Let $1 \leqslant p < \infty$ and $0 < \lambda < n$. Consider the function $f_\alpha \colon \mathbb R^n \to \mathbb R$ defined by
$$ f_\alpha(x) := \|x\|^{\frac{\lambda - n + \alpha}{p}} \chi_{B(0,1)}(x) + \| x \|^{\frac{\lambda - n}{p}} \chi_{\mathbb R^n \setminus B(0,1)}(x), \quad \forall x \in \mathbb R^n, $$
where $\alpha > 0$ is an arbitrarily fixed constant. My goal is to show that $f_\alpha$ satisfies the following property:
$$ \sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f_\alpha(y)|^p \, dy < \infty. $$
My attempt. I wasn't able to make any usefull progress but I believe I should start by analysing different cases. For context, yesterday I posted a question about this same function but proving another (similar) property (see here) and there I had to consider the cases $(i) \| x \| < 2r$ and $(ii) \| x \| \geqslant 2r$ to reach the result. Here we have two varying parameters so I believe the splitting is so easy to figure out.
Thanks for any help in advance.
Remark: I expect the ideas below work, although I did not check thoroughly the calculations, in particular, whether they are rigorous.
For any $x\in\mathbb R^n$, $r>0$, and $y\in B(x,r)$, we have $|f_\alpha(y)|^p\le \|y\|^{\lambda-n}$. So it suffices to show that $$I=r^{-\lambda}\int_{B(x,r)} \|y\|^{\lambda-n} dy<\infty$$ To bound $I$ we consider two cases which you suggested. Let $V_n$ be the volume of $B(0,1)$.
Suppose that $\|x\|\ge 2r$. Then $\|y\|\ge r$ for each $y\in B(x,r)$, so $$I\le r^{-\lambda}\int_{B(x,r)} r^{\lambda-n} dy\le r^{-\lambda}\cdot r^{\lambda-n}\cdot r^nV_n =V_n.$$
Suppose now that $\|x\|\le 2r$. Then $B(x,r)\subset B(0,3r)$, so $$I\le r^{-\lambda}\int_{B(0,3r)} \|y\|^{\lambda-n} dy= r^{-\lambda}\int_0^{3r} t^{\lambda-n} V_n nt^{n-1} dt=$$ $$nV_n r^{-\lambda}\int_0^{3r} t^{\lambda-1} dt= nV_n r^{-\lambda}\frac {t^{\lambda}}{\lambda} {\Huge|}_0^{3r}= nV_n r^{-\lambda}\frac {(3r)^\lambda}{\lambda} = \frac {nV_n3^\lambda}{\lambda}.$$