⊆ (−N,N ] for all p.Since ∩p∈NCp ⊆ ∩p∈NBp = ∅ it follows that, for some n, ∩1≤p≤nCp = ∅


by a compactness argument. Consequently

µ0(Bn) = µ0(Bn − ∩1≤p≤nCp)= µ0(∪1≤p≤nBn − Cp)≤ µ0(∪1≤p≤nBp − Cp) ≤

∑

1≤p≤nǫ/2p < ǫ

Since (Bp)p∈N is a decreasing sequence and ǫ > 0 was arbitrary thissays µ0(Bp) ց 0 as p → ∞ and this verifies µ0 is countably additive,and so a premeasure, in the case F (x) is constant outside (−N,N ] forsome N .

Consider now the general case. For each N ∈ N let FN(x) = F (x)if |x| ≤ N and FN(x) = F (−N) if x ≤ −N and FN(x) = F (N) ifx ≥ N . Then FN is a distribution function and the correspondingfinitely additive set function µN is in fact a premeasure by the firstpart of the proof and µN(A) = µ0(A) if A ∈ F0 and A ⊆ (−N,N ].Moreover, for any A ∈ F0, limN∈N µN(A) = µ(A) by the discussionpreceding the statement of this result. Let (Ap)p∈N be a sequence ofpairwise disjoint sets such that A = ∪p∈NAp is in F0. Then, simplybecause µ0 is finitely additive we have, for any n ∈ N

µ0(A) ≥ µ0(∪1≤p≤nAp) =∑

1≤p≤nµ0(Ap) so that µ0(A) ≥

∑

p∈Nµ0(Ap)

Conversely, by the special case

µ0(A) = limN∈N

µN(A) = limN∈N

∑

p∈NµN(Ap) ≤ lim

N∈N

∑

p∈Nµ0(Ap) =

∑

p∈Nµ0(Ap)

because µN(B) ≤ µ0(B) for any B ∈ F0. Combining these two in-equalities we have µ0(A) =

∑

p∈N µ0(Ap) and this verifies the µ0 is apremeasure and completes the proof. 2

Thus every distribution function corresponds to a premeasure (andindeed, we shall see, to a measure). The following result provides apartial converse.

Definition: We shall say that µ is a Lebesgue-Stieltjes measure ifµ is a measure on the Borel subsets B of R and µ(K) < ∞ if K is acompact subset of R

Proposition 3.8. Suppose that µ is a Lebesgue-Stieltjes mea-sure and C is any real constant. Define F (x) = µ((0, x]) + C if x ≥ 0and F (x) = C − µ((x, 0]) if x < 0. (Here (0, 0] = ∅ by convention.)Then F is a distribution function and µ((a, b]) = F (b)−F (a) whenevera < b.


Therefore every Lebesgue-Stieltjes measure corresponds to a distri-bution function which is unique up to an additive constant.

Proof: Since µ(A) ≤ µ(B) whenever A ⊆ B are sets in B, µ((0, x])is increasing x ≥ 0 and µ((x, 0]) is decreasing for x ≤ 0 and so F isincreasing on R.

Suppose a ≥ 0. Thenlim

x→a+F (x) = C + µ(∩x>a(0, x]) = C + µ((0, a]) = F (a)

so that F is right continuous on [0,∞). Next suppose a < 0. Thenlim

x→a+F (x) = C − µ(∪x>a(x, 0]) = C − µ((a, 0]) = F (a)

so that in fact F is right continuous everwhere and is therefore a dis-tribution function.

Next check that F (b) − F (a) = µ((a, b]) whenever a < b. If 0 ≤ athen this is obvious and similarly if b < 0. Suppose therefore that a < 0and b ≥ 0. Then F (b)−F (a) = C+µ((0, b])−(C−µ((a, 0])) = µ((a, b]).

The uniqueness, up to additive constants, of F with the propertyF (b) − F (a) = µ((a, b]) follows because if G is another distributionfunction and if G(0) = F (0) then G(b)−G(0) = µ((a, b]) = F (b)−F (0)implies G(b) = F (b) for b ≥ 0. Similarly G(0) − G(a) = µ((a, 0]) =F (0) − F (a) so that G(a) = F (a) for a < 0. Thus G = F . In generalG = F −F (0)+G(0) so that the choice of G is completely determinedby the choice of G(0). 2

4. EXTENSIONS AND MEASURES: 47

4. Extensions and Measures:

In this Section we show that certain premeasures extend to mea-sures and that a measure may be “completed.” As an application weare able to construct Lebesgue Stieltjes measures. Recall the definition:

Definition: A real valued function F defined on R is a distributionfunction if

(1) F is increasing which means, whenever x < y, F (x) ≤ F (y)(2) F is right continuous, which means that limx→a+ F (x) = F (a),

for every real a .

We begin by considering a premeasure µ0 defined on an algebra F0.Suppose that (Ap)p∈N is an increasing sequence in F0 and Ap ր A sothat A1 ⊆ A2 ⊆ A3 ⊆ . . . and A = ∪p∈NAn but A may not be itself inF0. We would like to extend µ0 to µ0 as µ0(A) = limp∈N µ0(Ap). Thelimit exists in R+ but we should verify that it does not depend on theparticular sequence and that is the object of the Lemma below.

Lemma 4.1. Suppose that µ0 is a premeasure defined on an al-gebra F0 and (Ap)p∈N and (Bp)p∈N are two increasing sequences of setsin F0 with limits, A = ∪p∈NAp and B = ∪p∈NBp. If A ⊆ B thenlimp∈N µ(Ap) ≤ limp∈N µ(Bp)

Proof: Because µ0 is a premeasure we have for any q ∈ Nµ0(Aq) = lim

p∈Nµ0(Aq ∩ Bp) ≤ lim

p∈Nµ0(Bp)

Taking the limit in q gives the result. 2Define therefore G to consist of all A ⊆ Ω such that there exists an

increasing sequence (Ap)p∈N ∈ F0, so that Ap ր A and define also aset function µ0 on G by

µ0(A) = limp∈N

µ0(Ap).

The preceding Lemma assures that µ0 is well defined because it does notdepend on the particular sequence (Ap)p∈N ∈ F0 which approximatesA. We see further that µ0 extends µ0 because any A ∈ F0 can bewritten as the limit of a constant sequence. An = A.

