Uniqueness of a measure that satisfies a property. Is that true that there exists a unique positive measure $\nu(dx)$ on the interval $[0, \infty)$ such that it holds
$$\int_0^\infty e^{-sx}\nu(dx)=\frac{1}{(1-e^{-s})^c}?$$
It should be a kind of consequence of the uniqueness of the inverse Laplace's transform?
I can prove that $$\nu(dx)=\sum_{j=0}^\infty C_j\delta(x-j)dx$$ where $C_j$ are positive coefficients that can be computed explicitly.
Since the function $$f(x):=\sum_{j=0}^\infty C_j\delta(x-j)$$ is pieceweise continuous in $[0, \infty)$ I can conclude (by the uniqueness of the inverse Laplace's transform) that it doesn't exisit another function piecewise continuous $g(x)$ such that
$\nu(dx)=g(x)dx$.
Is that correct?
Thank you
 A: Hint:
Consider the measure $\mu(dx)=\sum^\infty_{n=0}\delta_n(dx)$ where for any $p\in\mathbb{R}$, $\delta_p(dx)$ is the measure that gives has $1$ to any set that contains $p$ and $0$ otherwise. Then
$$\mathcal{L}(\mu)(s)=\int^\infty_0 e^{-sx}\,\mu(dx)=\frac{1}{1-e^{-s}}$$
Recall that for any pair of measures $\mu$ and $\nu$ in $[0,\infty)$,
$$\mathcal{L}(\mu*\nu)(s)=\big(\mathcal{L}(\mu)(s)\big)\,\big(\mathcal{L}(\nu)(s)\big)$$
for
$$\begin{align}\mathcal{L}(\mu*\nu)(s)&=\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-s(x+y)}\mathbb{1}_{[0,\infty)}(x)\mathbb{1}_{[0,\infty)}(y)\,\mu\otimes\nu(dx,dy)\\
&=\int_{\mathbb{R}}e^{-sx}\mathbb{1}_{[0,\infty)}(x)\Big(\int_{\mathbb{R}}e^{-sy}\mathbb{1}_{[0,\infty)}(y)\,\nu(dy)\Big)\,\mu(dx)
\end{align}$$
by Fubini's theorem and the definition of the Laplace transform of a measure in $[0,\infty)$.
For $c\in\mathbb{Z}_+$,
$$\mathcal{L}(\mu^{*c})(s)=\big(\mathcal{L}(\mu(s)\big)^c=\frac{1}{(1-e^{-s})^c}$$
where $\mu^{*c}$ is the convolution of $\mu$ with itself $c$ times.
