TITLE: Remark 11 - Theorem 9.12 (Morrey) - Brezis - Pag 282
QUESTION [5 upvotes]: I'm not sure about Remark 11.
(i) Why is $\mathbb{R}^{N}\setminus A$ dense in $\mathbb{R}^{N}$???
(ii) How to ensure that the ongoing representative he defined is unique?

REPLY [6 votes]: For (i), since $A$ has measure zero, it cannot contain any open ball, so its complement is dense.
For (ii), $u|(\mathbb R^N\setminus A)$ is continuous, and it's straightforward to define an extension $\hat u$ by $\hat u(x) = \lim_{y\to x,y\in \mathbb R^N\setminus A}u(y)$. The continuity and uniqueness of $\hat u$ should be simple to check.