integral of lebesgue function is continuous Let F be a lebesgue integrable function on $(0,\infty)$. For $0 \le t < \infty$, define $g(t)=\int_{0}^{\infty} e^{-tx}F(x)dx$. Can someone explain why $g$ and $g'$ are continuous over $(0,\infty)$?
 A: *

*Let $t_n \rightarrow t$, we show $\lim_n g(t_n) = g(t)$, basically showing $$\lim_n \int_0^\infty e^{-t_n x}F(x) dx =  \int_0^\infty \lim_n e^{-t_n x}F(x) dx.$$


Clearly $e^{-t_n x}F(x) \rightarrow  e^{-t x}F(x)$ pointwise a.e. then we just need to use Lebesgue dominated convergence theorem to pass the limit. This is easy since $e^{-t_n x} < 1$.


*For the derivative, we show 
$$g'(t) = \frac{d}{dt}\int_0^\infty e^{-t x}F(x) dx = \int_0^\infty \frac{\partial}{\partial t}e^{-t x}F(x) dx$$
The above is the same as showing
$$\lim_{h\rightarrow 0} \int_0^\infty \frac{e^{-(t+h) x} - e^{-(t) x}}{h}F(x) dx =  \int_0^\infty \lim_{h\rightarrow 0} \frac{e^{-(t+h) x} - e^{-(t) x}}{h}F(x) dx \\= \int_0^\infty  -x e^{-t x}F(x) dx$$


Then as $h\rightarrow 0$ we have $\frac{e^{-(t+h) x} - e^{-(t) x}}{h}F(x) \rightarrow -x e^{-t x}F(x)$ pointwise a.e. Now observe that the function $-x e^{-t x}$ is actually bounded uniformly in $x$ for each $t\in (0,\infty)$, so coming up with a bounding function shouldn't be too hard.
Using mean value theorem for the numerator $e^{-(t+h) x} - e^{-(t) x}$ and adding absolute value on both sides
we have 
$$|e^{-(t+h) x} - e^{-(t) x}| = xe^{-cx} |h| \quad\text{ for some } c\in(t,t+h)$$
and using monotonicity, since $t<c$ we have
$$xe^{-cx} \leq xe^{-tx}$$
then 
$$\frac{e^{-(t+h) x} - e^{-(t) x}}{h} \leq \frac{xe^{-tx}|h|}{|h|} = xe^{-tx}.$$
Since $xe^{-tx}$ is bounded, then $xe^{-tx}|F(x)|$ is integrable, and it will be our bounding function.
