Volume order of the difference of 2 large balls

Question

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?

Answer 1

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 $$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|}=0.$$

Answer 2

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 $$\lim\limits_{R\to\infty}\frac{|B(x,R)\setminus B(y, R)|}{|B(x, R)|} \leq \lim\limits_{R\to\infty}\frac{(R+|x-y|)^d-R^d}{R^d}=\lim\limits_{R\to\infty} \big(1+\frac{|x-y|}{R}\big)^d-1=0.$$