limln?x?n?dx?lim?gn?x?dx??limgn?x???0dx?0
n???n??n??n?0,????0,????0,????0,???n??由勒贝格控制收敛定理lim?0,????fn?x?dx?0
故lim
ln?x?n??xecosxdx?0.
n???n?0,n?
七、证明题
1.证明 对任何正数??0，由于
|(fn(x)?gn(x))?(f(x)?g(x))|?|fn(x)?f(x)|?|gn(x)?g(x)|
所以E[x|(fn(x)?gn(x))?(f(x)?g(x))|??]
?E[x|fn(x)?f(x)|??2]?E[x|gn(x)?g(x)|??2]
于是mE[x|(fn(x)?gn(x))?(f(x)?g(x))|??]
?mE[x|fn(x)?f(x)|??2]?mE[x|gn(x)?g(x)|??2]?0(n??)
故fn(x)?gn(x)?f(x)?g(x)
2.证明 因f(x),g(x)是E上L?可积，所以|f(x)|,|g(x)|在E上L?可积，从而
|f(x)|?|g(x)|L?可积，
又故f2(x)?g2(x)?(|f(x)|?|g(x)|)2?|f(x)|?|g(x)| f2(x)?g2(x)在E上L?可积
3.证明 反证，令A?E[x|f(x)?0]，则由f(x)的可测性知，A是可测集.下证mA?0，若不然，则mA?0
由于A?E[x|f(x)?0]?1E[x|f(x)?]，所以存在N?1，使 ?nn?11]?d?0 N?mE[x|f(x)? 于是
?Ef(x)dx??E[x|f(x)?1]Nf(x)dx??111ddx?mE[x|f(x)?]??0 1E[x|f(x)?]NNNNN因此4.证明
?Ef(x)dx?0，矛盾，故f(x)?0a.e于E
A\\(B?C)?A?(B?C)c?A?(Bc?Cc)?(A?Bc)?(A?Cc)?(A\\B)?(A\\C)