リーマン・ルベーグの定理

区分的に連続なので有限個の連続区間の和で表される。
ハイネ・カントールの定理より、有界閉区間で連続なので一様連続にある。
連続区間として\(\left[a',b'\right]\)をとる。
\begin{align*} \left|\lim_{k\rightarrow\infty}\int_{a'}^{b'}f\left(x\right)e^{ia_{k}x}dx\right| & =\left|\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\int_{x_{j}}^{x_{j+1}}f\left(x\right)e^{ia_{k}x}dx\right|\cmt{x_{j}=a'+\frac{b'-a'}{n}j}\\ & \leq\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left|\int_{x_{j}}^{x_{j+1}}f\left(x\right)e^{ia_{k}x}dx\right|\\ & =\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left|\int_{x_{j}}^{x_{j+1}}\left\{ \left(f\left(x\right)-f\left(x_{j}\right)\right)e^{ia_{k}x}+f\left(x_{j}\right)e^{ia_{k}x}\right\} dx\right|\\ & \leq\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \int_{x_{j}}^{x_{j+1}}\left|\left(f\left(x\right)-f\left(x_{j}\right)\right)e^{ia_{k}x}\right|dx+\left|f\left(x_{j}\right)\right|\left|\int_{x_{j}}^{x_{j+1}}e^{ia_{k}x}dx\right|\right\} \\ & \leq\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \int_{x_{j}}^{x_{j+1}}\left|f\left(x\right)-f\left(x_{j}\right)\right|dx+\max\left\{ \left|f\left(x\right)\right|;a'\leq x\le b'\right\} \left|\left[\frac{1}{ia_{k}}e^{ia_{k}x}\right]_{x_{j}}^{x_{j+1}}\right|\right\} \\ & =\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \int_{x_{j}}^{x_{j+1}}\left|f\left(x\right)-f\left(x_{j}\right)\right|dx+\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} \frac{1}{\left|a_{k}\right|}\left|e^{ia_{k}x_{j+1}}-e^{ia_{k}x_{j}}\right|\right\} \\ & \leq\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \int_{x_{j}}^{x_{j+1}}\left|f\left(x\right)-f\left(x_{j}\right)\right|dx+\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} \frac{1}{\left|a_{k}\right|}\left(\left|e^{ia_{k}x_{j+1}}\right|+\left|e^{ia_{k}x_{j}}\right|\right)\right\} \\ & =\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \int_{x_{j}}^{x_{j+1}}\left|f\left(x\right)-f\left(x_{j}\right)\right|dx+\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} \frac{2}{\left|a_{k}\right|}\right\} \\ & =\lim_{k\rightarrow\infty}\sum_{j=0}^{n-1}\left\{ \epsilon'\frac{b'-a'}{n}+\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} \frac{2}{\left|a_{k}\right|}\right\} \cmt{\text{一様連続より、}\frac{b'-a'}{n}\leq\left|x-x_{j}\right|<\delta\Rightarrow\left|f\left(x\right)-f\left(x_{j}\right)\right|<\epsilon'}\\ & =\lim_{k\rightarrow\infty}\left\{ \frac{\epsilon}{2}+\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} \frac{2n}{\left|a_{k}\right|}\right\} \cmt{\epsilon=2\epsilon'\left(b'-a'\right)}\\ & \le\lim_{k\rightarrow\infty}\left\{ \frac{\epsilon}{2}+\frac{\epsilon}{2}\right\} \cmt{\left|a_{k}\right|\geq\frac{4n\max\left\{ \left|f\left(x\right)\right|;a\leq x\le b\right\} }{\epsilon}}\\ & =\epsilon \end{align*} これより、左辺はいくらでも小さくできるので与式は成り立つ。