Does the convergence of $e^{ita_n}$ for all $t \in \mathbb{R}$ imply the convergence of $a_n$? Let $\{a_n\}_{n=1}^\infty \subseteq \mathbb{R}$ be a real sequence.
Is it true that the convergence of $\{e^{ita_n}\}_{n=1}^\infty$ for all $t \in \mathbb{R}$ implies the convergence of the sequence $\{a_n\}_{n=1}^\infty$?
 A: If the function $\varphi(t)=\lim_{n\to\infty}\exp(ita_n)$ is continuous, or even just continuous at $t=0$, then the $a_n$ converge, by (and this is ham-fisted) application of Levy's continuity theorem applied to the sequence of "random" variables $X_n$ for which $P(X_n=a_n)=1$.   Conversely, if $\varphi$ is not continuous, the sequence of distributions of $X_n$ is not tight, which means in this case that the $a_n$ are unbounded.
A: Here is a (simple?) proof that was also suggested by Conrad:
It is enough to show that $a_n$ is bounded, since then $a_n$ will have a convergent subsequence $a_{n_k} \to a$ and if $a_{m_k} \to b$ is any other converging subsequence, then $e^{it(a-b)} = 1$ for all $t \in \mathbb{R}$, which is only possible if $a = b$. 
Suppose therefore that $a_n$ is unbounded, i.e. there exists a subsequence $a_{n_k}$ such that $|a_{n_k}| \to \infty$. 
Let $g(t) = \lim_{k \to \infty} e^{ita_{n_k}}$. 
Then $g$ is measurable and $|g(t)| = 1$ for all $t \in \mathbb{R}$. 
By the Riemann-Lebesgue lemma and the dominated convergence theorem we have 
$$
\int_{\mathbb{R}} f(t)g(t) dt = \lim_{k \to \infty} \int_{\mathbb{R}} f(t)e^{ita_{nk}} dt = 0
$$
for any $f \in L^1$. 
Taking e.g. $f = \chi_{[0,1]}\bar{g}$ yields the contradiction.    
A: I don't think so, you could jump around any whole number of rotations on the circle ($\frac{2\pi k_n}{t}, k_n \in \mathbb Z$). 
It would make big difference for the angle $\{\phi + \frac{2\pi i n} {t}\}$ but $0$ difference for position on circle $\{e^{2\pi i n}\}$
