The "lower part" of $BV$ function is always lower semi-continuous. Let $u\in BV(\Omega)$ be a function of bounded variation, where $\Omega\subset\mathbb R^N$ is open bounded smooth boundary. Define 
$$
u^-(x):=\sup\left\{t\in\mathbb R:\,\lim_{r\to 0}\frac{\mathcal L^N(B(x,r)\cap\{u<t\})}{r^N}=0\right\}.
$$
Then $u^-$ is $\mathcal H^{N-1}$ a.e. well defined. 
My question: do we have $u^-$ is lower semi-continuous? In one dimension this is true. But what about multi-dimensions?
PS: by lower semi-continuous we mean for any $x_n\to x$, we have 
$$
\liminf_{n\to \infty} u^-(x_n)\geq u^-(x)
$$

Not necessarily. Consider $u(x)=\chi_E(x)$ where $E$ is a set which has an inner cusp, for instance the subset of $\mathbb R^2$ given by
$$E=\left\{(x_1,x_2):\,x_2< \sqrt{|x_1|}\right\}. $$
Then $u$, restricted to a bounded open set, is BV,  and $u^-(x)=u(x)$ for every $x$ except for the origin, in which $u^-(0)=1$, therefore lower semicontinuity fails for example along the sequence $(0,\frac1n)$.
