An Introduction to Measure-Theoretic Probability , Second Edition [2nd Ed] (Solutions) (Instructor's Solution Manual)

130 Pages·2014·1.061 MB·English

by  George G. Roussas

Revised Answers Manual to
an Introduction to
Measure-Theoretic
Probability
GeorgeG.Roussas
UniversityofCalifornia,Davis,UnitedStates
Chapter 1
Certain Classes of Sets, Measurability, Pointwise Approximation
1. (i) x ∈ limn→∞An if and only if x ∈ ∪n≥1∩j≥nAj, so that x ∈ ∩j≥n0Aj
forsomen0 ≥ 1,andthen x ∈ Aj forall j ≥ n0,or x ∈ ∪j≥nAj forall
(cid:2)n ≥1,sotha(cid:3)tx ∈∩(cid:2)n≥1∪j≥1Ajlim(cid:3)n→∞An.
(ii) (cid:2)limn→∞An(cid:3)c =(cid:2) ∪n≥1∩j≥nA(cid:3)j c = ∩n≥1∪j≥nAcj = limn→∞Acn,
limn→∞An c = ∩n≥1∪j≥nAj c =∪n≥1∩j≥nAcj =(cid:2)limn→∞Acn.(cid:3)
L(cid:2)et limn→∞(cid:3)An = A. Then limn→∞Acn = (cid:2)limn→∞An(cid:3)c =
(cid:2)limn→∞An(cid:3)c = Ac, and limn→∞An = limn→∞An c =
limn→∞An c = Ac,sothatlimn→∞Acn existsandis Ac.
(iii) To show that limn→∞(An ∩ Bn) = (limn→∞An) ∩ (limn→∞Bn).
Equivalently,
(cid:4) (cid:5) (cid:4) (cid:5)
∞ ∞ ∞ ∞ ∞ ∞
∪ ∩ (A ∩B )= ∪ ∩ A ∩ ∪ ∩ B .
j j j j
n=1j=n n=1j=n n=1j=n
Indeed,let x belongtotheleft-handside.Then x ∈ ∩∞ (A ∩ B )for
somen ≥ 1,hence x ∈ (A ∩ B )forall j ≥ n ,andj=tnh0en xj ∈ Aj and
0 j j 0 j
x ∈ B forall j ≥n .Hencex ∈∩∞ A andx ∈∩∞ B ,sothatx ∈
∪∞ ∩j∞ A andx ∈0 ∪∞ ∩∞ B j;=i.ne0.,xjbelongstotjh=en0rigjht-handside.
n=1 j=n j n=1 j=n j
Next,letx belongtotheright-handside.Thenx ∈∪∞ ∩∞ A andx ∈
n=1 j=n j
∪∞ ∩∞ B ,sothatx ∈∩∞ A andx ∈∩∞ B forsomen ,n ≥1.
Thn=en1xj∈=n∩∞j A andx ∈j∩=∞n1 jB wherenj==n2mjax(n ,n ),a1nd2hence
x ∈ A andj=xn∈0 Bj forall j ≥j=nn0 .jThus, x ∈0(A ∩ B 1)fo2rall j ≥ n ,
j j 0 j j 0
sothat x ∈ ∩∞ (A ∩ B )andhence x ∈ ∪∞ ∩∞ (A ∩ B );i.e., x
j=n0 j j n=1 j=n j j
belongstotheleft-handside.
Next, limn→∞(An ∪ Bn) = limn→∞(Acn ∩ Bnc)c = [limn→∞(Acn ∩
Bnc)]c(bypart(ii)),andthisequalsto[(limn→∞Acn)∩(limn→∞Bnc)]c(by
whatwejustproved),andthisequals[(limn→∞An)c∩(limn→∞Bn)c]c =
(limn→∞An)∪(limn→∞Bn),aswastobeseen.
AnIntroductiontoMeasure-TheoreticProbability.http://dx.doi.org/10.1016/B978-0-12-800042-7.00028-1 e1
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e2 Revised Answers Manual to an Introduction
(iv) To show that: limn→∞(An ∩ Bn) ⊆ (limn→∞An) ∩ (limn→∞Bn) and
limn→∞(An ∪Bn)⊇(limn→∞An)∪(limn→∞Bn). (cid:6) (cid:7)
Suffices to show: ∩∞ ∪∞ (A ∩ B ) ⊆ ∩∞ ∪∞ A ∩
(cid:6) (cid:7) n=1 j=n j j n=1 j=n j
∩∞ ∪∞ B .
n=1 j=n j
Indeed,letx belongtotheleft-handside.Thenx ∈∪∞ (A ∩B )for
j=n j j
all n ≥ 1, so that x ∈ (A ∩ B ) for some j ≥ n and all n ≥ 1. Then
j j
x ∈ A andx ∈ B forsome j ≥nandalln ≥1,hencex ∈∪∞ A and
j j j=n j
x ∈∪∞ B foralln ≥1,sothatx ∈∩∞ ∪∞ A andx ∈∩∞ ∪∞ B ,
j=n j (cid:6) (cid:7) (cid:6) n=1 j=n (cid:7)j n=1 j=n j
and hence x ∈ ∩∞ ∪∞ A ∩ ∩∞ ∪∞ B ;i.e., x belongs tothe
n=1 j=n j n=1 j=n j
right-handside.So,the(cid:6)aboveinclusion(cid:7)isc(cid:6)orrect. (cid:7)
Also,toshowthat: ∪∞ ∩∞ A ∪ ∪∞ ∩∞ B ⊆ ∪∞ ∩∞
n=1 j=n j n=1 j=n j n=1 j=n
(A ∪B ).
j j
Indeed, let x belong to the left-hand side. Then x ∈ ∪∞ ∩∞ A or
n=1 j=n j
x ∈ ∪∞ ∩∞ B ortoboth.Let x ∈ ∪∞ ∩∞ A .Then x ∈ ∩∞ A
forsomn=e1n j=≥n1,jhence x ∈ A forall jn=≥1 nj=,nandj then x ∈ (Aj=∪n0B )j
0 j 0 j j
for all j ≥ n , so that x ∈ ∪∞ ∩∞ (A ∪ B ); i.e., x belongs to the
0 n=1 j=n j j
right-handside.Similarlyifx ∈∪∞ ∩∞ B .