Lemma 4.2. Suppose that µ0 is a finite premeasure on an algebraF0 and µ0 is its extension to G as described above. Then G is closedunder finite intersections and countable unions and

a. If G1, G2 are in G thenµ0(G1 ∪G2) + µ0(G1 ∩G2) = µ0(G1) + µ0(G2)


b. If (Gp)p∈N is an increasing sequence in G then G = ∪p∈NGn isin G and

limp∈N

µ0(Gn) = µ0(G)

c. If Gp is a sequence in G thenµ0(∪p∈NGp) ≤

∑

p∈Nµ0(Gp)

Proof: Suppose that Ap, Bp are increasing sequences of sets inF0 and Ap ր G1 and Bp ր G2 so that G1, G2 ∈ G. It follows thatAp ∩ Bp ր G1 ∩G2 and Ap ∪ Bp ր G1 ∪G2 so that G is closed underfinite unions and intersections. Moreover since µ0 is a premeasure wecan take the limit in p ∈ N in the relation µ0(Ap∪Bp)+µ0(Ap∩Bp) =µ0(Ap) + µ0(Bp) to verify Part a.

To verify Part b, suppose that Gp is an increasing sequence in G,Gp ր G. For each p, suppose Apq is an increasing sequence in F0convergent to Gp = ∪q∈NApq.

A11 A12 . . . A1q . . . ր G1A21 A22 . . . A2q . . . ր G2...

......

......

Ap1 Ap2 . . . Apq . . . ր Gp...

......

......

Define Bq = ∪1≤p≤qApq. Then Bq is an increasing sequence in F0 andit is convergent to G so that G ∈ G and so µ0(G) = limq∈N µ0(Bq).Certainly µ0(G) ≥ µ0(Gp) for any p by the preceding Lemma. On theother hand µ0(Gp) ≥ µ0(Bp) so that limp∈N µ0(Gp) ≥ limp∈N µ0(Bp) =µ0(G). This establishes Part b.

To verify Part c, it suffices to check that, for any n ∈ Nµ0(∪1≤p≤nGp) ≤

∑

1≤p≤nµ0(Gp)

because we may take the limit in n ∈ N and applying Part b above. Inthe case n = 2 we have µ0(G1∪G2) ≤ µ0(G1)+µ0(G2) by Part a. Thecase of general n follows by induction. The proof is complete. 2.

Thus extension µ0 is a countably additive set function and G ⊇ F0is closed under countable unions, but A ∈ G does not imply Ac ∈ G.Also unions of sets in G may not be expressible as unions of disjointsets in G.

Definition: A mapping µ∗ on the power set P(Ω) of a set Ω to R+is said to be an outer measure if

(1) µ∗(∅) = 0


(2) If A ⊆ B ⊆ Ω then µ∗(A) ≤ µ∗(B).(3) If Aα∈I is a countable collection of sets in Ω then

µ∗(∪α∈IAα) ≤∑

α∈Iµ∗(Aα)

Frequently an outer measure will not be a measure because it is notadditive but it is possible that a restriction to a σ-algebra of subsets ofΩ may indeed be a measure.

We now suppose that µ0 is a finite premeasure on an algebra F0and that µ0 is the extension of µ0 to the G of the preceding Lemma.Later in this Section we weaken the assumption of finite to “σ-finite”to be defined below. Define

(4.1) µ∗(A) = inf{µ0(G) : G ⊇ A,G ∈ G}for any A ⊆ Ω. Alternatively, in terms of µ0 itself, we define

λ(A) = inf

{

∑

p∈Nµ0(Ap) : Ap ∈ F0,∪p∈NAp ⊇ A

}

We claim µ∗(A) = λ(A). For suppose that Ap, p ∈ N is a sequence inF0 and ∪p∈NAp ⊇ A, Define G = ∪p∈NAp so that G ∈ G and µ0(G) =limn µ0(∪1≤p≤nAp) ≤

∑

p∈N µ0(Ap). Consequently µ∗(A) ≤ λ(A). Con-

versely suppose G ∈ G so that there is a sequence Ap ∈ F0 so thatAp ր G and we can form a sequence Bp ∈ F0 which is pairwise dis-joint and ∪p∈NBp = G so that µ0(G) =

∑

p∈N µ0(Bp). This shows thatµ∗(A) ≥ λ(A) and complete the verification that λ = µ∗.

Now let us check that µ∗ is indeed an outer measure and recordsome of its properties.

Lemma 4.3. Suppose that µ0 is a finite premeasure on an algebraF0 of subsets of a set Ω and that µ∗ is defined as indicated above; (see(4.1)). Then µ∗ is an outer measure and agrees with µ0 on F0 and µ0on G. Moreover

(1) µ∗(A∪B)+µ∗(A∩B) ≤ µ∗(A)+µ∗(B) for any sets A,B ⊆ Ω.In particular µ∗(A) + µ∗(Ac) ≥ µ0(Ω).

(2) If Ap, p ∈ N is an increasing sequence of sets in Ω thenlimp∈N µ∗(Ap) = µ∗(∪p∈NAp).

Proof: It is clear that µ∗(G) = µ0(G) if G ∈ G, by the definition(4.1) of µ∗ and because µ0(G) ≤ µ0(G′) if G ⊆ G′ and G′ ∈ G. Similarlyone sees that, if A ⊆ B then µ∗(A) ⊆ µ∗(B), by the definition (4.1) ofµ∗.


Therefore, to complete the check that µ∗ is an outer measure, wesuppose that Ap, p ∈ N is a sequence of subsets. of Ω. We chooseGp ∈ G, Gp ⊇ Ap so that µ0(Gp) ≤ µ∗(Ap)+2−pǫ for each p ∈ N. Then

µ∗(∪p∈NAp) ≤ µ0(∪p∈NGp) ≤∑

p∈Nµ0(Gp) ≤

∑

p∈Nµ∗(Ap) + ǫ/2

p

where we have applied Part c of the preceding Lemma. Therefore µ∗

is an outer measure.Check next Part 1. We suppose that A,B ⊆ Ω and ǫ > 0 is given.

