4
$\begingroup$

Prove that $$\lim_{n \to \infty} \int_{[-n,n]}{f} = \int_{\mathbb{R}}f.$$

We're given $f$ a nonnegative measurable function on $\mathbb{R}$.

So far I have:

Let $f_n = 1_{[-n,n]}f$ then $\{f_n\}$ is nonnegative and monotone and $ f \to f_n$ pointwise.

By MCT, $$\lim_{n \to \infty} \int_{[-n,n]}{f} = \lim_{n \to \infty} \int_{[-n,n]}{f_n} = \lim_{n \to \infty} \int_{\mathbb{R}}{f} = \int_{\mathbb{R}}{f}.$$

Is this right? I'm a little concerned about my last line

$\endgroup$
5
$\begingroup$

What you are doing is basically correct, but you are not writing it properly.

You have that $f_n\nearrow f$ (it is essential that the convergence is monotone).

Then $$ \lim_n\int_{[-n,n]}f=\lim_n\int_{\mathbb R} f_n=\int_{\mathbb R}\lim_n f_n =\int_{\mathbb R}f, $$ where the Monotone Convergence Theorem is used in the second equality.

$\endgroup$
  • $\begingroup$ Thanks how do I know {$f_n$} is increasing? $\endgroup$ – Vinny Chase Nov 2 '17 at 20:01
  • $\begingroup$ Also could you explain why Dominated Convergence Theorem doesn't work in this case? I read somewhere that it doesn't apply but there was no explanation $\endgroup$ – Vinny Chase Nov 2 '17 at 20:02
  • $\begingroup$ You can prove that, for any $x\in\mathbb R$, $f_{n+1}(x)\geq f_n(x)$. And you cannot apply DCT because you have no integrable bound. $\endgroup$ – Martin Argerami Nov 2 '17 at 20:35
0
$\begingroup$

You have the right idea. Be sure to demonstrate that $f_n$ is a sequence of measurable functions, and as @MartinArgerami says, demonstrate that you have a pointwise increasing sequence of functions.

I would describe that last line as follows:

$$\lim_{n\to\infty} \int_{[-n,n]} f dm = \lim_{n\to\infty} \int_{\mathbb{R}} f_n dm = \int_{\mathbb{R}} \lim_{n\to\infty} f_n dm = \int_{\mathbb{R}} f dm,$$ where the exchange of limits follows from the MCT.

$\endgroup$
  • $\begingroup$ thanks. How do I know that $f_n$ is increasing? $\endgroup$ – Vinny Chase Nov 2 '17 at 19:48
  • $\begingroup$ @and why doesnt Dominated Convergence Theorem apply in this case? $\endgroup$ – Vinny Chase Nov 2 '17 at 19:48
  • $\begingroup$ @VinnyChase, You need $\int_\mathbb{R} f$ to be finite for DCT, as Martin said. $\endgroup$ – Joel Nov 2 '17 at 20:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.