江泽坚的实变函数课后习题答案

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rih2(HG_THeFuliSchoolofMathematicsScienceandComputingTechnologyCentralSouthUniversityC0(Æ$&June,2010};iA_}5‘Xb\1§1.1℄K^PwÆ..................................1§1.2℄KY...................................4§1.3℄K.....................................5§1.4\/Y℄..................................7(3.........................................8Fn%tgE‘12§2.1IU{Bolzano−Weierstrass...........12§2.2℄R℄%M℄..............................13§2.3V)℄R℄%M℄7{.......................15§2.4℄i..................................16(3.........................................186Cx~19§3.1$b......................................19§3.2b℄K.....................................20§3.3℄bJ..................................22r6T24§4.1bF℄^PjJ........................242E13§4.2Egoroff..................................29§4.3bFy7Lusin.........................29§4.4Wb~’...................................30(3.........................................31*^Jx~32§5.1#)F[$................................33§5.2[F.....................................37§5.3Fubini...................................42§5.4&$l\[$................................44(3.........................................45*1...........................................474p’050?6AaY℄zvWF4fLTC℄$1RFJ%(MO~}l*℄K4YswO~T}l*℄K℄KwÆ℄KY℄Ie℄K$zlK[|℄Ki4Y^7{E℄KT$zIK[YC℄KwÆv9fjzv$1fHV+F\:4T(V+℄K℄K4YO~}l*℄+j\:I6\:,LPob29℄^PYzT(/℄KYX|2℄np#(℄KimX|!dBernsteinnp#(/mUY!dCCJ(f/℄[℄da|℄)O~|)I3vr*C!#NPob+2℄J[J\$℄I\℄§1.1‘Xb=1-1-1(B−A)∪A=BmTo|A⊂B.STJ8a{vjA(B℄_A⊂B.m$JfA⊂B^B−A⊂BY(B−A)∪A⊂B.if(B−A)∪A⊃BMj#j|Æ1-1-2A−B=A∩BC.∀x∈A−B_x∈AWx∈B[_x∈AWx∈BCj|x∈A∩BC.yx∈A∩BC_x∈AWx∈BC.[_x∈AWx∈BV|x∈A−B.j1V~℄K^PY2Æ1-1-34(3)(4)6yI9.4(3)∀x∈Tλ∈ΛBλ|∃λ0∈Λ,s.t.x∈Bλ0yx∈Aλ0agx∈Tλ∈ΛAλ.XH8aj#ÆQÆ4(4)∀x∈Sλ∈Λ(Aλ∪Bλ)S{zλ0∈Λ,xx∈Aλ0∪Bλ0.fx∈Aλ0agx∈(Sλ∈ΛAλ)∪(Sλ∈ΛBλ)fx∈Bλ0aUgx∈(Sλ∈ΛAλ)∪(Sλ∈ΛBλ)..V!∀x∈(Sλ∈ΛAλ)∪(Sλ∈ΛBλ)|x∈Sλ∈ΛAλXx∈Sλ∈ΛBλ,\nx∈Sλ∈ΛAλ,|{zCλ0∈Λxx∈Aλ0yx∈Aλ0∪Bλ0:x∈Sλ∈Λ(Aλ∪Bλ).jAÆ6yV!∀x∈Tλ∈ΛAλC_x∈Tλ∈ΛAλS{zCλ0∈Λxx∈Aλ0_x∈ACλ0yx∈Sλ∈ΛACλ08aj#Æ.V!∀x∈Sλ∈ΛACλ0{zCλ0∈Λxx∈ACλ0_x∈Aλ0a5x∈Tλ∈ΛAλ_gx∈Tλ∈ΛAλC.jÆ9.℄+{An}Æjptliminfn→∞An=∞Sn=1∞Tm=nAm=∞Sn=1An_wlimsupn→∞An=liminfn→∞An=∞Sn=1AnYlimn→∞An=∞Sn=1An.