A Function of a Convolution (Laplace) A paper I am reading makes the following claim:
Assume that $a_n$ is a series of of positive, distinct, real numbers. Assume that $\epsilon_n$ are independent random standard exponential variables. They state that, "the partial fraction expansion of the Laplace transform" is, where $\Pi_{n,N}$ are defined in the standard partial fractions way:
$$E\left[\exp(-\lambda\sum_{n=1}^N a_n \epsilon_n)\right] = \prod_{n=1}^N \frac{1}{1+\lambda a_n} = \sum_{n=1}^N \frac{1}{\Pi_{n,N}}\frac{1}{1+\lambda a_n}$$
So far, so good. They then make the following claim that I don't understand: "This implies that for every non-negative measurable function $g$ such that 
$E[g(a_n \epsilon_1)]$ is finite for every $n$, then"
$$E\left[g\left(\sum_{n=1}^N a_n \epsilon_n\right)\right] = \sum_{n=1}^N \frac{1}{\Pi_{n,N}} E[g(a_n \epsilon_n)]$$
It seems to be right, e.g. checking with some functions by simulation, but I'm confused as to why it follows for $any$ function of the required type. Is it something special about the exponential distribution or would it work for, say, a gamma random variable?
p.s. The paper is really interesting (http://www.ams.org/journals/bull/2001-38-04/S0273-0979-01-00912-0/S0273-0979-01-00912-0.pdf) and I'm talking about section 5.
 A: I think I've figured it out. I'm not sure what the norms about answering your own question are, but I post this in case it's helpful to others.
1) Define $v_n = a_n \epsilon_n = Expo(1/a_n)$.
2) Define $X = \sum_n v_n$
As $X$ is a sum of independent random variables, its PDF $f(x)$ is a convolution of those random variables, thus the Laplace transform of $X$ is, i.e. the product of the individual Laplace transforms of the $v_i$. Because all $a_n$ are distinct (?), we can write this using the partial fractions notation.
$$\mathcal{L}\{f(x)\} = F^X(s) = \Pi_n F^{V_n}(s) = \sum_n \frac{1}{\Pi_{n,N}}F^{V_n}(s)$$
The RHS by definition is thus:
$$\sum_n \frac{1}{\Pi_{n,N}}\int_0^\infty f^{V_n}(t) e^{-st} dt$$
Because everything is nice and convergent, I think we can switch the order of summing and integrating to get:
$$\int_0^\infty \left[\sum_n \frac{1}{\Pi_{n,N}}f^{V_n}(t)\right] e^{-st} dt$$
The LHS is:
$$\int_0^\infty f^X(t) e^{-st} dt$$
Now, imagine I wanted to know the Laplace transform of $g(t)f^X(t)$, i.e.
$$\int_0^\infty g(t) f^X(t) e^{-st} dt$$
I think, and this is the part I'm not sure about, that is equivalent to writing the following on the RHS:
$$\int_0^\infty g(t) \left[\sum_n \frac{1}{\Pi_{n,N}}f^{V_n}(t)\right] e^{-st} dt$$
Because we assume that $E(g(v_n) < \infty$ for all $i$, I think we can again switch the order of integration / summation to get:
$$\sum_n \frac{1}{\Pi_{n,N}} \int_0^\infty g(t) f^{V_n}(t) e^{-st} dt$$
To complete the result, note that at $s=0$ we recover the expectation.
$$E(g(X)) = \sum_n \frac{1}{\Pi_{n,N}} E(g(a_n \epsilon_1))$$
