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在闭区间上$f$可积,则$|f|$可积

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hbghlyj Posted at 2022-10-16 22:20:58 |Read mode
If f is integrable, why is |f| integrable?
如果 $U(f,P) - L(f,P) < \epsilon$,那么我们也有 $U(|f|,P) - L(|f|,P) < \epsilon$.
因为,在每个子区间,令 $M_i=\sup f,m_i=\inf f$,则 $\sup|f|=\max(|M_i|, |m_i|)$, $\inf|f|=\min(|M_i|, |m_i|)$, 所以 $\sup|f|-\inf|f|\le\sup f-\inf f$.

直观地说,这意味着在每个子区间上,取 $f$ 的绝对值具有减小(或不改变)振幅(Oscillation)的效果,因而仍然可积。(如果取绝对值,上下达布和只能变得“更接近”,而不是更远。)

条件“在闭区间上”是必要的。这个命题对于无穷区间是不成立的,参见Absolutely convergent improper integral(EoM | Wikipedia)。
The existence of the improper integral does not guarantee its absolute convergence, as for instance is the case for the function $\sin x/x$ on $]0,∞[$.

但是反过来的“$|f|$可积,则$f$可积”总是对的.

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 Author| hbghlyj Posted at 2022-10-16 22:29:49
hw5第6题
Suppose that $f$ is integrable on $[a, b]$. Show that $|f |$ is integrable on $[a, b]$.

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2025-4-23 02:53 GMT+8

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