Define $f$ on $\mathbb{R}$ by $f(x)=\sqrt{n}$ if $n\le x \le n+(1/n^2))$ for $n=1,2,\ldots$ and $f(x)=0$ otherwise.

Question

Define $f$ on $\mathbb{R}$ by $$f(x)=\begin{cases}\sqrt{n} & \text{if $n\le x \le n+\dfrac{1}{n^2}$ where $n \ge 1$} \\ 0 & \text{otherwise} \end{cases}.$$ Show that $f$ is integrable with respect to Lebesgue measure but $f^2$ is not.

Answer 1

Try to use the definition of Lebesgue integrals to show that the integral of $f$ is equal to $\sum_{n=1}^{\infty}\frac{\sqrt{n}}{n^2}$ and similarly the integral of $f^2$ is $\sum_{n=1}^{\infty}\frac{n}{n^2}=\sum_{n=1}^{\infty}\frac{1}{n}$. Which one of these converges? Which one doesn't? And why?