Consider a finite Borel measure $\mu$ on $\mathbb R$. The Fourier transform $\hat{\mu}$ of $\mu$ is defined by $\hat{\mu} (\xi)= \int _{\mathbb R} e^{-ix\xi} d\mu(x)$. I would like to prove the following identity:
$$\lim_{R\to \infty}\frac{1}{2R}\int_{-R}^R {|\hat{\mu}(\xi)|}^2d\xi=\sum_{x\in \mathbb R} \mu({\{x\}}) ^2$$.
My attempt:
By Jensen's inequality,
$$\int_{-R}^R {|\hat{\mu}(\xi)|}^2d\xi=\int_{-R}^R|\int _{\mathbb R} e^{-ix\xi} d\mu(x)|^2d\xi\leq\int_{-R}^R\int _{\mathbb R} e^{-2ix\xi} d\mu(x)d\xi$$
By Fubini,
$$\int_{-R}^R\int _{\mathbb R} e^{-2ix\xi} d\mu(x)d\xi=\int _{\mathbb R}\int_{-R}^R\ e^{-2ix\xi} d\xi d\mu(x)$$ But I don't know how to proceed from here. Any advice will be appreciated.
Using Fubini's theorem is a good idea, but Jensen's inequality not so much if you want to establish an equality. Instead, you should use $$ |\hat\mu(\xi)|^2=\hat\mu(\xi)\overline{\hat\mu(\xi)}=\left(\int_{\mathbb{R}}e^{-ix\xi}\,d\mu(x)\right)\left(\int_{\mathbb{R}}e^{iy\xi}\,d\mu(y)\right)=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{i(y-x)\xi}\,d\mu(x)\,d\mu(y). $$ Now you can use Fubini's theorem to get $$ \int_{-R}^R|\hat\mu(\xi)|^2\,d\xi=\int_{\mathbb{R}^2}\int_{-R}^Re^{i(y-x)\xi}\,d\xi\,d(\mu\otimes\mu)(x,y). $$ If $y\neq x$, the inner integral is $$ \int_{-R}^Re^{i(y-x)\xi}\,d\xi=\frac{2\sin(y-x)R}{y-x}. $$ If $y=x$, we simply get $2R$. Thus $$ \frac 1{2R}\int_{-R}^R|\hat\mu(\xi)|^2\,d\xi=\int_{\mathbb{R}^2}\frac{\sin(y-x)R}{(y-x)R}1_{\mathbb{R}^2\setminus\Delta}(x,y)\,d(\mu\otimes \mu)(x,y)+\int_{\mathbb{R}^2}1_{\Delta}(x,y)\,d(\mu\otimes \mu)(x,y) $$ where $\Delta$ denotes the diagonal of $\mathbb{R}^2$, i.e., $\Delta=\{(x,y)\in\mathbb{R}^2\mid x=y\}$.
By the dominated convergence theorem, the first term goes to zero. For the second term we can once again apply Fubini's theorem: $$ \int_{\mathbb{R}^2}1_{\Delta}(x,y)\,d(\mu\otimes\mu)(x,y)=\int_{\mathbb{R}}\int_{\{x\}}d\mu(y)\,d\mu(x)=\int_{\mathbb{R}}\mu(\{x\})\,d\mu(x)=\sum_{x\in\mathbb{R}}\mu(\{x\})^2. $$ Note that $\{x\in\mathbb{R}\mid \mu(\{x\})\neq 0\}$ is countable for a finite measure, that's why one can replace the integral by a sum in the last step.