Is the function $f = \sum_{n=0}^{\infty} 2^{-n}\chi_{[n,n+1)}$ Lebesgue integrable on $\mathbb{R}$?

Question

Is the function $f = \sum_{n=0}^{\infty} 2^{-n}\chi_{[n,n+1)}$ Lebesgue integrable on $\mathbb{R}$? Justify your answer.

I came across this question on a past exam paper for a measure theory course I'm taking and I can't find anything similar in my professors notes to help me work through it. I have a feeling that I should be using a convergence theorem but I'm not quite sure which one. A push in the right direction would be appreciated! Thanks in advance.

Answer 1

Observe that

\begin{align*} \int_{\mathbb{R}}|f| &=\int_{\mathbb{R}} \left|\sum_{n=0}^{\infty} 2^{-n}\chi_{[n,n+1)} \right|\\ &= \int_{\mathbb{R}} \sum_{n=0}^{\infty}2^{-n}|\chi_{[n,n+1)}|\\ &=\sum_{n=0}^{\infty}2^{-n} \int_{\mathbb{R}}\chi_{[n,n+1)}&\because\text{Monotone Convergence Theorem}\\ &=\sum_{n=0}^{\infty}2^{-n}\\ &=\frac{1}{1-\frac{1}{2}}\\ &=2\\&<\infty \end{align*}

Therefore, $f$ is integrable.

Answer 2

Yes it is. Here you can compute the value of the integral directly using monotone convergence:

$$\int|f|=\int\lim_{n\rightarrow\infty} \sum^N_{k=10}2^{-k}\mathbb{1}_{[k,k+1})=\lim_{n\rightarrow\infty}\int\sum^n_{k=1}2^{-k}\mathbb{1}_{[k,k+1)}=\lim_n\sum^n_{k=0}2^{-k}=2<\infty$$