Generalized Riemann – Lebesgue

Let $f , g:\mathbb{R} \rightarrow \mathbb{R}$ be $1$ periodic and continuous functions. Prove that

$\displaystyle \lim_{n \rightarrow +\infty} \int_{0}^{1} f(x) g(nx) \, {\rm d}x = \int_{0}^{1} f(x) \, {\rm d}x \int_{0}^{1} g(x) \, {\rm d}x$

Solution

Since our functions are continuous and $1$ periodic this means that they are both uniform continuous and bounded. Thus,

$\begin{align*} \int_{0}^{1} f(x) g(nx) \, {\rm d}x &= \frac{1}{n} \int_{0}^{n} f \left ( \frac{x}{n} \right ) g(x) \, {\rm d}x \\ &= \frac{1}{n} \sum_{k=0}^{n-1} \int_{k}^{k+1} f \left ( \frac{x}{n} \right ) g(x) \, {\rm d}x\\ &= \frac{1}{n} \sum_{k=0}^{n-1} \int_{0}^{1} f \left ( \frac{x+k}{n} \right ) g(x) \, {\rm d}x\\ &= \frac{1}{n} \sum_{k=0}^{n-1} \int_{0}^{1} f \left ( \frac{k}{n} \right ) g(x) \, {\rm d}x\\ &= \frac{1}{n} \sum_{k=0}^{n-1} f \left ( \frac{k}{n} \right )\int_{0}^{1} g(x) \, {\rm d}x \end{align*}$

because for large enough $n$ by uniform continuity of $f$ and boundness of $g$ we can get $\displaystyle f\left(\frac{x+k}{n} \right)g(x)$ in an $\epsilon$ region of $\displaystyle f \left(\frac {k}{n}\right)g(x)$ . Taking limits the result follows.