Choose G1, G2 ∈ G A ⊆ G1 and B ⊆ G2 and µ0(G1) < µ∗(A) + ǫ andµ0(G2) < µ

∗(B) + ǫ. Then

µ∗(A)+µ∗(B) > µ0(G1)+µ0(G2)−2ǫ = µ0(G1∪G2)+µ0(G1∩G2)−2ǫby Part a of the preceding Lemma. Since G1 ∪ G2 ⊇ A ∪ B andG1 ∩G2 ⊇ A ∩B and G1 ∪G2 ∈ G and G1 ∪G2 ∈ G, we haveµ∗(A) + µ∗(B) > µ0(G1) + µ0(G2)− 2ǫ ≥ µ∗(A ∪B) + µ∗(A ∩B)− 2ǫand, since ǫ > 0 was arbitrary this implies Part 1.

It remains to check Part 2. Certainly µ∗(An) ≤ µ∗(∪p∈NAp) for anyn ∈ N so that limp∈N µ∗(Ap) ≤ µ∗(∪p∈NAp). To check the converse,suppose that ǫ > 0 is arbitrary, and that, for each p ∈ N, Gp ∈ Gis chosen so that Gp ⊇ Ap and µ0(Gp) ≤ µ∗(Ap) + 2−pǫ. We shallshow that there is a increasing sequence Hp ∈ G so that Hp ⊇ Apµ0(Hp) ≤ µ∗(Ap) +

∑

1≤q≤p 2−qǫ. The existence of such a sequence

would imply that

µ∗(∪p∈NAp) ≤ µ0(∪p∈NHp) = limp∈N

µ0(Hp) ≤ limp∈N

µ∗(Ap) +∑

1≤q≤p2−qǫ.

so that µ∗(∪p∈NAp) ≤ limp∈N µ∗(Ap) + ǫ and since ǫ is arbitrary thiswould complete the proof. To construct the sequence Hp we proceed byinduction on n supposing H1 ⊆ H2 ⊆ . . . ⊆ Hn have been constructedin G and Hp ⊇ Ap and µ0(Hp) ≤ µ∗(Ap) +

∑

1≤q≤p 2−qǫ for 1 ≤ p ≤ n.

Define Hn+1 = Hn ∪Gn+1 so that Hn+1 ∈ G andµ0(Hn+1) = µ0(Hn ∪Gn+1)

= µ0(Hn) + µ0(Gn+1)− µ0(Hn ∩Gn+1)≤ µ∗(An) +

∑

1≤p≤n2−pǫ+ µ∗(An+1) + ǫ/2

n+1 − µ∗(An)

since An ⊆ Hn ∩ Gn+1. Thus µ0(Hn+1) ≤ µ∗(An+1) +∑

1≤p≤n+1 2−pǫ

and that completes the construction and therefore the proof. 2Thus every finite premeasure µ0 extends to an outer measure µ

∗.One of the less desirable features of µ∗ is that there may exist sets A ⊆


Ω so that µ∗(A) + µ∗(Ac) > µ0(Ω) (but of course 0 ≤ µ∗(A) ≤ µ0(Ω)).However if we omit such sets then we have a measure as the followingTheorem asserts.

Theorem 4.4. Suppose that µ0 is a finite premeasure definedon an algebra F0 of subsets of a set Ω and that µ∗ is the outer measuredefined by (4.1). Define

F = {A ⊆ Ω : µ∗(A) + µ∗(Ac) = µ0(Ω)}Then F is a σ-algebra, F ⊇ F0 and indeed F ⊇ G and the restrictionof µ∗ to F is a measure that extends µ0 and µ0.

Proof: It is immediately clear that ∅ ∈ F and that A ∈ F impliesAc ∈ F . Also F0 ⊆ F because µ∗ = µ0 on F0 and µ0 is finitely additive.

Suppose that A,B ∈ F . By Part 1 of the preceding Lemma wehave

µ∗(A ∪ B) + µ∗(A ∩ B) ≤ µ∗(A) + µ∗(B)(4.2)µ∗(Ac ∪ Bc) + µ∗(Ac ∩ Bc) ≤ µ∗(Ac) + µ∗(Bc).

We add the two relations and recall that A,B ∈ F .µ∗(A ∪ B) + µ∗(A ∩ B) + µ∗(Ac ∪ Bc) + µ∗(Ac ∩ Bc) ≤ 2µ∗(Ω)

[µ∗(A ∪ B) + µ∗((A ∪ B)c)] + [µ∗(A ∩ B) + µ∗((A ∩ B)c)] ≤ 2µ∗(Ω)by rearranging. But we also know by Part 1 of the preceding lemmathat µ∗(A ∪ B) + µ∗((A ∪ B)c) ≥ µ∗(Ω) and µ∗(A ∩ B) + µ∗((A ∩B)c) ≥ µ∗(Ω) and so the above bound proves we must have equalityµ∗(A∪B)+µ∗((A∪B)c) = µ∗(Ω) and µ∗(A∩B)+µ∗((A∩B)c) = µ∗(Ω)which says that F is closed under union and intersection. Thus F is analgebra. We can say more because we must have equality in equation(4.2) and so, in the case A∩B = ∅, we have µ∗(A∪B) = µ∗(A)+µ∗(B)so that µ∗ is finitely additive on F .

Suppose now that Ap, p ∈ N is an increasing sequence in F ; wewant to show that A = ∪p∈NAp is also in F . By the preceding Lemmaµ∗(A) = limp∈N µ∗(Ap) and so µ∗(Ap)+µ∗(Acp) = µ

∗(Ω), for all p impliesµ∗(A)+limp∈N µ∗(Acp) = µ

∗(Ω). But Acp is a decreasing sequence of setsAcp ⊇ Ac and so this implies µ∗(A)+µ∗(Ac) ≤ µ∗(Ω). But by Part 1 ofthe previous Lemma µ∗(A) + µ∗(Ac) ≥ µ∗(Ω) and so we have A ∈ F .This proves that F is a σ-algebra and since limp∈N µ∗(Ap) = µ∗(A), bythe preceding Lemma, µ∗ is countably additive on F and therefore ameasure when restricted to F . Finally we recall that G consists of thelimits of all increasing sequences in F0 and of course these limits arein any σ-algebra that contains F0 and so F ⊇ G and µ∗ agrees with µ0by the preceding Lemma. 2


The next step is to generalize to not necessarily finite measures.

Definition: A premeasure µ0 on an algebra F0 of subsets of a setΩ is said to be σ-finite if there is a sequence sets Ap ∈ F0 so thatµ0(Ap) < ∞ and Ω = ∪p∈NAp.

