Measure of the boundary of the exceptional sets in the Egorov's theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:27:08Z http://mathoverflow.net/feeds/question/111229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111229/measure-of-the-boundary-of-the-exceptional-sets-in-the-egorovs-theorem Measure of the boundary of the exceptional sets in the Egorov's theorem Ana 2012-11-02T00:04:15Z 2012-11-02T19:54:08Z <p>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). </p> <p>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). </p> <p>Question: does it hold </p> <p>$m(\partial (E-A_{\varepsilon}))=O(g(\varepsilon))$ when $\varepsilon\to0$, where $g(x)\to0$ as $x\to0$?</p> <p>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 $L^2(E)$.</p> <p>Thank you for any comments.</p>