MHB Is the Limit of an Integrable Function Equal to Its Pointwise Limit?

[evinda]

Hey again! (Blush)

I am looking at the following exercise:
Let $f:[0,+\infty) \to \mathbb{R} $ integrable on each closed subinterval of $[0,+\infty)$.We suppose that $\lim_{x \to +\infty} f(x)=l$.Show that $\lim_{x \to +\infty} \frac{1}{x} \int_{0}^{x} f = l$.

The solution is the following:
Let $\epsilon>0$.As $\lim_{x \to +\infty} f(x)=l$, $\exists c_1>0$ such that $|f(x)-l|<\epsilon, \forall x \geq c_1$.
Also,$\exists c_2>0$ such that $\frac{1}{x} \int_{0}^{c_1} |f-l|< \epsilon, \forall x \geq c_2$
Therefore, $ \forall x \geq \max\{c_1,c_2\}$ we have $|\frac{1}{x} \int_0^x f - l|=\frac{1}{x} |\int_0^x (f-l)|\leq \frac{1}{x} \int_0^{c_1} |f-l|+\frac{1}{x} \int_{c_1}^x |f-l| \leq \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon$

Could you explain me the red part?? How do we conclude to that??

 

[Evgeny.Makarov]

Re: Why is it like that?

Let $f:[0,+\infty) \to \mathbb{R} $ integrable on each closed subinterval of $[0,+\infty)$...

Also,$\exists c_2>0$ such that $\frac{1}{x} \int_{0}^{c_1} |f-l|< \epsilon, \forall x \geq c_2$ ...

Could you explain me the red part??

This is because once we have chosen $c_1$, $\int_{0}^{c_1} |f-l|$ is a fixed number, which exists because $f$, and therefore $|f-l|$, is integrable. Meanwhile, $x$ increases unboundedly.

 

[evinda]

Re: Why is it like that?

This is because once we have chosen $c_1$, $\int_{0}^{c_1} |f-l|$ is a fixed number, which exists because $f$, and therefore $|f-l|$, is integrable. Meanwhile, $x$ increases unboundedly.

A ok!Thanks a lot! :)

 

[evinda]

Therefore, $ \forall x \geq \max\{c_1,c_2\}$ we have $|\frac{1}{x} \int_0^x f - l|=\frac{1}{x} |\int_0^x (f-l)|\leq \frac{1}{x} \int_0^{c_1} |f-l|+\frac{1}{x} \int_{c_1}^x |f-l| \leq \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon$

Could you also explain me why it stands that :

$$ \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon $$

? (Thinking) (Thinking)

 

[Evgeny.Makarov]

Could you also explain me why it stands that :

$$ \epsilon + \frac{x-c_1}{x} \epsilon \leq \epsilon $$

?

Well, obviously $\epsilon$ plus a positive number cannot be $\le\epsilon$, but it is $\le2\epsilon$.

 

[evinda]

Well, obviously $\epsilon$ plus a positive number cannot be $\le\epsilon$, but it is $\le2\epsilon$.

Nice! Thank you very much! (Smile)