A_Brief_Introduction_to_Measure_Theory_and_Integration-Richard_Bass.pdf

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ABriefIntroductionto
MeasureTheoryandIntegration
RichardF.Bass
DepartmentofMathematics
UniversityofConnecticut
September18,1998
Thesenotesarec1998byRichardBass.Theymaybeusedforpersonaluseorclassuse,butnotfor
commercialpurposes.
1.Measures.
LetXbeaset.Wewillusethenotation:A c ={x2X:x/2A}andA−B=A\B c .
Definition.AnalgebraorafieldisacollectionAofsubsetsofXsuchthat
(a);,X2A;
(b)ifA2A,thenA c 2A;
(c)ifA 1 ,...,A n 2A,then[ n i=1 A i and\ n i=1 A i areinA.
Aisa-algebraor-fieldifinaddition
(d)ifA 1 ,A 2 ,...areinA,then[ 1 i=1 A i and\ 1 i=1 A i areinA.
In(d)weallowcountableunionsandintersectionsonly;wedonotallowuncountableunionsandintersections.
Example.LetX=RandAbethecollectionofallsubsetsofR.
Example.LetX=RandletA={AR:AiscountableorA c iscountable}.
Definition.Ameasureon(X,A)isafunctionµ:A![0,1]suchthat
(a)µ(A)0forallA2A;
(b)µ(;)=0;
(c)ifA i 2Aaredisjoint,then
µ([ 1 i=1 A i )=
1 X
µ(A i ).
i=1
Example.Xisanyset,Aisthecollectionofallsubsets,andµ(A)isthenumberofelementsinA.
Example.X=R,Athecollectionofallsubsets,x 1 ,x 2 ,...2R,a 1 ,a 2 ,...>0,andµ(A)= P {i:x i 2A} a i .
Example. x (A)=1ifx2Aand0otherwise.Thismeasureiscalledpointmassatx.
Proposition1.1.Thefollowinghold:
(a)IfA,B2AwithAB,thenµ(A)µ(B).
(b)IfA i 2AandA=[ 1 i=1 A i ,thenµ(A) P 1 i=1 µ(A i ).
(c)IfA i 2A,A 1 A 2 ···,andA=[ 1 i=1 A i ,thenµ(A)=lim n!1 µ(A n ).
(d)IfA i 2A,A 1 A 2 ···,µ(A 1 )<1,andA=\ 1 i=1 A i ,thenwehaveµ(A)=lim n!1 µ(A n ).
Proof.(a)LetA 1 =A,A 2 =B−A,andA 3 =A 4 =···=;.Nowusepart(c)ofthedefinitionofmeasure.
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(b)LetB 1 =A 1 ,B 2 =A 2 −B 1 ,B 3 =A 3 −(B 1 [B 2 ),andsoon.TheB i aredisjointand
[ 1 i=1 B i =[ 1 i=1 A i .Soµ(A)= P µ(B i ) P µ(A i ).
(c)DefinetheB i asin(b).Since[ n i=1 B i =[ n i=1 A i ,then
µ(A)=µ([ 1 i=1 A i )=µ([ 1 i=1 B i )=
1 X
µ(B i )
i=1
=lim
n!1
n X
n!1 µ([ n i=1 B i )=lim
n!1 µ([ n i=1 A i ).
i=1
(d)Apply(c)tothesetsA 1 −A i ,i=1,2,....
Definition.Aprobabilityorprobabilitymeasureisameasuresuchthatµ(X)=1.Inthiscaseweusually
write(,F,P)insteadof(X,A,µ).
2.ConstructionofLebesguemeasure.
Definem((a,b))=b−a.IfGisanopensetandGR,thenG=[ 1 i=1 (a i ,b i )withtheintervals
m (A)=inf{m(G):Gopen,AG}.
Wewillshowthefollowing.
(1)m isnotameasureonthecollectionofallsubsetsofR.
(2)m isameasureonthe-algebraconsistingofwhatareknownasm -measurablesets.
(3)LetA 0 bethealgebra(not-algebra)consistingofallfiniteunionsofsetsoftheform[a i ,b i ).IfAis
thesmallest-algebracontainingA 0 ,thenm isameasureon(R,A).
Wewillprovethesethreefacts(andabitmore)inamoment,butlet’sfirstmakesomeremarksabout
theconsequencesof(1)-(3).
Ifyoutakeanycollectionof-algebrasandtaketheirintersection,itiseasytoseethatthiswillagain
bea-algebra.Thesmallest-algebracontainingA 0 willbetheintersectionofall-algebrascontaining
A 0 .
