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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?
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.
Observe $$\int_0^{\infty}te^{-tx}dx=[-e^{-tx}]_0^{\infty}=1 $$
so that $$ t\int_0^{\infty}e^{-tx}f(x)dx-1=t\int_0^{\infty}e^{-tx}f(x)dx-\int_0^{\infty}te^{-tx}dx=\int_0^{\infty}te^{-tx}[f(x)-1]dx$$
By $\lim_{x \to \infty}f(x)=1$, there exists $A>0$ such that $x>A \implies |f(x)-1|<\frac{\epsilon}{2}$. We then have $$|t\int_0^{\infty}e^{-tx}f(x)dx-1|=|\int_0^A te^{-tx}[f(x)-1]dx+\int_A^{\infty} te^{-tx}[f(x)-1]dx|$$ $$ \le t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}\int_A^{\infty} te^{-tx}dx=t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}e^{-tA}$$ Since $$\lim_{t \to 0} (t\int_0^A |f(x)-1|dx+\frac{\epsilon}{2}e^{-tA})=\frac{\epsilon}{2} < \epsilon $$ Then, there exsits $\delta>0$ such that $$0<t<\delta \implies |t\int_0^{\infty}e^{-tx}f(x)dx-1|< \epsilon $$
