1 Prove that ifAandBare subsets ofRand|B|=0, then|A∪B|=|A|. 2 SupposeA⊂Randt∈R. LettA={ta:a∈A}. Prove that|tA|=|t| |A|.
[Assume that0·∞is defined to be0.]
3 Prove that ifA,B⊂Rand|A|<∞, then|B\A| ≥ |B| − |A|.
4 SupposeFis a subset ofRwith the property that every open cover ofFhas a finite subcover. Prove thatFis closed and bounded.
5 SupposeAis a set of closed subsets ofRsuch thatTF∈AF=∅. Prove that ifA contains at least one bounded set, then there existn∈Z+andF1, . . . ,Fn∈ A such thatF1∩ · · · ∩Fn=∅.
6 Prove that ifa,b∈Randa<b, then
|(a,b)|=|[a,b)|=|(a,b]|=b−a.
7 Supposea,b,c,dare real numbers witha<bandc<d. Prove that
|(a,b)∪(c,d)|= (b−a) + (d−c)if and only if(a,b)∩(c,d) =∅.
8 Prove that ifA⊂Randt>0, then|A|=|A∩(−t,t)|+A∩ R\(−t,t). 9 Prove that|A|= lim
t→∞|A∩(−t,t)|for allA⊂R.
10 Prove that|[0, 1]\Q|=1.
11 Prove that if I1,I2, . . . is a disjoint sequence of open intervals, then
[∞ k=1
Ik=
∑
∞ k=1ℓ(Ik).
12 Supposer1,r2, . . .is a sequence that contains every rational number. Let F=R\
[∞ k=1
rk−21k,rk+ 1 2k
.
(a) Show thatFis a closed subset ofR.
(b) Prove that ifIis an interval contained inF, thenIcontains at most one element.
(c) Prove that|F|=∞.
13 Supposeε>0. Prove that there exists a subsetFof[0, 1]such thatFis closed, every element ofFis an irrational number, and|F|>1−ε.
14 Consider the following figure, which is drawn accurately to scale.
(a) Show that the right triangle whose vertices are(0, 0),(20, 0), and(20, 9) has area90.
[We have not defined area yet, but just use the elementary formulas for the areas of triangles and rectangles that you learned long ago.]
(b) Show that the yellow (lower) right triangle has area27.5.
(c) Show that the red rectangle has area45.
(d) Show that the blue (upper) right triangle has area18.
(e) Add the results of parts (b), (c), and (d), showing that the area of the colored region is90.5.
(f) Seeing the figure above, most people expect parts (a) and (e) to have the same result. Yet in part (a) we found area90, and in part (e) we found area 90.5. Explain why these results differ.
[You may be tempted to think that what we have here is a two-dimensional example similar to the result about the nonadditivity of outer measure (2.18). However, genuine examples of nonadditivity require much more complicated sets than in this example.]
2B Measurable Spaces and Functions
The last result in the previous section showed that outer measure is not additive.
Could this disappointing result perhaps be fixed by using some notion other than outer measure for the size of a subset ofR? The next result answers this question by showing that there does not exist a notion of size, called the Greek letter mu (µ) in the result below, that has all the desirable properties.
Property (c) in the result below is calledcountable additivity. Countable additivity is a highly desirable property because we want to be able to prove theorems about limits (the heart of analysis!), which requires countable additivity.
2.22 nonexistence of extension of length to all subsets of R There does not exist a functionµwith all the following properties:
(a) µis a function from the set of subsets ofRto[0,∞]. (b) µ(I) =ℓ(I)for every open intervalIofR.
(c) µ[∞
k=1
Ak
=
∑
∞ k=1µ(Ak)for every disjoint sequenceA1,A2, . . .of subsets ofR.
(d) µ(t+A) =µ(A)for everyA⊂Rand everyt∈R.
We will show thatµhas all the properties of outer measure that were used in the proof of2.18.
Proof Suppose there exists a functionµ with all the properties listed in the state- ment of this result.
Observe thatµ(∅) = 0, as follows
from (b) because the empty set is an open interval with length0.
