实变函数习题三答案.pdf

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1、1nÙÿ¼êù24,5¢C¼êؼ©Û6(1nþþ),2011.1Snno1.f´Eþÿ¼ê,é?¿¢êa,8Üx∈Ef(x)=a´ÿ8.nonoy²µnÏǑf3Eþoÿ,né?¿¢êao,8nÜx∈Ef(x)≥oax∈Ef(x)≤aÑ´ÿ8.l,x∈Ef(x)=a=x∈Ef(x)≥a∩x∈Ef(x)≤aǑÿ8.no2.f´Eþ¼ê,f3Eþÿ=éknêr,x∈Ef(x)>r´ÿ8.y²µ“⇒”w,.ey“⇐”.é?¿¢êt,3knê{rk},rk≥tklim→∞rk=t.·kn

2、o[∞nox∈Ef(x)>t=x∈Ef(x)>rk.k=1nonod^,:8x∈Ef(x)>rkÑ´ÿ8,x∈Ef(x)>tǑ´ÿ8,=f3Eþÿ.3.f´Rþÿ¼ê,é?¿~êa,f(ax)E´Rþÿ¼ê.y²µea=0,f(ax)=f(0)Ǒ~¼ê,ÿ.ea,0,é?¿t∈R,non1o1nox∈Rf(ax)>t=x∈Rf(x)>t=x∈Rf(x)>t.aano1f3Rþnÿ,Et=xo∈Rf(x)>tǑÿ8.l,dþÙS120aEtǑǑÿ8,=x∈Rf(ax)>tǑÿ8.Ïd

3、,f(ax)3Rþÿ.4.3f´Eþÿ¼ê,fǑ´Eþÿ¼ê.non1o33yn²µ5¿éo?¿t∈R,x∈nEf(x)>t=ox∈Ef(x)>t,f3Eþÿ,¤±,8Ü1x∈E333f(x)>tǑÿ8.l,x∈Ef(x)>tǑÿ8,=f3Eþÿ.5.f3?¿«m[α,β]⊂[a,b]þÿ,f3«m[a,b]þǑÿ.no[∞no11y²µ5¿é?¿t∈R,x∈(a,b)f(x)>t=x∈a+k,b−kf(x)>t,d^,nok=1no11é?¿k≥1,x∈a+k,b−kf(x)

4、>tÑ´ÿ8,x∈(a,b)f(x)>tǑÿ8,=f3(a,b)þÿ.l,f3[a,b]þÿ.6.231f´Rþÿ¼ê,f(x),f(x),f(x)ǑÑ´Rþÿ¼ê.no
2−1−1y²µg(x)=x,g(E)=xg(x)∈E,m(E)=0,mg(E)=0.fǑRþÿ−1−1¼ê,éR¥
?¿m8G,f(G)Ǒÿ8.f(G)=HZ
,Ù¥HǑGδ.8
m(Z)=0.−1f−1−1−1−1−1−1f−1u´g(G)=g(H
)g(Z),g(H)ǑǑGδ.8mg(Z)=0

5、,g(G)Ǒÿ8.2lf(x)=fg(x)ǑRþÿ¼ê.no7.2nnnf3Rþÿ,x∈Rf(x)>0´ÿ8,fǑ3Rþÿ.y²µy{.é?¿t∈R,knoxf(x)>0,t=0,nononoEt=xf(x)>t=xf(x)>0∩xf2>t2,t>0,nonoxf(x)>0∪xf20,f(x)=

6、f(x)

7、·χA(x)−χAc(x).ÏǑAÿ,¤±Aÿ,n2

8、nnnχA(x)−χAc(x)3Rþÿ.f(x)3Rþÿ,

9、f(x)

10、Ǒ3Rþÿ.Ïdf(x)3Rþÿ.8.f´ÿ8E⊂Rþ¼ê.eé?¿«m[a,b],f3E∩[a,b]þÿ,fǑ3Eþÿ.no[∞noy²µ5¿,∀t∈R,x∈Ef(x)>t=x∈E∩[k,k+1]f(x)>t,d^,é?¿nok=−∞nox∈Ek∈Z,x∈E∩[k,k+1]f(x)>tÑ´ÿ8,f(x)>tǑÿ8,=f3Eþÿ.9.′f3[a,b]þ,¼êf3[a,b]þÿ.1′f(x+h)−f(x)f(x+n

11、)−f(x)y²µÏǑf3[a,b]þ,f(x)=limh→0h=nlim→+∞1=nlim→+∞gn(x),nf(x+1)−f(x)n1Ù¥gn(x)=1.f(x)3[a,b]þëY,f(x)f(x+n)Ñ3[a,b]þÿ.l,gn(x)ǑÑn′(x)3[a,b]þÿ.Ïd,f3[a,b]þÿ.10.−1−1f3Eþÿ,(i)é?¿m8G⊂R,f(G)´ÿ8;(ii)é?¿48F⊂R,f(F)´−1ÿ8;(iii)é?¿Gδ8½Fσ8M⊂R,f(M)´ÿ8.[y²µ(i)dm8(½n,G

12、=(ak,bk),Ù¥(ak,bk)´õêpØm«nk≥1o
−1m.,f(x)3Eþÿ,∀t∈R,Et=x∈Ef(x)>t=f(t,+∞)´ÿ8.u´[[
[

−1−1−1−1−1f(G)=f(a,b)=f(a,b)=f(a,∞)f[b,∞)Ǒÿ8.kkkkkkk≥1k≥1k≥1−1−1cc(ii)df(F)=f(F)±9(i)=.\−1−1(iii)kM´Gδ8,=M=Gk