Theorem   Let $\mu$ be a finite Borel measure on $R$, then 

$$\lim\limits_{T\to \infty}\frac{1}{2T}\int_{-T}^T|\widehat{\mu}(\xi)|^2d\xi=\sum_{x\in R}\mu(\{x\})^2,$$

where $\widehat{\mu}(\xi)$ is the Fourier transformation of the measure $\mu,$ i.e., $\widehat{\mu}(\xi)=\int e^{-2\pi i\xi x}d\mu(x).$

Proof. Observe that $|\widehat{\mu}(\xi)|^2=\widehat{\mu}(\xi)\cdot \overline{\widehat{\mu}(\xi)}=\int_{R^2}e^{-2\pi i \xi (x-y)}d\mu(x)d\mu(y).$

Let $K_T(x,y)=\frac{1}{2T}\int_{-T}^Te^{-2\pi i \xi (x-y)}d\xi,$ then 

$$ K_T(x,y)= \begin{cases}
\frac{e^{2\pi iT(x-y)}-e^{-2\pi iT(x-y)}}{4Ti(x-y)} ,& x\not=y\\
1, & x=y
\end{cases} $$

Moreover, $\lim\limits_{T\to \infty}K_T(x,y)=\begin{cases}
0 ,& x\not=y\\
1, & x=y
\end{cases}$

Since $|K_T(x,y)|\le 1$, by the dominated convergence theorem we have that for all $x$,

$$\lim\limits_{T\to \infty}\int_RK_T(x,y) d\mu(y)=\mu(\{x\}).$$

Clearly, 

$$\frac{1}{2T}\int_{-T}^T|\widehat{\mu}(\xi)|^2d\xi=\int_{R^2}K_T(x,y)d\mu(x)d\mu(y).$$

It follows from Fubini‘s theorem,

$$\int_{R^2}K_T(x,y)d\mu(x)d\mu(y)=\int_R\mu(\{x\})d\mu(x)=\sum_{x\in R}\mu(\{x\})^2.$$

On a variant of Wiener's lemma