If $A\subset[0,2\pi]$ is measurable, prove that
$$\lim_{n\to\infty}\int_A \cos (nx)\ dx=\lim_{n\to\infty}\int_A \sin(nx) \ dx=0$$
Please, any suggestions are welcome.
If $A\subset[0,2\pi]$ is measurable, prove that
$$\lim_{n\to\infty}\int_A \cos (nx)\ dx=\lim_{n\to\infty}\int_A \sin(nx) \ dx=0$$
Please, any suggestions are welcome.
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This an application of the Riemann Lebesgue Lemma to the indicator funciton $1_A(x)$. You can however prove it from scratch. Prove the result holds true for $A$ being an interval, then a union of disjoint intervals. Then use the fact that any measurable set can be approximated by open/closed sets from outside/inside.