TITLE: How to prove: Moment Generating Function Uniqueness Theorem
QUESTION [24 upvotes]: Many results are based on the fact of the Moment Generating Function (MGF) Uniqueness Theorem, that says: 

If $X$ and $Y$ are two random variables and equality holds for their MGF's: $m_X(t) = m_Y(t)$ then $X$ and $Y$ have the same probability distribution: $F_X(x) = F_Y(y)$.

The proof of this theorem is never shown in textbooks, and I cannot seem to find it online or in any book I have access to.
Can someone show me the proof or tell me where to look it up?
Thanks for your time.

REPLY [6 votes]: The following is a proof with two steps. 
It requires familiarity with Cauchy-Riemann equations and the identity theorem from complex analysis. Dominated convergence theorem is also applied. Thus, it  fits for any advanced undergraduate or graduate student who is familiar with these topics.  
Step 1:
Assume that $X$ is a random variable such that there exists $\delta\in(0,\infty]$ for which
$$M_X(s)=Ee^{sX}<\infty\ , \ \forall s\in(-\delta,\delta)\,.$$
Now, let's denote $\Omega\equiv\left\{z\in\mathbb{C};\text{Re}(z)\in(-\delta,\delta)\right\}$ and define 
$$\phi_X(z)\equiv Ee^{zX}\ ,\ \forall z\in\Omega\,.$$
Let $z=s+it\in\Omega$ and observe that
$$\phi_X(z)=E\ \overset{u(s,t;X)}{\overbrace{e^{sX}\cos (tX)}}+iE\ \overset{v(s,t;X)}{\overbrace{e^{sX}\sin (tX)}}\,.$$
It is possible to apply dominated convergence theorem in order to show that $Eu(\cdot;X)$ and $Ev(\cdot;X)$ are differentiable on $(-\delta,\delta)\times\mathbb{R}$. To see this, notice that for every $(s,t)\in(-\delta,\delta)\times\mathbb{R}$
$$|\frac{\partial u(s,t;X)}{\partial t}|\ ,\ |\frac{\partial u(s,t;X)}{\partial s}|\ ,\ |\frac{\partial v(s,t;X)}{\partial t}|\ ,\ |\frac{\partial u(s,t;X)}{\partial s}|$$
are all dominated by the random variable $Y=|X|e^{sX}$. In addition notice that for every $\epsilon>0$
$$|X|=\epsilon^{-1}|\epsilon X|\leq\epsilon^{-1}\sum_{k=0}^\infty\frac{|\epsilon X|^k}{k!}$$
$$=\epsilon^{-1}e^{\epsilon |X|}\leq\epsilon^{-1}\left(e^{-\epsilon X}+e^{\epsilon X}\right)$$
and hence 
$$|Y|\leq \frac{e^{sX}}{\epsilon}\left(e^{-\epsilon X}+e^{\epsilon X}\right)=\epsilon^{-1}\left(e^{(s-\epsilon) X}+e^{(s+\epsilon)X}\right)\,.$$
Therefore, since $M_X(\cdot)$ is finite on $(-\delta,\delta)$, then by taking small enough $\epsilon>0$ deduce that 
$$E|Y|\leq\epsilon^{-1}\left[M_X(s-\epsilon)+M_X(s+\epsilon)\right]<\infty\,.$$ 
Thus dominated convergence theorem can be carried out to show that $Eu(\cdot;X)$ and $Ev(\cdot;X)$ have partial derivatives with respect to both of their coordinates. Moreover, these partial derivatives are obtained by differentiating inside the expectation. In addition, dominated convergence theorem can be applied once more (with the same dominating random variable) in order to show that these partial derivatives are continuous on $(-\delta,\delta)\times\mathbb{R}$, i.e., $Eu(\cdot;X)$ and $Ev(\cdot;X)$ are differentiable on $(-\delta,\delta)\times\mathbb{R}$. Now, it is straightforward to verify that the Cauchy-Riemann equations are satisfied on $(-\delta,\delta)\times\mathbb{R}$ and hence $\phi_X(\cdot)$ is holomorphic on $\Omega$. 
Step 2:
Assume that $X$ and $Y$ are two random variables such that there exists $\delta\in(0,\infty)$ for which $M_X(s)=M_Y(s)<\infty$ for every $s\in(-\delta,\delta)$. Then, define
$$H(z)\equiv\phi_X(z)-\phi_Y(z)\ , \ \forall z\in\Omega$$ 
and by the result of Step 1, $H(\cdot)$ is holomorphic on $\Omega$ which is an open connected subset of $\mathbb{C}$. Now, denote a sequence $z_n=\frac{\delta}{2n}\in\Omega,\forall n\in\mathbb{N}$ and note that $z_n\rightarrow 0\in\Omega$ as $n\to\infty$. In addition, $H(z_n)=0,\forall n\in\mathbb{N}$ and hence using the identity theorem deduce that $H(z)=0,\forall z\in\Omega$. Then, as a special case, for every $t\in\mathbb{R}$ set $z=it$ and get
$$Ee^{itX}=Ee^{itY}$$ 
which means that $F_X(u)=F_Y(u),\forall u\in\mathbb{R}$ due to the uniqueness property of characteristic function (There are plenty of sources for the proof of this property and as mentioned by others its proof requires no advanced staff but dominated convergence and Fubini's theorems).