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