By the Egorov's theorem, if
Let $E$ is a measurable E\subset\mathbb{R}^n$ be an open set and with a zero-measure boundary. Let ${f_k}$ is f_k$ be a sequence of measurable functions defined on $E$ such that $f_k\rightharpoonup f$ weakly in $H^1(E)$ and $f_k\to f$ a.e. on $E$, then 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). Suppose that
Question: does it hold
$E\subset\mathbb{R}^n$ is an open set with a zero-measure boundary. Moreover, m(\partial (E-A_{\varepsilon}))=O(g(\varepsilon))$ when $f_k-f$ is uniformly bounded in \varepsilon\to0$, where $H^1(E)$ g(x)\to0$ as $x\to0$?
I know that $m(\partial (but not pointwise a.e. bounded).
Question: are E-A_{\varepsilon}))=0$ if the sets $E-A_{\varepsilon}$ are Jordan-measurable? Actually, for my purpose it would be enough but I don't know if their boundaries had these sets are such. Also, I don't know how to use (if anyhow) the Lebesgue measure fact that $0$, or at least \nabla f_k$ are uniformly bounded in $m(\partial(E-A_{\varepsilon}))\leq C\varepsilon$.L^2(E)$.
Thank you for any comments.