IfA⊂B⊂R, thenµ(A)≤µ(B), as follows from (c) because we can writeB as the union of the disjoint sequenceA,B\A,∅,∅, . . .; thus
µ(B) =µ(A) +µ(B\A) +0+0+· · ·=µ(A) +µ(B\A)≥µ(A). Ifa,b ∈ Rwitha < b, then(a,b) ⊂ [a,b] ⊂ (a−ε,b+ε)for everyε> 0.
Thusb−a≤µ([a,b])≤b−a+2εfor everyε>0. Henceµ([a,b]) =b−a.
IfA1,A2, . . .is a sequence of subsets ofR, thenA1,A2\A1,A3\(A1∪A2), . . . is a disjoint sequence of subsets ofRwhose union isS∞k=1Ak. Thus
µ[∞
k=1
Ak
=µ
A1∪(A2\A1)∪ A3\(A1∪A2)∪ · · ·
=µ(A1) +µ(A2\A1) +µ A3\(A1∪A2)+· · ·
≤
∑
∞k=1
µ(Ak),
where the second equality follows from the countable additivity ofµ.
We have shown thatµhas all the properties of outer measure that were used in the proof of2.18. Repeating the proof of2.18, we see that there exist disjoint subsetsA,BofRsuch thatµ(A∪B)̸=µ(A) +µ(B). Thus the disjoint sequence A,B,∅,∅, . . .does not satisfy the countable additivity property required by (c). This contradiction completes the proof.
σ-Algebras
The last result shows that we need to give up one of the desirable properties in our goal of extending the notion of size from intervals to more general subsets ofR. We cannot give up2.22(b) because the size of an interval needs to be its length. We cannot give up2.22(c) because countable additivity is needed to prove theorems about limits. We cannot give up2.22(d) because a size that is not translation invariant does not satisfy our intuitive notion of size as a generalization of length.
Thus we are forced to relax the requirement in2.22(a) that the size is defined for all subsets ofR. Experience shows that to have a viable theory that allows for taking limits, the collection of subsets for which the size is defined should be closed under complementation and closed under countable unions. Thus we make the following definition.
2.23 Definition σ-algebra
SupposeXis a set andS is a set of subsets ofX. ThenSis called aσ-algebra onXif the following three conditions are satisfied:
• ∅∈ S;
• ifE∈ S, thenX\E∈ S;
• ifE1,E2, . . .is a sequence of elements ofS, then [∞ k=1
Ek∈ S.
Make sure you verify that the examples in all three bullet points below are indeed σ-algebras. The verification is obvious for the first two bullet points. For the third bullet point, you need to use the result that the countable union of countable sets is countable (see the proof of2.8for an example of how a doubly indexed list can be converted to a singly indexed sequence). The exercises contain some additional examples ofσ-algebras.
2.24 Example σ-algebras
• SupposeXis a set. Then clearly{∅,X}is aσ-algebra onX.
• SupposeXis a set. Then clearly the set of all subsets ofXis aσ-algebra onX.
• SupposeXis a set. Then the set of all subsetsEofXsuch thatEis countable or X\Eis countable is aσ-algebra onX.
Now we come to some easy but important properties ofσ-algebras.
2.25 σ-algebras are closed under countable intersection SupposeS is aσ-algebra on a setX. Then
(a) X∈ S;
(b) ifD,E∈ S, thenD∪E∈ SandD∩E∈ SandD\E∈ S; (c) ifE1,E2, . . .is a sequence of elements ofS, then
\∞ k=1
Ek∈ S.
Proof Because∅∈ SandX=X\∅, the first two bullet points in the definition ofσ-algebra (2.23) imply thatX∈ S, proving (a).
SupposeD,E∈ S. ThenD∪Eis the union of the sequenceD,E,∅,∅, . . .of elements ofS. Thus the third bullet point in the definition ofσ-algebra (2.23) implies thatD∪E∈ S.
De Morgan’s Laws tell us that
X\(D∩E) = (X\D)∪(X\E).
IfD,E∈ S, then the right side of the equation above is inS; henceX\(D∩E)∈ S; thus the complement inXofX\(D∩E)is inS; in other words,D∩E∈ S.
BecauseD\E = D∩(X\E), we see that ifD,E ∈ S, then D\E ∈ S, De Morgan’s Laws also show that if a collection of subsets of Xcontains the empty set, is closed under complementation, and is closed under countable intersections, then the collection is aσ-algebra.
completing the proof of (b).