n=1 j=n j
Analternativeproofofthesecondpartisasfollows:
(cid:10) (cid:11)
(cid:8)∞ (cid:9)∞ (cid:9)∞ (cid:8)∞ c
lim(A ∪B )= (A ∪B )= (Ac ∩Bc)
n n k k k k
(cid:12)n=1k=n (cid:13) (cid:12)(cid:2)n=1k=(cid:3)n (cid:2) (cid:3)(cid:13)
= lim(Ac ∩Bc) c ⊇ limAc ∩ limBc c
k k k k
(bythepreviouspart)
(cid:14) (cid:15) (cid:14) (cid:15)
(cid:9)∞ (cid:8)∞ c (cid:9)∞ (cid:8)∞ c
= ∪ Bc
k
(cid:14)n=1k=n (cid:15) n(cid:14)=1k=n (cid:15)
(cid:8)∞ (cid:9)∞ (cid:8)∞ (cid:9)∞
= A ∪ B =(limA )∪(limB ).
k k n n
n=1k=n n=1k=n
(v) That the inverse inclusions in part (iv) need not hold is demonstrated by
thefollowing
Counterexample:
Let A2j−1 = A,A2j = A0 and B2j−1 = B, B2j = B0, j ≥ 1,forsome
events A,A0,B and B0. Then: limn→∞An = A ∩ A0, limn→∞An =
A∪A0, limn→∞Bn = B ∩ B0, limn→∞Bn = B ∪ B0, limn→∞
(An∩Bn)=(A∩B)∪(A0∩B0),limn→∞(An∪Bn)=(A∪B)∩(A0∪B0).
Therefore(A∪B)∩(A ∪B )neednotcontain(A∪A )∩(B∪B ),and
0 0 0 0
(A∩ A )∪(B∩B )neednotcontain(A∪B)∩(A ∪B ).
0 0 0 0
As a concrete example, take (cid:2) = (cid:10),A = (0,1], A = [2,3],
0
B =[1,2],B =[3,4].Then:(A∪B)∩(A ∪B )=(0,2],(A∪A )∩
0 0 0 0
Revised Answers Manual to an Introduction e3
(B ∪ B ) = ((0,1]∪[2,3])∩([1,2]∪[3,4]) = {1}∪{3} = {1,3} (cid:2)
0
(0,2],and(A∩ A )∪(B∩B )=(cid:11)∪(cid:11)=(cid:11),(A∪B)∩(A ∪B )=
0 0 0 0
(0,2]∩[2,4]={2}notcontainedin(cid:11).
(vi) If limn→∞An = A and limn→∞Bn = B, then by parts (iii) and (iv):
limn→∞(An ∩ Bn) ⊆ A ∩ B and limn→∞(An ∩ Bn) = A ∩ B. Thus,
A ∩ B = limn→∞(An ∩ Bn) ⊆ limn→∞(An ∩ Bn) ⊆ A ∩ B, so that
limn→∞(An ∩ Bn) = A∩ B.Likewise: A∪ B ⊆ limn→∞(An ∪ Bn) ⊆
limn→∞(An ∪Bn)= A∪B,sothatlimn→∞(An ∪Bn)= A∪B.
(vii) Since A (cid:12)B = (A − B)+(B − A ) = (A ∩ Bc)+(B ∩ Ac), we
n n n n n
havelimn→∞(An∩Bc)=(limn→∞An)∩Bc = A∩Bcbypart(vi),and
limn→∞(B ∩ Acn) = B ∩(limn→∞Acn) = B ∩ Ac byparts(vi)and(ii).
Therefore,bypart(vi)again,limn→∞(An(cid:12)B) = limn→∞[(An ∩ Bc)+
(B ∩ Acn)] = limn→∞(An ∩ Bc) + limn→∞(B ∩ Acn) = (A ∩ Bc) +
(B∩ Ac)= A(cid:12)B.
(viii) A2j−1 = B,A2j =C, j ≥1.Then,asinpart(v),limn→∞An = B∩Cand
limn→∞An = B∪C.Thelimn→∞AnexistsifandonlyifB∩C = B∪C,
orB∪C =(B∩Cc)+(Bc∩C)+(B∩C)= B∩C.Then,bythepairwise
disjointnessofB∩Cc,Bc∩CandB∩C,wehaveB∩Cc = Bc∩C =(cid:11).
From B ∩Cc = (cid:11), it follows that B ⊆ C, and from Bc ∩C = (cid:11), it
follows that C ⊆ B. Therefore B = C. Thus, limn→∞An exists if and
onlyif B =C.#
2. (i) AllthreesetsA,A,andA(ifitexists)areinA,becausetheyareexpressed
intermsof A ,n ≥1,bymeansofcountableoperations.
n
(ii) Let An ↑. Then limn→∞An = ∪∞n=1∩∞j=nAj = ∪∞n=1An, and limn→∞
An =∩∞n=1∪∞j=nAj =∪∞j=nAj =∪∞j=1Aj =∪∞n=1An,sothatlimn→∞
A =∪∞ A .
n n=1 n
If A ↓, then Ac ↑ and hence ∩∞ ∪∞ Ac = ∪∞ ∩∞ Ac =
n n n=1 j=n j n=1 j=n j
∪∞ Ac,sothat,bytakingthecomplements,∪∞ ∩∞ A =∩∞ ∪∞
n=1 n n=1 j=n j n=1 j=n
Aj =∩∞n=1An,sothatlimn→∞An =∩∞n=1An.#
3. (i) ∩j∈IFj (cid:15)= (cid:11) since, e.g., (cid:2) ∈ Fj, j ∈ I. Next, if A ∈ ∩j∈IFj for all
j ∈ I,andhence Ac ∈Fj forall j ∈ I,sothat Ac ∈∩j∈IFj.Finally,if
A,B ∈ ∩j∈IFj,then A,B ∈ Fj forall j ∈ I,andhence A∪ B ∈ Fj
forall j ∈ I,sothat A∪B ∈∩j∈IFj.
