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:
$$ \lim_{r \to 0} \, \sup_{x \in \mathbb R^n} \, r^{-\lambda} \int_{B(x,r)} |f_\alpha(y)|^p \, dy = 0. $$
My attempt. Let $x \in \mathbb R^n$ and $0 < r < 1/3$ be arbitrary elements. In order to prove what I want I decided to split in the following cases: $(i) \, \| x \| < 2r $ and $(ii) \, \| x \| \geqslant 2r.$
For case $(i)$, It is obvious that $B(x,r) \subset B(0,3r) \subset B(0,1)$, which implies that
$$ r^{-\lambda} \int_{B(x,r)} |f_\alpha(y)|^p \, dy \leqslant r^{-\lambda}\int_{B(0,3r)} |f_\alpha(y)|^p \, dy = r^{-\lambda} \int_{B(0,3r)} \| y \|^{\lambda - n + \alpha} \, dy. $$
Hence, passing to polar coordinates, we obtain
$$ r^{-\lambda} \int_{B(0,3r)} \| y \|^{\lambda - n + \alpha} \, dy = c(n) r^{-\lambda} \int_0^{3r} t^{\lambda - n + \alpha + n -1} \, dt = c(n)r^{-\lambda} \int_0^{3r} t^{\lambda + \alpha - 1 } \, dt = c(n) \frac{3^{\lambda + \alpha}}{\lambda + \alpha} r^\alpha. $$
This is obviously a good upper bound since $r^\alpha \to 0$ as $r \to 0$.
Now I should find a similar bound for the case $ \| x \| \geqslant 2r$ but this is where I am stuck.
Remark. I have already asked a similar question to this one in MSE (see here). The answer provided there uses complex results that I would like to avoid applying, since I believe this can be solved with less machinery.
Thanks for any help in advance.