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


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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