Suppose $f\in \mathscr{R}$ on $[0,A]$ for all $A<\infty$, and $f(x)\to 1$ as $x\to +\infty$. Prove that $$\lim \limits_{t\to 0}t\int_{0}^{\infty}e^{-tx}f(x)dx=1 \quad (t>0).$$
Proof: Let's define $F(t)=t\int_{0}^{\infty}e^{-tx}(f(x)-1)dx$ for $t>0$. Let $\epsilon>0$ be given then $\exists A=A(\epsilon)>0$ such that for any $x\geqslant A$ we have $|f(x)-1|<{\epsilon}/{2}.$ Then $$|F(t)|=\left|t\int_{0}^{\infty}e^{-tx}(f(x)-1)dx\right|\leqslant t\int_{0}^{\infty}e^{-tx}|f(x)-1|dx=t\left(\int_{0}^{A}+\int_{A}^{\infty} \right)\leqslant$$$$\leqslant t\int_{0}^{A}e^{-tx}|f(x)-1|dx+t\frac{\epsilon}{2}\int_{A}^{\infty}e^{-tx}dx.$$
Since $|f-1|\in \mathscr{R}$ on $[0,A]$ then $|f-1|$ is bounded on $[0,A]$ (Rudin's assumption on Chapter 6) and let $C=\sup \limits_{[0,A]}|f(x)-1|$ then we get that: $$|F(t)|\le Ct\int_{0}^{A}e^{-tx}dx+t\frac{\epsilon}{2}\int_{A}^{\infty}e^{-tx}dx=C(1-e^{-At})+\dfrac{\epsilon}{2}e^{-At}<ACt+\dfrac{\epsilon}{2}$$ since $0<e^{-At}<1$ and $e^{-At}>1-At$. Taking $\delta=\dfrac{\epsilon}{2AC}$ then for any $t\in (0,\delta)$ we get $|F(t)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ which is equivalent to $\lim \limits_{t\to 0+}F(t)=1$ and we get our desired result.
Can anyone check my proof please? I would be very grateful for your help!
Your proof is basically correct. The last step can be like this. Since $$ |F(t)|\leqslant C(1-e^{-At})+\dfrac{\epsilon}{2}e^{-At} $$ There is $$ \varlimsup_{t\to0}|F(t)|\leqslant \varlimsup_{t\to0}C(1-e^{-At})+\dfrac{\epsilon}{2}=\dfrac{\epsilon}{2} $$ Thus $$ \varlimsup_{t\to0}|F(t)|=0\quad\text{and }\quad \lim_{t\to0}F(t)=0 $$ which is equivalent to $$ \lim_{t\to 0^+}t\int_{0}^{\infty}e^{-tx}f(x)dx=1 \quad $$