1 Show thatS ={Sn∈K(n,n+1]:K⊂Z}is aσ-algebra onR.
2 Verify both bullet points in Example2.28.
3 SupposeSis the smallestσ-algebra onRcontaining{(r,s]:r,s∈Q}. Prove thatSis the collection of Borel subsets ofR.
4 SupposeS is the smallestσ-algebra onRcontaining{(r,n]:r∈Q,n∈Z}. Prove thatSis the collection of Borel subsets ofR.
5 SupposeS is the smallestσ-algebra onRcontaining {(r,r+1) : r ∈ Q}. Prove thatSis the collection of Borel subsets ofR.
6 SupposeSis the smallestσ-algebra onRcontaining{[r,∞):r∈Q}. Prove thatSis the collection of Borel subsets ofR.
7 Prove that the collection of Borel subsets ofRis translation invariant. More precisely, prove that ifB⊂Ris a Borel set andt∈R, thent+Bis a Borel set.
8 Prove that the collection of Borel subsets ofR is dilation invariant. More precisely, prove that ifB ⊂ Ris a Borel set andt ∈ R, thentB(which is defined to be{tb:b∈B}) is a Borel set.
9 Give an example of a measurable space(X,S)and a function f:X→Rsuch that|f|isS-measurable butf is notS-measurable.
10 Show that the set of real numbers that have a decimal expansion with the digit5 appearing infinitely often is a Borel set.
11 SupposeT is aσ-algebra on a setYandX∈ T. LetS ={E∈ T :E⊂X}. (a) Show thatS ={F∩X:F∈ T }.
(b) Show thatS is aσ-algebra onX.
12 Suppose f:R→Ris a function.
(a) Fork∈Z+, let
Gk ={a∈R:there existsδ>0such that|f(b)−f(c)|< 1k for allb,c∈(a−δ,a+δ)}.
Prove thatGkis an open subset ofRfor eachk∈Z+.
(b) Prove that the set of points at which f is continuous equalsT∞k=1Gk. (c) Conclude that the set of points at whichf is continuous is a Borel set.
13 Suppose(X,S)is a measurable space,E1, . . . ,Enare disjoint subsets ofX, and c1, . . . ,cnare distinct nonzero real numbers. Prove thatc1χE
1+· · ·+cnχE
nis anS-measurable function if and only ifE1, . . . ,En ∈ S.
14 (a) Supposef1,f2, . . .is a sequence of functions from a setXtoR. Explain why
{x ∈X:the sequence f1(x),f2(x), . . .has a limit inR}
=
\∞ n=1
[∞ j=1
\∞ k=j
(fj−fk)−1 (−1n,1n).
(b) Suppose(X,S)is a measurable space and f1,f2, . . .is a sequence ofS- measurable functions fromXtoR. Prove that
{x ∈X:the sequence f1(x),f2(x), . . .has a limit inR} is anS-measurable subset ofX.
15 SupposeXis a set andE1,E2, . . .is a disjoint sequence of subsets ofXsuch thatS∞k=1Ek=X. LetS ={Sk∈KEk:K⊂Z+}.
(a) Show thatS is aσ-algebra onX.
(b) Prove that a function fromXtoRisS-measurable if and only if the function is constant onEkfor everyk∈Z+.
16 SupposeS is aσ-algebra on a setXandA⊂X. Let SA={E∈ S :A⊂EorA∩E=∅}. (a) Prove thatSAis aσ-algebra onX.
(b) Supposef: X→Ris a function. Prove that f is measurable with respect toSA if and only if f is measurable with respect toS and f is constant onA.
17 SupposeX is a Borel subset of R and f: X → R is a function such that {x ∈ X : f is not continuous atx} is a countable set. Prove f is a Borel measurable function.
18 Supposef: R→Ris differentiable at every element ofR. Prove that f′is a Borel measurable function fromRtoR.
19 SupposeXis a nonempty set andS is theσ-algebra onXconsisting of all subsets ofXthat are either countable or have a countable complement inX.
Give a characterization of theS-measurable real-valued functions onX.
20 Suppose(X,S)is a measurable space and f,g: X → RareS-measurable functions. Prove that iff(x)>0for allx∈X, then fg(which is the function whose value atx∈Xequals f(x)g(x)) is anS-measurable function.
21 Prove2.52.
22 SupposeB ⊂ Rand f: B → Ris an increasing function. Prove that f is continuous at every element ofBexcept for a countable subset ofB.
23 Suppose f: R → Ris a strictly increasing function. Prove that the inverse functionf−1: f(R)→Ris a continuous function.
