Given $u\in BV(R^N)$, we say $u$ is approximate continuous 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 $x_0\in R^N$ 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 $x_0\in R^N$,
$$ \lim_{r\to 0}\frac{\mathcal{L}^N(B(x_0,r)\cap \{u>l+\epsilon\})}{r^N} =0, \,\,\text{ for all }\epsilon>0\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
$$ \lim_{r\to 0}\frac{\mathcal{L}^N(B(x_0,r)\cap \{u>l\})}{r^N} =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
Clearly we have condition $(2)$ implies condition $(1)$. But can condition $(1)$ imply condition $(2)$ as well?
Thanks so much!