an equality related to the point measure 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..

Try to show that $\int_{-\delta}^\delta e^{in\theta}d\rho(\theta)\rightarrow \rho(\{0\})$, a kind of generalized Riemann Lebesgue Lemma. Then your result will follow by the fact that you are taking a Cesaro average of a sequence that converges. I believe you need some kind of sigma finite condition on your $\rho$ for this to work though.
