设 ${\displaystyle I_1={\int }_a^{b}f(x)\mathrm{d}x,I_2={\int }_a^{b}g(x)\mathrm{d}x}$, 则 ${\displaystyle \mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,\mathrm{\forall }\pi ,\xi (\pi )}$, 当 ${\displaystyle ||\pi ||<\delta }$:

$$|\sum _{\pi }f({\xi }_i)\mathrm{\Delta }{x}_i-I_1|<\epsilon ,|\sum _{\pi }g({\xi }_i)\mathrm{\Delta }{x}_i-I_2|<\epsilon .$$

则

$$\begin{array}{rl}& |\sum _{\pi }[\alpha f({\xi }_i)+\beta g({\xi }_i)]\mathrm{\Delta }{x}_i-(\alpha I_1+\beta I_2)|\\ \le & |\alpha ||\sum _{\pi }f({\xi }_i)\mathrm{\Delta }{x}_i-I_1|+|\beta ||\sum _{\pi }g({\xi }_i)\mathrm{\Delta }{x}_i-I_2|<(|\alpha |+|\beta |)\epsilon .\end{array}$$

由 定义 可得结论.

1. ${\displaystyle f(x)\in R[a,b]\Rightarrow \mathrm{\forall }\epsilon >0,\mathrm{\exists }{\pi }_{[a,b]}:\sum _{{\pi }_{[a,b]}}{\omega }_i\mathrm{\Delta }{x}_i<\epsilon }$.
   则 ${\displaystyle \mathrm{\forall }[\alpha ,\beta ]\subset [a,b]}$, 令 ${\displaystyle {\pi }_{[a,b]}^{\ast }={\pi }_{[a,b]}\cup \{\alpha ,\beta \}}$, 再令 ${\displaystyle {\pi }_{[\alpha ,\beta ]}={\pi }_{[a,b]}^{\ast }{|}_{[\alpha ,\beta ]}}$ 是 ${\displaystyle [\alpha ,\beta ]}$ 上的分割, 那么

$$\sum _{{\pi }_{[\alpha ,\beta ]}}{\omega }_i\mathrm{\Delta }{x}_i\le \sum _{{\pi }_{[a,b]}^{\ast }}{\omega }_i\mathrm{\Delta }{x}_i\le \sum _{{\pi }_{[a,b]}}{\omega }_i\mathrm{\Delta }{x}_i<\epsilon .$$

从而 ${\displaystyle f(x)\in R[\alpha ,\beta ]}$.
2. 由条件, ${\displaystyle \mathrm{\forall }\epsilon >0,\mathrm{\exists }{\pi }_{[a,c]},{\pi }_{[c,b]}:\sum _{{\pi }_{[a,c]}}{\omega }_i\mathrm{\Delta }{x}_i<\epsilon ,\sum _{{\pi }_{[c,b]}}{\omega }_i\mathrm{\Delta }{x}_i<\epsilon }$. 设 ${\displaystyle \pi ={\pi }_{[a,c]}\cup {\pi }_{[c,b]}}$, 由于 ${\displaystyle {\pi }_{[a,c]},{\pi }_{[c,b]}}$ 均在 ${\displaystyle c}$ 处存在分点, 因此 ${\displaystyle \pi }$ 没有加入新分点, 从而

$$\sum _{{\pi }_{[a,b]}}{\omega }_i\mathrm{\Delta }{x}_i=\sum _{{\pi }_{[a,c]}}{\omega }_i\mathrm{\Delta }{x}_i+\sum _{{\pi }_{[c,b]}}{\omega }_i\mathrm{\Delta }{x}_i<2\epsilon \Rightarrow f(x)\in R[a,b].$$

而 ${\displaystyle \mathrm{\forall }\delta >0,\mathrm{\exists }{\pi }_{[a,c]},{\pi }_{[c,b]},\xi ({\pi }_{[a,c]}),\xi ({\pi }_{[c,b]})}$, 且 ${\displaystyle ||{\pi }_{[a,c]}||,||{\pi }_{[c,b]}||<\delta }$:

$${\int }_a^{c}f(x)\mathrm{d}x=\underset{||{\pi }_{[a,c]}||\to 0}{lim}\sum _{{\pi }_{[a,c]}}f({\xi }_i)\mathrm{\Delta }{x}_i,{\int }_c^{b}f(x)\mathrm{d}x=\underset{||{\pi }_{[c,b]}||\to 0}{lim}\sum _{{\pi }_{[c,b]}}f({\xi }_i)\mathrm{\Delta }{x}_i.$$

