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1、⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯实变函数论课后答案第五章2第五章第二节习题1．设mE，f(x)在E上可测且几乎处处有限EnE[x;n1f(x)n]，n0,1,2,证明：f(x)在E上可积的充要条件是nmEn证明f(x)在E上可积f在E上可积fdx，显然En可测（由Ef可测）fdxfdxfdxf(x)dxf(x)dxE00EnEnEnEnn1nn1n0f(x)dxf(x)dxn1nEnEn若fdx，则E00fdx(n1)mEnnmEnnm
2、EnmEnnmEnEn1nn1n1n0nmEnm(En)nmEnnmEnmEn1n1n则从mE知nmEn。反过来，若nmEn，则00fdxf(x)dxf(x)dxnmEn(n1)mEnEn1EnEn1nnn0nmEn(n)mEnmEnnmEnmEn1nn所以此时，f可积，从而f(x)可积。证毕1⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯sinx12．证明，分别在(0,)和(0,1)上不可积。xxsinx证明f(x)显然在(0,)上连续，从而非负可
3、测。xsinxLf(x)dxLdx(0,)[2,)xsinxsinx[LdxLdx]k1[2k,(2k1))x[(2k1),(2k2))xsinxsinxLdxLdx（P142Th2）k1[2k,2k1]x[2k1,2k2]x2k1sinx2k2sinxRdxRdx（R积分不分开区间还是闭2kx2k1xk1区间）12k112k2sinxdxsinxdx2k12k2k22k1k112k112k2cosxcosxk12k12k2k22k122211k12k12k2k12k12k221n3nsinx所以在
4、(0,)上不可积。（P142Th1f可积于Ef也可积于E）x11dxdxk1dxLdxLdxLR0x1xxxkxk1[k,k1)k1[k,k1)k111k1k1n2n1则在0,上不可积。x2⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯111dxdxkLdxLR1k0xxx11k1[,]k1k11令k知Ldx0x1则在0,1上不可积。xb3.设f(x)在Riemann意义下的广义积分f(x)dx是绝对收敛的，证明abf(x)在[a,b]上可积，且f(
5、x)dxf(x)dxa[a,b]证明1）f(x)在[a,b]上可测。1事实上，f在(a,b)上广义Riemann可积n充分大，f在[a,b]上n1Riemann可积,故f在[a,b]上有界，且Riemann可积。由P156Th8f(x)n11在[a,b]上几乎处处连续，且可测（P157：m,n(x)f(x)a.e.于n11[a,b]，m,n(x)为简单函数，可测）n由n的任意性，知f在(a,b]上可测1(x)f(x)f(x)a.e.于[a,b][a,b]n2）f(x)在[a,b]上可积。我们只用证f
6、(x)dx。[a,b]bnN充分大，由f(x)dx作为广义Riemann绝对收敛，知f(x)在a1bb[a,b]上Riemann（有界）可积，且limR1f(x)dxRf(x)dxnnaan由1）已知f(x)在[a,b]上可测，从而f(x)ff也可测于[a,b]，再11由P142定理已知f(x)在[a,b]上Riemann可积知f(x)在[a,b]上nnL可积且bbf(x)dx1f(x)dx且Lf(x)dxR1f(x)dxaa1n1n[a,b][a,b]nn令gn(x)1(x)f(x)[a,b]n则
7、0gn(x)gn1(x)于[a,b]。gn(x)f(x)，gn(a)0，x(a,b]3⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯最新资料推荐⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯由Levi定理则f(x)dxf(x)dxlimgn(x)dxn[a,b](a,b](a,b]bblimLf(x)dxlimR1f(x)dxf(x)dxnnaa1n[a,b]n则f(x)在[a,b]上可积。b3）f(x)dxf(x)dxa[a,b]b从Lf(x)dxR1f(x)dx（前已证）a1n[a,b]n只用证limf
8、(x)dxf(x)dxn1[a,b][a,b]nf(x)dx1(x)f(x)dx[a,b]1[a,b]n[a,b]n11(x)f(x)f(x)，1(x)f(x)f(x)a.e.L于[a,b]。[a,b][a,b]nn由控制收敛定理，知bbf(x)dxlim1f(x)dxlimf(x)dxf(x)dxanann1[a,b][a,b]n4.设mE，证明如果fn(x)，n1,2,3,都是E上的可积函数，且在E上一致收敛于f(x)，则f(x)也在E上可积且f(x)dxlimf