Since(a,b]isinA 0 forallaandb,then(a,b)=[ 1 i=i 0 (a,b−1/i]2A,wherewechoosei 0 sothat
1/i 0 <b−a.Thensetsoftheform[ 1 i=1 (a i ,b i )willbeinA,henceallopensets.Thereforeallclosedsets
areinAaswell.
Thesmallest-algebracontainingtheopensetsiscalledtheBorel-algebra.ItisoftenwrittenB.
AsetNisanullsetifm (N)=0.LetLbethesmallest-algebracontainingBandallthenullsets.
LiscalledtheLebesgue-algebra,andsetsinLarecalledLebesguemeasurable.
Aspartofourproofsof(2)and(3)wewillshowthatm isameasureonL.Lebesguemeasureis
themeasurem onL.(1)showsthatLisstrictlysmallerthanthecollectionofallsubsetsofR.
Proofof(1).Definexyifx−yisrational.Thisisanequivalencerelationshipon[0,1].Foreach
equivalenceclass,pickanelementoutofthatclass(bytheaxiomofchoice)Callthecollectionofsuchpoints
A.GivenasetB,defineB+x={y+x:y2B}.Notem (A+q)=m (A)sincethistranslationinvariance
holdsforintervals,henceforopensets,henceforallsets.Moreover,thesetsA+qaredisjointfordierent
rationalsq.
2
µ(B i )=lim
disjoint.Definem(G)= P 1 i=1 (b i −a i ).IfAR,define
Now
[0,1][ q2[−2,2] (A+q),
wherethesumisonlyoverrationalq,so1 P q2[−2,2] m (A+q),andthereforem (A)>0.But
[ q2[−2,2] (A+q)[−6,6],
whereagainthesumisonlyoverrationalq,so12 P q2[−2,2] m (A+q),whichimpliesm (A)=0,a
contradiction.
Proposition2.1.Thefollowinghold:
(a)m (;)=0;
(b)ifAB,thenm (A)m (B);
(c)m ([ 1 i=1 A i ) P 1 i=1 m (A i ).
Proof.(a)and(b)areobvious.Toprove(c),let">0.ForeachithereexistintervalsI i1 ,I i2 ,...suchthat
A i [ 1 j=1 I ij and P j m(I ij )m (A i )+"/2 i .Then[ 1 i=1 A i [ i,j I ij and
X
m(I ij ) X
i
m (A i )+ X
i
"/2 i = X
i
m (A i )+".
i,j
Since"isarbitrary,m ([ 1 i=1 A i ) P 1 i=1 m (A i ).
Afunctiononthecollectionofallsubsetssatisfying(a),(b),and(c)iscalledanoutermeasure.
Definition.Letm beanoutermeasure.AsetAXism -measurableif
m (E)=m (E\A)+m (E\A c ) (2.1)
forallEX.
Theorem2.2.Ifm isanoutermeasureonX,thenthecollectionAofm measurablesetsisa-algebra
andtherestrictionofm toAisameasure.Moreover,Acontainsallthenullsets.
Proof.ByProposition2.1(c),
m (E)m (E\A)+m (E\A c )
forallEX.Sotocheck(2.1)itisenoughtoshowm (E)m (E\A)+m (E\A c ).Thiswillbetrivial
inthecasem (E)=1.
IfA2A,thenA c 2AbysymmetryandthedefinitionofA.SupposeA,B2AandEX.Then
m (E)=m (E\A)+m (E\A c )
=(m (E\A\B)+m (E\A\B c ))+(m (E\A c \B)+m (E\A c \B c )
Thefirstthreetermsontherighthaveasumgreaterthanorequaltom (E\(A[B))becauseA[B
(A\B)[(A\B c )[(A c \B).Therefore
m (E)m (E\(A[B))+m (E\(A[B) c ),
whichshowsA[B2A.ThereforeAisanalgebra.
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LetA i bedisjointsetsinA,letB n =[ n i=1 A i ,andB=[ 1 i=1 A i .IfEX,
m (E\B n )=m (E\B n \A n )+m (E\B n \A c n )
=m (E\A n )+m (E\B n−1 ).
Repeatingform (E\B n−1 ),weobtain
m (E\B n )=
n X
m (E\A i ).
i=1
So
n X
m (E)=m (E\B n )+m (E\B c n )
m (E\A i )+m (E\B c ).
i=1
Letn!1.Then
X
m (E)
m (E\A i )+m (E\B c )
i=1
m ([ 1 i=1 (E\A i ))+m (E\B c )
=m (E\B)+m(E\B c )
m (E).