Thus a finite premeasure is σ-finite and a Lebesgue-Stieltjes mea-sure is also σ-finite since it is bounded on every bounded interval.

Theorem 4.5. Carathéodory Extension Theorem: Sup-pose that µ0 is a σ-finite premeasure defined on an algebra F0 of sub-sets of a set Ω. Then µ0 has an unique extension to a measure on thesmallest σ-algebra, σ(F0) that contains F0.

Remark: The hypothesis that µ0 be σ-finite is only necessary toprove the uniqueness of the extension of µ0 and its not required toprove existence. See Problem 3, page 22 of Ash’s book.

Proof: Let Ep be a sequence of disjoint subsets in F0 of finiteµ0 measure such that Ω = ∪p∈NEp. Then Ep ∩ F0 is an algebra ofsubsets of Ep (by an Exercise in §1.1) and µp(A) = µ0(A ∩Ep) definesa premeasure on Ep∩F0 for each p ∈ N (check this). Since µp is a finitepremeasure, it extends as to a measure on a σ-algebra that containsEp ∩F0 and hence σ(Ep ∩F0) = Ep ∩ σ(F0). Denote the extension byµp as well. Define µ on σ(F0) by

µ(A) =∑

p∈Nµp(A ∩ Ep)

Then µ extends µ0 because if A ∈ F0 thenµ(A) =

∑

p∈Nµ0(A ∩ Ep) = µ0(A)

because µ0 is countably additive. Moreover if Aq is a sequence of dis-joint subsets in σ(F0) thenµ(∪q∈NAq) =

∑

p∈Nµ0(∪q∈NAq ∩ Ep)

=∑

p∈N

∑

q∈Nµ0(Aq ∩ Ep)

=∑

q∈N

∑

p∈Nµ0(Aq ∩ Ep) =

∑

q∈Nµp(Aq ∩ Ep) =

∑

q∈Nµ(Aq)

because the order of summation does not matter if all terms are non-negative. Therefore µ is σ-additive and therefore a measure on σ(F0).

We have shown that µ0 has an extension µ defined on σ(F0) andit remains to show µ is unique. Suppose therefore that ν is another


measure defined on σ(F0) so that µ0(A) = ν(A) for all A ∈ F0. Define,for fixed p ∈ N , M = {A ∩ σ(F0) : µ(A∩Ep) = ν(A∩Ep)}. Then Mcontains F0; moreover it is a monotone class because the limit of anyincreasing or decreasing sequence of sets in M is also in M. By theMonotone Class Lemma the smallest monotone class that contains F0is σ(F0) and so µ(A∩Ep) = ν(A∩Ep) for every A ∈ σ(F0). But p ∈ Nwas arbitrary and so

µ(A) =∑

p

µ(Ep ∩ A) =∑

µ(Ep ∩ A) =∑

p

ν(Ep ∩ A) = ν(A)

if A ∈ σ(F0) by countable additivity and so µ = ν which verifiesuniqueness. 2.

Corollary 4.6. If F is a distribution function defined on Rthen there is one and only measure µ on the Borel subsets B(R) so thatµ((a, b]) = F (b)− F (a) for every half closed interval (a, b].

Proof: This follows from the Carathéodory Extension Theorem byway of Lemma 3.7.

Exercise Construct a set which is unbounded (and measurable)but has finite Lebesgue measure.

Example: Consider the rationals Q and define an algebra F0 asfollows. All intervals of the form (a, b] where a < b are rationals aswell as intervals of the form (a,∞) and (−∞, b] with a, b ∈ Q belongto F0 as well as all finite disjoint unions of such intervals. Finally wesuppose the full space Q and ∅ belong to F0. It is not difficult to checkthat F0 is closed under finite intersections and complements and so itis an algebra. Define µ0 on F0 by µ(A) = ∞ if A 6= ∅ and µ(∅) = 0.Then µ0 is a premeasure. It is however not σ-finite. It can be shownthat every premeasure defined on an algebra has an extension to thesmallest σ-algebra containing the original algebra but, as we shall seethe extension need not be unique. Indeed in the present case σ(F0)obviously contains the singleton sets because {b} = ∩n(b− 1/n, b] andso σ(F0) contains all countable sets and so σ(F0) = P(Q) since Q iscountable. Define µ1 on P(Q), µ1(A) = ∞ if A 6= ∅ and µ1(∅) = 0.Then µ1 is a measure which extends µ0. Define µ2 to be the countingmeasure on P(Q). Again µ2 extends µ0 and yet µ2 6= µ1. Observefurther that µ2 is σ-finite even though it extends a premeasure that isnot σ-finite.

Proposition 4.7. Assume the hypotheses of the CarathéodoryExtension Theorem 4.5 so that µ0 is a σ-finite premeasure defined onan algebra F0 and µ is its extension to the σ-algebra F = σ(F0). Then,


for any set A ∈ F such that µ(A) < ∞ and any ǫ > 0 there is B ∈ F0so that µ(A∆B) < ǫ.

Proof: In the case that µ is finite we recall the construction ofLemma 4.2 of G ⊇ F0 and of µ0(G) = sup{µ0(Ap) : Ap ր G}. Recallfurther that if A ∈ F then µ(A) = µ∗(A) = inf{µ0(G) : G ∈ G, G ⊇A}. It follows that A∆Ap = (A−Ap)∪ (Ap−A) ⊆ (G−Ap)∪ (G−A)has arbitrarily small measure provided G ∈ G is chosen appropriatelyand then Ap ∈ F0 approximates G closely enough. This proves theresult in the case µ is finite.

In the general case let Ep be a sequence of disjoint subsets in F0of finite µ0 measure such that Ω = ∪p∈NEp. Define µp by µp(A) =µ(A∩Ep) so that µp is a finite measure on Ep. By the first part of theproof, if A ∈ F , and µ(A) < ∞ and ǫ > 0, then there exist sets Ap ⊆ Epwith Ap ∈ F0 with µp((A ∩ Ep)∆Ap) < ǫ2−p. Consider B = ∪p∈NAp.Then

µ(A∆B) =∑

p∈Nµ((A∆B) ∩ Ep) =

∑

p∈Nµ((A ∩ Ep)∆Ap) ∩ Ep) < ǫ

Here we have used the observations that B∩Ep = Ap and that (C∆D)∩E = (C ∩ E)∆(D ∩ E), for any sets C,D and EOf course B may not itself be in F0 but it can be approximated byBN = ∪1≤p≤NAp which is in F0 and BN ր B We havelim

N→∞µ(A∆BN) = lim

N→∞µ((A−BN) ∪ (BN − A))

= limN→∞

µ(A− BN) + µ(BN − A) = µ(A∆B) < ǫ.

by Theorem 3.5 and because µ(A) < ∞. Therefore for N large enoughwe have µ(A∆BN) < 2ǫ say. Since ǫ > 0 is arbitray the proof iscomplete. 2

Exercise: Show that, for any Borel subset A of the real line withfinite Lebesgue measure and for any ǫ > 0, there is a compact (that isclosed and bounded) set K so that the Lebesgue measure of A∆K isat most ǫ.

