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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 $\mathbb{R}^n$.

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    $\begingroup$ I think the statement as written is false. WLOG $x=0$ and $y=e_1$. Then we're measuring the set of points which are a distance $r$ from $0$ but not from $e_1$. This set contains a half ball of radius $1$ centred at $-(r-1)e_1$, so it has measure at least half the measure of the unit ball. $\endgroup$ – Sean Eberhard Apr 24 '15 at 14:12
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    $\begingroup$ I think the correct version should have $$\frac{\lambda_n(B_r(x)\setminus B_r(y))}{r^n}$$ $\endgroup$ – user228113 Apr 24 '15 at 14:14
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    $\begingroup$ @G.Sassatelli You're probably right. If so then after rescaling this is the same as the limit of $\lambda_n(B_1(x)\setminus B_1(y))$ as $y\to x$, and now it's clear. $\endgroup$ – Sean Eberhard Apr 24 '15 at 14:17
  • $\begingroup$ @G.Sassatelli Sorry, I've updated my question. $\endgroup$ – 0xbadf00d Apr 24 '15 at 14:43
  • $\begingroup$ @SeanEberhard Could you please provide an answer? It's still not that clear to me. $\endgroup$ – 0xbadf00d Apr 24 '15 at 17:33
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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.

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  • $\begingroup$ Since $y\in\mathbb{R}^n$ is a vector, $1-y/r$ is not a "radius". Have you made a typo? $\endgroup$ – 0xbadf00d Apr 25 '15 at 21:58
  • $\begingroup$ Thanks, yes, I meant the modulus of $y$. It's fixed now. $\endgroup$ – Sean Eberhard Apr 26 '15 at 0:26

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