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I'm trying to prove the following

Let $A$ be a measurable subset of $[0,2\pi]$

$$\lim_{n\to \infty} \int_A e^{inx} \, dx=0$$

There is a hint "this is the special case of the Riemann Lebesgue lemma"

But I have no idea how to approach this problem using the hint. Since the problem is in Hilbert space chapter, I have considered $e^{inx}$ as a element of orthnormal basis of $L^2$ space. But it's of no use I think.

Can I get some hints? Thank you:)

Michael Hardy
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user277793
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  • To apply the Riemann Lebesgue lemma, you just need that $f\in L^1(\mathbb{R})$ where the integrand is $f(x)e^{inx}$. In this case, $f(x) = \chi_A(x)$, which is $L^1$ for this $A$. – TrivialCase Dec 05 '15 at 16:55

1 Answers1

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The Riemann–Lebesgue lemma says that if $\displaystyle\int_{\mathbb R} |f(x)| \, dx < \infty$ then $\displaystyle\lim_{n\to\infty} \int_{\mathbb R} f(x) e^{inx}\,dx = 0$.

Since the set $A$ is bounded, we have $\displaystyle\int_{\mathbb R} 1_A(x)\,dx = (\text{the Lebesgue measure of A}) <\infty$ (where $1_A$ is the indicator function of $A$, taking the value $1$ at arguments in $A$ and $0$ elsewhere). Therefore, by the Riemann–Lebesgue lemma, $\displaystyle \lim_{n\to\infty} \int_{\mathbb R} 1_A(x) e^{inx}\,dx=0$, or, in other words $\displaystyle\lim_{n\to\infty} \int_A e^{inx} \, dx = 0$.

Michael Hardy
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  • Right. Thank you so much !! :) Then I'm wandering why this problem is in hilbert space chapter. Is Riemann Lebesgue lemma related to hilbert space? – user277793 Dec 05 '15 at 20:43