Show that $\lim\limits_{n\to\infty}\int_{A}\sin(nx)dx = 0$ for any Lebesgue measurable $A \subseteq [0,1]$ I have to show that
$$\lim_{n\to\infty}\int_{A}\sin(nx)dx = 0$$
for any Lebesgue Measurable $A \subseteq [0,1]$
I have seen many solutions which invoke several theorems from real analysis that I do not study on my course (Riemann-Lebesgue lemma for example). I want to see different approaches to understand better the problem and try to solve this with my methods. I covered the Royden book up to chapter 5.
 A: HINT: Let $A$ measurable, $\mu(A) < \infty$.  Then for every $\epsilon>0$ there exists $B$ a finite union of intervals such that 
$$\mu(A \Delta B) < \epsilon$$
Indeed, first take $K$ compact so that $\mu(A\Delta K) < \epsilon/2$, then $U$ open, $U\supset K$, so that $\mu(K\Delta U) < \epsilon/2$. Now $U$ is a union of intervals, and finitely many of them will cover $K$, so take $B$ that finite cover.
Now one can reduce the problem  to the case $A$ a  finite interval, which is simple.
$\bf{Added:}$ One can show in this way that $\int_A \sin(\lambda x) dx \to 0$ as $\lambda \to \infty$.
A: Here is a proof avoiding analysis (basically reproving Bessel's inequality as mentioned by user363464). I will use these facts:


*

*There is an inner product
$$\langle f,g\rangle=\frac{1}{\pi}\int_{0}^{2\pi} f g$$
defined on $L_2([0,2 \pi])$.

*The functions $s_n$ defined by $s_n(x)=\sin nx$ for $n\geq 1$ are orthonormal: $$\langle s_n, s_m\rangle=\begin{cases}
0\text{ if }n\neq m\\
1\text{ if }n=m
\end{cases}$$
(Actually I won't use that $\langle.,.\rangle$ is positive definite, and it only has to operate on a vector space including $\chi_A$ and the functions $s_n$.)
Define $a_n=\langle \chi_A, s_n\rangle$. We need to show $|a_n|\to 0$.
For each $N$ we can define $g_N=\chi_A-\sum_{n=1}^Na_ns_n$. Then
$$\chi_A=a_1s_1+\dots+a_Ns_N + g_N$$
but these terms are all orthogonal, so
$$\langle\chi_A,\chi_A\rangle=\langle a_1s_1,a_1s_1\rangle+\dots+\langle a_Ns_N, a_Ns_N\rangle + \langle g_N,g_N\rangle$$
Using $\langle g_N,g_N\rangle\geq 0$ and $\langle s_n,s_n\rangle=1$ we get
$$\langle\chi_A,\chi_A\rangle\geq a_1^2+\dots+a_N^2$$
so $\sum_{n=1}^{\infty} a_n^2$ converges, which implies $a_n\to 0$.
