# Prove that $\lim_{n\rightarrow \infty} \int_{[-n,n]} f\,d\lambda= \int f\,d\lambda.$

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?

• Yes, and you're pretty much done. – T. Bongers Nov 19 '18 at 0:54
• @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 $\bigcup_{n=1}^\infty E_n=(-\infty,\infty)$? – Thy Art is Math Nov 19 '18 at 0:59
• It is $(-\infty, \infty)$, but don't you want it to be? You're integrating over $\mathbb{R}$. – T. Bongers Nov 19 '18 at 1:28
• Ah, that makes sense! Thanks! – Thy Art is Math Nov 19 '18 at 1:40

## 1 Answer

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.