Prove $\int_A {\cos nxdx} \to 0$ as $n \to \infty $ on any measurable subset $A$ of $[0,2\pi]$

Question

Let $A$ be a measurable subset of $[0,2\pi]$, show that $\int_A {\cos nxdx} \to 0$ as $n \to \infty $.

My attempt is that since $A$ is measurable, then $\forall \epsilon>0$ there exists an open set $G$ such that $A\subseteq G$ and $|G-A|<\epsilon$ where $|.|$ denotes Lebesgue measure.

Now $G$ can be written as countable union of disjoint open intervals and for each such open interval $(a,b)$ we have $\int_a^b {\cos nxdx} = \frac{1}{n}[\sin (nb) - \sin (na)] \to 0$ as $n \rightarrow \infty$, thus it is proved that $\int_A {\cos nxdx} \to 0$ as $n \to \infty $ on set $G$.

Then I got stuck. There is a set $G-A$ left with arbitrary small measure $\epsilon$. I also feel that my attempt might not be correct... I just realized the sum of countable infinitesimal might not be zero...

Hope someone can help. Thank you!

Answer 1

The problem can be solved along those lines: find a finite union $G' \subset G$ of (say) $N$ intervals such that $|G-G'| < \epsilon$, and then $\left|\int_A \cos nx \, dx\right| < 2\epsilon + 2N/n$, which is $<3 \epsilon$ once $n > 2N/\epsilon$; since $\epsilon$ was an arbitrary positive number we're done.

A classic alternative approach is to use the orthogonality of the functions $f_n := \cos nx$ with respect to the inner product $\langle f, g \rangle := \int_0^{2\pi} f(x) g(x) \, dx$. Let $\chi$ be the characteristic function of $A$; we're to prove that $\langle \chi, f_n \rangle \rightarrow 0$ as $n \rightarrow \infty$. Since $\langle \chi, \chi \rangle \leq 2\pi$, while $\langle f_n, f_n \rangle = \pi$ for each $n>0$, we have $\sum_{n=1}^N \langle \chi, f_n \rangle^2 \leq 2\pi^2$ for every $N$. Hence for every $\epsilon>0$ there are at most $2\pi^2/\epsilon^2$ choices of $n$ for which $\langle \chi, f_n \rangle \geq \epsilon$; therefore, $|\langle \chi, f_n \rangle| < \epsilon$ once $n$ is large enough, QED.