The infinite set of $SBV$ function?

Question

Let $u\in SBV(\Omega)$ where by $SBV$ we denote the special bounded variation function and $\Omega\subset \mathbb R^N$ is open bounded.

Let's identify $u$ by its approximation representative (see page 213 in Evans & Gariepy). Hence in this way $u$ is determined $\mathcal H^{N-1}$ a.e.. I am trying to study the following set $$ S_\infty:=\{x\in\Omega:\,\, u(x)=+\infty\} $$

Clearly, if $N=1$, then the set $S_\infty$ is empty since an one dimensional $BV$ function belongs to $L^\infty$.

Now let $N>1$. By Evans & gariepy, page 211, Theorem 2, we have $\mathcal H^{N-1}(S_\infty)=0$. (Note it is $\mathcal H^{N-1}$ measure but not $\mathcal L^N$ measure). Then, can we show that $\mathcal H^{N-1}(\overline{S_\infty})<+\infty$ too? (where by $\overline{S_\infty}$ we denote the closure of $S_\infty$)

If, in addition, say we do have $\mathcal H^{N-1}(\overline{S_\infty})<+\infty$ hold, Then can we deduce that $u\in L^\infty(\Omega\setminus (\overline {S_\infty})_\epsilon)$ for any $\epsilon>0$? where we define $$ (\overline {S_\infty})_\epsilon:=\{x\in\Omega:\,\,\operatorname{dist}(x,\overline {S_\infty})<\epsilon)\}. $$

Thank you!

Answer 1

No. It is a basic fact (see for example: Section 5.2.2, Example 4 in Evans, Partial Differential Equations, 1998) that there are Sobolev functions in $W^{1,1}(\Omega)\subset\mathrm{SBV}(\Omega)$ which are locally unbounded everywhere in $\Omega$. For example, consider $u(x)=\sum_{k=1}^\infty 2^{-k}|x-r_k|^{-\alpha}$ with $\alpha=\frac{N-1}{2}$, where $r_k$ is a countable dense subset of the unit ball $\Omega$. So $u$ has approximate limit $\infty$ at every $r_k$, and hence $\{r_k : k\in\mathbb{N}\}\subset S_\infty$. Therefore $\Omega\subset\overline{S_\infty}$.

Also the second question has a negative answer since $u$ can be smooth, unbounded and $\mathrm{SBV}$.