Let $\mu$ be a finite, non-negative Borel measure on $[0,\infty)$ and let
$f(r) = \int_0^{\infty} e^{-rt} d\mu(t).$
A book I am reading states that "differentiation under the integral sign is justified because the measure is finite". That is,
$-\frac{df}{dr} = -\frac{d}{dr} \int_0^{\infty} e^{-rt} d\mu(t) dt = \int_0^{\infty} t e^{-rt} d\mu(t)$
I'm not exactly sure how this follows. I have the following theorem from my real analysis textbook by Folland: if $\frac{\partial }{\partial r}e^{-rt}$ exists and if there is a $g \in L^1([0,\infty)$ such that
$\left|\frac{\partial }{\partial r}e^{-rt}\right| \leq g(t)$
for all $r$ and $t$, then we may interchange derivatives. However, for fixed $r$, I can do some differentiation and find a relative extrema at $t = \frac{1}{r}$ where the derivative of the integrand is $\frac{1}{re}$ at this point. As $r$ becomes very small, this blows up to infinity, so finding a $g(t)$ that bounds $te^{-rt}$ (the derivative of the integrand) seems difficult to me, as $g$ cannot change with $r$. I don't seem to be using the finiteness of the measure, so I know I'm not on the right track. Any suggestions would be greatly appreciated.