This is a textbook corollary in my Real Analysis text book. I'm slightly paraphrasing the first part that I fully understand:
\begin{align*} f(x) &= \begin{cases} \frac{1}{|x|^{d+1}} & \text{if $x \neq 0$} \\ 0 & \text{otherwise} \\ \end{cases} \\ \end{align*} We will prove that $f$ is integrable outside any ball $|x| \ge \epsilon$. Let: \begin{align*} A_k &= \{ x \in \mathbb{R}^d : 2^k \epsilon < |x| < 2^{k+1} \epsilon \} \\ a_k(x) &= \frac{1}{(2^k \epsilon)^{d+1}} \chi_{A_k}(x) \\ g(x) &= \sum\limits_{k=0}^\infty a_k(x) \\ \end{align*} We can see that $A_k$ are disjoint, $g$ is a rounded version of $f$ such that $f(x) \le g(x)$ so $\int f \le \int g$. For any measurable $E$ on $\mathbb{R}^d$, the dilation property says that $m(\delta E) = \delta^d m(E)$. Since $A_k$ is $\mathcal{A} = \{1 < |x| < 2 \}$ dilated by a factor of $2^k \epsilon$, then $m(A_k) = m(2^k \epsilon \mathcal{A}) = (2^k \epsilon)^d m(\mathcal{A})$. Using an earlier theorem that says that the integral of a series of non-negative functions equals the series of the integrals: \begin{align*} \int \sum_{k=0}^\infty a_k(x) \, dx &= \sum_{k=0}^\infty \int a_k(x) \, dx \\ \end{align*} we can use that, plug in $a_k$ and get: \begin{align*} \int g &= \sum_{k=0}^\infty \frac{m(A_k)}{(2^k \epsilon)^{d+1}} \\ &= m(\mathcal{A}) \sum_{k=0}^\infty \frac{(2^k \epsilon)^d}{(2^k \epsilon)^{d+1}} \\ &= \frac{2 m(\mathcal{A})}{\epsilon} \\ \end{align*} Since $g$ is integrable, by monotonicity, so is $f$.
I'm having trouble understanding this next piece. My textbook literally continues from the above with:
Note that the same dilation-invariance property in fact shows that \begin{align*} \int_{|x| \ge \epsilon} \frac{dx}{|x|^{d+1}} &= \frac{1}{\epsilon} \int_{|x| \ge 1} \frac{dx}{|x|^{d+1}} \\ \end{align*}
Why is this so? How can I see this final step and conclusion from the earlier steps?
Let $x=\epsilon y$ and then $y=\frac{x}{\epsilon}$ maps $|x|\ge \epsilon$ to $|y|\ge 1$. Since $dx=\epsilon^ddy$, $$ \int_{|x| \ge \epsilon} \frac{dx}{|x|^{d+1}} =\int_{|y| \ge 1} \frac{\epsilon^ddy}{|\epsilon y|^{d+1}}=\frac1\epsilon\int_{|y|\ge1}\frac{dy}{|y|^{d+1}}. $$