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设f∈L([0,1]),fn∈L([0,1])(n∈N).若有,|fn(x)|≥1,a.e.x∈[0,1],试问是否有|f(x)|≥1,a.e.x∈[0,1]?
设f∈L([0,1]),fn∈L([0,1])(n∈N).若有,|fn(x)|≥1,a.e.x∈[0,1],试问是否有|f(x)|≥1,a.e.x∈[0,1]?
![](https://static.youtibao.com/asksite/comm/h5/images/m_q_a.png)
是.因为依题设知,存在{fnk(x)},使得,a.e.x∈[0,1],且有|fnk(x)|≥1,a.e.x∈[0,1],所以当k→∞时即得|f(x)|≥1,a.e.x∈[0,1].
![](https://static.youtibao.com/asksite/comm/h5/images/solist_ts.png)