(ii) IfAi ∈∩j∈IAj,i =1,2,...,thenAi ∈Aj,i =1,2,...,forall j ∈ I,
andhence∪i∞=1Ai ∈Aj forall j ∈ I,sothat∪i∞=1Ai ∈∩j∈IAj.#
4. Let(cid:2)=(cid:10),F ={A ⊆(cid:10); either Aor Ac isfinite},andlet A ={1,2,..., j},
j
j ≥ 1. Then F is a field and A ∈ F, j ≥ 1, but ∪∞ A = {1,2,...} ∈/ F,
j j=1 j
becauseneitherthissetnoritscomplementisfinite.
Also,if B = {j +1, j +2,...},then B ∈ F since Bc isfinite,whereas
j (cid:6) (cid:7) j j j
c
∩∞ B =∩∞ Ac = ∪∞ A ∈/ F,asithasbeenseenalready.#
j=1 j j=1 j j=1 j
e4 Revised Answers Manual to an Introduction
5. Clearly, C is (cid:15)= (cid:11), every member of C is a countable union of members of P,
and C is the smallest σ-field containing P, if indeed, is a σ-field. If B ∈ C,
then B =∪i∈IAi forsome I ⊆N={1,2,...},andthen Bc =∪j∈JAj,where
wJh=erNe −IjI⊆,soNthaantdBIci∈∩CI.jF=ina(cid:11)lly.,TifheBnj∪∈∞jC=,1Bj =j =1,∪2,∞j=..1.∪,it∈hIejnABjij,=the∪ui∈niIojnAjoif,
membersofP,sothat∪∞ B belongsinC.#
j=1 j
6. SinceC andC(cid:16) ⊆C , j =1,...,8,itfollowsthatσ(C )andσ(C(cid:16))⊆σ(C )=
j j 0 j j 0
B,sothatitsufficestoshowthatB ⊆σ(C )andB ⊆σ(C(cid:16)),whichareimplied,
j j
respectively, by C ⊆ σ(C ) and C ⊆ σ(C(cid:16)), j = 1,...,8. As an example,
0 j 0 j
considertheclassesmentionedinthehint.
So,toshowthatC ⊆ σ(C ).Inallthatfollows,alllimitsaretakenasn → ∞.
0 1
Indeed, for y ↓ y, we have (x,y ) ∈ C and ∩∞ (x,y ) = (x,y] ∈ σ(C ).
n n 1 n=1 n 1
Likewise,forx ↑ x,wehave(x ,y) ∈ C and∩∞ (x ,y) = [x,y) ∈ σ(C ).
n n 1 n=1 n 1
Next,withx andy asabove,(x ,y )∈C and∩∞ (x ,y )=[x,y]∈σ(C ).
n n n n 1 n=1 n n 1
Also, for x ↓ −∞, we have (x ,a) ∈ C and ∩∞ (x ,a) = (−∞,a) ∈
n n 1 n=1 n
σ(C ),andlikewise(x ,a] ∈ C and∪∞ (x ,a] = (−∞,a] ∈ σ(C ).Finally,
1 n 1 n=1 n 1
(b,∞) = (−∞,b]c ∈ σ(C ),and[b,∞)=(−∞,b)c ∈ σ(C ).Itfollowsthat
1 1
C ⊆σ(C ).
0 1
ThatC ∈σ(C(cid:16))isseenasfollows.For(x,y),thereexistx and y rationals
0 1 n n
with x ↓ x and y ↑ y, so that (x,y) = ∪∞ ∈ σ(C(cid:16)). Also, for y ↓ y,
n n n=1 j n
wehave(x,y ) ∈ σ(C(cid:16)),aswasjustproved,andthen∩∞ (x,y ) = (x,y] ∈
n 1 n=1 n
σ(C(cid:16)).Likewise,withx ↑ x,wehave(x ,y)∈σ(C(cid:16))andthen∩∞ (x ,y)=
1 n n 1 n=1 n
[x,y) ∈ σ(C(cid:16)).Also,with x ↑ x and y ↓ y,wehave(x ,y ) ∈ σ(C(cid:16)),and
1 n n n n 1
∩∞ (x ,y ) = [x,y] ∈ σ(C(cid:16)). Likewise, with x ↓ −∞, we have (x ,a) ∈
n=1 n n 1 n n
σ(C(cid:16))and ∪∞ (x ,a) = (−∞,a) ∈ σ(C(cid:16)),whereas (x ,a] ∈ σ(C(cid:16)),sothat
1 n=1 n 1 n 1
∪∞ (x ,a] = (−∞,a] ∈ σ(C(cid:16)). Finally, (b,∞) = (−∞,b]c ∈ σ(C(cid:16)) since
n=1 n 1 1
(−∞,b] ∈σ(C(cid:16)),and[b,∞) =(−∞,b)c ∈σ(C(cid:16))since(−∞,b) ∈σ(C(cid:16)).It
1 1 1
followsthatC ⊆σ(C(cid:16)).
0 1
Aslightlyalternativeversionoftheprooffollows.Wewillshow(a)σ(C )=B
1
and(b)σ(C(cid:16))=B.
1
(a) σ(C )=B.
