# Showing $1/x^2$ is Lebesgue Integrable on $\mathbb{R}_{\ge 1}$

I am trying to show that $\int_{\mathbb{R}_{\ge 1}} 1/x^2 < \infty$.

(1) By definition, $\int_{\mathbb{R}_{\ge 1}} 1/x^2 = \underset{0 \le \phi \le 1/x^2}{sup} \int_{\mathbb{R}_{\ge 1}} \phi$ for $\phi$ a simple function.

(2) If we let $\psi_n$ be a simple function s.t. $\psi_n = \Sigma_{k=2}^n 1/k^2 \chi_{[k,k-1]}$ then since the p-series $\Sigma_{n=2}^\infty 1/n^2 < \infty$ we have that $\forall n \in \mathbb{N}$, $\int_{\mathbb{R}_{\ge 1}} \psi_n < \infty.$

From this it seems clear that the supremum of the $\int_{\mathbb{R}_{\ge 1}} \phi$ couldn't possibly be greater than the supremum of the $\int_{\mathbb{R}_{\ge 1}} \psi_n$ so that $\int_{\mathbb{R}_{\ge 1}} 1/x^2 < \infty$, yet how could I show this last step rigorously?

-

To show the last step rigorously you just need to show that for every $\phi$ bounded by $1/x^2$ there is an $n$ such that $\phi$ is also bounded by $\psi_n$. To do this, you want to (re)define the sequence $(\psi_n)$ so that these functions are above the graph of $1/x^2$ whenever they are not $0$. Then, since $\phi$ is always below the graph, you get the inequality.