Let $\Omega \subset \mathbb{R}^n$ be a bounded and convex open set. Let $u \in W^{1,p}(\Omega)$ for some $p \ge 1$ and suppose we are able to prove:
"If $x,y \in \Omega$ are two distinct Lebesgue points of $u$, then:
$$ |u(x)-u(y)| \le K |x-y| $$
where $K$ is a positive constant."
Can I conclude that $u$ admits a Lipschitz representative? Intuitively I think that the answer is yes, but I can't really get why. By the well-known Lebesgue's theorem, almost every point of $\Omega$ is a Lebesgue point for $u$, then it is natural to try and define $\overline{u} \in W^{1,p}$ as equal to $u$ at every Lebesgue point of $u$. Following the same idea behind a Lipschitz extension to the boundary, if $z \in \Omega$ is not a Lebesgue point for $u$ I would be tempted to write:
$$ \overline{u}(z) := \lim_{x \rightarrow z} u(x) $$
Before talking about the Lipschitz condition, how can I even prove that such limit exists for every $z$ since none guarantees that $z$ is approached along Lebesgue points of $u$?
You always have Kirszbraun's theorem to rely on: if $E \subset \mathbb R^n$ and $f : E \to \mathbb R$ is Lipschitz, then $f$ has an extension to the whole space $\mathbb R^n$ with the same Lipschitz constant.
In this case take $E$ to be the set of Lebesgue points of $f$.