1
Thatσ(C1) ⊆ B isclear;toshowB ⊆ σ(C1)itsufficestosh(cid:2)owthatC0(cid:3) ⊆
σ(C ). To this end, we show that (x,y] ∈ σ(C ). Indeed, x,y+ 1 ∈
1 (cid:16) (cid:2) (cid:3) 1 (cid:2) (cid:3) n
C , so that ∞ x,y+ 1 = (x,y] ∈ σ(C ). Next, x − 1,y ∈ C ,
1 (cid:16) (cid:2)n=1 (cid:3) n 1 (cid:2) n (cid:3) 1
so that ∞ x − 1,y = [x,y) ∈ σ(C ). Also, x − 1,y+ 1 ∈ C ,
(cid:16)n=1(cid:2) n (cid:3) 1 n n 1
so that ∞ x − 1,y+ 1 = [x,y] ∈ σ(C ). Next, (−n,x) ∈ C , so
(cid:17) n=1 n n 1 (cid:2) (cid:13) 1
that ∞ (−n,x) = (−∞,x) ∈ σ(C ). Also, −∞,x + 1 ∈ C , so
(cid:16)n=1(cid:2) (cid:13) 1 n 1
that ∞ −∞,x + 1 = (−∞,x] ∈ σ(C ). Likewise, (x,n) ∈ C , so
(cid:17)n=1 n (cid:2)1 (cid:3) 1
that ∞ (x,n) = (x,∞) ∈ σ(C ); and x − 1,∞ ∈ σ(C ), so that
(cid:16) (cid:2)n=1 (cid:3) 1 n 1
∞ x − 1,∞ =[x,∞)∈σ(C ).Theproofiscomplete.
n=1 n 1
(b) σ(C(cid:16))=B.
1
Revised Answers Manual to an Introduction e5
Since, clearly, σ(C(cid:16)) ⊆ σ(C ), it suffices to show that σ(C ) ⊆ σ(C(cid:16)).
1 1 1 1
For x,y ∈ (cid:10) with x < y, there exist x ↓ x and y ↑ y with x ,y
n n n n
r(cid:17)at∞iona(lxn,umy b)e=rs(axn,dyx)n∈<σy(nCf(cid:16)o).rSeoacChn⊆.Sσin(cCe(cid:16))(,xann,dynh)en∈ceCσ1(cid:16),(iCtfo)l⊆lowσs(Cth(cid:16)a)t.
n=1 n n 1 1 1 1 1
Theproofiscomplete.#
7. (i) Let A∈C.Thentherearethefollowingpossiblecases:
(cid:18)
(a) A= m I ,I =(α ,β ],i =1,...,m.
i=1 i i i i
( ] ( ] ( ] ( ]
α β α β ··· α β α β
1 1 2 2 m−1 m−1 m m
Then Ac =(−∞,α1]+(β1,α2]+...+(βm−1,αm]+(βm,∞)and
thisisinC.
(b) Aconsistsonlyofintervalsoftheform(−∞,α].Thentherecanbe
onlyonesuchinterval;i.e., A = (−∞,α]andhence Ac = (α,∞)
whichisinC.
(c) Aconsistsonlyofintervalsoftheform(β,∞).Thentherecanonly
beonesuchinterval;i.e., A=(β,∞)sothat Ac =(−∞,β]which
isinC.
(d) Aconsistsonlyofintervalsoftheform(−∞,α]and(β,∞).Then
A will be as follows: A = (−∞,α] + (β,∞) (α < β), so that
Ac =(α,∞)∩(−∞,β]=(α,β]whichisinC.
(e) Finally,let Aconsistofintervalsofallforms.Then Aisasbelow:
] ( ] ( ] ( ] ( ] (
−∞ α α β α β ··· α β α β β ∞
1 1 2 2 m−1 m−1 m m
Then,clearly,
Ac =(α,α1]+(β1,α2]+...+(βm−1,αm]+(βm,β]
which is in C. So, C is closed under complementation. It is also
closed under the union of two sets A and B in C, because, clearly,
theunionoftwosuchsetsisalsoamemberofC.Thus,C isafield.
Next, let C = {(α,β]; α,β ∈ (cid:10),α < β}. Then, by Exercise 6,
2
σ(C )=B.Also,C ⊂C,sothatB =σ(C )⊆σ(C).Furthermore,
2 2 2
C ⊆σ(C )=Bandhenceσ(C)⊆B.Itfollowsthatσ(C)=B.
0 (cid:18)
(ii) If A ∈ C,then A = m I ,where I sareoftheforms:(α,β),(α,β],
i=1 i i
[α,β),[α,β],(−∞,α),(−∞,α],(β,∞),[β,∞). But (α,β)c =
(−∞,α]+[β,∞),(α,β]c =(−∞,α]+(β,∞),[α,β)c =(−∞,α)+
(β,∞),[α,β]c =(−∞,α)+(β,∞),(−∞,α)c =[α,∞),(−∞,α]c =
(α,∞),(β,∞)c =(−∞,β],and[β,∞)c =(−∞,β).Then,consider-
ing all possibilities as in part (i)(cid:18), we conclude that Ac ∈ C in all cases.
Next,for A asaboveand B = n J with J beingfromamongthe
j=1 j j
aboveintervals,itfollowsthat A∪Bisafinitesumofintervalsasabove,
e6 Revised Answers Manual to an Introduction
andhence A∪ B ∈ C.Thus,C isafield.Finally,fromC ⊂ C ⊂ B,it
0
followsthatB =σ(C )⊆σ(C)⊆B,sothatσ(C)=B.#
0
8. Clearly,F is(cid:15)= (cid:11)since,forexample, A = A∩(cid:2)andhence A ∈ F .Next,
A A
forB ∈F ,itfollowsthatB = A∩C,C ∈F,andBc(=complementofBwith
A A
respectto A)=A∩Cc ∈F sinceCc ∈F.Finally,for B ,B ∈F ,itfollows
A 1 2 A
that B = A ∩C ,C ∈F,i =1,2,andthen B ∪B = A∩(C ∪C )∈F ,
i i i i 1 2 1 2 A
sinceC ∪C ∈F.#
1 2
9. That A (cid:15)= (cid:11) and that itis closed under complementation is as inExercise 8.
A
ForBi ∈AA,i =1,2,...,itfollows(cid:2)thatBi (cid:3)= A∩Ci forsomeCi ∈A,i ≥1,
and∪∞ B =∪∞ (A∩C )= A∩ ∪∞ C ∈A since∪∞ C ∈A.
i=1 i i=1 i i=1 i A i=1 i
Thus, A is a σ-field. Since F ⊆ A, it follows that F ⊆ A and hence
A A A
σ(FA)⊆AA.SinceforeveryF ⊆Ai,i ∈ I,itfollowsFA ⊆Ai,A,i ∈ I,then
σ(FA)⊆∩i∈IAi,A.Also,σ(FA)=∩j∈JA∗j forallσ-fieldsofsubsetsofAwith
A∗ ⊇F .Inordertoshowthatσ(F )=A ,itmustbeshownthatforeveryσ-
j A A A
fieldA∗ofsubsetsof AwithA∗ ⊇F ,wehaveA∗ ⊇A .Thatthisis,indeed,
A A
thecaseisseenasfollows.DefinetheclassMby:M={C ∈A; A∩C ∈A∗}.
