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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?

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    @Marc: Well, for the Lebesgue integral to even be defined you need the domain of integration to be Lebesgue measurable...2010-11-03
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    Can I proceed like this? if B = (a,b) then using the FTC we can easily see the limit is zero. Now assume B is the disjoint union of open intervals, so the integral over B is equal: $\sum_{j=1}^{\infty} \int_{(a_{j},b_{j})} f$ and then?2010-11-03

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