ThisshowsB2A.
IfwesetE=Binthislastequation,weobtain
m (B)=
1 X
m (A i ),
i=1
orm iscountablyadditiveonA.
Ifm (A)=0andEX,then
m (E\A)+m (E\A c )=m (E\A c )m (E),
whichshowsAcontainsallnullsets.
NoneofthisisusefulifAdoesnotcontaintheintervals.Therearetwomainstepsinshowingthis.
LetA 0 bethealgebraconsistingofallfiniteunionsofintervalsoftheform(a,b].Thefirststepis
Proposition2.3.IfA i 2A 0 aredisjointand[ 1 i=1 A i 2A 0 ,thenwehavem([ 1 i=1 A i )= P 1 i=1 m(A i ).
Proof.Since[ 1 i=1 A i isafiniteunionofintervals(a k ,b k ],wemaylookatA i \(a k ,b k ]foreachk.Sowe
mayassumethatA=[ 1 i=1 A i =(a,b].
First,
n X
m(A)=m([ n i=1 A i )+m(A−[ n i=1 A i )m([ n i=1 A i )=
m(A i ).
i=1
Lettingn!1,
X
m(A)
m(A i ).
i=1
Letusassumeaandbarefinite,theothercasebeingsimilar.Bylinearity,wemayassumeA i =
(a i ,b i ].Let">0.Thecollection{(a i ,b i +"/2 i )}covers[a+",b],andsothereexistsafinitesubcover.
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1
1
Discardinganyintervalcontainedinanotherone,andrelabeling,wemayassumea 1 <a 2 <···a N and
b i +"/2 i 2(a i+1 ,b i+1 +"/2 i+1 ).Then
m(A)=b−a=b−(a+")+"
N X
(b i +"/2 i −a i )+"
i=1
1 X
m(A i )+2".
i=1
Since"isarbitrary,m(A) P 1 i=1 m(A i ).
ThesecondstepistheCarath´eodoryextensiontheorem.Wesaythatameasuremis-finiteifthere
existE 1 ,E 2 ,...,suchthatm(E i )<1foralliandX[ 1 i=1 E i .
Theorem2.4.SupposeA 0 isanalgebraandmrestrictedtoA 0 isameasure.Define
n 1 X
o
m (E)=inf
m(A i ):A i 2A 0 ,E[ 1 i=1 A i
.
i=1
Then
(a)m (A)=m(A)ifA2A 0 ;
(b)everysetinA 0 ism -measurable;
(c)ifmis-finite,thenthereisauniqueextensiontothesmallest-fieldcontainingA 0 .
Proof.Westartwith(a).SupposeE2A 0 .Weknowm (E)m(E)sincewecantakeA 1 =Eand
A 2 ,A 3 ,...emptyinthedefinitionofm .IfE[ 1 i=1 A i withA i 2A 0 ,letB n =E\(A n −[ n−1
i=1 A i ).The
theB n aredisjoint,theyareeachinA 0 ,andtheirunionisE.Therefore
m(E)=
1 X
m(B i )
1 X
m(A i ).
i=1
i=1
Thusm(E)m (E).
Nextwelookat(b).SupposeA2A 0 .Let">0andletEX.PickB i 2A 0 suchthatE[ 1 i=1 B i
m (E)+"
1 X
1 X
1 X
m(B i \A c )
m(B i )=
m(B i \A)+
i=1
i=1
i=1
m (E\A)+m (E\A c ).
Since"isarbitrary,m (E)m (E\A)+m (E\A c ).SoAism -measurable.
Finally,supposewehavetwoextensionstothesmallest-fieldcontainingA 0 ;lettheotherextension
becalledn.WewillshowthatifEisinthissmallest-field,thenm (E)=n(E).
SinceEmustbem -measurable,m (E)=inf{ P 1 i=1 m(A i ):E[ 1 i=1 A i ,A i 2A 0 }.Butm=non
Let">0andchooseA i 2A 0 suchthatm (E)+" P i m(A i )andE[ i A i .LetA=[ i A i and
B k =[ k i=1 A i .Observem (E)+"m (A),hencem (A−E)<".Wehave
m (A)=lim
k!1 m (B k )=lim
k!1 n(B k )=n(A).
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and P i m(B i )m (E)+".Then
A 0 ,so P i m(A i )= P i n(A i ).Thereforen(E) P i n(A i ),whichimpliesn(E)m (E).

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