I want to prove the following statement:

Let $A \subseteq \mathbb{R}^d$ be an open set and $\alpha \in (0, \lambda^d(A))$, where $\lambda$ is the Lebesgue-measure on $\mathbb{R}^d$. Then there is some open subset $B$ of $A$ which is dense in $A$ and has measure $\lambda^d(B)=\alpha$.

So far every attempt of mine failed. I think that I might somehow have to extract some nowhere dense set from $A$. Thanks!

  • 1
    $\begingroup$ Do you know about fat Cantor sets? $\endgroup$ – Daniel Fischer Dec 11 '15 at 15:37

First, let $(q_n)_{n \in \Bbb {N}_0} $ be an enumeration of $\Bbb {Q}\cap A $. Define

$$ B_0 := \bigcup_n (B_{\alpha /(c \cdot 2^{n})}(q_n) \cap A) $$ for a suitable $c>0$ which we will choose below.

Note that $B_0 \subset A $ is open and dense with $$ \lambda (B_0) \leq \sum \lambda (B_{\alpha /(c \cdot 2^{n})}(q_n))= \lambda (B_1 (0)) \sum_n \frac {\alpha^d}{c^d 2^{dn}} = \lambda (B_1 (0)) \frac{\alpha^d}{c^d} \cdot \frac{1}{1-2^{-d}} <\alpha $$ for suitable $c $.

Now, consider the map $$ \Phi : [0,\infty) \to [0,\infty), t \mapsto \lambda ( [B_0 \cup (-t,t)^d] \cap A). $$ Show that this map is continuous with $\Phi (0)< \alpha$ and $\Phi (t) \to \lambda (A) >\alpha $ as $t\to \infty $. Now apply the intermediate value theorem.

  • 1
    $\begingroup$ omg great. how did you know how to prove this? $\endgroup$ – Joker123 Dec 11 '15 at 18:31
  • $\begingroup$ @Joker123: I knew the proof for showing that a set $B $ always contains a set of measure $A $ of measure $\alpha $, for arbitrary $0\leq \alpha <\lambda (B ) $. This is essentially the second part of the proof. I also knew how to construct dense open sets with small measure (a variant of the first part). Then I combined both arguments. $\endgroup$ – PhoemueX Dec 11 '15 at 18:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.