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!



Question regarding to approximate continuity
Based on the following observation, I believe that conditions $(1)$ and $(2)$ are not equivalent. 


Since the set 
$$B(r)=:\{B(x\_0,r)\cap \{u>l\}\}=\bigcup\_{n}\{B(x\_0,r)\cap \{u>l+\frac{1}{n}\}\}:=\bigcup\_{n}B\_n(r),$$
and $B\_n(r)\subset B\_{n+1}(r)\subset\cdots\to B(r)$, we know that 
$$\lim\_{n\to \infty}\mathcal{L}^N\big(B\_n(r)\big)=\mathcal{L}^N\big(B(r)\big)<\infty.$$
To prove that $(1)$ implies $(2)$, one necessarily needs to interchange the limits, namely
$$\lim\_{r\to 0}\lim\_{n\to \infty}\frac{\mathcal{L}^N\big(B\_n(r)\big)}{r^N}=\lim\_{n\to\infty}\lim\_{r\to \infty}\frac{\mathcal{L}^N\big(B\_n(r)\big)}{r^N},$$
for which there is no obvious reason why this should hold.


Let us do a counter-example along the following lines: suppose $N=2$, $x\_0=0$ and $l=0$, we may define a function $u:B(0,1)\to \mathbb{R}$ such that for each $r\in (0,1)$, the set $B\_n(r)$ is equal to the cusp area of degree $r\log^{-\frac{1}{n}}(1/r)$, namely 
$$B\_n(r)=\{(x,y):0<x<r,\ 0<y<x\log^{-\frac{1}{n}}(1/x)\}.$$
Then it is clear that 
$$\lim\_{r\to 0}\frac{\mathcal{L}^N\big(B\_n(r)\big)}{r^N}=0,$$
since $0$ is the tip of the cusp. On the other hand, $B\_n$ increases to the cone 
$$B(r)=\{(x,y):0<x<r,\ 0<y<x\},$$
for which 
$$\lim\_{r\to 0}\frac{\mathcal{L}^N\big(B(r)\big)}{r^N}=\frac{1}{4}>0.$$
So this means you probably need some other assumption on the function $u$.


Here is a counter-example to the first question.
Let $u:B(0,1)\to \mathbb{R}$ be defined as follows:


First, we set $f(x)=|x|$, this function is countinuous at $0$. We modify the function $f$ such that it is approximately continuous at $0$, but your first conclusion fails in the following way:


Fix a decreasing sequence $\{a\_n\}$ such that 
$$\lim\_{n\to \infty}a\_n=0.$$
Set $A\_k=\in B(0,2^{-k})\backslash B(0,2^{-k-1})$, choose a small ball $B\_k$ of radius $\alpha\_k=a\_k 2^{-k-1}$ and re-define the value of $f$ on $B\_k$ by setting $f=1+a\_k$ on $B\_k$, for $k\in \mathbb{N}$. Then $f$ is approximately continuous at $0$ since $\lim a\_k=0$, however, your first claim fails since at each scale, we have a hole $B\_k$ such that it is of size $a\_k2^{-k-1}$, which is of positive lebesgue measure.


