Integration estimate:$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy < \infty $ I want to show that for $0< \lambda < n$ and $c_1>0$ fixed, we have
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy \leq c < \infty $$
for any $x \in B_r(0) \subseteq\mathbb{R}^n, 0<r<c_1$. It is
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy \leq \int_{B_r(x)} \! |y|^{- n} dy \leq \int_{B_r(0)} \! |y|^{- n} dy \leq \int_{B_{c_1}(0)} \! |y|^{- n}$$
so the assertion would be true if $\frac{1}{|y|^n} \in L^1(B_r(0))$, but i think this is not true? I'm thankful for any suggestions on how to continue. 

First, Using polar coordinate for every $x\in\Bbb R $ we get 
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y-x|^{\lambda - n} \, dy= |\Bbb S^{n-1}|\frac{1}{r^\lambda}\int_{0}^r \! |\rho|^{\lambda - n} \rho^{n-1}\, d\rho \\=|\Bbb S^{n-1}|\frac{1}{r^\lambda}\int_{0}^r \! \rho^{\lambda - 1}  \, d\rho  = \frac{|\Bbb S^{n-1}|}{\lambda}\tag{I}$$


*

*If $0<\lambda< n$ then from (I) we have,
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy=\frac{1}{r^\lambda} \int_{B_r(x)\cap B_r(0)} \! |y|^{\lambda - n} \, dy+\frac{1}{r^\lambda} \int_{B_r(x)\cap B^c_r(0)} \! |y|^{\lambda - n} \, dy\\\le\frac{1}{r^\lambda} \int_{ B_r(0)} \! |y|^{\lambda - n} \, dy+\frac{1}{r^\lambda} \int_{B_r(x)\cap B^c_r(0)} \! |y|^{\lambda - n} \, dy \\=\frac{|\Bbb S^{n-1}|}{\lambda} +\frac{1}{r^\lambda} \int_{B_r(x)\cap B^c_r(0)} \! |y|^{\lambda - n} \, dy $$
However, Since $x\in B_r(x)$ we have,  $y\in B_r(x)\cap B^c_r(0) \implies r\le |y|\le |x|+|y-x|\le  2r \implies (2r)^{\lambda-n} \le |y|^{\lambda-n}\le r^{\lambda-n}$
Accordingly,
$$\frac{1}{r^\lambda} \int_{B_r(x)\cap B^c_r(0)} \! |y|^{\lambda - n} \, dy \le \frac{1}{r^\lambda} \int_{B_r(x)} \! r^{\lambda - n} \, dy \le r^{-n}|B_r(x)| = |B_1(0)| $$
That is, $$\color{blue}{\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy\le  |B_1(0)|+ \frac{|\Bbb S^{n-1}|}{\lambda}}$$


*Now If $\lambda> n$ then we proceed as follows.
Since,  $x\in B_r(0)$ i.e $|x|\le r$ we have, 
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |x|^{\lambda - n} \, dy\le r^{-n}|B_r(x)| = |B_1(0)|\tag{II} $$


*

*On other hand,  $|y| \le|x|+|y-x| $. And since $\lambda >n$ we have, $$\color{red}{|y|^{\lambda - n}\le (|x|+|y-x|)^{\lambda - n}\le 2^{\lambda -n}(|x|^{\lambda - n}+|y-x|^{\lambda - n})}\tag{III}$$


Therefore,  from I, II, III we have, 
$$\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy\le \frac{ 2^{\lambda -n}}{r^\lambda} \int_{B_r(x)} \! |x|^{\lambda - n} \, dy+\frac{ 2^{\lambda -n}}{r^\lambda} \int_{B_r(x)} \! |y-x|^{\lambda - n} \, dy\\\le 2^{\lambda -n}( |B_1(0)|+ \frac{|\Bbb S^{n-1}|}{\lambda})$$
That is, $$\color{blue}{\frac{1}{r^\lambda} \int_{B_r(x)} \! |y|^{\lambda - n} \, dy\le 2^{\lambda -n}( |B_1(0)|+ \frac{|\Bbb S^{n-1}|}{\lambda})}$$
