The Lebesgue-Borel measuref the difference between two open balls tends to $0$ as the radii tend to $\infty$

Question

Let $\lambda_n$ be the Lebesgue-Borel measure on the Borel-$\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ and $x,y\in\mathbb{R}^n$. What is the easiest way to prove $$\frac 1{c_nr^n}\lambda_n\left(B_r(x)\setminus B_r(y)\right)\to 0\;\;\;\text{for }r\to\infty ,$$ where $B_r(a)$ denotes the open ball around $a\in\mathbb{R}^n$ with radius $r>0$ and $c_n$ is the $n$-dimensional volume of the unit ball in $\mathbb{R}^n$.

Answer 1

By translating we may assume $x=0$ for convenience, and then by rescaling we have

$$\frac{1}{c_nr^n}\lambda_n(B_r(0)\setminus B_r(y)) = \frac{1}{c_n} \lambda_n(B_1(0)\setminus B_1(y/r)).$$

Now observe $y/r\to 0$ as $r\to\infty$ and say "dominated convergence theorem".

The dominated convergence theorem is overkill. To be more elementary, $B_1(0)\cap B_1(y/r)$ contains the ball of radius $1-|y|/r$ centred at $0$, so clearly $$\lambda(B_1(0)\cap B_1(y/r))\geq c_n(1-|y|/r)^n \to c_n$$ as $r\to\infty$.

It might be the very reason you asked the question, but this computation leads to a short proof of Liouville's theorem that a bounded harmonic function on $\mathbf{R}^n$ is constant. Suppose $f:\mathbf{R}^n\to\mathbf{R}$ is bounded and harmonic. Then $f(x)$ is equal to the average of $f$ over $B_r(x)$, and similarly $f(y)$ is the average of $f$ over $B_r(y)$, so by taking $r$ large compared to $|x-y|$ we see that the difference $f(x)-f(y)$ is bounded by $\|f\|_\infty \lambda_n(B_r(x)\triangle B_r(y))/(c_nr^n)$, which as we've seen tends to zero.

Nelson, Edward. A proof of Liouville's theorem. Proc. Amer. Math. Soc. 12 1961 995.