Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $E\subset\mathbb{R}^n$ be an open set with a zero-measure boundary. Let $f_k$ be a sequence of functions on $E$ such that $f_k\rightharpoonup f$ weakly in $H^1(E)$ and $f_k\to f$ a.e. on $E$ (but $(f_k)$ is not pointwise a.e. bounded).

By the Egorov's theorem, for any $\varepsilon>0$ there is a closed set $A_{\varepsilon}\subset E$ such that $m(E-A_{\varepsilon})\leq\varepsilon$ and $f_k\to f$ uniformly on $A_{\varepsilon}$ ($m$ is the Lebesgue measure).

Question: does it hold

$m(\partial (E-A_{\varepsilon}))=O(g(\varepsilon))$ when $\varepsilon\to0$, where $g(x)\to0$ as $x\to0$?

I know that $m(\partial (E-A_{\varepsilon}))=0$ if the sets $E-A_{\varepsilon}$ are Jordan-measurable, but I don't know if these sets are such. Also, I don't know how to use (if anyhow) the fact that $\nabla f_k$ are uniformly bounded in $L^2(E)$.

Thank you for any comments.

share|improve this question
1  
You can easily create an open set $E$ with the boundary of positive measure. How do you propose to get rid of it by removing a closed subset from $E$? (forget about functions, uniform convergence, etc., just think of this). –  fedja Nov 2 '12 at 2:37
    
Are you pointing out that the boundary of an opet set is contained in it's (closed) complement? This is not enough here. –  Ana Nov 2 '12 at 12:27
1  
Then you should state it this way from the beginning to avoid idiotic comments like mine :). OK, I'll think of this now when it makes more sense. –  fedja Nov 2 '12 at 13:26
2  
You are welcome. The best idea is to ask the question exactly as you have it not trying to generalize or sidestep (unless you have some really good reason to believe that your generalization is the right thing to do). So, if you edit your post and let us know what you are really after, somebody might be able to help :). –  fedja Nov 2 '12 at 15:46
1  
Unfortunately, $H^1$ does not prevent the functions from having dense sets of singularities in dimensions $n>1$, so my last counterexample still stands: $f_n=f+\frac 1n g$ where $g$ is some function with finite $H^1$ norm but a dense set of bad points. –  fedja Nov 9 '12 at 4:57
show 8 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.