How to argue that $X$ is a discrete variable?

Let $$X$$ be a random variable on the probabality space $$(\Omega, \mathcal{F}, P)$$. Let $$t_0 \in \mathbb{R} \setminus \{0\}$$ such that $$\varphi_X(t_0) = 1$$. Then I have to show that $$X$$ is discrete.

My attempt:

\begin{align*} 1 = \varphi_X(t_0) = \mathbb{E}[e^{it_0 X}] = \mathbb{E}[\cos(t_0 X)] + i \mathbb{E}[\sin(t_0 X)] = \mathbb{E}[\cos(t_0 X)] \end{align*}

Thus since $$\cos(t_0X)$$ at most can be $$1$$ then $$1 = \cos(t_0X)$$ almost sure. However, $$1 = \cos(t_0X) \iff Xt_0 \in \{2p \pi \mid p \in \mathbb{Z}\}$$. Is this enough to conclude that $$Xt_0$$ is discrete since $$Xt_0$$ is in a countable subset of $$\mathbb{R}$$? Thus, since $$t_0 \neq 0$$, $$X = \frac{1}{t_0} t_0 X$$ must also be discrete. Is this ok?

You arrive at the fact that $$\operatorname{img}(X)\subset \frac{2\pi}{t_0}\mathbb{Z}$$ and $$\frac{2\pi}{t_0}\mathbb{Z}$$ is a discrete set, so we are done.
• But what allows the asker to consider that $\mathbb{E}(\sin t_0X)=0$ ? Commented Jun 13, 2022 at 14:20
• @JeanMarie if you have $1=a+ib$, where $a,b\in \mathbb{R}$ then $1=a$ and $b=0$ Commented Jun 13, 2022 at 14:25
• Hi Masacrosco. Thanks for the comment. In the next task I am asked to prove that $X$ is discrete where I now know that $|\varphi_X(t_0)| = 1$. Now by definition there exists $\theta_0 \in (-\pi, \pi]$ such that $\varphi_X(t_0) = e^{-i \theta_0}$. Therefore, $$\varphi_{X + \theta_0/t_0}(t_0) = \mathbb{E}[e^{i t_0 (X + \theta_0 / t_0)}] = \varphi_X(t_0)e^{i \theta_0} = e^{-i \theta_0}e^{i\theta_0} = 1$$, thus $X + \theta_0/t_0$ is discrete from what I just showed. But how do I now conclude that $X$ is discrete? Commented Jun 13, 2022 at 14:28