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# 實變一題，可積分函數

If |E| = 0, then ∫_E f = 0

謝謝

Update:

怎麼這麼複雜？？課本只簡單的帶過…

沒想到這麼複雜！

### 1 Answer

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- Scharze spaceLv 71 decade agoFavorite Answer
只需考慮f≥0的情形 因為f =f^+-f^-

則∫_E f=(∫_E f^+)-(∫_E f^-)

If f=0, then ∫_{f=0} f=0, f∈L^1(E), f<+∞ a.e 即 |{x:f(x)=+∞}|=0

and ∫_{f=+∞}f=0

so it sufficies to show 0<f<+∞:

Set E_n={x:2^(n-1)<f(x)<=2^n} E_n∩E_m=∅

|E_n|=0 for all n

Note that {x:0<f(x)<+∞}=∪_(n=1~∞){x:2^(n-1)<f(x)<=2^n}

Hence Σ_(n=1~∞)2^(n-1)|x:2^(n-1)<f(x)<=2^n}|<=Σ_(n=1~∞)∫_{x:2^(n-1)<f(x)<=2^n}f =∫_E f <=Σ_(n=1~∞)2^n|{x2^(n-1)<f(x)<=2^n}

上面不等式左右兩邊都是零,兩邊一夾, ∫_Ef=0

2008-12-29 00:18:24 補充：

我漏了{ x:0<=1}的情形:

0<=∫_{x:0<=1} f<=|{x:0<=1}|=0

=>∫_{x0<=1} f=0

2008-12-29 00:19:01 補充：

0<=1 是 0<=1

2008-12-29 00:19:21 補充：

0<=1是0<=1

2008-12-29 00:22:45 補充：

0<=1是 0小於f(x)小於等於1

2008-12-29 00:34:36 補充：

我想應該還有更簡單的方法

目前只想到這樣作

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