1

I need some hints regarding this exercice.

if $f : [0, \infty)\rightarrow \mathbb{R}$ is a measurable function s.t $\lim_{x\rightarrow \infty} f(x) = a$, prove :

\begin{align} \lim_{x\rightarrow \infty} \frac{1}{x}\int_0^xf(t)dt = a. \end{align}

If I try to integrate the lim inside of the integral by the monotone convergence theorem, I end up with a pathological case $\frac{\infty}{\infty}$.

Thanks for any help !

user149705
  • 1,163
  • 1
    I think you need to assume $f$ is integrable over bounded sets. You should be able to use the ideas in the second answer here for a proof. – David Mitra Jun 13 '14 at 23:12
  • The usual epsilon-delta approach (or rather, epsilon-x_0) works perfectly here: let epsilon be positive, then there exists some x_0 such that... – Did Aug 07 '14 at 21:03

1 Answers1

0

If you have that $$\lim_{x\rightarrow \infty} \int_0^xf(t)dt=\infty$$ the simplest way is to use L'Hopital rule $$\begin{align} \lim_{x\rightarrow \infty} \frac{\int_0^xf(t)dt}{x} = \lim_{x\rightarrow \infty} f(x)=a. \end{align}$$

Dario
  • 5,749
  • 2
  • 24
  • 36
  • Is there any other proof using measure theory techniques ? I haven't seen l'Hopital rule in the book I'm reading and I'm guessing there is a measure theoric proof to it. – user149705 Jun 13 '14 at 22:29