Definition 4.8. A measure space (Ω,F , µ) is said to be com-plete (or µ is said to be complete) if, whenever A ∈ F has measureµ(A) = 0, then every subset B ⊆ A is also in F .

Therefore µ is complete if every subset of a set of measure zero isa set of measure zero.

Suppose now that (Ω,F , µ) is any measure space. DefineN = {N ⊆ Ω : there exists M ∈ F ,M ⊇ N and µ(M) = 0}.


Define further Fµ to consist of all sets of the form A∪N where A ∈ Fand N ∈ N .

Exercise Show that

F = {A∆N : A ∈ F and N ∈ N}is equal to Fµ. Suggestion: Begin by showing that, for N ⊆ M ,

A ∪N = (A−M)∆(M ∩ (A ∪N)A∆N = (A−M) ∪ [M ∩ (A∆N)]

Theorem 4.9. Suppose that (Ω,F , µ) is a measure space andFµ is defined as above. Define µ on Fµ by µ(A ∪ N) = µ(A), for allA ∪ N ∈ Fµ so that where A ∈ F and N ∈ N . Then Fµ is a σ-algebra, µ is well defined and a measure that extends µ and (Ω,Fµ, µ)is a complete measure space.

Proof: It is clear that Fµ contains ∅. Moreover if Ap ∪ Np is asequence in Fµ (so that AP ∈ F and Np ∈ N ) then ∪p∈NAp ∪ Np =(∪p∈NAp) ∪ (∪p∈NNp) is in Fµ also. Suppose next that A ∪ N ∈ Fµwith A ∈ F and N ∈ N and suppose that N ⊆ M where M ∈ F andµ(M) = 0. To show that Fµ contains the complement of A ∪ N wenote that (A ∪ N)c = Ac ∩ N c = (Ac ∩ M c) ∪ Ac ∩ (N c − M c). NowN c −M c = N c ∩M ⊆ M and so Ac ∩ (N c −M c) ∈ N and so (A∪N)cis in Fµ and we have Fµ is a σ-algebra.

Check next that µ is well defined. Suppose therefore that A1∪N1 =A2 ∪ N2 where A1, A2 ∈ F and N1, N2 ∈ N . Then A1 = (A1 ∩ A2) ∪(A1 − A2). and µ(A1 − A2) = 0 because A1 − A2 ⊆ (A1 ∪N1)− A2 =(A2 ∪ N2) − A2 ⊆ N2. Therefore µ(A1) = µ(A1 ∩ A2); and by asymmetric argument µ(A2) = µ(A1 ∩ A2). This shows µ(A1 ∪ N1) =µ(A1) = µ(A2) = µ(A2 ∪ N2) so that µ is well defined. Obviously µextends µ

Check next that µ is countably additive. Suppose that Ap∪Np is asequence of pairwise disjoint sets in Fµ (so that AP ∈ F and Np ∈ N )then

µ(∪p∈NAp ∪Np) = µ ((∪p∈NAp) ∪ (∪p∈NNp))= µ(∪p∈NAp) =

∑

p∈Nµ(Ap) =

∑

p∈Nµ(Ap ∪Np)

because Ap forms a sequence of pairwise disjoint sets in F . Thereforeµ is a measure. To check that it is complete, suppose that A∪N ∈ Fµwhere N ∈ N and A ∈ F and µ(A∪N) = µ(A) = 0. Then A∪N ∈ Nand so any subset of A ∪ N is also in N and so has zero µ measure.Thus µ is complete. 2


We have constructed Lebesgue-Stieltjes measure on the Borel setsB(R) corresponding to a distribution function F and of course Lebesguemeasure corresponds to the case F (x) = x. Each of these measures hasa (unique) completion; and usually “Lebesgue Stieltjes measure” refersto this completion.

Historical Note: Constantin Carathéodory (September 13,1873 to February 2, 1950) was born in Berlin of Greek parents and grewup in Brussels. He is known for his work in the calculus of variations,conformal representations and the foundations of thermodynamics be-sides measure theory; he was also an accomplished linguist, speakingsix modern and several ancient languages. He joined the faculty of theill-fated Greek University of Smyrna (in present day Izmir, Turkey) in1920 and stayed until the Greek expulsion in 1922. He was Professorof Mathematics at the University of Munich 1924-1950.

5. THE INTEGRAL 57

5. The Integral

In this section we introduce the integral and derive some of itselementary properties.

Suppose that (Ω,F , µ) is a measure space and S = S(Ω,F) is theset of all simple functions; see Definition 2.10. Define S0 to be thesubset of S which consists of all those f ∈ S so that µ({x ∈ Ω : f(x) 6=0}) < ∞. (Observe that {x ∈ Ω : f(x) 6= 0} is measurable because fis.) Consequently f =

∑

1≤j≤n λjχAj for some scalars λj and Aj ∈ Fwith µ(Aj) < ∞. We define the integral of f with respect to µ as

∫

f dµ =∑

1≤j≤nλjµ(Aj)

Since the representation of f is not unique it is necessary to check thatthe integral is well defined.