Then,clearly,F ⊆M⊆AandM (=M∩ A)⊆A∗.Thisissobecause,for
A
C ∈F,itfollowsthatC ∩ A ∈F andhenceC ∩ A ∈A∗ (⊇F ).Also,with
A A
M = {C ⊆ A;C = M ∩ A,M ∈ M}, it follows that M ⊆ A∗ from the
A A
definitionofM.WeassertthatMisamonotoneclass.Indeed,letC(cid:2)n ∈Mw(cid:3)ith
Cn ↑o(cid:2)rCn ↓.(cid:3)Then,forthecasethatCn ↑,A∩(limn→∞Cn)= A∩ ∪∞n=1Cn =
∪∞n=1 A∩Cn ∈ A∗ since A ∩Cn ∈ A∗,n ≥ 1, so that limn→∞Cn ∈ M.
Likewise,forCn ↓, A∩(limn→∞Cn)= A∩(∩∞n=1Cn)=∩∞n=1(A∩Cn)∈A∗
since A∩Cn ∈A∗,n ≥1,sothatlimn→∞Cn ∈M.SoMisamonotoneclass
⊇F,andhenceM⊇minimalmonotoneclassM ,say,⊇F.SinceFisafield,
0
it follows that M is a σ-field and indeed M = A (by Theorem 6). Finally,
0 0
A=M0 ⊆MimpliesAA =M0,A ⊆MA ⊆A∗,aswastobeseen.#
10. SetF = ∪∞ A ,andlet A ∈ F.Then A ∈ A forsomen,sothat Ac ∈ A
n=1 n n n
and hence A ∈ F. Next, let A,B ∈ F. Then A ∈ A ,B ∈ A for some n
n1 n2 1
andn ,andletn = max(n ,n ).Then A,B ∈ A ,sothat A∪ B ∈ A and
2 0 1 2 n0 n0
A∪B ∈F.Then, Ac ∈F and A∪B ∈F,sothatF isafield.
Itneednotbeaσ-field.
Counterexample: Let (cid:2) = (cid:10) and let A = {A ⊆ [−n,n]; either A or Ac is
n
countable},n ≥ 1. Then A is a σ-field (by Example 8) and A ↑. However,
n n
F isnotaσ-fieldbecause,if A = {rationalsin[−n,n]},n ≥ 1,andifweset
n
A=∪∞ A ,then A∈/ F,becauseotherwise A∈A forsomen,whichcannot
n=1 n n
happen.#
11. Set M∩j∈IMj and let An ∈ M,n ≥ 1, where the Ans form a monotone
sequence.Then An ∈ Mj foreach j ∈ I andalln ≥ 1,sothatlimn→∞An is
alsoinMj.Sincethisistrueforall j ∈ I,itfollowsthatlimn→∞An isinM,
andMisamonotoneclass.#
12. Let(cid:2) = {1,2,...},M = {(cid:11),{1,...,n},{n,n +1,...},n ≥ 1,(cid:2)}.ThenM
is a monotone class, but not a field, because, e.g., if A = {1,...,n} and B =
{n−2,n−1,...}(n ≥3),then A,B ∈M,but A∩B ={n−2,n−1,n}∈/ M.
Revised Answers Manual to an Introduction e7
Asanotherexample,let(cid:2) = (0,1)andM = {(0,1− 1],n ≥ 1,(cid:2)}.Then
n
Misamonotoneclassand(0,1]∈M,but(0,1]c =(1,1)∈/ M.
2 2 2
Still as a third example, let (cid:2) = (cid:10) and let M = {(cid:11),(0,n),(−n,0),n ≥
1,(0,∞),(−∞,0)}. Then M is a monotone class, but not a field since, for
A=(−1,0)andB =(0,1),wehave A,B,∈M,but A∪B =(−1,1)∈/ M.#
13. (i) Forω=(ω ,ω )∈ Ec,wehaveω∈/ E = A×B,sothateitherω ∈/ A
1 2 1
orω ∈/ B orboth.Letω ∈/ A.Thenω ∈ Ac and(ω ,ω )∈ Ac×(cid:2) ,
2 1 1 1 2 2
whetherornotω ∈ B.Hence Ec ⊆(A×Bc)+(Ac×(cid:2) ).Ifω ∈ A,
2 2 1
thenω ∈/ B,sothat(ω ,ω )∈ A×BcandEc ⊆(A×Bc)+(Ac×(cid:2) ).
2 1 2 2
Next,if(ω ,ω )∈ A×Bc,thenω ∈ Aandω ∈/ B,sothat(ω ,ω )∈/ E
1 2 1 2 1 2
andhence(ω ,ω )∈ Ec.If(ω ,ω )∈ Ac×(cid:2) ,thenω ∈/ Aandhence
1 2 1 2 2 1
(ω ,ω ) ∈/ A× B = E whether or not ω ∈ B. Thus (ω ,ω ) ∈ Ec.
1 2 2 1 2
In both cases, (A× Bc)+(Ac ×(cid:2) ) ⊇ Ec and equality follows. The
2
secondequalityisentirelysymmetric.
