Volume order of the difference of 2 large balls Let $x, y\in \Bbb R^d$, I would like to show that
$$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}=0.$$
My feeling is that we must also show that
$$|B(x,R)\setminus B(y, R)|\simeq R^{d-1}, \quad R>>1.$$
Is this equivalence true? if yes, how to prove it?
 A: One has
$$\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}=\frac{|B(x,R)\setminus (B(x,R)\cap B(y, R))|}{|B(x, R)|}.$$
Therefore,
$$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}=1-\lim\limits_{R\to\infty}\frac{|B(x,R)\cap B(y, R)|}{|B(x, R)|}.$$
But, for all $\varepsilon>0$, there exists $M>0$ such that for all $R\geq M$,
$$B(\frac{x+y}{2},R-\varepsilon)\subset B(x,R)\cap B(y, R).$$
Thus, for all $\varepsilon>0$, there exists $M>0$ such that for all $R\geq M$,
$$\left|1-\frac{|B(x,R)\cap B(y, R)|}{|B(x, R)|}\right|=1-\frac{|B(x,R)\cap B(y, R)|}{|B(x, R)|}\leq 1-\frac{(R-\varepsilon)^d}{R^d}.$$
Hence, $$\lim\limits_{R\to\infty}\frac{|B(x,R)\cap B(y, R)|}{|B(x, R)|}=1$$ and $$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}=0.$$
A: I have found an answer
We have that $B(x, R)\subset B(y, R+|x-y|)$ and $$|B(y, R+|x-y|)\setminus B(y,R)|= c_d (R+|x-y)^d-c_dR^d\sim R^{d-1}$$
we get
$$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}
\leq \lim\limits_{R\to\infty}\frac{(R+|x-y|)^d-R^d}{R^d}=\lim\limits_{R\to\infty} \big(1+\frac{|x-y|}{R}\big)^d-1=0.$$
