# Open and dense is full Lebesgue measure?

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 $${\displaystyle . Then the image $$f(X)$$ has Lebesgue measure $$0$$ in $$\mathbb {R} ^{m}$$.

Thank you in advance.

• The complement of a fat Cantor set is an open dense set in he real line which does not have full measure. Jun 8, 2020 at 10:09
• Thanks a lot for your example. And do you know any sufficient conditions for full measure set? @KaviRamaMurthy Jun 8, 2020 at 10:13

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 $$\Bbb R^n$$.