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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!

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1 Answer 1

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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.

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  • $\begingroup$ Yes. You counter-example indeed works. What do you think about my first question? $\endgroup$
    – JumpJump
    Dec 4, 2014 at 15:33
  • $\begingroup$ I think there should be a simple counter-example to your first question, but I cannot write down it explicitly yet. I will add it later. All the concepts of approximate (limits,continuity,differentiability, ...) just means that in the limiting case, the set where f fails the property has measure zero. In principle, this means at arbitrary small scale, bad sets can have smaller portion. $\endgroup$ Dec 4, 2014 at 16:05
  • $\begingroup$ Well... I mentioned this problem because I saw a fairly similar statement in Evans & Gariepy's book, page 214, equation $(\ast)$...It states something similar, i.e., without take $r\to 0$... $\endgroup$
    – JumpJump
    Dec 4, 2014 at 16:07
  • $\begingroup$ so you are saying that it is a typo in the book? $\endgroup$
    – JumpJump
    Dec 4, 2014 at 16:11
  • $\begingroup$ Thx for your answer and comment! I'll keep work on this to try to find a counterexample myself before you offer me one :) $\endgroup$
    – JumpJump
    Dec 4, 2014 at 16:13

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