→ keroro321 :no , f≦0 ,let fn =f*χ_[0, (-1/n)+π/2] 05/02 10:11
→ keroro321 :Lebes(f,[0,∞]) ≦ Lebes(fn) for all n 05/02 10:12
→ keroro321 :Lebes(fn) = Riema(fn)->-∞ 05/02 10:12
→ bineapple :所以我們是不是可以推出 如果函數measurable且恆正或 05/02 10:39
→ bineapple :恆負 則可推得Riemann可積若且惟若Lebesgue可積呢? 05/02 10:40
→ keroro321 :no , 若函數measurable且恆正或恆負 R=>L 反過來不對 05/02 15:04
→ keroro321 :f=χ_([0, 1]的有理數) f:L可積 ,但不是R可積 05/02 15:08
→ keroro321 :他們關係 , 許多書上都有. 05/02 15:09
→ bineapple :寫錯了 不是measurable 應該是conti a.e. 05/02 15:48
→ bineapple :不過還是感謝你喔 thx 05/02 15:49
→ keroro321 :f is continuous a.e.[m] =>f is Riemann integrable 05/02 16:00
→ bineapple :應該還要bounded吧 05/02 16:57
→ keroro321 :嗯嗯 05/02 19:18