$U \subset \mathbb{R}^d$ open and $D \subset U$ open and dense $\implies \lambda(D) = \lambda(U)$

Question

This is not a homework exercise!

Let $U \subset \mathbb{R}^d$ be and open subset and $D \subset U$ open and dense in $U$. Can we conclude that $\lambda(D) = \lambda(U)$?

Here, $\lambda$ denotes the $d$-dimensional Lebesgue measure.

This condition is trivially true for any $\lambda$-null set $U$. For the general case, I think this boils down to the question whether a $\lambda$-non null set $A$ such that $\lambda(A) > 0$ has a non-empty interior: $\mathring{A} \neq \emptyset$. Since then, we could argue that $\lambda(U \setminus D) = 0$, which implies the highlighted proposition.

I´d be grateful for a proof or counterexample for any of the two statements.

Thanks in advance :)

Answer 1

First, let $\tau_d := \frac{\pi^{\frac{d}{2}}}{\Gamma\left(\frac{n}{2} + 1\right)}$ the volume of the $d$-dimensional unit sphere and $\Gamma(x)$ the gamma function. Then, $\lambda(B(x,r)) = \tau_d r^d$.

As @copperhat suggests in the comments, a counterexample for the first statement goes as follows:

Let $L > 0$ be arbitrary. Let $U$ be the open unit sphere in $\mathbb{R}^d$ and $(q_k)_{k \in \mathbb{N}} \subset \mathbb{Q} \cap U$ an enumeration of all rational points in $U$. Then, let $$D := \bigcup_{k \in \mathbb{N}} B(q_k, \varepsilon_k),$$ where for each $k \in \mathbb{N}$ the number $\varepsilon_k > 0$ is chosen so that $$ B(q_k, \varepsilon_k) \subset U \qquad \text{and} \qquad \varepsilon_k < \sqrt[d]{\frac{L}{2^{k - 1} \cdot \tau_d}}. $$ Then, $$ \lambda(D) \le \sum_{n \in \mathbb{N}} \lambda\left(B(q_n, \varepsilon)\right) = \sum_{n \in \mathbb{N}} \tau_d \varepsilon_{k}^{d} < \sum_{n \in \mathbb{N}} \tau_d \cdot \frac{L}{2^{k - 1} \cdot \tau_{d}} = \sum_{n \in \mathbb{N}} \frac{L}{2^{k - 1}} = L. $$

Therefore, we can get $\lambda(D)$ to be as small as we want, and it will be still open as union of open sets an dense since it contain all $q \in \mathbb{Q} \cap U$.

@Will_M. gives a great counterexample for the second proposition: the fat cantor set.