Suppose therefore that f ∈ S0 and so f takes on finitely many nonzero distinct values {fi : 1 ≤ i ≤ n}. If Ai = {x ∈ Ω : f(x) = fi}then f =

∑

1≤i≤n fiχAi . We want to show that if there is any otherrepresentation f =

∑

1≤j≤m λjχBj then

∑

1≤i≤nfiµ(Ai) =

∑

1≤j≤mλjµ(Bj)

It is convenient to define f0 = 0 and A0 = (∪1≤j≤mBj) − (∪1≤i≤nAi).Then, for arbitrary i, 0 ≤ i ≤ n

fiχAi =∑

1≤j≤mλjχBj∩Ai

Since the Ai are disjoint, if one shows that fiµ(Ai) =∑

1≤j≤m λjµ(Bj∩Ai) then that suffices (for we can sum over i.) We also note that∪1≤j≤mBj ∩ Ai = Ai. For i > 0 this is because fi 6= 0 and for i = 0it is by the choice of A0. We now subdivide Ai into disjoint subsets asfollows. For each subset I of {1 ≤ j ≤ m} let

CI = [(∩j∈IBj)− (∪j /∈IBj)] ∩ Ai =[

(∩j∈IBj) ∩(

∩j /∈IBcj)]

∩ AiIf J is another such subset of {1 ≤ j ≤ m} then CJ is disjoint fromCI . Some of the CI may be empty but their union (over all 2

m − 1choices of I which are nonempty) is Ai. Suppose now that x ∈ CI .Then we must have fi =

∑

j∈I λj. But we also have Ai = ∪ICI wherethe sum is over all 2m − 1 nonvoid subsets I of {1 ≤ j ≤ m}. Take theintersection with Bj to get Bj ∩ Ai = ∪IBj ∩ CI = ∩I∋jCI where now


the intersection is over all subsets I which contain j. Therefore∑

1≤j≤mλjµ(Bj ∩ Ai) =

∑

1≤j≤m

∑

I∋jλjµ(CI)

=∑

I

∑

j∈Iλjµ(CI) =

∑

I

fiµ(CI) = fiµ(Ai)

where the interchange of summation is justified by checking that indexset {(j, I) : 1 ≤ j ≤ m, I ∋ j} is the same set as {(j, I) : I 6= ∅, j ∈ I}.

Theorem 5.1. Suppose f, g ∈ S0(Ω,F , µ) and λ is a real. Thena.∫

Ωλf dµ = λ

∫

Ωf dµ;

b.∫

Ωf + g dµ =

∫

Ωf dµ+

∫

Ωg dµ;

c.∫

Ωf dµ ≥ 0 if f ≥ 0;

d.∫

Ωf dµ ≤

∫

Ωg dµ if f ≤ g;

e. |∫

Ωf dµ| ≤

∫

Ω|f | dµ.

Proof. The proof of Parts a and c are left as elementary exercises.For Part b we have f =

∑

1≤i≤n fiχAi and g =∑

1≤j≤m gjχBj . Thenf + g =

∑

1≤k≤m+n hkχCk where hk = fk and Ck = Ak, if 1 ≤ k ≤ nand hk = gk−n and Ck = Bk−n, if n < k ≤ n+m. Therefore∫

Ω

f + g dµ =∑

1≤k≤n+mhkµ(Ck)

=∑

1≤k≤nfkµ(Ak) +

∑

n ǫ}) = 0.

5. THE INTEGRAL 59

If (Ω,F , µ) is a probability space and A = Ω then fp is said to convergein probability.

Proposition 5.2. Suppose that fp is a sequence of real valued,measurable functions defined on a measure space (Ω,F , µ) and conver-gent pointwise to f . Then f is measurable and, for every A ∈ F offinite measure, fp converges in measure to f on A.

Proof. We have already seen in Proposition 2.11 that the point-wise limit of a sequence of real valued, measurable functions is mea-surable. We may suppose that f = 0 because otherwise we replace fpby fp − f . For ǫ > 0, we claim that

∪m∈N ∩p≥m {x ∈ A : |fp(x)| ≤ ǫ} = AIn fact, for any x ∈ A, limp fp(x) = 0 implies that there is some m forwhich |fp(x)| ≤ ǫ for every p ≥ m, that is there is m so that x ∈ Amwhere Am = ∩p≥m{x ∈ A : |fp(x)| ≤ ǫ}. Since x ∈ A was arbitrary thisverifies the claim. Therefore A−Am is a decreasing sequence of sets offinite measure converging to the empty set and so limm∈N µ(A−Am) = 0by Theorem 3.5. But

A− Am = {x ∈ A : |fp(x)| > ǫ, for some p ≥ m}and the result follows. �

Thus pointwise convergence implies convergence in measure on setsof finite measure.

Exercise: Construct a sequence of measurable functions on a prob-ability space which converges to 0 in probability but does not convergepointwise.

Theorem 5.3. Let fp be a decreasing sequence of nonnegativefunctions in S0(Ω,F , µ) which converges pointwise to 0. Then

limp∈N

∫

Ω

fp dµ = 0.

This result will allow us to show later that the integral can beextended to a wider class of functions.

Proof. We have f1 ≥ f2 ≥ . . . ≥ fp ≥ . . . ≥ 0. Since f1 is finitevalued, there is a constant M > 0 so that f1 ≤ M and hence fp ≤ Mfor all p. Also, the set A = {x : f1(x) > 0} has finite measure andfp(x) = 0, for every p ∈ N if x ∈ Ac. Suppose the ǫ > 0. Define

Ap = {x : fp(x) > ǫ}Then Ap is a decreasing sequence of sets of finite measure; indeed Ap ⊆A. We know from the preceding Proposition that limp∈N µ(Ap) = 0.


For any p we have∫

Ω

fp dµ =

∫

Ω

χA−Apfp dµ+

∫

Ω

χApfp dµ

≤∫

Ω

χA−Apǫ dµ+

∫

Ω

χApM dµ

≤ ǫµ(A− Ap) +Mµ(Ap) ≤ ǫµ(A) +Mµ(Ap)Since limp∈N µ(Ap) = 0, it follows that, provided p is large enough,∫

Ωfp dµ ≤ ǫ(µ(A)+M), for example. Since ǫ > 0 was arbitrary

∫

Ωfp dµ

must converge to zero. �

Corollary 5.4. Let fp be an increasing sequence of functionsin S0(Ω,F , µ) which converges pointwise to a function f also in S0.Then

limp∈N

∫

Ω

fp dµ =

∫

Ω

f dµ

Proof. Apply the Theorem to the sequence f − fp. �We are now prepared to extend the integral to a broader family

of functions. It will be convenient to focus first on nonnegative func-tions. Let S+0 = S+0 (Ω,F , µ) be the subset of S0(Ω,F , µ) consistingof nonnegative functions. Thus S+0 is the set of all nonegative simplefunctions f such that {x : f(x) 6= 0} has finite measure. Define furtherD+(Ω,F , µ) = D+ by

D+ = {f : Ω → R+ : f = supH for some countable H ⊆ S+0 }.Later we shall see that for f ∈ D+, then we can extend the concept ofintegral as

∫ ∗

Ω

f dµ = sup{∫

Ω

h dµ : h ∈ S+0 , h ≤ f}

but first we explore the properties of D+. A cautionary observation isthat elements of D+ may take on the value ∞ unlike those in S+0 .

