Inner product of function of period $2\pi$ with exponential

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous with period $2\pi$. Prove that $$\lim_{N\rightarrow\infty}\dfrac{1}{N}\sum_{j=1}^Nf\left(\dfrac{2\pi j}{N}\right)e^{-2\pi ijn/N}=\dfrac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx.$$

The right-hand side is the inner product $\langle f,e^{inx}\rangle$. What can we use to relate it to the left-hand side?

Answer 1

Let $g(x)=f(x)e^{-inx}$. Multiply both sides by $2\pi$. The statement becomes $$ \lim_{N\rightarrow\infty}\dfrac{2\pi}{N}\sum_{j=1}^N g\left(\dfrac{2\pi j}{N}\right) = \int_0^{2\pi}g(x) dx $$ which comes straight out of the definition of Riemann integral. The function $g$ is Riemann integrable because it's continuous.

The periodicity of $f$ does not matter.