Open and dense is full Lebesgue measure?

Question

Consider the Euclidean space $\mathbb{R}^n$. Is an open and dense subset of $\mathbb{R}^n$ is full Lebesgue measure? Also, I want to know the sufficient conditions for a subset in $\mathbb{R}^n$ is full Lebesgue measure or a null set. Moreover, I want to know if there is any other theorem like Sard's theorem:

Let $f\colon \mathbb{R}^n \rightarrow \mathbb{R}^m$ be $C^{k}$, (that is, $k$ times continuously differentiable), where $k\geq \max\{n-m+1, 1\}$ and $X$ denote the critical set of $f$, which is the set of points $x\in \mathbb {R} ^{n}$ at which the Jacobian matrix of $f$ has rank ${\displaystyle <m}$. Then the image $f(X)$ has Lebesgue measure $0$ in $\mathbb {R} ^{m}$.

Thank you in advance.

Answer 1

If $C$ is a "fat Cantor set" (see here e.g. ) then $\Bbb R\setminus C$ is open and dense but does not have full measure. E.g. We could take one copy of such a set in each $[n,n+1]$ ($n \in \Bbb Z$) interval and the complement would be open and dense and have "half" the measure of $\Bbb R$ (and both sets, the closed and nowhere dense set and its complement would have infinite Lebesgue measure. We can extend such examples to any $\Bbb R^n$.

So no an open and dense set is not of full measure.

Answer 2

No. Here's an example much simpler than a fat Cantor set. Let $\{a_1, a_2, \cdots, a_n, \cdots\}$ be the set of all rational numbers in $(0,1)$. For each $i$, take a neighborhood of length $\frac{1}{2^{i+1}}$ near $a_i$ and denote it by $B_i$. Let $U=\cup B_i$. Then $|U| \le \Sigma |B_i| \le \frac{1}{2}$. Theore, $U$ is an open dense subset of the unit interval (since it contains all rational numbers), but has measure less than $1$.