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I met a problem like following:

If $E\subseteq \mathbb{R}$ is Lebesgue measurable and its measure is finite. Show that $$\lim_{n\to\infty}\int_Ee^{inx}\,\mathrm{d}x=0.$$ A brutal attempt of using the Dominated Convergence Theorem fails as we do not have pointwise convergence. I also tried the change of variable $y=nx$, but this only gives an upper bound of the norm of the integral which is strictly positive (not useful).

Can anyone give me some hint or suggestion on this problem? Thank you very much!

OnoL
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2 Answers2

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If $E$ is Lebesgue measurable, then for any $\epsilon\gt0$, we can find a finite, disjoint union of open intervals $U$ such that $|U\setminus E|+|E\setminus U|\le\epsilon$.

For each interval $(a,b)$, $$ \begin{align} \left|\,\int_a^be^{inx}\,\mathrm{d}x\,\right| &=\frac1n\left|\,e^{inb}-e^{ina}\,\right|\\ &\le\frac2n\tag{1} \end{align} $$ If $N$ is the number of intervals in $U$, then $$ \left|\,\int_Ee^{inx}\,\mathrm{d}x\,\right| \le\frac{2N}{n}+\epsilon\tag{2} $$ Thus, $$ \limsup_{n\to\infty}\left|\,\int_Ee^{inx}\,\mathrm{d}x\,\right|\le\epsilon\tag{3} $$ and since $(3)$ is true for arbitrary $\epsilon\gt0$, $$ \lim_{n\to\infty}\int_Ee^{inx}\,\mathrm{d}x=0\tag{4} $$

robjohn
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This is a special case of the Riemann-Lebesgue lemma, the proof of which is contained in the linked-to article.

Igor Rivin
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  • While the linked article proves the Riemann-Lebesgue Lemma, links can go stale, and this renders link-only answers less useful. It is better to describe a bit about the contents of the link in order to make the answer more self-contained. – robjohn Dec 10 '13 at 07:42
  • @robjohn fair enough, but I would have thought that "Riemann-Lebesgue Lemma" will bring up useful info in perpetuity... – Igor Rivin Dec 10 '13 at 13:06
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    That is most likely. However, even stating the Riemann-Lebesgue Lemma would make your answer more self-contained, and might attract more upvotes :-) – robjohn Dec 10 '13 at 14:08