For some measure space $(X, M, \mu)$ such that $\mu(X)<\infty$. For any measurable function $f: X \to [0,\infty]$ with the property that
$$\exists C>0, \exists \alpha < -1, \forall \epsilon>0, \mu(\{x:f(x)>\epsilon\}) \leq C\epsilon^\alpha$$
Then, the following is true
$$\int f d\mu < \infty$$
$\forall r<1, a \in \mathbb{R}$ then $f(x) = \frac{1}{|x-a|^r} \in L^1([0,1],m)$ where $m$ is the Lebesgue measure
I don't know how this can be proven. I had the idea of finding another function from $f$ that is greater than it but still finite, but I don't know how to go about that.
By Fubini-Tonelli's theorem, it holds that $$\begin{align*} \int_0^\infty \mu\left(\{x:f(x)>\epsilon\}\right)\mathrm d\epsilon&=\int_0^\infty\int_X 1_{\{f(x)>\epsilon\}}\mathrm d\mu\mathrm d\epsilon\\ &= \int_X\int_0^\infty 1_{\{f(x)>\epsilon\}}\mathrm d\epsilon\mathrm d\mu\\&=\int_X \int_0^{f(x)}\mathrm d\epsilon \mathrm d\mu =\int_X f(x) \mathrm d\mu. \end{align*}$$ Since for all $\epsilon>0$, $$ \mu\left(\{x:f(x)>\epsilon\}\right)\le \mu(X)<\infty, $$ we have $$\begin{align*} \int_0^\infty \mu\left(\{x:f(x)>\epsilon\}\right)\mathrm d\epsilon&=\int_0^1 \mu\left(\{x:f(x)>\epsilon\}\right)\mathrm d\epsilon+\int_1^\infty \mu\left(\{x:f(x)>\epsilon\}\right)\mathrm d\epsilon\\&\le \mu(X) +C\int_1^\infty \epsilon^\alpha \mathrm d\epsilon \\&=\mu(X) -\frac{C}{1+\alpha}<\infty \end{align*}$$ as wanted. Since $$ m\left(\{x:|x-a|^{-r} >\epsilon\}\right)\le 2\epsilon^{-\frac1{r}}, $$ this criterion shows that $\int_0^1 \frac{1}{|x-a|^r}\mathrm dx <\infty$ for $r<1$.