# Lebesgue Integrable in $\mathbb{R}$ Implies Convergence of Integral on $[-n,n]$

$$f: \mathbb{R}\rightarrow [0,\infty)$$ is integrable on $$\mathbb{R}$$ with Lebesgue measure $$\lambda$$, need to show that $$f$$ is Lebesgue integrable on $$[-n,n]$$ for each integer $$n$$, and the sequence $$u_n = \int_{-n}^n f d\lambda$$ converges.

The part to show that $$f$$ is Lebesgue integrable on $$[-n,n]$$ is easy, but how to show that $$u_n$$ converges? I am currently here: $$\int_{\mathbb{R}} f d\lambda = \int_{(-\infty,-n]}f d\lambda + \int_{[-n,n]}f d\lambda + \int_{[n,\infty)}f d\lambda$$ using the definition of Lebesgue integral of nonnegative functions. If $$u_n =\int_{[-n,n]} f d\lambda$$ converges, then $$\int_{(-\infty,-n]}f d\lambda + \int_{[n,\infty)}f d\lambda$$ needs to go to $$0$$, and I am not sure how to show that. There is nothing I can say about $$f$$, and the measure $$\lambda([n,\infty))$$ is always infinite (?).

Thanks for any help.

Let $$f:\mathbb{R}\to \mathbb{R}$$ be Lebesgue integrable. We have that $$\mathbb{I}_{[-n,n]}|f|\leq |f| \in \mathcal{L}^1(\lambda)$$ therefore $$\mathbb{I}_{[-n,n]}f \in \mathcal{L}^1(\lambda)\, \forall n \in \mathbb{N}$$. This also satisfies one of the requirements of the DCT. Pointwise, $$\lim_{n \to \infty}\mathbb{I}_{[-n,n]}f(x)=f(x)$$ because eventually the set $$[-n,n]$$ of the indicator function will contain any fixed $$x \in \mathbb{R}$$. Therefore we can use the DCT $$\lim_{n \to \infty}\int_{[-n,n]}f\,d\lambda=\lim_{n \to \infty}\int_{\mathbb{R}}\mathbb{I}_{[-n,n]}f\,d\lambda=\int_{\mathbb{R}}f \,d\lambda$$
this is a direct consequence of the Monotone Convergence Theorem, by letting $$f_n(x)=1_{[-n,n]}f(x)$$.
• The limit is finite because $f$ is integrable by assumption. So we can show that it is an usual convergence by applying the dominated convergence theorem.