# Measure of the boundary of the exceptional sets in the Egorov's theorem

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 $L^2(E)$.

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
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