[Note that this exercise does not have as a hypothesis that f is continuous.]
24 Supposef:R→Ris a strictly increasing function andB⊂Ris a Borel set.
Prove thatf(B)is a Borel set.
25 SupposeB⊂Randf: B→Ris an increasing function. Prove that there exists a sequence f1,f2, . . .of strictly increasing functions fromBtoRsuch that
f(x) = lim
k→∞fk(x) for everyx ∈B.
26 SupposeB⊂Rand f:B →Ris a bounded increasing function. Prove that there exists an increasing functiong:R →Rsuch thatg(x) = f(x)for all x∈B.
27 Prove or give a counterexample: If(X,S)is a measurable space and f:X→[−∞,∞]
is a function such that f−1 (a,∞) ∈ S for every a ∈ R, then f is an S-measurable function.
28 Supposef: B→Ris a Borel measurable function. Defineg:R→Rby g(x) =
(f(x) ifx∈B,
0 ifx∈R\B.
Prove thatgis a Borel measurable function.
29 Give an example of a measurable space(X,S) and a family{ft}t∈R such that each ft is anS-measurable function from Xto [0, 1], but the function
f:X→[0, 1]defined by
f(x) =sup{ft(x):t∈R} is notS-measurable.
[Compare this exercise to2.53, where the index set isZ+rather thanR.]
30 Show that
j→∞lim lim
k→∞ cos(j!πx)2k=
(1 ifxis rational, 0 ifxis irrational for everyx ∈R.
[This example is due to Henri Lebesgue.]
2C Measures and Their Properties
Definition and Examples of Measures
The original motivation for the next definition came from trying to extend the notion of the length of an interval. However, the definition below allows us to discuss size in many more contexts. For example, we will see later that the area of a set in the plane or the volume of a set in higher dimensions fits into this structure. The word measureallows us to use a single word instead of repeating theorems forlength,area, andvolume.
2.54 Definition measure
SupposeXis a set andSis aσ-algebra onX. Ameasureon(X,S)is a function µ:S →[0,∞]such thatµ(∅) =0and
µ[∞
k=1
Ek
=
∑
∞ k=1µ(Ek) for every disjoint sequenceE1,E2, . . .of sets inS.
In the mathematical literature, sometimes a measure on(X,S) is just called a measure onXif theσ-algebraS is clear from the context.
The concept of a measure, as defined here, is sometimes called apositive measure (although the phrasenonnegative measure would be more accurate). The countable additivity that forms the key
part of the definition above allows us to prove good limit theorems. Note that countable ad- ditivity implies finite additivity: ifµis a mea- sure on(X,S) and E1, . . . ,En are disjoint sets inS, then
µ(E1∪ · · · ∪En) =µ(E1) +· · ·+µ(En), as follows from applying the equation µ(∅) =0and countable additivity to the dis- joint sequence E1, . . . ,En,∅,∅, . . . of sets inS.
2.55 Example measures
• IfXis a set, thencounting measureis the measureµdefined on theσ-algebra of all subsets ofXby settingµ(E) =nifEis a finite set containing exactlyn elements andµ(E) =∞ifEis not a finite set.
• SupposeXis a set,S is aσ-algebra onX, andc∈X. Define theDiracmeasure δcon(X,S)by
δc(E) =
(1 ifc∈E, 0 ifc∈/E.
This measure is named in honor of mathematician and physicist Paul Dirac (1902–
1984), who won the Nobel Prize for Physics in 1933 for his work combining relativity and quantum mechanics at the atomic level.
• SupposeXis a set,S is aσ-algebra onX, andw: X →[0,∞]is a function.
Define a measureµon(X,S)by
µ(E) =
∑
x∈E
w(x)
forE ∈ S. [Here the sum is defined as the supremum of all finite subsums
∑x∈Dw(x)asDranges over all finite subsets ofE.]
• SupposeXis a set andS is theσ-algebra onXconsisting of all subsets ofX that are either countable or have a countable complement inX. Define a measure µon(X,S)by
µ(E) =
(0 ifEis countable, 3 ifEis uncountable.
• SupposeSis theσ-algebra onRconsisting of all subsets ofR. Then the function that takes a setE⊂Rto|E|(the outer measure ofE) is not a measure because it is not finitely additive (see2.18).
• SupposeBis theσ-algebra onRconsisting of all Borel subsets ofR. We will see in the next section that outer measure is a measure on(R,B).
The following terminology is frequently useful.
2.56 Definition measure space
Ameasure spaceis an ordered triple(X,S,µ), whereXis a set,S is aσ-algebra onX, andµis a measure on(X,S).