取定 ${\displaystyle \pi ={\pi }_{[a,c]}\cup {\pi }_{[c,b]}}$, 则 ${\displaystyle ||\pi ||<\delta }$, 故

$$\begin{array}{rl}& {\int }_a^{b}f(x)\mathrm{d}x=\underset{||\pi ||\to 0}{lim}\sum _{\pi }f({\xi }_i)\mathrm{\Delta }{x}_i\\ =& \underset{||{\pi }_{[a,c]}||\to 0}{lim}\sum _{{\pi }_{[a,c]}}f({\xi }_i)\mathrm{\Delta }{x}_i+\underset{||{\pi }_{[c,b]}||\to 0}{lim}\sum _{{\pi }_{[c,b]}}f({\xi }_i)\mathrm{\Delta }{x}_i={\int }_a^{c}f(x)\mathrm{d}x+{\int }_c^{b}f(x)\mathrm{d}x.\end{array}$$

首先证明

现在我们用 闭区间套定理 证明结论.
由引理, 对 ${\displaystyle \epsilon =1}$, 取 ${\displaystyle [a_1,b_1]\subset [a,b]:{\omega }_{[a_1,b_1]}\le 1}$. 令 ${\displaystyle \epsilon =\frac{1}{2}}$, 由 对区间可加性 (1), 有 ${\displaystyle f(x)\in R[a_1,b_1]}$. 故可取 ${\displaystyle [a_2,b_2]\subset [a_1,b_1]:{\omega }_{[a_2,b_2]}\le \frac{1}{2}}$. 从而可得闭区间列 ${\displaystyle \{[a_n,b_n]\}}$ 满足:

1. ${\displaystyle [a_n,b_n]\supset [a_{n+1},b_{n+1}]}$.
2. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(b_n-a_n)=0}$.
3. ${\displaystyle {\omega }_{[a_n,b_n]}\le \frac{1}{n}}$.

由闭区间套定理, 唯一存在 ${\displaystyle \xi \in \bigcap _{n=1}^{\mathrm{\infty }}[a_n,b_n]}$. 那么

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }{n}_0\in \mathbb{N},\frac{1}{n_0}<\epsilon ,\mathrm{\exists }\delta >0:o(\xi ,\delta )\subset [a_{n_0},b_{n_0}]$$

于是 ${\displaystyle \mathrm{\forall }x\in o(\xi ,\delta )}$:

$$|f(x)-f(\xi )|\le {\omega }_{[a_{n_0},b_{n_0}]}<\frac{1}{n_0}<\epsilon .$$

从而 ${\displaystyle f(x)}$ 在 ${\displaystyle \xi }$ 处连续.

记 ${\displaystyle F(x)=f(x)-g(x)}$, 则除了有限个点外, ${\displaystyle F(x)\equiv 0}$. 因此 ${\displaystyle F(x)}$ 有界, 且只有有限个间断点, 则由 命题 (1), ${\displaystyle F(x)\in R[a,b]}$, 且由定积分的定义不难算得

$${\int }_a^{b}F(x)\mathrm{d}x={\int }_a^b[f(x)-g(x)]\mathrm{d}x=0.$$

从而由 线性性 得 ${\displaystyle g(x)=f(x)-F(x)\in R[a,b]}$, 且

$$\begin{array}{rl}& {\int }_a^{b}g(x)\mathrm{d}x={\int }_a^b[f(x)-F(x)]\mathrm{d}x\\ =& {\int }_a^{b}f(x)\mathrm{d}x-{\int }_a^{b}F(x)\mathrm{d}x={\int }_a^{b}f(x)\mathrm{d}x.\end{array}$$

因此结论得证.

