$f: \mathbb{R}\rightarrow [0,\infty)$ 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.
Best Answer
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 $$\lim_{n \to \infty}\int_{[-n,n]}f\,d\lambda=\lim_{n \to \infty}\int_{\mathbb{R}}\mathbb{I}_{[-n,n]}f\,d\lambda=\int_{\mathbb{R}}f \,d\lambda$$