Riemann-Lebesgue's Lemma

The content is directly from SiZhengJun.1 and SiZhengJun.2.

Riemann-Lebesgue’s Lemma. If $f\in \mathcal R([a,b])$ and $g\in \mathcal R([0,T])$ with a period $T>0$, then

$$\lim_{n\to\infty}\int_a^b f(x)g(nx)\,\mathrm dx=\dfrac 1T\int^b_a f(x)\,\mathrm dx\int^T_0 g(x)\,\mathrm dx.$$

Proof. Consider that $g\geq 0$ first. Extend $f$ from $[a,b]$ to $[-mT,mT]$ by padding zeros, where $m$ is sufficient to satisfy $[a,b]\subset[-mT,mT]$, then

$$\int^b_af(x)g(nx)\,\mathrm dx=\int^{mT}_{-mT}f(x)g(nx)\,\mathrm dx=\sum^{mn-1}_{k=-mn}\int^{\frac {k+1}{n}T}_{\frac{k}{n}T}f(x)g(nx)\,\mathrm dx=\dfrac 1n\sum^{mn-1}_{k=-mn}f(\xi_k)\int^{T}_0g(x)\,\mathrm dx=\dfrac 1T\int^b_a f(x)\,\mathrm dx\int^T_0 g(x)\,\mathrm dx,$$

where $\xi_k\in[\frac{k}{n}T,\frac{k+1}{n}T]$ by the mean value theorem. Since the operation is linear, $g\geq 0$ is not necessary. The proof is complete.