(ii) Let (ω ,ω ) ∈ E ∩ E , so that (ω ,ω ) ∈ E and (ω ,ω ) ∈ E
1 2 1 2 1 2 1 1 2 2
and hence ω ∈ A ,ω ∈ B , and ω ∈ A , ω ∈ B . It follows that
1 1 2 1 1 2 2 2
ω ∈ A ∩A ,ω ∈ B ∩B andhence(ω ,ω )∈(A ∩A )×(B ∩B ).
1 1 2 2 1 2 1 2 1 2 1 2
Next, (ω ,ω ) ∈ (A ∩ A )×(B ∩ B ), so that ω ∈ A ∩ A and
1 2 1 2 1 2 1 1 2
ω ∈ B ∩ B .Thus,ω ∈ A ,ω ∈ A andω ∈ B ,ω ∈ B ,sothat
2 1 2 1 1 1 2 2 1 2 2
(ω ,ω )∈ A ∩B and(ω ,ω )∈ A ∩B ,or(ω ,ω )∈ E ∩E ,so
1 2 1 1 1 2 2 2 1 2 1 2
thatequalityoccurs.Thesecondconclusionisimmediate.
(iii) Indeed, E ∩F = (A ∩A(cid:16))×(B ∩B(cid:16))and E ∩F = (A ∩A(cid:16))×
1 1 1 1 1 1 2 2 2 2
(B ∩B(cid:16)),bypart(ii),andthefirstequalityfollows.Next,againbypart(ii),
2 2
andreplacingE by(A ∩A(cid:16))×(B ∩B(cid:16))andE by(A ∩A(cid:16))×(B ∩B(cid:16)),
1 1 1 1 1 2 2 2 2 2
we obtain thesecond equality. The thirdequality isimmediate. Finally,
thelastconclusionisimmediate.#
14. (i) Eitherbytheinclusionprocessorasfollows:
(A ×B )−(A ×B )
1 1 2 2
=(A ×B )∩(A ×B )c
1 1 2 2
=(A ×B )∩[(A ×Bc)+(Ac ×(cid:2) )](byLemma2)
1 1 2 2 2 2
=(A ×B )∩(A ×Bc)+(A ×B )∩(Ac ×(cid:2) )
1 1 2 2 1 1 2 2
=(A ∩ A )×(B ∩Bc)+(A ∩ Ac)×(B ∩(cid:2) )(clearly)
1 2 1 2 1 2 1 2
=(A ∩ A )×(B −B )+(A − A )×B .
1 2 1 2 1 2 1
(ii) Let A× B = (cid:11).Then(x,y) ∈ A× B,sothatx ∈ Aand y ∈ B.Also,
(x,y)∈(cid:11)andthiscanhappenonlyifatleastoneof Aor B is=(cid:11).On
theotherhand,ifatleastoneof AorBis=(cid:11),then,clearly, A×B =(cid:11).
(iii) Let A × B ⊆ A × B . Then (x,y) ∈ A × B , so that x ∈ A
1 1 2 2 1 1 1
and y ∈ B . Also, (x,y) ∈ A × B implies x ∈ A and y ∈ B .
1 2 2 2 2
Thus, A ⊆ A and B ⊆ B . Next, let A ⊆ A and B ⊆ B . Then
1 2 1 2 1 2 1 2
A × B ⊆ A × B since(x,y) ∈ A × B ifandonlyif x ∈ A and
1 1 2 2 1 1 1
y ∈ B .Hence,x ∈ A and y ∈ B or(x,y)∈ A ×B .
1 2 2 2 2
e8 Revised Answers Manual to an Introduction
(iv) A ×B (cid:15)=(cid:11)andA ×B (cid:15)=(cid:11).ThenA ×B = A ×B orA ×B ⊆
1 1 2 2 1 1 2 2 1 1
A × B andthen(by(iii)), A ⊆ A and B ⊆ B .Also, A × B =
2 2 1 2 1 2 2 2
A ×B or A ×B ⊆ A ×B ,andthen(by(iii)again), A ⊆ A and
1 1 2 2 1 1 2 1
B ⊆ B .
2 1
So,both A ⊆ A and A ⊆ A ,andtherefore A = A .Likewise,
1 2 2 1 1 2
B ⊆ B and B ⊆ B sothat B = B .
1 2 2 1 1 2
(v)
A×B =(A ×B )+(A ×B ) (*)
1 1 2 2
From(cid:11)=(A ×B )∩(A ×B )=(A ∩A )×(B ∩B )andpart(ii),
1 1 2 2 1 2 1 2
wehavethatatleastoneofA ∩A ,B ∩B is(cid:11).LetA ∩A =(cid:11).Then
1 2 1 2 1 2
theclaimisthat A = A + A .Infact,(x,y) ∈ A× B implies x ∈ A
1 2
(and y ∈ B).Also,(x,y)belongingtotheright-handsideof(*)implies
(x,y)∈ A ×B or(x,y)∈ A ×B .Let(x,y)∈ A ×B .Thenx ∈ A
1 1 2 2 1 1 1
(andy ∈ B ),sothatA⊆ A .Ontheotherhand,(x,y)∈ A ×B implies
1 2 2 2
x ∈ A (and y ∈ B ), so that A ⊆ A . Thus, A ⊆ A + A . Next, let
2 2 2 1 2
again(x,y)belongtotheright-handsideof(*).Then(x,y)∈ A ×B
1 1
or (x,y) ∈ A × B . Now (x,y) ∈ A × B implies that x ∈ A
2 2 1 1 1
(and y ∈ B ).Also,(x,y)belongingtotheleft-handsideof(*)implies
1
(x,y) ∈ A× B,sothat x ∈ A (and y ∈ B).Hence A ⊆ A.Likewise,
1
(x,y) ∈ A × B implies A ⊆ A, so that A + A ⊆ A, and hence
2 2 2 1 2
A= A +A .Next,let A= A +A .Then A×B =(A +A )×B =
1 2 1 2 1 2
(A × B)+(A × B).Also, A× B = (A × B )+(A × B ).Thus,
1 2 1 1 2 2
(A ×B)+(A ×B)=(A ×B )+(A ×B ).(x,y)belongingtothe
1 2 1 1 2 2
left-handsideof(*)implies(x,y)∈ A ×Bor(x,y)∈ A ×B.(x,y)∈
1 2
A × B yields y ∈ B (and x ∈ A ). Same if (x,y) ∈ A × B. Also,
1 1 2
(x,y)belongingtotheright-handsideof(*)implies(x,y)∈ A ×B or
1 1
(x,y)∈ A ×B .For(x,y)∈ A ×B ,wehavey ∈ B (andx ∈ A ),so
2 2 1 1 1 1
that B ⊆ B .For(x,y)∈ A ×B ,wehave B ⊆ B likewise.Next,let
1 2 2 2
again(x,y)belongtotheright-handsideof(*).Then(x,y)∈ A ×B
1 1
or(x,y)∈ A ×B .For(x,y)∈ A ×B ,wehavey ∈ B (andx ∈ A ).
