# Reduction of open set to dense open set with smaller measure

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!

• Do you know about fat Cantor sets? – 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.
• @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. – PhoemueX Dec 11 '15 at 18:53