A functional problem on page 268 in the book GTM 95 probability 
I have noticed that if 
$$
\lim_{c\downarrow 0}\int_{-\infty}^{\infty} e^{c|x|}P(dx) < \infty
$$
the system $\{1, x, x^2, \cdots \}$ is complete in $L^2$. 
So I want to know which theorem or result was used here?
 A: This seems to be more of a problem in complex analysis (+ a bit of
Fourier analysis) than in functional analysis.
Your assumption on $\mathbb{P}$ implies that $\int e^{c\left|x\right|}d\mathbb{P}\left(x\right)<\infty$
for some $c>0$. This easily implies $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$. 
Now, let $f\in L^{2}\left(\mathbb{P}\right)$ be arbitrary. We first
show that the function
$$
G:\left\{ z\in\mathbb{C}\mid{\rm Re}\left(z\right)<\frac{c}{4}\right\} \to\mathbb{C},z\mapsto\int f\left(x\right)\cdot e^{zx}d\mathbb{P}\left(x\right)
$$
is well-defined and holomorphic.
The integral is well-defined because of 
$$
\left|f\left(x\right)\cdot e^{zx}\right|=\left|f\left(x\right)\right|\cdot e^{{\rm Re}\left(zx\right)}\leq\left|f\left(x\right)\right|\cdot e^{{\rm Re}\left(z\right)\cdot\left|x\right|}\leq\left|f\left(x\right)\right|\cdot e^{\frac{c}{2}\left|x\right|}\in L^{1}\left(\mathbb{P}\right),
$$
by Cauchy Schwarz, since $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$.
Since the integrand is obviously complex differentiable, we can use
differentiation under the integral sign. We have
\begin{eqnarray*}
\left|\frac{\partial}{\partial z}\left[f\left(x\right)\cdot e^{zx}\right]\right| & = & \left|f\left(x\right)\right|\cdot\left|x\right|\cdot\left|e^{zx}\right|\\
 & = & \left|f\left(x\right)\right|\cdot\left|x\right|\cdot e^{{\rm Re}\left(zx\right)}\\
 & \leq & \frac{4}{c}\cdot\left|f\left(x\right)\right|\cdot\frac{c}{4}\left|x\right|\cdot e^{{\rm Re}\left(z\right)\cdot\left|x\right|}\\
 & \leq & \frac{4}{c}\cdot\left|f\left(x\right)\right|\cdot e^{\frac{c}{4}\left|x\right|}\cdot e^{\frac{c}{4}\left|x\right|}\in L^{1}\left(\mathbb{P}\right),
\end{eqnarray*}
by Cauchy Schwarz, since $e^{\frac{c}{2}\left|x\right|}\in L^{2}\left(\mathbb{P}\right)$.
All in all, this implies that $G$ is holomorphic.
Now assume that $\left\langle f,x^{n}\right\rangle _{L^{2}\left(\mathbb{P}\right)}=0$
for all $n\in\mathbb{N}_{0}$. For $\left|z\right|<\frac{c}{4}$,
we have
\begin{eqnarray*}
\int\left|f\left(x\right)\right|\cdot\sum_{n=0}^{\infty}\left|\frac{\left(zx\right)^{n}}{n!}\right|d\mathbb{P}\left(x\right) & = & \int\left|f\left(x\right)\right|\cdot e^{\left|zx\right|}d\mathbb{P}\left(x\right)\\
 & \leq & \int\left|f\left(x\right)\right|\cdot e^{\frac{c}{4}\left|x\right|}d\mathbb{P}\left(x\right)<\infty,
\end{eqnarray*}
which justifies the interchange of integration and summation in
$$
G\left(z\right)=\int f\left(x\right)\cdot\sum_{n=0}^{\infty}\frac{\left(zx\right)^{n}}{n!}d\mathbb{P}\left(x\right)=\sum_{n=0}^{\infty}\left[\frac{z^{n}}{n!}\cdot\int f\left(x\right)\cdot x^{n}d\mathbb{P}\left(x\right)\right]=0.
$$
But since $G$ is holomorphic, this implies $G\equiv0$.
In particular, we get
$$
0=G\left(-i\xi\right)=\int f\left(x\right)\cdot e^{-ix\xi}d\mathbb{P}\left(x\right)=\widehat{f\cdot d\mathbb{P}}\left(\xi\right),
$$
where $f\cdot d\mathbb{P}$ is a finite, complex measure since $f\in L^{2}\left(\mathbb{P}\right)\subset L^{1}\left(\mathbb{P}\right)$.
But this implies $f\cdot d\mathbb{P}\equiv0$ and thus $f=0$ almost
everywhere.
