For any positive measure $\rho$ on $[-\pi, \pi]$, prove the following equality:
$\lim_{N\to\infty}\int_{-\pi}^{\pi}\frac{\sum_{n=1}^Ne^{in\theta}}{N}d\rho(\theta)=\rho(\{0\}).$
Remark:
It is easy to check that for any fixed positive number $0<\delta<\pi$, then $|\int_{\delta}^{\pi}\frac{\sum_{n=1}^Ne^{in\theta}}{N}d\rho(\theta)|\leq \int_{\delta}^{\pi}\frac{2}{2sin(\frac{\delta}{2})N}d\rho(\theta)\to 0\text{ as}\;N\to\infty,$
so I think we have to show that
$\int_{-\delta}^{\delta}\frac{\sum_{n=1}^Ne^{in\theta}}{N}d\rho(\theta)\sim \int_{-\delta}^{\delta}d\rho(\theta)\sim \rho(\{0\})?$
Maybe we also have to choose $\delta=\delta(N)$ etc..