Properties of Measures
The hypothesis thatµ(D) < ∞is needed in part (b) of the next result to avoid undefined expressions of the form∞−∞.
2.57 measure preserves order; measure of a set difference
Suppose(X,S,µ)is a measure space andD,E∈ Sare such thatD⊂E. Then (a) µ(D)≤µ(E);
(b) µ(E\D) =µ(E)−µ(D)provided thatµ(D)<∞.
Proof BecauseE=D∪(E\D)and this is a disjoint union, we have µ(E) =µ(D) +µ(E\D)≥µ(D),
which proves (a). Ifµ(D) < ∞, then subtractingµ(D)from both sides of the equation above proves (b).
The countable additivity property of measures applies to disjoint countable unions.
The following countable subadditivity property applies to countable unions that may not be disjoint unions.
2.58 countable subadditivity
Suppose(X,S,µ)is a measure space andE1,E2, . . .∈ S. Then µ[∞
k=1
Ek
≤
∑
∞k=1
µ(Ek).
Proof LetD1=∅andDk =E1∪ · · · ∪Ek−1fork≥2. Then E1\D1,E2\D2,E3\D3, . . .
is a disjoint sequence of subsets ofXwhose union equalsS∞k=1Ek. Thus µ[∞
k=1
Ek
=µ[∞
k=1
(Ek\Dk)
=
∑
∞ k=1µ(Ek\Dk)
≤
∑
∞k=1
µ(Ek),
where the second line above follows from the countable additivity ofµand the last line above follows from2.57(a).
Note that countable subadditivity implies finite subadditivity: ifµis a measure on (X,S)andE1, . . . ,Enare sets inS, then
µ(E1∪ · · · ∪En)≤µ(E1) +· · ·+µ(En),
as follows from applying the equationµ(∅) =0and countable subadditivity to the sequenceE1, . . . ,En,∅,∅, . . .of sets inS.
The next result shows that measures behave well with increasing unions.
2.59 measure of an increasing union
Suppose(X,S,µ) is a measure space and E1 ⊂ E2 ⊂ · · · is an increasing sequence of sets inS. Then
µ[∞
k=1
Ek
= lim
k→∞µ(Ek).
Proof Ifµ(Ek) = ∞for somek ∈ Z+, then the equation above holds because both sides equal∞. Hence we can consider only the case whereµ(Ek)<∞for all k∈Z+.
For convenience, letE0=∅. Then [∞ k=1
Ek= [∞ j=1
(Ej\Ej−1), where the union on the right side is a disjoint union. Thus µ[∞
k=1
Ek
=
∑
∞ j=1µ(Ej\Ej−1)
= lim
k→∞
∑
k j=1µ(Ej\Ej−1)
= lim
k→∞
∑
k j=1µ(Ej)−µ(Ej−1)
= lim
k→∞µ(Ek),
Another mew.
as desired.
Measures also behave well with respect to decreasing intersections (but see Exer- cise10, which shows that the hypothesisµ(E1)<∞below cannot be deleted).
2.60 measure of a decreasing intersection
Suppose(X,S,µ)is a measure space and E1 ⊃ E2 ⊃ · · · is a decreasing sequence of sets inS, withµ(E1)<∞. Then
µ\∞
k=1
Ek
= lim
k→∞µ(Ek). Proof One of De Morgan’s Laws tells us that
E1\
\∞ k=1
Ek= [∞ k=1
(E1\Ek).
NowE1\E1 ⊂ E1\E2 ⊂ E1\E3 ⊂ · · · is an increasing sequence of sets inS. Thus2.59, applied to the equation above, implies that
µ E1\
\∞ k=1
Ek
= lim
k→∞µ(E1\Ek). Use2.57(b) to rewrite the equation above as
µ(E1)−µ
\∞ k=1
Ek
=µ(E1)− lim
k→∞µ(Ek), which implies our desired result.
The next result is intuitively plausible—we expect that the measure of the union of two sets equals the measure of the first set plus the measure of the second set minus the measure of the set that has been counted twice.
2.61 measure of a union
Suppose(X,S,µ)is a measure space andD,E∈ S, withµ(D∩E)<∞. Then µ(D∪E) =µ(D) +µ(E)−µ(D∩E).
Proof We have
D∪E= D\(D∩E)∪ E\(D∩E)∪ D∩E . The right side of the equation above is a disjoint union. Thus
µ(D∪E) =µ D\(D∩E)+µ E\(D∩E)+µ D∩E
= µ(D)−µ(D∩E)+ µ(E)−µ(D∩E)+µ(D∩E)
=µ(D) +µ(E)−µ(D∩E), as desired.