Prove that $g_a \in L^1(\mathbb{R}^d) \iff a<-d.$ Where $$ g_a(x)=\begin{cases} |x|^a & \text{ if } |x|>1 \\ 0 & \text{ otherwise} \\ \end{cases}$$
We will be using the following lemma
Lemma 4
We have $f\in L^1(\mathbb{R}^d) \iff$ $$ \sum_{k\in \mathbb{Z}}2^km(\{x\in \mathbb{R}^d: 2^k\leq |f(x)|\leq 2^{k+1}\})<\infty $$
My proof attempt:
Proof. If $a\geq 0$, then $|x|^a> 1$ since $|x|>1$. Therefore, $$ \int 1_{|x|>1}<\int |x|^a\cdot 1_{|x|>1}$$ Since $\int 1_{|x|>1}=\infty \implies g_a\notin L^1(\mathbb{R}^d)$.
So let $a<0$. for $k\in \mathbb{Z}$, let \begin{align*} E_k&=\{x\in \mathbb{R}^d: 2^k\leq |x|^a\leq 2^{k+1}\} \\ &=\{x:2^{k/a}\geq |x|\geq 2^{(k+1)/a} \} \\ \end{align*} If $k\geq 0$, then $2^{k/a}\leq1 \implies |x|\leq 1$. But $g_a$ vanishes over that part of the domain. So we may exclude these sets.
If $k<0 \implies 2^{(k+1)/a}\geq 1 \implies |x|\geq 1$. Hence, we care about these sets. For every $x\in E_k$, $E_k \subset B_{2^{k/a}}(0)$. Thus, $E_k$ is contained in a cube of radius $2^{k/a} \implies m(E_k)\leq 2^{dk/a}$. By Lemma 4, $g_a\in L^1(\mathbb{R}^d) \iff \sum_{k<0}2^{k}m(E_k)<\infty$. Observe that \begin{align*} \sum_{k<0}2^{k}m(E_k)&\leq \sum_{k<0}2^{k}2^{dk/a}\\ &=\sum_{k<0}2^{k(1+d/a)} \end{align*} Therefore, the sum converges $\iff 1+d/a>0$. Therefore, $1>-d/a\implies a<-d$. This completes the proof.
Any critique of the proof or comments on style are much appreciated. My main concern in my proof is whether my reasoning for $g_a\notin L^1(\mathbb{R}^d)$ was correct and whether or not my reasoning for leaving out $E_k$ for $k\geq 0$ was correct.