Proposition 5.5. Properties of D+(Ω,F , µ)(1) S+0 ⊆ D+.(2) A function f : Ω → R+ is in D+ if and only if f is the

pointwise limit of an increasing sequence gp ∈ S+0 .(3) A function f : Ω → R+ is in D+ if and only if f is nonnegative

and measurable and {x ∈ Ω : f(x) > 0} is the union of at mostcountably many sets of finite measure.

(4) Suppose f : Ω → R+ is measurable and f ≤ g where g ∈ D+.Then f ∈ D+.

(5) If f, g ∈ D+ then f + g ∈ D+.

5. THE INTEGRAL 61

(6) If f, g ∈ D+ then fg ∈ D+. Here it is understood the ∞·0 = 0,that is if f(x) = ∞ and g(x) = 0 then fg(x) = 0.

(7) If f ∈ D+ and λ ≥ 0 then λf ∈ D+.(8) If f, g ∈ D+ then sup{f, g} and inf{f, g} are also in D+.

Proof: If f ∈ S+0 then f ∈ D+ because f = supH where H = {f},that is H consists of only one element. Thus S+0 ⊆ D+ which is Part(a).

Suppose f ∈ D+ so that f = supH where H is a countable set,and suppose H = {hp : p ∈ N} is an enumeration of H. Defineg1 = h1, g2 = sup{g1, h2}, and in general gp = sup{gp−1, hp} for p ≥ 2so that gp is an increasing sequence in S+0 . Since hp ≤ gp ≤ f we havef = sup{gp}. Conversely if f is the pointwise limit of an increasingsequence gp in S+0 then f ∈ D+ because we can choose H = {gp}.

If f ∈ D+ then f ≥ 0 and f is the pointwise limit of an increasingsequence gp of functions in S+0 . As the limit of measurable functions fis measurable by Proposition 2.11. Moreover

{x ∈ Ω : f(x) > 0} = ∪p∈N{x ∈ Ω : gp(x) > 0}

where {x ∈ Ω : gp(x) > 0} is an increasing sequence of sets of finitemeasure.

Conversely suppose that f is a nonnegative measurable functionand {x ∈ Ω : f(x) > 0} = ∪p∈NAp where Ap is a sequence of setsof finite measure. We may assume that Ap is an increasing sequence.Because f ≥ 0 is measurable there is an increasing sequence gp ∈ Sof nonnegative functions which converges pointwise to f by Theorem2.15. The sequence χApgp belongs to S+0 , is increasing and convergespointwise to f because, if x /∈ ∪pAp then f(x) = 0 = gp(x) and ifx ∈ Aq, for some q then x ∈ Ap if p ≥ q and f(x) = limp gp(x) =limp χAp(x)gp(x). This shows that f ∈ D+ and verifies Part 3.

If f is measurable and 0 ≤ f ≤ g where g ∈ D+ then {x : f(x) >0} ⊆ {x : g(x) > 0}. It follows that {x : f(x) > 0} can be written asthe countable union of sets of finite measure because {x : g(x) > 0}can, Therefore by Part 3. f ∈ D+.

Let f and g be in D+ so that, by Part 2, there are increasingsequences fp and gp, p ∈ N in S+0 so that f = limp fp and g = limp gp.Consequently f + g = limp fp + gp and this implies that f + g ∈ D+,again by Part 2. This verifies Part (5). The proof of Part (6) is similarbecause fg = limp fpgp. (Here we use the convention 0 · ∞ = 0.) Thisverifies Part 6.

Exercise: Verify Properties 7 and 8 of D+. 2


For f in D+, define Sf = {g ∈ S+0 : g ≤ f} and∫ ∗

Ω

f dµ = sup{∫

Ω

g dµ : g ∈ Sf}

Of course Sf is a very large set but we could actually use a much smallerset as the following result shows.

Theorem 5.6. If f ∈ D+ then there is an increasing sequencegp in S+0 so that f = supp gp. Moreover for any such sequence,

∫ ∗

Ω

f dµ = supp

∫ ∗

Ω

gp dµ

Proof. We have already seen that there exists an increasing se-quence gp in S+0 so that f = sup gp: this is Property 2 of D+. For anyp,

∫ ∗

Ω

gp dµ ≤ sup{∫

Ω

h dµ : h ∈ Sf}

simply because gp ∈ Sf . We may take the supremum over p to get

sup

{∫

Ω

gp dµ : p ∈ N}

≤ sup{∫

Ω

h dµ : h ∈ Sf}

=

∫ ∗

Ω

f dµ

To prove the reverse inequality, we shall show that if h ∈ Sf then∫

Ωh dµ ≤ supp{

∫

Ωgp dµ. Taking the supremum over all such h will

complete the argument. Observe that h = supp inf{h, gp} so that h isthe pointwise limit of an increasing sequence in S+0 . By Corollary 5.4,

∫

Ω

h dµ = limp∈N

∫

Ω

inf{h, gp} dµ ≤ limp∈N

∫

Ω

gp dµ

This completes the proof. �

Notation: We have S+0 ⊆ D+ (Property 1 of D+). Indeed anyh ∈ S+0 is the supremum of the constant sequence hp = h and

∫ ∗

Ω

h dµ = supp

∫

Ω

hp dµ =

∫

Ω

h dµ

Therefore the two integrals∫ ∗Ωh dµ =

∫

Ωh dµ coincide on the intersec-

tion S+0 of their domains. Later we shall construct an extension of theintegral on S0 and that integral (which will be finite valued on its do-main) will be denoted

∫

Ω. The extension

∫ ∗Ωdefined on D+ is distinct

in that it may take on the value ∞.Proposition 5.7. If f, g ∈ D+(Ω,F , µ) and λ > 0. Then

a.

∫ ∗

Ω

λf dµ = λ

∫ ∗

Ω

f dµ;

5. THE INTEGRAL 63

b.

