TITLE: Prove that $\lim_{t \rightarrow 0} t \int_{0}^{\infty} e^{-tx} f(x) dx =1$
QUESTION [6 upvotes]: I am trying to solve Rudin 8.11:
Suppose $f$ is Riemann-integrable on $[0,A]$ for all $A<\infty$, and $f(x) \rightarrow 1$ as $x \rightarrow \infty$. 
Prove that 
$$\lim_{t \rightarrow 0} \;\int_{0}^{\infty}t e^{-tx} f(x) dx =1.$$
This is easy if one assumes $f$ is differentiable (it is a special case of the Final Value Theorem for Laplace transforms), but I'm apparently not even allowed to assume continuity here. Anyone have any ideas?

REPLY [7 votes]: Assuming you left out the $t$, split the interval $[0,\infty)$ into $[0,A]$ and $[A,\infty)$ and choose $A$ so that $|f(x)-1|<\epsilon$ for all $x>A$. Write the integral as the sum of integrals over either interval. On the first interval, you can use dominated convergence theorem and the fact that $f$ is integrable on $[0,A]$. On the infinite interval, use the fact that $f$ is close to 1.