Prove that if the measure P is discrete, then its characteristic function $\varphi(\lambda)$ does not tend to zero as $\lambda \to \infty$ Prove that if the measure P is discrete, then its characteristic function
$\varphi(\lambda)$ does not tend to zero as $\lambda \to \infty$
I don't know how to give a proof. But I try some discrete random variables and I find the characteristic function does not go to 0. For example, $X=1$ or $X=-1$ with probabilities $1/2$. Then I find the characteristic function us $\cos t$, which is periodic. Thus it cannot go to 0. However, I don't know how to prove the general statement above.
 A: Suppose $X$ has characteristic function $\phi$.
Claim. Let $U_n\sim\mathcal U([-n,n])$. Then $\mathbb P(X=0)=\lim_{n\to\infty}\mathbb E[e^{iU_nX}]$.
Proof. Indeed, first see that
$$\mathbb E[e^{iU_nX}\mid X]=\int e^{iuX}\frac{1}{2n}\mathbf1_{[-n,n]}(u)\mathop{}\!\mathrm{d}u=\begin{cases}
\frac{e^{inX}-e^{-inX}}{2niX} & \text{if }X\neq0, \\
1 & \text{if }X=0.
\end{cases}$$
Let $Y_n=\frac{e^{inX}-e^{-inX}}{2niX}\cdot\mathbf1_{\{X\neq0\}}$. Then $Y_n\to0$ a.s. so we can use DCT to get
$$\mathbb E[e^{iU_nX}]=\mathbb E\left[\mathbb E[e^{iU_nX}\mid X]\right]=\mathbb E[Y_n]+\mathbb P(X=0)\to\mathbb P(X=0),$$
as claimed. $\square$
Now by translation, equivalently we have for any $a\in\mathbb R$ that
$$\mathbb P(X=a)=\lim_{n\to\infty}\frac{1}{2n}\int_{-n}^ne^{-iua}\phi(u)\mathop{}\!\mathrm{d}u.$$
It suffices to show that assuming $\phi(u)\to0$ as $|u|\to\infty$ implies that the RHS is $0$ (and so $X$ cannot be discrete). Given $\epsilon>0$, pick $R$ s.t. $|\phi(u)|<\epsilon$ for $|u|>R$. Then
\begin{align*}
\left|\int_{-n}^ne^{-iua}\phi(u)\mathop{}\!\mathrm{d}u\right|\leq\int_{-n}^n|\phi(u)|\mathop{}\!\mathrm{d}u\leq2R+2\epsilon|n-R|.
\end{align*}
So dividing by $2n$ and taking the limit,
$$\lim_{n\to\infty}\left|\frac{1}{2n}\int_{-n}^ne^{-iua}\phi(u)\mathop{}\!\mathrm{d}u\right|\leq\lim_{n\to\infty}\frac{2R+2\epsilon|n-R|}{2n}=\epsilon.$$
Now take $\epsilon\to0$ and we are done.