Finally, suppose E1,E2, . . . is a se- quence of elements ofS. De Morgan’s Laws tell us that
X\
\∞ k=1
Ek = [∞ k=1
(X\Ek).
The right side of the equation above is inS. Hence the left side is inS, which implies thatX\(X\T∞k=1Ek)∈ S. In other words,T∞k=1Ek ∈ S, proving (c).
The wordmeasurableis used in the terminology below because in the next section we introduce a size function, called a measure, defined on measurable sets.
2.26 Definition measurable space; measurable set
• Ameasurable spaceis an ordered pair(X,S), whereXis a set andS is a σ-algebra onX.
• An element ofS is called anS-measurable set, or just ameasurable setifS is clear from the context.
For example, ifX=RandS is the set of all subsets ofRthat are countable or have a countable complement, then the set of rational numbers isS-measurable but the set of positive real numbers is notS-measurable.
Borel Subsets of R
The next result guarantees that there is a smallestσ-algebra on a setXcontaining a given setAof subsets ofX.
2.27 smallestσ-algebra containing a collection of subsets
SupposeXis a set andAis a set of subsets ofX. Then the intersection of all σ-algebras onXthat containAis aσ-algebra onX.
Proof There is at least oneσ-algebra onXthat containsAbecause theσ-algebra consisting of all subsets ofXcontainsA.
LetS be the intersection of allσ-algebras onXthat containA. Then∅ ∈ S because∅is an element of eachσ-algebra onXthat containsA.
SupposeE∈ S. ThusEis in everyσ-algebra onXthat containsA. ThusX\E is in everyσ-algebra onXthat containsA. HenceX\E∈ S.
SupposeE1,E2, . . .is a sequence of elements ofS. Thus eachEkis in everyσ- algebra onXthat containsA. ThusS∞k=1Ekis in everyσ-algebra onXthat contains A. HenceS∞k=1Ek ∈ S, which completes the proof thatSis aσ-algebra onX.
Using the terminologysmallestfor the intersection of allσ-algebras that contain a setAof subsets ofXmakes sense because the intersection of thoseσ-algebras is contained in everyσ-algebra that containsA.
2.28 Example smallestσ-algebra
• SupposeXis a set andAis the set of subsets ofXthat consist of exactly one element:
A={x}:x∈X .
Then the smallestσ-algebra onXcontainingAis the set of all subsetsEofX such thatEis countable orX\Eis countable, as you should verify.
• SupposeA= {(0, 1),(0,∞)}. Then the smallestσ-algebra onRcontaining Ais{∅,(0, 1),(0,∞),(−∞, 0]∪[1,∞),(−∞, 0],[1,∞),(−∞, 1),R}, as you should verify.
Now we come to a crucial definition.
2.29 Definition Borel set
The smallest σ-algebra on Rcontaining all open subsets of R is called the collection ofBorel subsetsofR. An element of thisσ-algebra is called aBorel set.
We have defined the collection of Borel subsets ofRto be the smallestσ-algebra onRcontaining all the open subsets ofR. We could have defined the collection of Borel subsets ofRto be the smallestσ-algebra onRcontaining all the open intervals (because every open subset ofRis the union of a sequence of open intervals).
2.30 Example Borel sets
• Every closed subset ofRis a Borel set because every closed subset ofRis the complement of an open subset ofR.
• Every countable subset ofRis a Borel set because ifB = {x1,x2, . . .}, then B=S∞k=1{xk}, which is a Borel set because each{xk}is a closed set.
• Every half-open interval[a,b)(wherea,b∈R) is a Borel set because[a,b) = T∞
k=1(a−1k,b).
• Iff: R→Ris a function, then the set of points at which f is continuous is the intersection of a sequence of open sets (see Exercise12in this section) and thus is a Borel set.
The intersection of every sequence of open subsets ofRis a Borel set. However, the set of all such intersections is not the set of Borel sets (this is not obvious, but it is not closed under countable unions). The set of all countable unions of countable intersections of open subsets ofRis also not the set of Borel sets (again, this is not obvious, but it is not closed under countable intersections). And so onad infinitum—
there is no finite procedure involving countable unions, countable intersections, and complements for constructing the collection of Borel sets.
