Show that $\lim\limits_{n\to\infty}\int_{A}\sin(nx)dx = 0$ for any Lebesgue measurable $A \subseteq [0,1]$

Question

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.

Answer 1

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 $\lambda \to \infty$.

Answer 2

Here is a proof avoiding analysis (basically reproving Bessel's inequality as mentioned by user363464). I will use these facts:

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

2. 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 $a_n\to 0$.