Let $A\subset [-\pi, \pi]$ be a measurable set. Show that $\lim_{k\to \infty}\int_A \cos(kt) \text{d}t=0. $

Question

This is a homework problem, however I don't know how to start with. Let $A\subset [-\pi, \pi]$ be a measurable set. Show that $$\lim_{k\to \infty}\int_A \cos(kt) \text{d}t=0. $$ I was thinking theorems like dominated convergence theorem etc. but don't know how to apply them. I read somewhere that this can be shown with Riemann-Lebesgue lemma, but we haven't proved that in class. What we recently proved was that the functions $\{\frac{1}{\sqrt{2\pi}}e^{ikt}:k\in\mathbb{Z}\}$ form a Hilbert basis in the space of square-integrable $2\pi$-periodic functions $L^2_{2\pi, \mathbb{C}}([-\pi, \pi])$. Any hints or help is appreciated.

Answer 1

Here is the (detailed) solution to the question. The integral can be written as $$\int_A\cos(kt)\text{d}t=\sqrt{\pi}\int_{[-\pi, \pi]}\chi_A\frac{\cos(kt)}{\sqrt{\pi}}\text{d}t=\sqrt{\pi}\Big\langle\chi_A, \frac{\cos(kt)}{\sqrt{\pi}}\Big\rangle. $$ Let's recap the Bessel's inequality $$\sum_{e\in E}|\langle x, e\rangle|^2\leq \lVert x\rVert^2$$ for any orthonormal set $E$. We know that the set $$\Big\{\frac{1}{\sqrt{\pi}}\cos(kt):k\in\mathbb{N}\Big\}\cup\Big\{\frac{1}{\sqrt{\pi}}\sin(kt):k\in\mathbb{N}\Big\}\cup\Big\{\frac{1}{\sqrt{2\pi}}\Big\}$$ is an orthonormal set in $L^2_{\mathbb{R}}([-\pi, \pi])$. Let $\epsilon>0$. Then $$\Big|\Big\langle \chi_A, \frac{\cos(kt)}{\sqrt{\pi}}\Big\rangle\Big|\geq\frac{\epsilon}{\sqrt{\pi}}$$ for only finite number of $k$'s, because otherwise Bessel's inequality would give $\lVert\chi_A\rVert^2=\infty$. Therefore there exists $N\in\mathbb{N}$ s.t. $$\Big|\Big\langle \chi_A, \frac{\cos(kt)}{\sqrt{\pi}}\Big\rangle\Big|<\frac{\epsilon}{\sqrt{\pi}} \quad \forall k\geq N.$$ Thus we have $$\Big|\int_A\cos(kt)\text{d}t\Big|=\sqrt{\pi}\Big|\Big\langle \chi_A, \frac{\cos(kt)}{\sqrt{\pi}}\Big\rangle\Big|<\epsilon \quad \forall k\geq N $$ i.e. $$\lim_{k\to\infty}\int_A\cos(kt)\text{d}t=0. $$