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I am working on the following exercise:

Let $\lambda$ denote Lebesgue measure on $\mathbf{R}$. Suppose $f:\mathbf{R}\rightarrow \mathbf{R}$ is a Borel measurable function such that $\int|f|<\infty$. Prove that $$\lim_{n\rightarrow \infty} \int_{[-n,n]} f\,d\lambda= \int f\,d\lambda.$$

Now, I have some ideas as to things that could help me here, but I can't really put any of it together. Here is what I have so far:


  1. I know that, given a measure space $(X,\mathcal{S},\mu)$, a set $E\in \mathcal{S}$, and an $\mathcal{S}$-measurable function $f$, $$\int_E f\,d\mu=\int f\chi_{E}\,d\mu$$ if the RHS is defined. In this case, I believe it is, since $\int |f|<\infty$.

  2. Then this question got me thinking: what if I write $[-n,n]$ as the limit of an increasing sequence of sets? That is, can I write $E=[-n,n]=\bigcup_{n=1}^\infty E_n$ where $E_n=[-n,n]$ for $n\in\mathbf{N}$?

  3. If (2) is true, then I can define $f_n=f\chi_{E_n}$ and $$\lim_{n\rightarrow\infty}f_n=f\chi_E.$$ Then I think I would have my answer by the Dominated Convergence Theorem.


Am I on the right track here?

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    $\begingroup$ Yes, and you're pretty much done. $\endgroup$ – T. Bongers Nov 19 '18 at 0:54
  • $\begingroup$ @T.Bongers Really? Well that is certainly a relief. If I may ask an additional question in regards to my second step. Did I set that up right? For whatever reason I am having a hard time convincing myself that what I wrote is true, i.e. that $\bigcup_{n=1}^\infty E_n=[-n,n]$. I tried drawing a picture, but it made me think -- shouldn't $\bigcup_{n=1}^\infty E_n=(-\infty,\infty)$? $\endgroup$ – Thy Art is Math Nov 19 '18 at 0:59
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    $\begingroup$ It is $(-\infty, \infty)$, but don't you want it to be? You're integrating over $\mathbb{R}$. $\endgroup$ – T. Bongers Nov 19 '18 at 1:28
  • $\begingroup$ Ah, that makes sense! Thanks! $\endgroup$ – Thy Art is Math Nov 19 '18 at 1:40
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Your proof is fine as-is.


For an alternative proof, you could apply the monotone convergence theorem to the sequences $f^+ \chi_{E_n}$ and $f^{-} \chi_{E_n}$ separately.

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