Why $\bigcap_{t>0}\left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)$ is a singleton? I had to prove that if the characteristic function $\chi_X$ of a r.v. $X$ is s.t. $|\chi(t)|=1$ for all $t>0$, then $X$ is constant. So the proof goes as follow :
We have that $$|\mathbb E e^{itX}|=1,$$
and thus that $$\mathbb E[e^{itX}]=e^{i\theta (t)},$$
for some function $\theta$. Moreover, we can prove that $$e^{itX-i\theta (t)}=1,\quad a.s.$$
and thus  $$tX-\theta (t)\in 2\pi \mathbb Z,\quad a.s.$$
for all $t$, i.e. $$X=\frac{\theta (t)}{t}+\frac{2\pi \mathbb Z}{t},$$
for all $t$.
Until here, I agree. But I don't understand the rest. They say, take $t_1,t_2$ rationally independent, i.e. there are no $q\in\mathbb Q$ s.t. $t_1=qt_2$. Since $$\frac{2\pi}{t_1}\mathbb Z\cap \frac{2\pi}{t_2}\mathbb Z=\emptyset,$$
(which I agree), we have that $$\left(\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z\right)\cap\left(\frac{\theta (t_2)}{t_2}+\frac{2\pi }{t_2}\mathbb Z\right),$$
has at most one element. Therefore, there is $k\in\mathbb Z$ s.t. $$X=\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z=\theta (1)+2\pi k$$
for some $k\in\mathbb Z$.

Question I don't understand the conclusion : why $$\left(\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z\right)\cap\left(\frac{\theta (t_2)}{t_2}+\frac{2\pi }{t_2}\mathbb Z\right)$$
has at most one element ? And why it implies that $$\bigcap_{t>0}\left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)=\{\theta (1)+2\pi k\}$$
for some $k$ ? (They took $t=1$). Because it could be also be empty, no ?
 A: Let $x_1=\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}k_1=\frac{\theta (t_2)}{t_2}+\frac{2\pi}{t_2}k_2$ and $x_2=\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\ell_1=\frac{\theta (t_2)}{t_2}+\frac{2\pi}{t_2}\ell_2$ be two elements of
$$\left(\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z\right)\cap\left(\frac{\theta (t_2)}{t_2}+\frac{2\pi }{t_2}\mathbb Z\right).$$
Then
$$
x_2-x_1=\frac{2\pi}{t_1}(k_1-\ell_1)=\frac{2\pi}{t_2}(k_2-\ell_2)
$$
hence $(k_1-\ell_1)t_2=(k_2-\ell_2)t_1$. The "rational independence" between $t_1$ and $t_2$ thus implies that $k_1=\ell_1$ and $k_2=\ell_2$, and hence $x_1=x_2$. This implies the first statement. 
From now on we fix $t_1=1$ and $t_2>0$ irrational. We have, almost surely
$$
X\in \left(\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z\right)\cap\left(\frac{\theta (t_2)}{t_2}+\frac{2\pi }{t_2}\mathbb Z\right).
$$
This implies that the above intersection is non-empty (there exists at least one $\omega$ such that $X(\omega)$ is in the intersection). Since it also contains at most one element, it contains exactly one element, say $x_0$.
For all $t>0$, almost surely, we have $X\in \left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)$. Hence, almost surely, 
$$
X\in \left(\frac{\theta (t_1)}{t_1}+\frac{2\pi}{t_1}\mathbb Z\right)\cap\left(\frac{\theta (t_2)}{t_2}+\frac{2\pi }{t_2}\mathbb Z\right)\cap \left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)=\lbrace x_0\rbrace \cap \left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right).
$$
In particular, the above intersection is non-empty, and hence  $x_0\in \left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)$ for all $t>0$.
We conclude that the intersection of the $\left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)$ is non-empty and contains at most one element. You can pick the unique element in any of the $\left(\frac{\theta (t)}{t}+\frac{2\pi}{t}\mathbb Z\right)$, for instance $t=1$.
