In the following proof (screenshot is attached):
How exactly to prove that $\dfrac{V(D_r)}{V(B(0, r))}$ tends to $0$ as $r \to \infty$?
In $\Bbb {R}$ it is very easy since the balls are precisely intervals and we can explicitly calculate their volumes and check that that is indeed true.
But for $\Bbb {R}^n, ~ n \gt 1$ do we really need to calculate explicitly the volume $V(D_r)$ and $V(B(0, r)$ to prove that conclusion??
Any suggestion is highly appreciated. thanks.
One strategy is to investigate the size of $I_r = B(0, r) \cap B(x, r)$. If we can show that it is sufficiently large as $r \to \infty$, then we may be able to show that $\frac{V(I_r)}{ V(B(0, r))} \to 1$ as $r \to \infty$, which is equivalent to the claim (details below).
Let's find the largest sphere that fits inside $I_r$, and since we're taking $r \to \infty$, let's assume $r$ is large enough that $I_r \neq \emptyset$. The center of this large sphere is the midpoint $x/2$ between $x$ and $0$. Let's define $m = x/2$, and $d = |x/2|$, the distance between $m$ and $0$. It is a straightforward geometric exercise to show that $I_r$ contains the sphere $B(m, r - d)$.
Thus, we have that $V(I_r) > V(B(m, r - d))$. Hence $$ \frac{V(I_r)}{V(B(0, r))} > \frac{V(B(m, r-d))}{V(B(0, r))} = \frac{(r-d)^n}{r^n} $$
Note that $d$ is a constant with respect to $r$. Thus, as we take $r \to \infty$, this quantity tends to $1$.
Why is this equivalent to the original claim?
We can break up the symmetric difference into two pieces: $B(0, r) \setminus I_r$ and $B(x, r) \setminus I_r$. If we show that $\frac{V(B(0,r)) - V(I_r)}{V(B(0, r))} \to 0$ as $r\to \infty$ (and likewise for $V(B(x, r)) - V(I_r)$), then we're done. But this is an immediate corollary of the above proof.