Prove $ \lim\limits_{n\to\infty}\int_0^1 f(x)g(nx)\,dx=\int_0^1 f(x)\,dx\int_0^1 g(x) \, dx $ Let $f$ and $g$ be a real valued continuous functions on $\mathbb{R}$ such that $f(x+1)=f(x)$ and $g(x+1)=g(x)$ for all $x\in \mathbb{R}$. Prove that
$$
\lim_{n\to\infty}\int_0^1 f(x)g(nx)\,dx=\int_0^1 f(x)\,dx\int_0^1 g(x)\,dx.
$$
 A: Let 
$$
R_n=\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{i}{n})g(x)dx
$$
First we prove that
$$
\lim\limits_{n\to\infty}R_n=\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx
$$
Since $f(x)$ is uniform continuous on $[0,1]$, $\forall \epsilon>0,\space\exists N>0, \forall n>N$, let $\delta=\dfrac{1}{n},\space \forall x_1,x_2\in [0,1], |x_1-x_2|<\delta$, that $|f(x_1)-f(x_2)|<\epsilon. \space$ So there is
\begin{align}
\left|R_n-\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx\right|&=\left|\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{i}{n})g(x)dx-\int_{0}^{1}f(x)dx\int_{0}^{1}g(x)dx\right|
\\
&=\left|\dfrac{1}{n}\sum\limits_{i=0}^{n-1}f(\dfrac{i}{n})\int_{i}^{i+1}g(x)dx-\dfrac{1}{n}\int_{0}^{n}f(\dfrac{x}{n})dx\int_{0}^{1}g(x)dx\right|
\\
&=\left|\dfrac{1}{n}\sum\limits_{i=0}^{n-1}f(\dfrac{i}{n})\int_{0}^{1}g(x+i)dx-\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{x}{n})dx\int_{0}^{1}g(x)dx\right|
\\
&=\left|\dfrac{1}{n}\left(\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}\left(f(\dfrac{i}{n})-f(\dfrac{x}{n})\right)dx\right)\int_{0}^{1}g(x)dx\right|
\\
&\leqslant\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}\left|f(\dfrac{i}{n})-f(\dfrac{x}{n})\right|dx\int_{0}^{1}|g(x)|dx
\\
&\leqslant M\epsilon \hspace{20 mm} \left(M=\int_{0}^{1}|g(x)|dx\right)
\end{align}
Next
\begin{align}
\left|\int_{0}^{1}f(x)g(nx)dx-R_n\right|&=\left|\dfrac{1}{n}\int_{0}^{n}f(\dfrac{x}{n})g(x)dx-\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\frac{i}{n})g(x)dx\right|
\\
&=\left|\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\dfrac{x}{n})g(x)dx-\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}f(\frac{i}{n})g(x)dx\right|
\\
&=\left|\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}\left(f(\dfrac{x}{n})-f(\frac{i}{n})\right)g(x)dx\right|
\\
&\leqslant\dfrac{1}{n}\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}\left|f(\dfrac{x}{n})-f(\frac{i}{n})\right||g(x)|dx
\\
&\leqslant M\epsilon \hspace{20 mm} \left(M=\sum\limits_{i=0}^{n-1}\int_{i}^{i+1}|g(x)|dx=\int_{0}^{1}|g(x)|dx\right)
\end{align}
So we have
$$
\lim_{n\to\infty}\int_0^1 f(x)g(nx)\,dx=\int_0^1 f(x)\,dx\int_0^1 g(x)\,dx.
$$
