Let $A$ be a subset of $R^d$ such that $0 < \mu(A) < \infty$. Given $0 < \alpha < 1$, there exists a cube $Q$ such that $\mu(A ∩ Q) \geq α |Q|$,
where $\mu$ denotes the exterior Lebesgue measure. I think Caratheodory's Criterion will work here, but I could not get it.
Maybe this will work:
Suppose not. Cover $A$ with cubes $Q_i$ such that distinct cubes have disjoint interior and $$\mu(\cup _i Q_i -A)< (1 - \alpha ) \mu (A) $$
But we also have
$$\mu (\cup Q_i -A) = \Sigma _i \mu (Q_i - A \cap Q_i ) > \Sigma _i \mu (Q_i)(1- \alpha) = \mu (\cup Q_i)(1- \alpha) \geq \mu (A) (1- \alpha)$$
Contradiction.