Let $B \subset [0,2\pi]$ be a Lebesgue measurable set. Prove that:
$\displaystyle \lim_{n \to \infty} \int_{B} \cos(nx) dx = 0$
OK I did this assuming B is an open interval, this is pretty easy using the fact that the sine function is bounded by 1. Now I'm stuck in the general case, I'm somewhat confused by "Lebesgue measurable" I know this means that the measure is given by the outer measure i.e the infimum of the sum of all measures of open covers of B. But I'm having trouble writing it, I get confused when working with Lebesgue measurable sets. Can you please help me?
Best Answer
hint
Re-write the integral as the following
$$ \int_0^{2\pi} \chi_B(x) \cos(nx) dx $$
where $\chi_B$ is the characteristic function of the set $B$ (so that it equals 1 on $B$ and 0 else where).
Now, you've already proven the case where $B$ is an open interval. Now take a sequence of decreasing coverings for $B$ by finitely many disjoint open intervals. For each covering, the result is true by what you've already shown. Take the limit using dominated convergence theorem (if $B \subset C$, then $|\chi_B \cos| \leq |\chi_C \cos|$).