> 
> From Folland "Real Analysis":
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> 
> "It is instructive to consider following simple examples on $(0, \infty)$ wtih Lebesgue measure. Let $f\_a(x) = x^{-a}$, where $a>0.$ Elementary calculus shows that $f\_a\chi\_{(0,1)}\in L^p$ iff $p<a^{-1}$, and $f\_a\chi\_{(1, \infty)} \in L^P$ iff $p > a^{-1}$."
> 
> 
> 

I don't quite understand what the author did here.

I know that $f \in L^p$ if $|f|^p$ is Lebesgue integrable. And $$\int|x^{-a}\chi\_{(0,1)}|^p = \int|x^{-a}|^p\chi\_{(0,1)}.$$

So, for example, if $p = 1/a^2<1/a$,
$$\int|x^{-1/a}|\chi\_{(0,1)},$$
and this is integrable. But if $p=2/a\ge 1/a,$ $$\int|x^{-2}|\chi\_{(0,1)},$$ and this is not integrable.

Could you explain why?