℄+{An}Æ6rtlimsupn→∞An=∞Tn=1∞Sm=nAm=∞Tn=1An_wliminfn→∞An=limsupn→∞An=∞Tn=1AnYlimn→∞An=∞Tn=1An.1-1-4(A−B)SB=(ASB)−BmTo|B=∅.m$J8a9STJdgB6=∅|/x∈Byx∈(ASB)−BWx∈(A−B)SB.ln6ÆQÆ1-1-5nS={1,2,3,4},A={{1,2},{3,4}}[F(A)ifBS={1n|n=1,2,···},A0={{1n|nisodd}},A1={{1},{13},···,{12i+1},···},F(A0)IF(A1)|u7V~℄K^PY3Solution.F(A)={∅,{1,2},{3,4},S}F(A0)={∅,{1n|nisodd},{1n|niseven},S}F(A1)={E|E⊂S}.1-1-6jS℄A,℄AF(ϕA(x)=(1,ifx∈A0,ifx∈A.fBA1,A2,···,An,···|S℄N+|ϕliminfn→∞An(x)=liminfn→∞ϕAn(x),ϕlimsupn→∞An(x)=limsupn→∞ϕAn(x).Proof.1).x∈liminfn→∞Ant$Uj1wtf6\:℄∃N,nNtx∈An,_ϕAn(x)=1.YhUUj1.x∈liminfn→∞Ant$Uj0wtf6\:℄∀N,{z∃nNx∈An,_ϕAn(x)=0.YhUUj0.2)./yÆ1-1-7nf(x)|℄zEjvFa(VfE[fa]=∞Sn=1E[f≥a+1n],E[f≥a]=∞Tn=1E[fa−1n].Proof.7VyÆV!L8a∀n,E[fa]⊃E[f≥a+1n]_wE[fa]⊃∞Sn=1E[f≥a+1n]..V!∀x∈E[fa],_x∈E,f(x)aF7nm$|ta+1nf(x)U|nm$|tx∈E[fa+1n]Yx∈∞Sn=1E[f≥a+1n]:E[fa]⊂∞Sn=1E[f≥a+1n].VyÆyyÆV!L8aE[f≥a]⊂E[fa−1n],∀n_wE[f≥a]⊂∞Tn=1E[fa−1n]..V!∀x∈∞Tn=1E[fa−1n],_∀n,x∈E[fa−1n]F7x∈EW∀n,a−1nf(x)fn\Jff(x)≥a,Yx∈E[f≥a]:E[f≥a]⊃∞Sn=1E[fa−1n].yÆV~℄K^PY41-1-8fBvFN+{fn(x)}zEj~’jf(x)|j\fagE[f≤a]=∞Tk=1liminfnE[fn≤a+1k]=∞Tk=1liminfnE[fna+1k].Proof.V!∀x∈E[f≤a],fjfn(x)→f(x)(n→∞),Y∀k,∃N,nNt|fn(x)−f(x)|1k_fn(x)≤f(x)+1k,Yfn(x)≤a+1k._wnNtx∈E[fn≤a+1k],E[f≤a]⊂∞Tk=1liminfnE[fn≤a+1k]..V!∀x∈∞Tk=1liminfnE[fn≤a+1k],Y∀k,x∈liminfnE[fn≤a+1k],_∃N,nNtx∈E[fn≤a+1k],YnNtx∈EWfn(x)≤a+1k,fjfn(x)→f(x)(n→∞),Yf(x)≤a+1kfk\Jf(x)≤a.Yx∈E[f≤a].:∞Tk=1liminfnE[fn≤a+1k]⊂E[f≤a].jV/yj#Æ/yƧ1.2‘X\1-2-1dz1y1q(−1,1)I(−∞,+∞)iV/1-1bSolution.y=tanπ2x,Px∈(−1,1).1-2-2Tabg(a,b)∼(0,1).Proof./y=a+(b−a)xPx∈(−1,1).|(a,b)l(0,1)p#1-1bÆ:(a,b)∼(0,1).1-2-3NjJ\~qqj%j℄|I/Nj%j℄NjJ#℄I/Nj%j℄ÆProof.7\~q+qIR2Æ{vjf31-2-1+q8VlNjC;Æy\\~qqI\~q+qC0ÆnC(Fz(x0,y0)V~℄K^PY5J(r\~qqÆ%mφ:C→C0,(x,y)→1r(x−x0,y−y0).8aφ(yCC0VVm_wC∼C0.Yf4uJV/y4j#ÆnE(℄Æf℄℄{zV\~qqC⊂EYc=C≤E≤R2=cYfBernstainEI/Nj%j℄Ƨ1.3r‘X1-3-1Nj&X(g7jV℄KÆnNj&X(g7j℄(A.g℄Q={r1,···,rn,···}.[An={(rn,r)|r∈Q},n=1,2,···|A=∞Sn=1An.8aAn_wA1-3-2Y;jQ\s\i(o\℄KDg/oz8V\i℄Vgfn;jCgQ^Cgj℄(A|_g℄B℄fjB(℄YAU(℄daAl\i1-1bj|QÆ1-3-3g4(g?yjV℄K\n4(?y(nv?ypn(x)=a0xn+a1xn−1+···+an−1x+an8a(n+1/g4(a0,a1,···,an)_4(gnv?y_℄Pn={pn(x)}ln+1)iRn+1&X(g℄KAn={(a0,a1,···,an)}gJAn|℄FBPnÆ9AndGgn=0|An={a0}a(℄gnn−1tAn−1(℄\nAn

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