Let $f \in L^2[0,1].$ The $n$th Fourier coefficient of $f$ is given by $$\hat f(n) = \int_0^1 f(t)e^{-2\pi int} \, dt$$ The Fourier exponentials $u_n(t) = e^{2\pi int}$ are orthonormal in $L^2[0,1].$
Using Bessel's inequality, we have $$\sum_n|\hat f(n)|^2=\sum_n\lvert \langle f,u_n\rangle \rvert^2 \leq \lVert f\rVert^2=\int_0^1 \lvert f \rvert^2 < \infty$$ so $\lvert \hat{f}(n)\rvert ^2 \to 0$ as $\lvert n \rvert \to \infty$ by Cauchy's criterion, which in turn implies that the $\hat{f}(n)$ go to zero.
I additionally want to show that for measurable $A \subset [0,1]$, we have that $$\int_Ae^{2\pi int} \to 0 \text{ as } n \to \infty$$ What is the correct way to approach this? Do we treat this as the Fourier transform of the characteristic function $\chi_A?$
Thank you!
\begin{align} & \int_Ae^{2\pi int} \, dt = \int_0^1 f(t) e^{2\pi int} \, dt \quad \text{if } f(t) = \begin{cases} 1 & \text{if } t\in A, \\ 0 & \text{if } t\notin A. \end{cases} \\[10pt] & \text{And } f\in L^2[0,1]. \end{align}