Here's my proof that moment generation function (if exists) uniquely determines characteristic function.
Can you please see how to make it more rigorous or improve in either way (e.g. by citing relevant well-known theorems). Thank you very much - any suggestions are welcome!
Theorem. Let $F(X)$ be a probability distribution of $X$ and suppose the MGF, $M(t)$ exists in an open ball centered at $t = 0$. Then, the characteristic function is uniquely determined by $M(t)$ alone. In other words, if $M(t)=\sum_{x=0}^{\infty} \frac{t^n \mu_n}{n!}$ has a positive radius of convergence about $0$, there is a unique characteristic function associated with it. \newline
Proof: The theorem can be easily proven by using analytical continuation technique. Let's extend MGF on a ball $R = \{x | |x| <r\} , r \in \mathbb{R}$ to a strip $S =\{z | |Re(z)| < r\}$ on complex plain. it is easy to see that $M(z)$ is bounded on $S$:
$|M(z)| = | \int\limits_{\mathbb{R}} e^{zt} dF(t) | < \int\limits_{\mathbb{R}} |e^{zt}| dF(t) = M(x)$,
so using Fubini's theorem we can see that
$\int\limits_{\partial B} M(z) dz = \int\limits_{\partial B} \int\limits_{\mathbb{R}} e^{zt} dt dz = \int\limits_{\mathbb{R}}( \int\limits_{\partial B} e^{zt} dz ) dt = 0 $
on every open ball $B \subset S$, so $M(z)$ is analytical continuation on $S$.
Because the uniqueness of analytical continuation, if $M_1(x) = M_2(x), x \in R$, then $C_1(x) = M_1(-ix) = M_2(-ix) = C_2(x)$. Q.E.D.