We will see later that there exist subsets ofRthat are not Borel sets. However, any subset ofRthat you can write down in a concrete fashion is a Borel set.
Inverse Images
The next definition is used frequently in the rest of this chapter.
2.31 Definition inverse image; f−1(A)
If f:X→Yis a function andA⊂Y, then the setf−1(A)is defined by f−1(A) ={x∈X: f(x)∈A}.
2.32 Example inverse images
Supposef:[0, 4π]→Ris defined by f(x) =sinx. Then f−1 (0,∞)= (0,π)∪(2π, 3π),
f−1([0, 1]) = [0,π]∪[2π, 3π]∪ {4π}, f−1({−1}) ={3π2 ,7π2 },
f−1 (2, 3)=∅, as you should verify.
Inverse images have good algebraic properties, as is shown in the next two results.
2.33 algebra of inverse images Supposef: X→Yis a function. Then
(a) f−1(Y\A) =X\f−1(A)for everyA⊂Y;
(b) f−1(SA∈AA) =SA∈Af−1(A)for every setAof subsets ofY;
(c) f−1(TA∈AA) =TA∈Af−1(A)for every setAof subsets ofY. Proof SupposeA⊂Y. Forx∈Xwe have
x∈ f−1(Y\A) ⇐⇒ f(x)∈Y\A
⇐⇒ f(x)∈/ A
⇐⇒ x∈/ f−1(A)
⇐⇒ x∈X\f−1(A). Thus f−1(Y\A) =X\f−1(A), which proves (a).
To prove (b), supposeAis a set of subsets ofY. Then x∈ f−1( [
A∈A
A) ⇐⇒ f(x)∈ [
A∈A
A
⇐⇒ f(x)∈Afor someA∈ A
⇐⇒ x∈ f−1(A)for someA∈ A
⇐⇒ x∈ [
A∈A
f−1(A). Thus f−1(SA∈AA) =SA∈Af−1(A), which proves (b).
Part (c) is proved in the same fashion as (b), with unions replaced by intersections andfor somereplaced byfor every.
2.34 inverse image of a composition
Supposef: X→Yandg:Y→Ware functions. Then (g◦f)−1(A) = f−1 g−1(A) for everyA⊂W.
Proof SupposeA⊂W. Forx∈Xwe have
x∈(g◦f)−1(A) ⇐⇒ (g◦f)(x)∈A ⇐⇒ g f(x)∈ A
⇐⇒ f(x)∈g−1(A)
⇐⇒ x ∈ f−1 g−1(A). Thus(g◦f)−1(A) = f−1 g−1(A).
Measurable Functions
The next definition tells us which real-valued functions behave reasonably with respect to aσ-algebra on their domain.
2.35 Definition measurable function
Suppose (X,S) is a measurable space. A function f: X → R is called S-measurable(or justmeasurableifSis clear from the context) if
f−1(B)∈ S for every Borel setB⊂R.
2.36 Example measurable functions
• IfS = {∅,X}, then the onlyS-measurable functions fromX toRare the constant functions.
• If S is the set of all subsets of X, then every function from X toR is S- measurable.
• IfS = {∅,(−∞, 0),[0,∞),R}(which is aσ-algebra onR), then a function f:R →RisS-measurable if and only if f is constant on(−∞, 0)and f is constant on[0,∞).
Another class of examples comes from characteristic functions, which are defined below. The Greek letter chi (χ) is traditionally used to denote a characteristic function.
2.37 Definition characteristic function;χE
SupposeEis a subset of a setX. Thecharacteristic functionofEis the function χE:X→Rdefined by
χE(x) =
(1 ifx∈ E, 0 ifx∈/ E.
The setXthat containsEis not explicitly included in the notationχEbecauseXwill always be clear from the context.
2.38 Example inverse image with respect to a characteristic function Suppose(X,S)is a measurable space,E⊂X, andB⊂R. Then
χE−1(B) =
E if0 /∈Band1∈B, X\E if0∈Band1 /∈B, X if0∈Band1∈B,
∅ if0 /∈Band1 /∈B.
Thus we see thatχEis anS-measurable function if and only ifE∈ S.
