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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? |
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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. |
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