Lower semicontinuous representative of positive Sobolev function? For a function u in the Sobolev space $W_0^{1,p} (\mathcal O )$, ($p \in [ 1, n ]$), having $u > 0$ inside $\mathcal O$, where $\mathcal O$ is an open bounded connected set in $\mathbb R^n$, can one always find a precise reprentative  of $u$ which is lower semicontinuous ? 
 A: For $n = 1$, this is obviously true.
I don't think it holds for $n > 1$. Let me sketch an example for $n = 2$. W.l.o.g. we assume that the unit square belongs to $O$. Let $\tilde F \subset \mathbb{R}$ be the fat cantor set. Then, $F = \tilde F \times (0,1)$ has positive measure. The set $U = (0,1)^2 \setminus F$ is open and dense in $[0,1]^2$, since $[0,1] \setminus \tilde F$ is dense in $[0,1]$. Hence, we find a countable set of points $\{x_n\} \subset U$, which are dense in $[0,1]^2$.
Since a point as capacity $0$, we can construct functions $f_n \in H_0^1(O) \cap C(\bar O)$ with


*

*$f_n \equiv 1$ on a small ball around $x_n$

*$0 \le f_n \le 1$ on $O$

*$\mathrm{supp} (f_n)$ lies in $U$

*$\| f_n \|_{H_0^1} \le 2^{-n}$


Then, for $f = \sum_{n=1}^\infty f_n$, we have


*

*$f \in H_0^1(O)$

*$f \ge 0$ on $O$

*$f \ge 1$ on a small ball around $x_n$

*$\mathrm{supp} (f)$ lies in $U$


Then, for any representative $\tilde f$ of $f$, and any $x \in [0,1]^2$, we find a sequence $\{y_n\}$, $y_n \to x$ and $\tilde f(y_n) \ge 1$. But $\tilde f$ is zero on $F$ a.e. This is a contradiction.
Edit: ups, this is actually lower semicontinuous. But you can consider something like $g - \min(f,1)$, for $g \in H_0^1(O) \cap C(\bar O)$ large enough.