Note that if f:X→Ris a function and a∈R, then
f−1 (a,∞)={x∈X: f(x)>a}. The definition of anS-measurable
function requires the inverse image of every Borel subset ofRto be in S. The next result shows that to ver- ify that a function isS-measurable, we can check the inverse images of a much smaller collection of subsets ofR.
2.39 condition for measurable function
Suppose(X,S)is a measurable space and f:X→Ris a function such that f−1 (a,∞)∈ S
for alla∈R. Then f is anS-measurable function.
Proof Let
T ={A⊂R: f−1(A)∈ S}.
We want to show that every Borel subset ofRis inT. To do this, we will first show thatT is aσ-algebra onR.
Certainly∅∈ T, because f−1(∅) =∅∈ S. IfA∈ T, then f−1(A)∈ S; hence
f−1(R\A) =X\f−1(A)∈ S
by2.33(a), and thusR\A∈ T. In other words,T is closed under complementation.
IfA1,A2, . . .∈ T, then f−1(A1),f−1(A2), . . .∈ S; hence f−1[∞
k=1
Ak
= [∞ k=1
f−1(Ak)∈ S
by2.33(b), and thusS∞k=1Ak ∈ T. In other words, T is closed under countable unions. ThusT is aσ-algebra onR.
By hypothesis, T contains {(a,∞) : a ∈ R}. Because T is closed under complementation,T also contains{(−∞,b]:b∈R}. Because theσ-algebraT is closed under finite intersections (by2.25), we see thatT contains{(a,b]:a,b∈R}. Because(a,b) =S∞k=1(a,b−1k]and(−∞,b) =S∞k=1(−k,b−1k]andT is closed under countable unions, we can conclude thatT contains every open subset ofR.
Thus theσ-algebraT contains the smallestσ-algebra onRthat contains all open subsets ofR. In other words,T contains every Borel subset ofR. Thus f is an S-measurable function.
In the result above, we could replace the collection of sets{(a,∞) : a ∈ R} by any collection of subsets ofRsuch that the smallestσ-algebra containing that collection contains the Borel subsets ofR. For specific examples of such collections of subsets ofR, see Exercises3–6.
We have been dealing withS-measurable functions fromXtoRin the context of an arbitrary setXand aσ-algebraS onX. An important special case of this setup is whenXis a Borel subset ofRandSis the set of Borel subsets ofRthat are contained inX(see Exercise11for another way of thinking about thisσ-algebra). In this special case, theS-measurable functions are called Borel measurable.
2.40 Definition Borel measurable function
SupposeX⊂R. A function f:X→Ris calledBorel measurableif f−1(B)is a Borel set for every Borel setB⊂R.
IfX⊂Rand there exists a Borel measurable function f:X→R, thenXmust be a Borel set [becauseX= f−1(R)].
IfX⊂Rand f:X→Ris a function, then f is a Borel measurable function if and only iff−1 (a,∞)is a Borel set for everya∈R(use2.39).
SupposeXis a set and f: X→Ris a function. The measurability of f depends upon the choice of aσ-algebra onX. If theσ-algebra is calledS, then we can discuss whether f is anS-measurable function. IfXis a Borel subset ofR, thenSmight be the set of Borel sets contained inX, in which case the phraseBorel measurable means the same asS-measurable. However, whether or notS is a collection of Borel sets, we consider inverse images of Borel subsets ofRwhen determining whether a function isS-measurable.
The next result states that continuity interacts well with the notion of Borel measurability.
2.41 every continuous function is Borel measurable
Every continuous real-valued function defined on a Borel subset ofRis a Borel measurable function.
Proof SupposeX⊂Ris a Borel set and f:X→Ris continuous. To prove that f is Borel measurable, fixa∈R.
Ifx∈Xand f(x)>a, then (by the continuity off) there existsδx>0such that f(y)>afor ally∈(x−δx,x+δx)∩X. Thus
f−1 (a,∞)= [
x∈f−1 (a,∞)(x−δx,x+δx)∩X.
The union inside the large parentheses above is an open subset of R; hence its intersection withXis a Borel set. Thus we can conclude thatf−1 (a,∞)is a Borel set.
Now2.39implies that f is a Borel measurable function.