2 2 1 1 1 1
Thus B ⊆ B.For(x,y) ∈ A × B ,wehave B ⊆ B.Itfollowsthat
1 2 2 2
B = B = B .
1 2
Tosummarize: A ∩ A =(cid:11)implies A = A + A and B = B = B .
1 2 1 2 1 2
Likewise, B ∩ B = (cid:11)implies B = B + B and A = A = A .Fur-
1 2 1 2 1 2
thermore,A ∩A =(cid:11)andB ∩B =(cid:11)cannothappensimultaneously.
1 2 1 2
Indeed, A ∩ A =(cid:11)implies A = A + A ,and B ∩ B =(cid:11)implies
1 2 1 2 1 2
B = B +B .ThenA×B =(A +A )×(B +B )=(A ×B )+(A ×
1 2 1 2 1 2 1 1 2
B )+(A ×B )+(A ×B ).Also,A×B =(A ×B )+(A ×B ),sothat:
2 1 2 2 1 1 1 2 2
(A ×B )+(A ×B )+(A ×B )+(A ×B )=(A ×B )+(A ×B ).
1 1 2 2 1 2 2 1 1 1 2 2
Then(A ×B )+(A ×B )=(cid:11)implies(A ×B )=(A ×B )=(cid:11),
1 2 2 1 1 2 2 1
sothatatleastoneof A , A ,B ,B =(cid:11)(bypart(ii)).However,thisis
1 2 1 2
notpossiblebythefactthat A ×B (cid:15)=(cid:11), A ×B (cid:15)=(cid:11).#
1 1 2 2
Revised Answers Manual to an Introduction e9
15. (i) Ifeither A or B = (cid:11),then,clearly, A× B = (cid:11).Next,if A× B = (cid:11),
and A (cid:15)= (cid:11) and B (cid:15)= (cid:11), then there exist ω ∈ A and ω ∈ B, so that
1 2
(ω ,ω )∈ A×B,acontradiction.
1 2
(ii) Bothdirectionsofthefirstassertionareimmediate.Withouttheassump-
tion E and E (cid:15)= (cid:11),theresultneednotbetrue.Indeed,let(cid:2) = (cid:2) ,
1 2 1 2
A (cid:15)=(cid:11),B = A = B =(cid:11).Then E = E =(cid:11),but A (cid:3) A .#
1 1 2 2 1 2 1 2
16. (i) Ifatleastoneof A ,...,A is=(cid:11),then,clearly, A ×...× A =(cid:11).
1 n 1 n
Next, let E = (cid:11) and suppose that A (cid:15)= (cid:11),i = 1,...,n. Then there
i
existsω ∈ A ,i =1,...,n,sothat(ω ,...,ω )∈ E,acontradiction.
i i 1 n
(ii) Letω = (ω ,...,ω ) ∈ E ∩ F,or(ω ,...,ω ) ∈ (A ×...× A )∩
1 n 1 n 1 n
(B ×...×B ).Then(ω ,...,ω )∈ A ×...×A and(ω ,...,ω )∈
1 n 1 n 1 n 1 n
B ×...×B .Itfollowsthatω ∈ A andω ∈ B ,i =1,...,n,sothat
1 n i i i i
ω ∈ A ∩B ,i =1,...,n,andhence(ω ,...,ω )∈(A ∩B )×...×
i i i 1 n 1 1
(A ∩B ).Next,let(ω ,...,ω )∈(A ∩B )×...×(A ∩B ).Then
n n 1 n 1 1 n n
ω ∈ A ∩ B ,i = 1,...,n,sothatω ∈ A andω ∈ B ,i = 1,...,n.
i i i i i i i
Itfollowsthat(ω ,...,ω )∈ A ×...× A and(ω ,...,ω )∈ B ×
1 n 1 n 1 n 1
...×B ,sothat(ω ,...,ω )∈(A ×...× A )∩(B ×...×B ).#
n 1 n 1 n 1 n
17. We have E = F +G and E,F,G are all (cid:15)= (cid:11). This implies that A ,B , and
i i
C ,i = 1,...,n are all (cid:15)= (cid:11); this is so by Exercise 16(i). Furthermore, by
i
Exercise16(ii):
F∩G =(B ×...×B )∩(C ×...×C )=(B ∩C )×...×(B ∩C ),
1 n 1 n 1 1 n n
whereas F∩G =(cid:11).Itfollowsthat B ∩C =(cid:11)foratleastone j,1≤ j ≤n.