∫ ∗

Ω

f + g dµ =

∫ ∗

Ω

f dµ+

∫ ∗

Ω

g dµ;

c.

∫

Ω

f dµ ≤∫

Ω

g dµ if f ≤ g.

Proof. Suppose that f ∈ D+ and f = supp fp where fp is anincreasing sequences in S+0 . Then, for λ > 0. λfp is an increasingsequence in S+0 convergent pointwise to λf so that λf ∈ D+ and

∫ ∗

Ω

λf dµ = supp

∫

Ω

λfp dµ = λ supp

∫

Ω

fp dµ = λ

∫ ∗

Ω

f dµ

by the Theorem 5.1. This proves Part a because∫ ∗

=∫

.Suppose now that g ∈ D+ also and g = supp gp where gp is an

increasing sequence in S+0 . Then f + g = sup fp + gp and∫ ∗

Ω

f + g dµ = supp

∫

Ω

fp + gp dµ = supp

∫

Ω

fp dµ+

∫

Ω

gp dµ

= supp

∫

Ω

fp dµ+ supp

∫

Ω

gp dµ

=

∫ ∗

Ω

f dµ+

∫ ∗

Ω

g dµ

again by Theorem 5.1 and this proves Part b.If we further assume that f ≤ g then fp ∈ Sg so that

∫

Ω

fp dµ ≤ suph∈Sg

∫

Ω

h dµ =

∫ ∗

Ω

g dµ

Taking the supremum over p gives Part c. �

Historical Notes. Henri Léon Lebesgue 1875-1941 was born inBeauvais France and received a teaching diploma from the École Nor-male Supérieure in Paris in 1897. From 1899-1902 he was a professor atthe Lycée Centrale at Nancy where he wrote several papers includinghis “Sur une généralisation de l’intégrale définie” in which he intro-duces his novel approach to integration on the real line. Lebesgue’sdoctoral dissertation, Intégrale, longueur, aire , presented to the Fac-ulty of Science in Paris in 1902, and the 130 page work was publishedin the Annali di Matematica in the same year. contains a chapter onBorel measure and measure theory and ends with a discussion of thePlateau problem.


6. Convergence of Integrals

In this section we extend the integral to real valued functions andshow that the integral of a sequence of pointwise convergent functionswill converge in many circumstances.

Theorem 6.1. Monotone Convergence Theorem: Let fpbe an increasing sequence in D+ and f = supp fp. Then f ∈ D+ and

∫

Ω

f dµ = sup

∫

Ω

fp dµ

Proof. For each p there is an increasing sequence h(p)n in S+0 so

that fp = supn h(p)n .

h(1)1 h

(1)2 . . . h

(1)p . . . ր f1

h(2)1 h

(2)2 . . . h

(2)p . . . ր f2

......

......

......

h(p)1 h

(p)2 . . . h

(p)p . . . ր fp

......

......

......

We define

gp = sup{h(1)p , h(2)p , . . . , h(p)p }so that, gp ∈ S+0 for all p ∈ N and

(1) gp is an increasing sequence;(2) gp ≤ fp;(3) f = limp gp.

Assuming these three for the moment we can now derive the result. Wehave f ∈ D+ as the pointwise limit of a sequence in S+0 and gp ≤ fp ≤ fso that

∫

Ω

gp dµ ≤∫ ∗

Ω

fp dµ ≤∫ ∗

Ω

f dµ

If we take the supremum over p, we see sup∫ ∗Ωfp dµ =

∫ ∗Ωf dµ.

Check the three properties. The sequence gp increases because h(n)p

increases as p does if n is fixed. We also have gp ≤ fp because h(n)p ≤ fnand fn ≤ fp if n ≤ p. Finally we show that, for an arbitrary x ∈ Ω,f(x) = supp gp(x). Certainly gp ≤ fp ≤ f and so we need only showthat if λ < f(x) then there is n so that gn(x) > λ. Choose p so large

that fp(x) > λ. Choose n ∈ N so large that h(p)n (x) > λ and n ≥ p.Then gn(x) ≥ h(p)n (x) > λ. This completes the proof. �

6. CONVERGENCE OF INTEGRALS 65

Example It is not true that, if fn is a decreasing sequence in D+which is pointwise convergent to f then

∫

Ωfn dµ converges to

∫

Ωf dµ.

For example fn to be the characteristic function of [−n, n]c convergespointwise to 0 but

∫

Rfn dµ = ∞. However, if we further suppose that

∫

Ωfn0 dµ < ∞ for some n0 then the integrals converge as can be seen by

applying the Monotone Convergence theorem to the sequence fn0 − fnwhich is in D+ by property 4 of Proposition 5.5 D+.

Corollary 6.2. If gn, n ∈ N is a sequence in D+ then f =∑

n∈N gn is also in D+ and∫

Ω

∑

n∈Ngn dµ =

∑

n∈N

∫

Ω

gn dµ

Proof. Define fk =∑

1≤n≤k gn so that fk ∈ D+ and apply theTheorem. Because

∫

Ωfk dµ =

∑

1≤n≤k∫

Ωgn dµ the result follows. �

We shall need the concept of lim infn an and lim supn an in the casethat an is a sequence in R. For such a sequence, define bk = infn≥k anfor each k ∈ N; bk is well defined and an increasing sequence and itslimit is lim infn an. Similarly, if ck = supn≥n ak then ck is a decreasingsequence with limit lim supn an.

Lemma 6.3. Suppose that fp : Ω → R, p ∈ N is a sequence ofreal valued functions in D+. Then lim infp fp ∈ D+.

lim infp

∫

Ω

fp dµ ≥∫

Ω

lim infp

fp dµ

Proof. For each n ∈ N, define gn = infp≥n fp. Then gn is measur-able as the limit of the measurable sequence hN = infp:N≥p≥n fp. andgn is in D+ by Property 4 of Proposition 5.5 because gn ≤ fn. The se-quence gn increases to lim infp fp which, by the Monotone ConvergenceTheorem is in D+, and

∫

Ω

lim infp∈N

fp dµ = limn∈N

∫

Ω

gn dµ.

On the other hand gn ≤ fp for all p ≥ n so that∫

Ω

gn dµ ≤∫

fp dµ for all p ≥ n

If we take the infimum over p and then the limit over

Measure Theory and Integration - Mathematics &...

Documents

Transcript of Measure Theory and Integration - Mathematics &...