Question in the proof of Set Point Difference by One from a Compact has Measure Zero

Question

I am reading the top voted proof here given in To show that the set point distant by 1 of a compact set has Lebesgue measure $0$. Granted a great proof, I can not prove the first line of inequality: $$m(B_\epsilon(x_0) \cap B) \geq \frac{1}{2^n} m(B_\epsilon(x_0)).$$ I can see this from a picture drawn in 2D as the quadrant of $B_\epsilon(x_0)$ is obviously contained in $B_\epsilon(x_0 \cap B)$. But how does one show this rigorously using properties of measure? How would one use, say, scaling factors of Lebesgue measures in $\mathbb{R}^n$? I see that this inequality is the same as $$ m(B_\epsilon(x_0) \cap B) \geq m(B_\frac{\epsilon}{2}(x_0)). $$ But then what?

Answer 1

The crucial claim here is that, since $x_0$ is on the boundary of $B_1(y_0)$, we can fit an orthant $\tilde{B}_\epsilon(x_0)$ of $B_\epsilon(x_0)$ inside $B_\epsilon(x_0) \cap B_1(y_0)$ (we'll look at this claim carefully in a bit). An orthant is an intersection of a quadrant with a sphere, so there are $2^n$ rotated copies of $\tilde{B}_\epsilon(x_0)$ inside $B_\epsilon(x_0)$. Using the translation- and rotation-invariance of the Lebesgue measure is what thus allows us to say that $$ m(\tilde{B}_\epsilon(x_0)) = \frac{1}{2^n} m(B_\epsilon(x_0)), $$ which renders $$ m(B_\epsilon(x_0)) \geq m(B_\epsilon(x_0) \cap B_1(y_0)) \geq m(\tilde{B}_\epsilon(x_0)) = \frac{1}{2^n} m(B_\epsilon(x_0)). $$

Now, we'll show that we can chose an orthant $\tilde{B}_\epsilon(x_0) \subseteq B_\epsilon(x_0) \cap B_1(y)$. By translation and rotation invariance, we can assume without loss of generality that $x_0 = (0, \dotsc, 0)$ and $y_0 = (1/\sqrt{n}, \dotsc, 1/\sqrt{n})$. We choose this so that we can use the positive orthant $$ \tilde{B}_\epsilon(0) = \{\mathbf{x} \in B_\epsilon(0) : x_i \geq 0 \text{ for all } i\}. $$ And any $\epsilon < 2/\sqrt{n}$ (in the original proof, they choose $\epsilon < 1/2$, but I think that this is actually a mistake - correct me if I'm wrong!).

One of the ways of showing that $\tilde{B}_\epsilon(0) \subseteq B_1(y)$ is to show that the $[0, \epsilon]^n \subseteq B_1(y)$, since $\tilde{B}_\epsilon(0) \subseteq [0, \epsilon]^n$. To to this, we need only show that $B_1(y)$ contains the vertices of $[0, \epsilon]^n$, each of which have $k$ entries with value $\epsilon$ and $n-k$ entries with value $0$s. Thus, for any such vertex $v$, we have \begin{align*} d(v, y_0)^2 &= k (\epsilon - 1/\sqrt{n})^2 + (n-k) \cdot (0- 1/\sqrt{n})^2 \\ &= k\epsilon^2 - \frac{2k\epsilon}{\sqrt{n}} + 1 \\ &\leq 1, \end{align*} where in the last line we use the fact that $\epsilon < 2/\sqrt{n}$. Hence, $v \in B_1(y_0)$, and so we are done.