Show that $\int_{\mathbb{R}} |\mathbb{E}(e^{i \xi X})|^2 \, d\xi < \infty$ implies $P(X=x)=0, \forall x\in\mathbb{R}$. 
Let $X$ be a real-valued random variable with $\varphi_X\in L^2(\mathbb{R})$, where $\varphi_X$ denotes the characteristic function of $X$. Prove that $P(X=x)=0, \forall x\in\mathbb{R}$. 

Since $\varphi_X\in L^2(\mathbb{R})$, I believe that implies that $X$ must be continuous, then we use that to conclude that $P(X=x)=0$ for all $x\in\mathbb{R}$. However, I am having a hard time formalizing this idea. Any help would be appreciated. 
 A: Hints: Show the following auxiliary statements:


*

*Let $\chi$ be the characteristic function of a real-valued random variable $Y$ with distribution $\mu$, i.e. $$\chi(\xi) = \mathbb{E}e^{i \xi Y} = \int_{\mathbb{R}} e^{i \xi y}  \mu(dy).$$ Apply Fubini's theorem to show that $$\frac{1}{2T} \int_{-T}^T \chi(\xi) e^{-ix \xi}  d\xi = \mu(\{x\}) + \int_{\mathbb{R} \backslash \{x\}} \frac{\sin(T(y-x))}{T(y-x)} \mu(dy).$$ Use the dominated convergence theorem to let $T \to \infty$: $$\lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T \chi(\xi) e^{-ix \xi} \, d\xi = \mu(\{x\}) = \mathbb{P}(Y=x).$$

*Let $\phi$ be the characteristic function of a real-valued random variable $X$. Show that $|\phi|^2$ is the characteristic function of the random variable $Y:=X-X'$ where $X' \sim X$ is an independent copy of $X$.


Now let $X$ be a random variable such that its characteristic function $\phi$ is square-integrable. By Step 2, we know that $|\phi|^2$ is the characteristic function of $Y=X-X'$ where $X' \sim X$ is an independent copy of $X$. Since
$$\lim_{T \to \infty} \left| \frac{1}{2T} \int_{-T}^T |\phi(\xi)|^2 e^{-ix \xi}  d\xi \right| \leq \lim_{T \to \infty} \frac{1}{2T} \int_{\mathbb{R}} |\phi(\xi)|^2 d \xi =0$$
it follows from Step 1 (applied to $\chi=|\phi|^2$) that
$$\mathbb{P}(Y=x) = 0$$
for all $x \in \mathbb{R}^d$. Since $Y=X-X'$ for $X \sim X'$ independent, this implies that $\mathbb{P}(X=x)=0$ for all $x \in \mathbb{R}^d$. 