设不可导的点为 ${\displaystyle a<x_1<\cdots <x_{n-1}<b}$. 记 ${\displaystyle a=x_0,b=x_n}$. 则在 ${\displaystyle [x_i,x_{i+1}](0\le i\le n-1)}$ 上由 Newton-Leibniz 公式得

$${\int }_{x_i}^{x_{i+1}}f(x)\mathrm{d}x={F(x)|}_{x_i}^{x_{i+1}}.$$

加和可得结论式.

注意到 ${\displaystyle {\int }_a^{b}f(x)\mathrm{d}x={\int }_a^{x_0}f(x)\mathrm{d}x+{\int }_{x_0}^{b}f(x)\mathrm{d}x}$. 考察 ${\displaystyle {\int }_a^{x_0}f(x)\mathrm{d}x}$ 的取值. 设

$${\int }_a^{x_0}f(x)\mathrm{d}x={\int }_a^{x_0-\epsilon }f(x)\mathrm{d}x+{\int }_{x_0-\epsilon }^{x_0}f(x)\mathrm{d}x={F(x)|}_a^{x_0-\epsilon }+I_{\epsilon },$$

其中 ${\displaystyle I_{\epsilon }={\int }_{x_0-\epsilon }^{x_0}f(x)\mathrm{d}x}$.
由 单侧极限的局部有界性 得 ${\displaystyle \mathrm{\exists }m\le M}$: $$m\epsilon \le {\int }_{x_0-\epsilon }^{x_0}m\mathrm{d}x\le I_{\epsilon }\le {\int }_{x_0-\epsilon }^{x_0}M\mathrm{d}x\le M\epsilon .$$ 令 ${\displaystyle \epsilon \to 0^{+}:I_{\epsilon }\to 0}$, 从而

$$\underset{\epsilon \to 0^{+}}{lim}{\int }_a^{x_0-\epsilon }f(x)\mathrm{d}x={\int }_a^{x_0}f(x)\mathrm{d}x=F(x_0^{-})-F(a).$$

同理

$${\int }_{x_0}^{b}f(x)\mathrm{d}x=F(b)-F(x_0^{+}).$$

因此证明了结论.

由 ${\displaystyle f(x)\in R[a,b]}$, ${\displaystyle \mathrm{\forall }\epsilon >0,\mathrm{\exists }\pi :\sum _{\pi }{\omega }_i(f)\mathrm{\Delta }{x}_i<\epsilon }$, 而

$${\omega }_i(|f|)=\underset{[x_{i-1},x_i]}{sup}||f(x)|-|f(y)||\le \underset{[x_{i-1},x_i]}{sup}|f(x)-f(y)|={\omega }_i(f).$$

故

$$\sum _{\pi }{\omega }_i(|f|)\mathrm{\Delta }{x}_i\le \sum _{\pi }{\omega }_i(f)\mathrm{\Delta }{x}_i<\epsilon ,$$

从而 ${\displaystyle |f(x)|\in R[a,b]}$. 又由于

$$|\sum _{\pi }f({\xi }_i)\mathrm{\Delta }{x}_i|\le \sum _{\pi }|f({\xi }_i)|\mathrm{\Delta }{x}_i.$$

令 ${\displaystyle ||\pi ||\to 0}$ 即得 ${\displaystyle |{\int }_a^{b}f(x)\mathrm{d}x|\le {\int }_a^b|f(x)|\mathrm{d}x}$.

由 ${\displaystyle f(x)\in R[a,b]}$ 及有界性, ${\displaystyle \mathrm{\exists }M>0:|f(x)|\le M}$. 则

$$\begin{array}{rl}& {\omega }_i(f^2)=\underset{[x_{i-1},x_i]}{sup}|f^2(x)-f^2(y)|\\ =& \underset{[x_{i-1},x_i]}{sup}|f(x)-f(y)||f(x)+f(y)|\le 2M{\omega }_i(f).\end{array}$$

得

$$\sum _{\pi }{\omega }_i(f^2)\mathrm{\Delta }{x}_i\le 2M\sum _{\pi }{\omega }_i(f)\mathrm{\Delta }{x}_i<2M\epsilon .$$

