Let $A\subset \mathbb{R}^n$ be a bounded set and $\alpha \in (0,1)$ such that: $\mu^*(A\cap Q)≤\alpha\cdot\mu(Q)$. Show that A is the null set.

Question

Let $A\subset \mathbb{R}^n$ be a bounded set and $\alpha \in (0,1)$ such that for all cuboids $Q\subset\mathbb{R}^n$: $$\mu^*(A\cap Q)≤\alpha\cdot\mu(Q)$$ Show that A is the null set.

How to begin? What conclusions can I draw from A being bounded?

Answer 1

Let $\{ Q_n \}$ be a countable cover for $A$, then

$\mu^*(A) = \mu^*(\bigcup_n Q_n \cap A) \leq \sum_n \mu^*(Q_n \cap A) \leq \alpha \sum_n \mu(Q_n)$.

Now use the definition of $\mu^*(A) = \inf \left\{ \sum_i \mu(Q_i) \ | \ A \subset \bigcup_i Q_i \right\}$.

Answer 2

Let $\beta< 1 $ a bit larger than $\alpha$.

Let's first agree on this: for every box $Q$ there exists a countable family of boxes $(Q_i)$ inside $Q$ that cover $Q\cap A$ and such that $$\sum \mu(Q_i) \le \beta\cdot \mu(Q)$$ Repeat this procedure several times ($N$ times) to get a countable family of boxes covering $Q\cap A$ with sum of volumes $\le \beta^N \cdot \mu(Q)$.