$\lim_{n \to \infty} [a_n \cos(nx) + b_n \sin(nx)] =0$ implies $\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = 0$ Assume $E \subset [0, 2\pi]$ has positive measure. Let $a_n$ and $b_n$ be real numbers such that$$\lim_{n \to \infty} [a_n \cos(nx) + b_n \sin(nx)]  =0$$for all $x \in E$. How do I see that$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = 0?$$
 A: First we show that $a_n$ and $b_n$ are bounded in order to apply the dominated convergence theorem below. Suppose $|a_n|$ is unbounded. For the sake of argument, take the subsequence where $|a_{n_k}| \to \infty$ and rename it $a_n$. By the density of $nx \mod 2 \pi$ in $[0, 2 \pi]$ for $x$ irrational, we can always pick an $n$ such that $|\cos nx -1 | < \epsilon$ and $|\sin nx| < 1/|b_n|$ as well. If $|b_n| \neq 0$ only finitely many times then we have $a_n \cos(n x) = a_n \cos(nx) + b_n \sin (nx)$ not converging to zero. Otherwise,
$$ a_{n_k} \cos(n_kx) + b_{n_k} \sin (n_k x) = a_{n_k} (1 + \epsilon_{n_k}) + b_{n_k} \epsilon^*_{n_k}$$
with $|\epsilon^*_{n_k} b_{n_k}| < 1$. From this we can conclude the expression doesn't converge to zero. A similar argument with show that $b_n$ must be bounded.
Square the expression, use $2 \sin x \cos x = \sin 2x$, and integrate over $[0, 2 \pi]$ to get
$$\int_0^{2 \pi} [a_n^2+b_n^2 + a_nb_n \sin (2n x)] dx = 2 \pi( a_n^2+b_n^2)$$
since $\int_0^{2 \pi} \sin (2n x) dx = 0 $. Now realize that taking the limit as $n \to \infty$ on both sides, we can pass the limit through the integral by the dominated convergence theorem. This means $a_n^2+b_n^2 \to 0$, from which it follows that $a_n,b_n \to 0$.
A: It is more convenient to use instead the condition:
$$f_n(x) = c_n e^{i n x} + d_n e^{-i n x}\to 0$$
For  $\epsilon > 0$ and $n$ natural define
$$E^{\epsilon}_n=\{ x \in E \ | \  |f_m(x)| < \epsilon \textrm{ for all } m \ge n \}$$
Note that for every $\epsilon> 0 $ $E$ is the increasing union of the subsets $E_n^{\epsilon}$.  Hence, for every $\epsilon > 0$ there exists $n_{\epsilon}$ so that 
$\mu ( E^{\epsilon}_{n_\epsilon}) > \frac{1}{2}\mu (E)$. We'll use this later
Recall that we have $|f_m(x) | < \epsilon$ for all $x \in E^{\epsilon}_n$ and $m \ge n$. Therefore, we also have 
$$|e^{\pm i m x} \cdot f_m (x)|< \epsilon$$ Integrating over $E^{\epsilon}_n$ we obtain
$$| \mu(E^{\epsilon}_n) \cdot c_m  + \int_{E^{\epsilon}_n} e^{-2 i m x} d x\cdot d_m| < \epsilon \mu(E^{\epsilon}_n) \\
|\int_{E^{\epsilon}_n} e^{2 i m x} d x \cdot c_m +   \mu(E^{\epsilon}_n) \cdot d_m   | < \epsilon \mu(E^{\epsilon}_n)$$
Fix now $\epsilon > 0$ and some $n \ge n_{\epsilon}$ ( so that $\mu(E^{\epsilon}_n) > \frac{1}{2} \mu(E)\ $).  By the  Riemann-Lebesgue lemma
we have 
$\int_{E_n^{\epsilon}} e^{\pm i m x} d x \to 0$ as $m \to \infty$
From the above we conclude $\limsup |c_m| \le \epsilon$, $\limsup |d_m| \le \epsilon$. 
Since $\epsilon> 0 $ was arbitrary we get $\lim c_n = \lim d_n = 0$
