The Problem is: If $x$ and $y$ are two distinct points in $\mathbb R^n$, then show that $$\frac {Vol(B_R(x)\setminus B_R(y)) + Vol(B_R(y)\setminus B_R(x))}{B_R(x)}\underset{R\to \infty}{\to} 0$$
$\mathbf {My \ approach}:$ I was trying to apply mean value theorem taking the indicator functions $\chi_{B_R(x)\setminus B_R(y)}$ and $\chi_{B_R(y)\setminus B_R(x)}$ but I can't proceed any further. Any hint is appreciated. Thanks in advance.
Let $\omega_n := \operatorname{Vol}(B_1)$ and $x,y\in \mathbb R^n$. Making the change of coordinates $\tilde z= \frac{z-x}R$ gives that $$\operatorname{Vol}(B_R(x)\setminus B_R(y)) = \int_{B_R(x)\setminus B_R(y)} \, dz = R^n \int_{B_1\setminus B_1((y-x)/R )} \, d\tilde z. $$ Since $\operatorname{Vol}(B_R(x)) = \omega_n R^n$, it follows that \begin{align*} \frac{\operatorname{Vol}(B_R(x)\setminus B_R(y))}{\operatorname{Vol}(B_R(x))} = \frac 1 {\omega_n} \int_{B_1\setminus B_1((y-x)/R )} \, d z = \frac 1 {\omega_n} \int_{B_1} \chi_{_{B_1\setminus B_1((y-x)/R )}}(z) \, d z \end{align*} where $\chi_A$ denotes the characteristic function of a set $A\subset \mathbb R^n$. Then $$ \vert \chi_{_{B_1\setminus B_1((y-x)/R )}}(z) \vert \leqslant 1 \in L^1(B_1) $$ and $$ \chi_{_{B_1\setminus B_1((y-x)/R )}}(z) \to 0 $$ pointwise as $R\to +\infty$, so, by the dominated convergence theorem, $$ \lim_{R\to +\infty} \frac{\operatorname{Vol}(B_R(x)\setminus B_R(y))}{\operatorname{Vol}(B_R(x))} =\frac 1 {\omega_n} \int_{B_1} \lim_{R\to+\infty} \chi_{_{B_1\setminus B_1((y-x)/R )}}(z) \, d z= 0 $$ as required.