• "${\displaystyle \Rightarrow }$" 由 ${\displaystyle |f(x)|\in R[a,b]}$ 及有界性, ${\displaystyle \mathrm{\exists }M>0:|f(x)|\le M}$, 则

$$\begin{array}{rl}& {\omega }_i(f^2)=\underset{[x_{i-1},x_i]}{sup}|f^2(x)-f^2(y)|\\ =& \underset{[x_{i-1},x_i]}{sup}||f(x)|-|f(y)||(|f(x)|+|f(y)|)\le 2M{\omega }_i(|f|).\end{array}$$

故 ${\displaystyle f^2(x)\in R[a,b]}$.

• "${\displaystyle \Leftarrow }$" 由已知, ${\displaystyle \mathrm{\forall }\epsilon >0,\mathrm{\exists }\pi :\sum _{\pi }{\omega }_i(f^2)\mathrm{\Delta }{x}_i<{\epsilon }^2}$. 又 ${\displaystyle \mathrm{\forall }x,y\in [x_{i-1},x_i]}$:

$$||f(x)|-|f(y)||\le \sqrt{|f^2(x)-f^2(y)|},$$

则 ${\displaystyle {\omega }_i^2(|f|)\le {\omega }_i(f^2)}$.

$$\begin{array}{rl}\sum _{\pi }{\omega }_i(|f|)\mathrm{\Delta }{x}_i& =\sum _{\pi }{\omega }_i(|f|)\sqrt{\mathrm{\Delta }{x}_i}\cdot \sqrt{\mathrm{\Delta }{x}_i}\\ & \le \sqrt{(\sum _{\pi }{\omega }_i^2(|f|)\mathrm{\Delta }{x}_i)\left(\sum _{\pi }\mathrm{\Delta }{x}_i\right)}\\ & \le \sqrt{(\sum _{\pi }{\omega }_i(f^2)\mathrm{\Delta }{x}_i)(b-a)}<\epsilon \sqrt{b-a}.\end{array}$$

设 ${\displaystyle |f(x)|,|g(x)|}$ 的上界分别为 ${\displaystyle M_f,M_g}$. 由于 ${\displaystyle f,g\in R[a,b]}$, ${\displaystyle \mathrm{\exists }\pi :\sum _{\pi }{\omega }_i(f)\mathrm{\Delta }{x}_i<\epsilon ,\sum _{\pi }{\omega }_i(g)\mathrm{\Delta }{x}_i<\epsilon }$.
而 ${\displaystyle \mathrm{\forall }x,y\in [a,b]}$:

$$\begin{array}{rl}|f(x)g(x)-f(y)g(y)|& =|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|\\ & \le |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|\\ & \le M_f|g(x)-g(y)|+M_g|f(x)-f(y)|.\end{array}$$

故 $${\omega }_i(fg)\le M_f{\omega }_i(g)+M_g{\omega }_i(f)$$

$$\Rightarrow \sum _{\pi }{\omega }_i(fg)<M_f\sum _{\pi }{\omega }_i(g)+M_g\sum _{\pi }{\omega }_i(f)<(M_f+M_g)\epsilon .$$

从而 ${\displaystyle fg\in R[a,b]}$.

不妨设 ${\displaystyle g(x)\ge 0}$. 当 ${\displaystyle g(x)\equiv 0}$, 式子总是成立的. 否则, 由 定积分的严格正性 得 ${\displaystyle {\int }_a^{b}g(x)\mathrm{d}x>0}$.
记 ${\displaystyle m=\underset{[a,b]}{min}f(x),M=\underset{[a,b]}{max}f(x)}$, 则

$$mg(x)\le f(x)g(x)\le Mg(x).$$$$m\le \frac{{\int }_a^{b}f(x)g(x)\mathrm{d}x}{{\int }_a^{b}g(x)\mathrm{d}x}\le M.$$

从而由介值性, ${\displaystyle \mathrm{\exists }\xi \in [a,b]:\frac{{\int }_a^{b}f(x)g(x)\mathrm{d}x}{{\int }_a^{b}g(x)\mathrm{d}x}=f(\xi )}$. 因而结论得证.