So, I have to show that if $E \subset [0, 2\pi]$ is a set of finite measure, then $\lim_{n \to \infty} \int_E \cos(nx) = \lim_{n \to \infty} \int_E \sin(nx) = 0$
My original plan of action was to observe the following:
$$\left|\int_E e^{inx} \,dx\right| = \left|\frac{e^{inx}}{in}\right| = \frac{1}{|n|}$$
The integral would then go to $0$ as $|n| \to \infty$, and since $\cos(nx), \sin(nx)$ are the real and imaginary parts of $e^{inx}$ we would be done. However, I realized that I didn't really use the hypothesis regarding $E$ being a set of finite measure, so I don't think my proof is correct. However, I would like some advice as to how to proceed in actually proving the above statements.
It's redundant to say $E$ has finite measure since its measure is no greater than the measure of $[0,2\pi]$, which is $2\pi$. In any case, since the indicator function $1_E$ is Lebesgue integrable on $[0,2\pi]$ you can fix a positive number $\epsilon$ and choose a smooth function $g$ on $[0,2\pi]$ for which $\sup_{x\in [0,2\pi]} \lvert 1_E(x) - g(x)\rvert < \frac{\epsilon}{2\pi}$. Let $h(x) = 1_E(x) - g(x)$ and show that $\int_0^{2\pi} h(x) e^{inx}\, dx$ is bounded by $\epsilon$. By integration by parts, show that $\int_0^{2\pi} g(x)e^{inx}\, dx = O(\frac{1}{n})$ as $n\to \infty$. Using those two estimates show that $\int_E e^{inx}\, dx = \int_0^{2\pi} 1_E(x)e^{inx}\, dx \to 0$ as $n\to \infty$. The result follows from taking real and imaginary parts.