For every $E \subseteq \mathbb{R}^d$ with $0 < |E|_e < \infty$ and $0 < a < 1$ there exists a cube Q such that $|E \cap Q|_e \geq a|Q|$

Question

For every $E \subseteq \mathbb{R}^d$ with $0 < |E|_e < \infty$ and $0 < a < 1$ there exists a cube Q such that $|E \cap Q|_e \geq a|Q|$

Here the exterior Lebesgue measure $|E|_e$ is defined as $inf \lbrace \sum vol(Q_k) : Q_k$ cover $E \rbrace$ where $Q_k$ indicates closed generalized rectangles (e.g. intervals, rectangles, rectangular prisms) whose sides are all parallel to coordinate axes of $\mathbb{R}^d$.

I've seen a solution to this problem solved in $\mathbb{R}$, but it appears to rely on the fact that open sets in $\mathbb{R}$ are disjoint countable unions of open intervals, which as far as I know doesn't generalize to higher-dimensional open sets being unions of open cubes.

The problem comes with a hint saying only "$\frac{1}{a} |E|_e$", which I can't really make sense of; I tried turning the guaranteed covering of E by generalized rectangles into a covering by cubes and showing that if none of those cubes has the desired property then $|E|_e = 0$, but my argument didn't really work because it involved a sum whose terms didn't actually go to zero at all.

Answer 1

Hint: Take a cover of $E$ by cubes $Q_k$ so that $$\sum |Q_k| \le \frac{1}{a} \cdot |E|_e$$

Now look at the pieces $|E\cap Q_k|_e$.