The behavior of BV functions at a point of approximate continuity Given $u\in BV(\mathbb R^N)$, we say $u$ is approximate continuos at $x$ and the approximation limit is $l\in R$ if
$$  \lim_{r\to 0}\frac{\mathcal{L}^N(B(x,r)\cap \{|u-l|>\epsilon\})}{r^N} =0 $$
for all $\epsilon>0$.  (I know the approximate continuity can be defined for even just a barely measurable function. But given I am studying $BV$, let's keep $u\in BV$ and maybe it is useful).
Now fix any $x_0\in \mathbb R^N$ such that $u$ is approximate continuous at this point $x_0$. My first question is: for a fixed $\epsilon_0>0$, would it be possible to have $r_{\epsilon_0}$ defined such that 
$$ \mathcal{L}^N(B(x,r)\cap \{|u-l|>\epsilon_0\})=0 $$
for all $r<r_{\epsilon_0}$
My second question is not related with above. Given two conditions: for a fixed $x_0\in R^N$, 
$$  \lim_{r\to 0}\frac{\mathcal{L}^N(B(x_0,r)\cap \{u>l+\epsilon\})}{r^N} =0, \,\,\text{ for all }\epsilon>0 \tag{1}$$

(I'll answer the first question; please post the second one separately.)
The answer is negative. In $N\ge 2$ dimensions, let $\xi$ be some unit vector and define 
$$
u(x) =\sum_{n=1}^\infty \left(1-4^n |x-2^{-n}\xi|\right)^+
$$
This function is in BV, because the BV norm of the $n$th term (i.e., $L^1$ norm of its gradient)  is of order $4^n4^{-nN} = 4^{-(N-1)n}$.  It is approximately continuous at the origin because the set $\{u\ne 0\}$ is the union of balls $B(2^{-n} \xi, 4^{-n})$ which has Lebesgue density $0$ at the origin. Yet, for every $r>0$ the set $\{u>1/2\}$ intersects $B(0,r)$ in a set of positive measure. 
