A problem similar to $L^2$ Fourier transform, but in the setting of Borel measure. Problem: Let $\mu$ be a finite Borel measure on the real axis, supported on a countable set $\mathbb{Q} \subset \mathbb{R}$ (I'm not sure whether here $\mathbb{Q}$ is all rational numbers ).
And let $$
F(t) = \int_{-\infty}^{\infty}\, e^{ixt}d \mu(x)$$ denote its Fourier transform. Show that
$$ \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T}\, |F(t)|^2dt = \sum_{q \in Q} |\mu (q)|^2
$$
My thought: First of all, the desired identity seems to do with $L^2(\mathbb{R})$ Fourier transform. In this original setting, we have a good space $\mathcal{S}(\mathbb{R})$  over which Fourier transform is very nice, and hence we are allowed to apply density argument. However, here we deal with Borel measure, so perhaps all things are different. 
 A: I am not sure what the author of the exam question meant by "supported on a countable set." A purely atomic finite Borel measure $\mathbb{R}$ necessarily has at most countably many atoms. Since $\mu$ is a finite Borel measure on a separable metric space, every atom of $\mu$ contains a singleton of positive measure. Whence, the set $Q:=\left\{x_{k}\in\mathbb{R}: a_{k}:=\mu(\left\{x_{k}\right\})>0\right\}$ is at most countable. If the author meant support as defined in the Wikipedia article, then this would exclude measures like $\sum_{k}2^{-k}\delta_{x_{k}}$ if $Q$ is a countable dense set in $\mathbb{R}$ (e.g. $Q=\mathbb{R}$).
Assuming that $\mu$ is purely atomic, we follow T.A.E.'s suggestion. Write $$\mu=\sum_{k}a_{k}\delta_{x_{k}},\quad Q:=\left\{x_{k}\in\mathbb{R}: a_{k}:=\mu(\left\{x_{k}\right\})>0\right\}\tag{1}$$
and observe that $\sum_{k}\left|a_{k}\right|<\infty$, since $\mu$ is finite. It follows from approximating $e^{ixt}$ by simple functions that $F(t)=\sum_{k}a_{k}e^{ix_{k}t}$, whence
$$\left|F(t)\right|^{2}=F(t)\overline{F(t)}=\left(\sum_{k}a_{k}e^{ix_{k}t}\right)\left(\sum_{j}\overline{a_{j}}e^{-ix_{j}t}\right)=\sum_{k=j}\left|a_{k}\right|^{2}+\sum_{k\neq j}a_{k}\overline{a_{j}}e^{i(x_{k}-x_{j})t}\tag{2}$$
Integrating over the interval $[-T,T]$ and scaling by $1/2T$, we obtain
$$\dfrac{1}{2T}\int_{-T}^{T}\left|F(t)\right|^{2}dt=\sum_{k}\left|a_{k}\right|^{2}+\sum_{k\neq j}\dfrac{a_{k}\overline{a_{j}}}{2T}\int_{-T}^{T}e^{i(x_{k}-x_{j})t}dt\tag{3}$$
where, by dominated convergence or Fubini, we can interchange summation and integration since $\sum_{k\neq j}\left|a_{k}a_{j}\right|<\infty$. Direct computation shows that
$$\lim_{T\rightarrow\infty}\dfrac{1}{2T}\int_{-T}^{T}e^{i(x_{k}-x_{j})t}dt=0,\qquad k\neq j\tag{4}$$
Appealing to the dominated convergence theorem, we can compute the limit as $T\rightarrow\infty$ term-by-term to obtain the desired conclusion.
