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时间:2019-03-10
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1、MeasureTheoryV.Liskevich19981IntroductionWealwaysdenotebyXouruniverse,i.e.allthesetsweshallconsideraresubsetsofX.Recallsomestandardnotation.2XeverywheredenotesthesetofallsubsetsofagivensetX.IfAB=?thenweoftenwriteAtBratherthanA[B,tounderlinethedisjointness.Thecomplement(inX)ofasetAisdenote
2、dbyAc.ByA4Bthesymmetricdi®erenceofAandBisdenoted,i.e.A4B=(AnB)[(BnA).Lettersi;j;kalwaysdenotepositiveintegers.Thesign¹isusedforrestrictionofafunction(operatoretc.)toasubset(subspace).1.1TheRiemannintegralRecallhowtoconstructtheRiemannianintegral.Letf:[a;b]!R:Considerapartition¼of[a;b]:a=x0
3、4、DarbouxsumsXn¡1Xn¡1s(f;¼)=mk¢xk;s¹(f;¼)=Mk¢xk:k=0k=0Onecanshow(theDarbouxtheorem)thatthefollowinglimitsexistZblims(f;¼)=sups(f;¼)=fdxj¼j!0¼aZblims5、¹(f;¼)=infs¹(f;¼)=fdx:j¼j!0¼a1Clearly,ZbZbs(f;¼)·fdx·fdx·s¹(f;¼)aaforanypartition¼.ThefunctionfissaidtobeRiemannintegrableon[a;b]iftheupperandlowerintegralsareequal.ThecommonvalueiscalledRiemannintegraloffon[a;b].Thefunctionscannothavealargesetofpointsofdiscontinuity.Morepresicelythiswillb6、estatedfurther.1.2TheLebesgueintegralItallowstointegratefunctionsfromamuchmoregeneralclass.First,consideraveryusefulexample.Forf;g2C[a;b],twocontinuousfunctionsonthesegment[a;b]=fx2R:a6x6bgput½1(f;g)=maxjf(x)¡g(x)j;a6x6bZb½2(f;g)=jf(x)¡g(x)jdx:aThen(C[a;b];½1)isacompletemetricspace,when(C[7、a;b];½2)isnot.Toprovethelatterstatement,considerafamilyoffunctionsf'g1asdrawnonFig.1.ThisisaCauchynn=1sequencewithrespectto½2.However,thelimitdoesnotbelongtoC[a;b].26¯¯L¯L¯L¯L¯L¯L¯L¯L¯LL-¡1¡1+11¡1122n2n2Figure1:Thefunction'n.2SystemsofSetsDe¯nition2.1Aringofsetsisanon-emptysubsetin2Xwhichi8、sclosedwithrespecttotheoperations[andn.Proposition.LetKbearingofsets.Then?2K.Proof.SinceK6=?,thereexistsA2K.SinceKcontainsthedi®erenceofeverytwoitselements,onehasAnA=?2K.¥Examples.1.ThetwoextremecasesareK=f?gandK=2X.2.LetX=RanddenotebyKall¯niteunionsofsemi-seg
4、DarbouxsumsXn¡1Xn¡1s(f;¼)=mk¢xk;s¹(f;¼)=Mk¢xk:k=0k=0Onecanshow(theDarbouxtheorem)thatthefollowinglimitsexistZblims(f;¼)=sups(f;¼)=fdxj¼j!0¼aZblims
5、¹(f;¼)=infs¹(f;¼)=fdx:j¼j!0¼a1Clearly,ZbZbs(f;¼)·fdx·fdx·s¹(f;¼)aaforanypartition¼.ThefunctionfissaidtobeRiemannintegrableon[a;b]iftheupperandlowerintegralsareequal.ThecommonvalueiscalledRiemannintegraloffon[a;b].Thefunctionscannothavealargesetofpointsofdiscontinuity.Morepresicelythiswillb
6、estatedfurther.1.2TheLebesgueintegralItallowstointegratefunctionsfromamuchmoregeneralclass.First,consideraveryusefulexample.Forf;g2C[a;b],twocontinuousfunctionsonthesegment[a;b]=fx2R:a6x6bgput½1(f;g)=maxjf(x)¡g(x)j;a6x6bZb½2(f;g)=jf(x)¡g(x)jdx:aThen(C[a;b];½1)isacompletemetricspace,when(C[
7、a;b];½2)isnot.Toprovethelatterstatement,considerafamilyoffunctionsf'g1asdrawnonFig.1.ThisisaCauchynn=1sequencewithrespectto½2.However,thelimitdoesnotbelongtoC[a;b].26¯¯L¯L¯L¯L¯L¯L¯L¯L¯LL-¡1¡1+11¡1122n2n2Figure1:Thefunction'n.2SystemsofSetsDe¯nition2.1Aringofsetsisanon-emptysubsetin2Xwhichi
8、sclosedwithrespecttotheoperations[andn.Proposition.LetKbearingofsets.Then?2K.Proof.SinceK6=?,thereexistsA2K.SinceKcontainsthedi®erenceofeverytwoitselements,onehasAnA=?2K.¥Examples.1.ThetwoextremecasesareK=f?gandK=2X.2.LetX=RanddenotebyKall¯niteunionsofsemi-seg
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