Exchanging integrals with series when the monotone convergence theorem does not apply Let $g$ and $h$  measurable functions on $X $, and assume $g\in L^{1}(X)$ and $h\geq 0$. Show that if $$F(t)=\int g(x)e^{-th(x)}\,d\mu(x)$$ then for $t >0$,
$$F(t)=\sum\limits_{n=0}^{\infty} \left[\int g(x)h^{n}(x)\,d\mu(x)\right] \frac{(-t)^{n}}{n!}$$ 
The question is, what convergence theorem can be used, as the monotone convergence theorem can not be used?
 A: So let $F_n(t)$ be the lower representation summed from $1$ to $n$, i.e. 
$$F_n(t) = \sum_{i=0}^n\int g(x)h^n(x)d\mu(x)\frac{(-t)^n}{n!}$$
When $n$ is finite we are able to rearrange the summation and integration to give an equivalent:
$$F_n(t) = \int g(x) \sum_{i=o}^{n}\frac{(-th(x))^n}{n!}\,d\mu(x)$$
So to use Lebesgue dominated convergence we need to find some integrable function $l(x)$ such that 
$$\left|g(x)\right|\left| \sum_{i=o}^{n}\frac{(-th(x))^n}{n!}\right| \le l(x)$$
For all $n$. 
But we also have that $\exp(-x) \le 1$  for all $x \in \Bbb{R}^+$. And that for every $\epsilon > 0$ there exists an $N$ such that $$\sum_{n>N}\frac{(-th(x))^n}{n!}<\epsilon$$
for fixed $t$ and all $x$.  We have 
$$\sum_{i=o}^{\infty}\frac{(-th(x))^n}{n!} \le 1$$
So for $n>N$ we have 
$$\sum_{i=o}^{n}\frac{(-th(x))^n}{n!} \le 1+\epsilon $$
Andthis gives us $l(x) = |g(x)|(1+\epsilon)$ as a bound if we reindex $\{F_n\}$ to $\{F_{n'}\}$ for $n > N$ this converges to
$$g(x)\exp(-th(x))$$
And LDCT shows that the integral converges to our first definition of $F$ as well, giving the desired result. 