Next we come to another class of Borel measurable functions. A similar definition could be made for decreasing functions, with a corresponding similar result.
2.42 Definition increasing function; strictly increasing SupposeX⊂Randf:X→Ris a function.
• f is calledincreasingif f(x)≤ f(y)for allx,y∈Xwithx<y.
• f is calledstrictly increasingif f(x)< f(y)for allx,y∈Xwithx<y.
2.43 every increasing function is Borel measurable
Every increasing function defined on a Borel subset ofRis a Borel measurable function.
Proof SupposeX⊂Ris a Borel set and f:X→Ris increasing. To prove that f is Borel measurable, fixa∈R.
Letb=inff−1 (a,∞). Then it is easy to see that
f−1 (a,∞)= (b,∞)∩X or f−1 (a,∞)= [b,∞)∩X.
Either way, we can conclude that f−1 (a,∞)is a Borel set.
Now2.39implies that f is a Borel measurable function.
The next result shows that measurability interacts well with composition.
2.44 composition of measurable functions
Suppose(X,S) is a measurable space and f: X → Ris an S-measurable function. Supposeg is a real-valued Borel measurable function defined on a subset ofRthat includes the range of f. Theng◦f:X→Ris anS-measurable function.
Proof SupposeB⊂Ris a Borel set. Then (see2.34) (g◦f)−1(B) = f−1 g−1(B).
Becausegis a Borel measurable function,g−1(B)is a Borel subset ofR. Because f is anS-measurable function, f−1 g−1(B)∈ S. Thus the equation above implies that(g◦f)−1(B)∈ S. Thusg◦f is anS-measurable function.
2.45 Example if f is measurable, then so are−f,12f,|f|,f2
Suppose(X,S)is a measurable space andf:X→RisS-measurable. Then2.44 implies that the functions−f,12f,|f|,f2are allS-measurable functions because each of these functions can be written as the composition off with a continuous (and thus Borel measurable) functiong.
Specifically, takeg(x) = −x, theng(x) = 12x, theng(x) = |x|, and then g(x) =x2.
Measurability also interacts well with algebraic operations, as shown in the next result.
2.46 algebraic operations with measurable functions
Suppose(X,S)is a measurable space andf,g:X→RareS-measurable. Then (a) f+g,f −g, and f gareS-measurable functions;
(b) ifg(x)̸=0for allx∈X, then gf is anS-measurable function.
Proof Supposea∈R. We will show that 2.47 (f +g)−1 (a,∞)= [
r∈Q
f−1 (r,∞)∩g−1 (a−r,∞),
which implies that(f+g)−1 (a,∞)∈ S. To prove2.47, first suppose
x ∈(f+g)−1 (a,∞).
Thusa< f(x) +g(x). Hence the open interval a−g(x),f(x)is nonempty, and thus it contains some rational numberr. This implies thatr< f(x), which means thatx ∈ f−1 (r,∞), anda−g(x)<r, which implies thatx∈g−1 (a−r,∞). Thusxis an element of the right side of2.47, completing the proof that the left side of2.47is contained in the right side.
The proof of the inclusion in the other direction is easier. Specifically, suppose x∈ f−1 (r,∞)∩g−1 (a−r,∞)for somer∈Q. Thus
r< f(x) and a−r<g(x).
Adding these two inequalities, we see thata< f(x) +g(x). Thusxis an element of the left side of2.47, completing the proof of2.47. Hence f+gis anS-measurable function.
Example2.45tells us that−gis anS-measurable function. Thus f −g, which equals f+ (−g)is anS-measurable function.
The easiest way to prove that f gis anS-measurable function uses the equation f g= (f +g)2− f2−g2
2 .
The operation of squaring an S-measurable function produces anS-measurable function (see Example2.45), as does the operation of multiplication by 12(again, see Example2.45). Thus the equation above implies thatf gis anS-measurable function, completing the proof of (a).
Supposeg(x)̸=0for allx∈X. The function defined onR\ {0}(a Borel subset ofR) that takesxto 1x is continuous and thus is a Borel measurable function (by 2.41). Now2.44implies that1gis anS-measurable function. Combining this result with what we have already proved about the product ofS-measurable functions, we conclude that gf is anS-measurable function, proving (b).