$f_a\chi_{(0,1)}$ is in $L^p$ iff $p
From Folland "Real Analysis":
"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?
 A: Take $f_a \chi_{(0,1)}$ as example. Since $f_a \chi _{(0,1)}$ is nonnegative and measurable, by [Beppo Levi's] monotonic convergence theorem we have
\begin{align*}
&\phantom{==}(\mathrm L)\int x^{-ap}\chi _{(0,1)}(x)\mathrm dx \\
&= \lim_n (\mathrm L)\int x^{-ap}\chi_{[1/n, 1)} \mathrm dx \\
&=\lim_n (\mathrm R)\int_{1/n}^1 x^{-ap}\mathrm dx \\
&=
\begin{cases}
\lim_n \dfrac {1 - (1/n)^{1-ap}}{1-ap}, & ap\neq 1,\\
\lim_n -\log(1/n) = +\infty, & ap = 1,
\end{cases}\\
&=
\begin{cases}
+\infty, & ap\geqslant 1,\\
\dfrac 1 {1-ap} < +\infty, & ap < 1,
\end{cases}
\end{align*}
hence the claim. 
A: Your computations just prove Folland's claim in the special cases $p = 1/a^2$ (if $a > 1$ otherwise $1/a^2 \geq 1/a$) and $p = 2/a$. The proof of the claim for general $a > 0$ is not much harder. So let us check for which $a >0$ the function $f_a \chi_{(0,1)} \in L^p(0, \infty)$:
$$
\int_0^\infty \lvert f_a(x) \chi_{(0,1)}(x)\rvert^p \, \text{d} x = \int_0^1 x^{-ap} \, \text{d}x = \frac{1}{1-ap}\Big[ x^{1-ap} \Big]_{x=0}^{x=1}
= \begin{cases}
\frac{1}{1-ap}& ap < 1 \\
+ \infty & ap > 1
\end{cases}\, .
$$ 
So $f_a \chi_{(0,1)} $ is in $L^p(0, \infty)$ if and only if $ap < 1$, which is the first part of the claim. The proof of the second part is similar. 
