Let $g\in L_{\infty}(\mathbb{T})$. For any $-\pi<a<b<\pi$ let $\chi_{[a,b]}$ be the characristic function of $[a,b]$. Prove that
$$\lim_{n\to\infty}\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}\chi_{[a,b]}(t)g(nt)dt=\dfrac{(b-a)\hat{g}(0)}{2\pi}.$$
I see that $(b-a)/2\pi$ is exactly $\widehat{\chi_{[a,b]}}(0)$, so I am trying to make left hand side $\widehat{(\chi_{[a,b]}*g)}(0)$. Can someone help me?
$\frac{1}{2\pi}\int_{-\pi}^{\pi}\chi_{[a,b]}(t)g(nt)dt=\frac{1}{2n\pi}\int_{-n\pi}^{n\pi}\chi_{[a,b]}(u/n)g(u)du=\frac{1}{2n{\pi}}\int_{-n\pi}^{n\pi}\chi_{[na.nb]}(u)g(u)du$ $=\frac{1}{2n{\pi}}\int_{na}^{nb}g(u)du=\frac{b-a}{2n\pi(b-a)}\int_{na}^{nb}g(u)du$.
Renormalizing the orthogonal functions $e^{-2\pi imx}$ for $m\in\mathbb{Z}$ on $[na,nb]$, so we have an orthonormal basis, and considering the above computation tells us that
$\frac{1}{2\pi}\int_{-\pi}^{\pi}\chi_{[a,b]}(t)g(nt)dt=\frac{b-a}{2\pi} \hat{g}(0)$. This is true for all n as n goes to infinity so
$\lim_{n\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}\chi_{[a,b]}(t)g(nt)dt=\frac{b-a}{2\pi} \hat{g}(0)$.