Show that $K$ has measure zero

Question

This is a problem from my measure theory book:

Let $K$ be a compact subset of $\mathbb{R}^d$ such that the intersection $H_r(K)\cap H_{r'}(K)$ of two homothetic images ($H_r(x)=rx$ for $x\in \mathbb{R}^d$ and $r\in\mathbb{R}$) of $K$ has Lebesgue-Borel measure zero whenever $0<r<r'<1$. Prove that $\lambda^d(K)=0$. Hint: $H_r(K) \subset \tilde{K}=\{tx:0\leq t\leq 1, x\in K \}$ which is a compact set. Hence $\lambda^d(\tilde{K})<\infty$.

I know that $\lambda^d(H_r(K))=|r|^d \lambda^d(K)$ approaches $\lambda^d(K)$ as $r$ approaches $1$, but not sure where I can go from there.

Any help is greatly appreciated.

Answer 1

Let's assume that $\lambda^d(K) > 0$. We consider the sets $H_{r_n}$, where $r_n = \frac{1}{2}+\frac{1}{n+2}$. We know that the $H_{r_n}$ are pairwise "almost disjoint", and included in $\overline{K}$.

Can you conclude from there, using what you have already noticed ?

Answer 2

Following FiMePr's suggestion, let $(r_n)\subset(0,1)$ be the sequence defined by $r_n=\frac{1}{2}+\frac{1}{n+2}$. For each $n$, let $A_n=H_{r_n}(K)$. The sequence $(A_n)$ is almost disjoint, in the sense that $\lambda^d(A_i \cap A_j)=0$ whenever $i\neq j$. Using this, one can show that

$$\lambda^d \bigg(\bigcup_{i=1}^n A_i \bigg)=\sum_{i=1}^n \lambda^d(A_i)$$ for each $n$. Then, using continuity of measure from below, we get

$$\lambda^d \bigg(\bigcup_{n=1}^\infty A_n \bigg)=\sum_{n=1}^\infty \lambda^d(A_n)$$

Now, since $\bigcup_{n=1}^\infty A_n \subset \tilde{K}$ we get

$$\sum_{n=1}^\infty \lambda^d(A_n)\leq \lambda^d(\tilde{K})<\infty$$

which implies $\lambda^d(A_n) \to 0$ as $n\to \infty$. But $\lambda^d(A_n)=|r_n|^d \lambda^d(K)$ and $r_n \to \frac{1}{2}$ as $n\to \infty$, so this is possible only if $\lambda^d(K)=0$.