Convergence of a sequence I have to prove the following statement: 
Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of real numbers, such that $\exp(itX_n)$ converges for every $t\in\mathbb{R}$. Show that the sequence $(X_n)$ converges.
The problem I see is, that the complex logarithm is not continous. Hence, I have to work around, but I dont know how. I hope someone can help.
 A: Since I've been chewing on this for a while now, allow me to first construct a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent for all $t\in\mathbb Q$, before I shall solve the original problem below.
Counterexample for $\mathbb Q$ (incomplete space)
Let $(q_n)_{n\in\mathbb N}$ be an enumeration of $\mathbb Q\setminus 0$.
Observe that $\bigcap_{k=1}^n q_k\mathbb Z=r_n\mathbb Z$ for some $r_n\in \mathbb Q_{>0}$.
By letting $y_1=0$, and selecting $y_n\in r_n\mathbb Q$ with $y_n>y_{n-1}+1$ we obtain a diverging sequence $(y_n)_{n\in\mathbb N}$.
If $t\in\mathbb Q$, then $ty_n\in\mathbb Z$ for almost all $n$: If $t=0$, this is trivial; if $t\ne 0$,  we find $N$ with $\frac1t=q_N$ and have $y_n\in q_N\mathbb Z$ for all $n\ge N$.
Therefore, by letting $X_n=2\pi y_n$ we find a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent (in fact, is eventually constant $=1$) for all $t\in \mathbb Q$.
Proof for the case $\mathbb R$ (complete space)
What's different with $\mathbb R$ instead of $\mathbb Q$?
First let's see that $(X_n)_{n\in\mathbb N}$ is bounded:
For $c>0$ and $N\in\mathbb N$ consider the set
$$A_{c,N}=\{t\in\mathbb R\mid\forall n,m\ge N\colon t(X_n-X_m)\in[-c,c]+2\pi\mathbb Z\}.$$
If $t\notin A_{c,N}$, then there are $n,m\ge N$, $k\in\mathbb Z$ with $2k\pi+c<t(X_n-X_m)<2(k+1)\pi-c$, which also holds for $t'$ if
$$|t'-t|<\frac{\min\{t(X_n-X_m)-2k\pi-c,2(k+1)\pi-c-t(X_n-X_m)\}}{|X_n-X_m|}.$$
Therefore the complements of the $A_{c,N}$ are open and the $A_{c,N}$ themselves are closed.
The convergence of $(e^{itX_n})_{n\in\mathbb N}$ for all $t\in\mathbb R$ implies that for any $c>0$ we have
$$\mathbb R=\bigcup_{N\in\mathbb N}A_{c,N}$$
Specifically, we can consider $c=\frac{2\pi}5$.
By the Baire category theorem, there exists an $N$ such that $A_{c,N}$ contains an open interval $(t_0-\epsilon,t_0+\epsilon)$ with $\epsilon>0$.
Without loss of generality, $t_0\ne0$ and $\epsilon< |t_0|$.
Let $$M=\max\left\{|X_1|,\ldots,|X_N|\right\}+\frac\pi{\epsilon}.$$
Then $(X_n)_{n\in\mathbb N}$ is bounded by $M$. To prove this, assume that $|X_n|>M$ for some $n$.
Then clearly $n>N$ and $|X_n-X_N|>\frac\pi{\epsilon}$.
Because $n>N$, there exists $k\in\mathbb Z$ such that $|t_0(X_n-X_N)-2k\pi|\le c$. Let $t_1=t_0+\frac\pi{(X_n-X_N)}$ so that $|t_1-t_0|<\epsilon$ and hence $t_1\in A_{c,N}$.
Then there is $k'\in\mathbb Z$ with $|t_1(X_n-X_N)-2k'\pi|\le c$.
But $|t_1(X_n-X_N)-t_0(X_n-X_N)|=\pi$ makes this impossible because $2c<\pi<2\pi-2c$: the left inequality rules out $k=k'$, the other rules out $|k-k'|\ge 1$.
We conclude that $(X_n)_{n\in\mathbb N}$ is bounded.
Now the final step is easy: If $|X_n|\le M$ for all $n$, then consider the case $0<t<\frac\pi{2M}$. For such $t$, all $tX_n$ are in $(-\frac\pi2,\frac\pi2)$.
Since the map $(-\frac\pi2,\frac\pi2)\to \mathbb C$, $x\mapsto e^{ix}$ is an embedding, convergence of $(e^{itX_n})_{n\in\mathbb N}$ implies convergence of $(tX_n)_{n\in\mathbb N}$ and finally convergence of $(X_n)_{n\in\mathbb N}$.$_\blacksquare$
A: Here's a solution that takes advantage of a little Fourier analysis.  
Set $y_n = \int_{-\infty}^\infty e^{-t^2/2} e^{i t x_n}\,dt$.  Since $|e^{i t x_n}| = 1$ and $e^{-t^2/2}$ is integrable, the dominated convergence theorem implies that $y_n$ converges to some finite value $y$.  On the other hand, computing the integral above shows that $y_n = \frac{1}{\sqrt{2 \pi}} e^{-x_n^2/2}$.  (This is the fact that a Gaussian is its own Fourier transform.)  If we define $f : [0,\infty) \to \mathbb{R}$ by $f(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, we have $y_n = f(|x_n|)$.  Now $f$ is continuous and strictly decreasing, so it has a continuous inverse.  Hence $|x_n| = f^{-1}(y_n) \to f^{-1}(y)$.  Call this limit $x$.
If $x=0$ then we have $x_n \to 0$.  Otherwise, let $s_n$ be the sign of $x_n$ and consider $z_n := \sin(\frac{\pi}{2} \frac{x_n}{x})$.  On the one hand $z_n$ is the imaginary part of $e^{i \pi x_n / 2x}$ and hence $z_n$ converges to some $z$.  On the other hand, since $\sin$ is an odd function we have
$$s_n = \frac{z_n}{\sin(\frac{\pi}{2} \frac{|x_n|}{x})}.$$
Since $|x_n| \to x$, the denominator converges to 1, and so $s_n$ converges to some $s = \pm 1$.  Thus $x_n = s_n |x_n| \to s x$.
A: Actually, we don't need to require the converge of $\{e^{itx_n}\}$ for each $t$ in the real line, but only for $t\in A$, where $A$ is of positive Lebesgue measure.Let $l(t):=\lim_{n \to +\infty}e^{itx_n}$ for $t\in A$.


*

*We show that $\{x_n\}$ is bounded. Suppose not. 
By the dominated convergence theorem, for all $A'\subset A$, $\int_{A'}e^{itx_n}dt\to \int_{A'}l(t)dt$. But it also converges to $0$ by Riemann-Lebesgue lemma. 
We deduce that $l=0$ on $A'$, a contradiction. 

*Assume that for a subsequences of $\{x_n\}$, say $\{x_{n'}\}$ and $\{x_{n''}\}$ converge respectively to $x'$ and $x''$. We have to show that $x'=x''$. We have by dominated converge theorem that $\int_{A'}e^{itx'}dt=\int_{A'}e^{itx''}dt$ for all $A'\subset A$ measurable. This gives that $e^{itx'}=e^{itx''}$ for all $t\in A$. If $x'\neq x''$, then defining $S:=\frac 1{x'-x''}A$, we would have $e^{is}=1$ for $s$ is a set of positive Lebesgue measure, a contradiction, as $S$ should contain points which are not of the form $2k\pi$, $k\in\Bbb Z$. 
