由于 ${\displaystyle f(x)\in R[a,b]}$, 则 ${\displaystyle \mathrm{\forall }\epsilon >0}$, 存在阶梯函数 ${\displaystyle p(x)=c_k,x\in [x_{k-1},x_k],k=1,\dots ,p}$, 使得

于是

由于 ${\displaystyle \underset{\lambda \to +\mathrm{\infty }}{lim}\sum _{k=1}^p|c_k|\frac{2}{\lambda }=0}$, 于是对于 ${\displaystyle \epsilon >0,\mathrm{\exists }X>0,\mathrm{\forall }\lambda >X:\sum _{k=1}^p|c_k|\frac{2}{\lambda }<\epsilon }$, 从而

也即 ${\displaystyle \underset{\lambda \to +\mathrm{\infty }}{lim}{\int }_a^{b}f(x)\sin (\lambda x)\mathrm{d}x=0}$. 同理 ${\displaystyle \underset{\lambda \to +\mathrm{\infty }}{lim}{\int }_a^{b}f(x)\cos (\lambda x)\mathrm{d}x=0}$.

分三步进行证明.

• 第一步: 设 ${\displaystyle f(x)\equiv 1}$. 由于我们考虑的是 ${\displaystyle \lambda \to +\mathrm{\infty }}$ 的极限, 故不妨设 ${\displaystyle \lambda >0}$. 令 ${\displaystyle n=\lfloor \frac{\lambda (b-a)}{T}\rfloor \in \mathbb{N}}$, 则

注意到

又 ${\displaystyle g(x)\in R[a,b]\Rightarrow g(x)}$ 有界 (设上界是 ${\displaystyle M_g}$), 故又有

于是

• 第二步: 设 ${\displaystyle f(x)}$ 是阶梯型函数, 即存在 ${\displaystyle [a,b]}$ 的分割 ${\displaystyle \pi :a=x_0<x_1<\cdots <x_m=b}$, 使得

则有

这里用到了第一步得出的结论.

• 第三步: 对一般的 ${\displaystyle f(x)\in R[a,b]}$, ${\displaystyle \mathrm{\forall }\epsilon >0}$, ${\displaystyle \mathrm{\exists }\pi \subset [a,b]}$: ${\displaystyle a=x_0<x_1<\cdots <x_m=b}$ 和对应值点下的 Riemann 和 ${\displaystyle S(\pi )}$, 使得 $$|{\int }_a^{b}f(x)\mathrm{d}x-S(\pi )|<\frac{\epsilon }{3M_g+1}.$$ 取 ${\displaystyle c_k=\underset{[x_{k-1},x_k]}{inf}f(x)}$, 记相应的阶梯形函数为 ${\displaystyle L(x)}$, 则 ${\displaystyle S(\pi )={\int }_a^{b}L(x)\mathrm{d}x}$, 且 $$|{\int }_a^{b}f(x)\mathrm{d}x-S(\pi )|=|{\int }_a^b[f(x)-L(x)]\mathrm{d}x|={\int }_a^b|f(x)-L(x)|\mathrm{d}x.$$ 由第二步的已知结论, 对上述 ${\displaystyle \epsilon >0,\mathrm{\exists }\mathrm{\Lambda }>0,\mathrm{\forall }\lambda >\mathrm{\Lambda }}$: $$|{\int }_a^{b}L(x)g(\lambda x)\mathrm{d}x-\frac{1}{T}{\int }_0^{T}g(x)\mathrm{d}x{\int }_a^{b}L(x)\mathrm{d}x|<\epsilon ,$$ 于是$$\begin{array}{rl}& |{\int }_a^{b}f(x)g(\lambda x)\mathrm{d}x-\frac{1}{T}{\int }_0^{T}g(x)\mathrm{d}x{\int }_a^{b}f(x)\mathrm{d}x|\\ \le & |{\int }_a^b[f(x)-L(x)]g(\lambda x)\mathrm{d}x|+|{\int }_a^{b}L(x)g(\lambda x)\mathrm{d}x-\frac{1}{T}{\int }_0^{T}g(x)\mathrm{d}x{\int }_a^{b}L(x)\mathrm{d}x|\\ & +|\frac{1}{T}{\int }_0^{T}g(x)\mathrm{d}x{\int }_a^b[L(x)-f(x)]\mathrm{d}x|\\ <& \frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon .\end{array}$$

由已知,

于是结论得证.

取分割 ${\displaystyle \pi :a=x_0<x_1<\cdot \cdot \cdot <x_n=b}$, 则

由于

因此在(*)中令 ${\displaystyle ||\pi ||\to 0}$, 得

注意到 ${\displaystyle g(x)}$ 的单调性, 不妨设其单调增. 令 ${\displaystyle F(x)={\int }_a^{x}f(t)\mathrm{d}t\in C[a,b]}$, 则 ${\displaystyle F(x_0)=F(a)=0}$. 应用 Abel 变换, 得

于是 $$\begin{array}{rl}& \sum _{k=1}^{n-1}F(x_k)[g(x_{k+1})-g(x_k)]=\sum _{k=0}^{n-1}F(x_k)[g(x_{k+1})-g(x_k)]\\ \le & M\sum _{k=1}^{n-1}[g(x_{k+1})-g(x_k)]=M[g(b)-g(a)],\end{array}$$ 同理 $$\sum _{k=1}^{n-1}F(x_k)[g(x_{k+1})-g(x_k)]\ge m[g(b)-g(a)].$$ 于是 $$m\le \frac{g(b)F(b)-\sum _{k=1}^n{\int }_{x_{k-1}}^{x_k}g(x_k)f(x)\mathrm{d}x}{g(b)-g(a)}\le M.$$ 令 ${\displaystyle ||\pi ||\to 0}$, 有 $$m\le \frac{g(b)F(b)-{\int }_a^{b}g(x)f(x)\mathrm{d}x}{g(b)-g(a)}\le M.$$ 从而 ${\displaystyle \mathrm{\exists }\xi \in [a,b]}$: $$\begin{array}{rl}& \frac{g(b)F(b)-{\int }_a^{b}g(x)f(x)\mathrm{d}x}{g(b)-g(a)}=F(\xi )={\int }_a^{\xi }f(t)\mathrm{d}t\\ \Rightarrow & g(b){\int }_a^{b}f(t)\mathrm{d}t-{\int }_a^{b}g(x)f(x)\mathrm{d}x=g(b){\int }_a^{\xi }f(t)\mathrm{d}t-g(a){\int }_a^{\xi }f(t)\mathrm{d}t.\end{array}$$ 整理即得结论形式.