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# JordanmeasurabilityMeasure of the openboundaryoftheexceptional sets appearing in the Egorov's theorem

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$.L^2(E)$.

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By the Egorov's theorem, if $E$ is a measurable set and ${f_k}$ is a sequence of measurable functions defined on $E$ such that $f_k\to f$ a.e. on $E$, then 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 $E\subset\mathbb{R}^n$ is an open set with a zero-measure boundary. Moreover, $f_k-f$ is uniformly bounded in $H^1(E)$ (but not pointwise a.e. bounded).

Question: are the sets $E-A_{\varepsilon}$ Jordan-measurable? Actually, for my purpose it would be enough if their boundaries had the Lebesgue measure $0$, or at least $m(\partial(E-A_{\varepsilon}))\leq C\varepsilon$.

Thank you.

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By the Egorov's theorem, if $E$ is a measurable set and ${f_k}$ is a sequence of measurable functions defined on $E$ such that $f_k\to f$ a.e. on $E$, then 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 $E\subset\mathbb{R}^n$ is an open set with a zero-measure boundary.

Question: are the sets $E-A_{\varepsilon}$ Jordan-measurable? Actually, for my purpose it would be enough if their boundaries had the Lebesgue measure $0$.0$, or at least$m(\partial(E-A_{\varepsilon}))\leq C\varepsilon\$.

Thank you.

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