j j
Withoutlossofgenerality,supposethat B ∩C =(cid:11).Thenweshallshowthat
1 1
A = B +C and A = B = C , i = 2,...,n. To this end, let ω ∈ A ,
1 1 1 i i i j j
j = 1,...,n. Then (ω ,...,ω ) ∈ A ×...× A or (ω ,...,ω ) ∈ E or
1 n 1 n 1 n
(ω ,...,ω ) ∈ (F +G). Hence (ω ,...,ω ) ∈ F or (ω ,...,ω ) ∈ G. Let
1 n 1 n 1 n
(ω ,...,ω ) ∈ F.Then(ω ,...,ω ) ∈ B ×...× B andhenceω ∈ B or
1 n 1 n 1 n 1 1
ω ∈(B ∪C ),sothat A ⊆ B ∪C .Likewiseif(ω ,...,ω )∈G.Next,let
1 1 1 1 1 1 1 n
ω ∈ B , j =1,...,n.Then(ω ,...,ω )∈ B ×...×B or(ω ,...,ω )∈ F
j j 1 n 1 n 1 n
or(ω ,...,ω )∈ E or(ω ,...,ω )∈(A ×...×A ),henceω ∈ A ,which
1 n 1 n 1 n 1 1
impliesthatB ⊆ A .Bytakingω ∈C , j =1,...,nandarguingasbefore,we
1 1 j j
concludethatC ⊆ A .From B ⊆ A andC ⊆ A ,weobtain B ∪C ⊆ A .
1 1 1 1 1 1 1 1 1
Sincealso A ⊆ B ∪C ,weget A = B ∪C .Since B ∩C =(cid:11),wehave
1 1 1 1 1 1 1 1
then A = B +C .
1 1 1
It remains for us to show that A = B = C , i = 2,...,n. Without loss
i i i
of generality, it suffices to show that A = B = C , the remaining cases
2 2 2
being treated symmetrically. As before, let ω ∈ A , j = 1,...,n. Then
j j
(ω ,...,ω )∈(A ×...×A )or(ω ,...,ω )∈ Eor(ω ,...,ω )∈(F+G).
1 n 1 n 1 n 1 n
Hence either (ω ,...,ω ) ∈ F or (ω ,...,ω ) ∈ G. Let (ω ,...,ω ) ∈ F.
1 n 1 n 1 n
Then (ω ,...,ω ) ∈ B ×...× B and hence ω ∈ B , so that A ⊆ B .
1 n 1 n 2 2 2 2
e10 Revised Answers Manual to an Introduction
Likewise A ⊆C if(ω ,...,ω )∈G.Next,let(ω ,...,ω )∈ B ×...×B
2 2 1 n 1 n 1 n
or (ω ,...,ω ) ∈ F or (ω ,...,ω ) ∈ (F + G) or (ω ,...,ω ) ∈ E or
1 n 1 n 1 n
(ω ,...,ω ) ∈ (A × ... × A ) and hence ω ∈ A , so that B ⊆ A . It
1 n 1 n 2 2 2 2
follows that A = B . We arrive at the same conclusion A = B if we take
2 2 2 2
(ω ,...,ω )∈G.So,tosumitup, A = B +C ,and A = B =C ,andby
1 n 1 1 1 2 2 2
symmetry, A = B =C ,i =3,...,n.
i i i
Avariationtotheaboveproofisasfollows.
LetE = F+Gor A ×...×A =(B ×...×B )+(C ×...×C ),and
1 n 1 n 1 n
let(ω ,...,ω )∈ E.Then(ω ,...,ω )∈ A ×...×A ,sothatω ∈ A ,i =
1 n 1 n 1 n i i
1,...,n.Thenω ∈ B ,i =1,...,norω ∈C ,i =1,...,n(butnotboth).So,
i i i i
A = B ∪C ,i =1,...,nandA = B +C foratleastone j.Considerthecase
i i i j j j
n =2,andwithoutlossofgeneralitysupposethatA = B +C ,A = B ∪C .
1 1 1 2 2 2
Then,clearly:
A × A =(B +C )×(B ∪C )
1 2 1 1 2 2
=(B ×B )∪(C ×C )∪(B ×C )∪(C ×B ).
1 2 1 2 1 2 1 2
However, A × A =(B ×B )+(C ×C ),andthisimpliesthat B ×C ⊆
1 2 1 2 1 2 1 2
B × B and C × B ⊆ B × C , hence C ⊆ B and B ⊆ C , so that
1 2 1 2 1 2 2 2 2 2
B =C (= A ).Next,assumetheassertiontobetruefornandconsider:
2 2 2
A1×...×An×An+1 =(B1×...×Bn×Bn+1)+(C1×...×Cn×Cn+1),
or An × An+1 = (Bn × Bn+1) = (Cn ×Cn+1),where An = A1 ×...× An,
Bn = B ×...×B andCn =C ×...×C .Applythereasoningusedinthe
1 n 1 n
casen = 2byreplacing A1 by An and A2 by An+1 (sothat B1,B2 andC1,C2
arereplaced,respectively,by Bn,Bn+1andCn,Cn+1)togetthat:
An = Bn +Cn, An+1 = Bn+1∪Cn+1.
The first union is a “+” by the induction hypothesis. The second union may or
maynotbea“+”asofnow.Then:
An × An+1 =(Bn ∪Cn)×(Bn+1∪Cn+1)
=(Bn ×Bn+1)∪(Cn ×Cn+1)∪(Bn ×Cn+1)∪(Cn ×Bn+1).
However, An×An+1 =(Bn×Bn+1)+(Cn×Cn+1).Therefore Bn×Cn+1 ⊆
Bn×Bn+1andCn×Bn+1 ⊆Cn×Cn+1,sothatCn+1 ⊆ Bn+1andBn+1 ⊆Cn+1,
andhence Bn+1 =Cn+1.Theproofiscompleted.#
18. Theonlypropertiesoftheσ-fieldsA andA usedintheproofofTheorem7is
1 2
thatA ,i = 1,2areclosedundertheintersectionoftwosetsinthemandalso
i
closedundercomplementations.Sincethesepropertiesholdalsoforthecasethat
A ,i =1,2arefields,F ,i =1,2,theproofiscompleted.#
i i
19. C asdefinedhereneednotbeaσ-field.Hereisa
Counterexample: (cid:2) = (cid:2) = [0,1]. For n ≥ 2, let I = [0, 1], I =
1 2 n1 n nj
(j−1, j], j = 2,...,n, and set E = I × I , j = 1,...,n. Also, let
n n nj nj nj