设 $f(x)\in R[0,1], F(x)=\int_0^x f(t)\mathrm{ d} t, DF(x)=\varlimsup_{h\to 0}\frac{F(x+h)-F(x)}{h}$. 求证: $$\begin{aligned} DF(x)\in R[0,1]\mbox{且} \int_0^1 DF(x)\mathrm{ d} x=\int_0^1 f(x)\mathrm{ d} x. \end{aligned}$$
设 $\displaystyle f(x)\in R, F(x)=\int_0^x f(t)\mathrm{ d} t, DF(x)=\varlimsup_{h\to 0}\frac{F(x+h)-F(x)}{h}$. 求证:$$\begin{aligned} DF(x)\in R\mbox{且} \int_0^1 DF(x)\mathrm{ d} x=\int_0^1 f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 我们要用实变函数的知识:$\displaystyle f\in R\Leftrightarrow f$ 的不连续点集 $\displaystyle D$ 是零测度集, 即 $\displaystyle f\mbox{a.e.}$ 连续. 由 $\displaystyle f\in R$ 知存在零测度集 $\displaystyle E\subset $, 使得
$$\begin{aligned} \forall\ x\in \backslash E, &f\mbox{在}x\mbox{处连续}\\ \Rightarrow& \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}= \lim_{h\to 0}\frac{1}{h}\int_x^{x+h}f(t)\mathrm{ d} t=f(x). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这就证明了除了零测度集 $\displaystyle E$ 外, $\displaystyle DF(x)=f(x)$ 是连续的, 而 $\displaystyle DF(x)\in R$, 且
$$\begin{aligned} \int_0^1 DF(x)\mathrm{ d} x=&\lim_{n\to\infty}\sum_{k=1}^n DF(\xi_k)\Delta x_k\\ &\left(\mbox{$$ 中必有 $\displaystyle f(x)$ 的连续点, 取其中一个为$\displaystyle \xi_k$}\right)\\ =&\lim_{n\to\infty}\sum_{k=1}^n f(\xi_k)\Delta x_k=\int_0^1 f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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