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Let $E,E_n \subset \mathbb{R}^N$ be measurable such that $E_n \subset E, E$ is bounded domain, $E_{n+1} \subset E$ and $\lim_{n\rightarrow \infty}L^N(E_n) = 0$. Are there $K_n$ compact such that $E_n \subset K_n \subset E$ and $\lim_{n\rightarrow \infty}L^N(K_n)=0$?

$L^N(A)$ denotes the $N$-dimensional Lebesgue measure of $A$.

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You are right. Is Ok now? –  user29999 Aug 23 '12 at 17:33

2 Answers 2

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No. For a counterexample, take $N=1$, enumerate the rationals in $(0,1)$ as $(q_k)_{k\geqslant1}$ and consider $E=(-3,+8)$, $A_n=\bigcup\limits_{k\geqslant1}\,(q_k-1/(2^kn),q_k+1/(2^kn))$ and $E_n=A_n\cap (0,1)$. Then each $E_n$ is dense in $[0,1]$ hence the only $K_n$ compact such that $E_n\subseteq K_n\subseteq E$ are such that $K_n\supseteq[0,1]$. And $\mathrm{Leb}(E_n)\leqslant1/n$ but $\mathrm{Leb}(K_n)\geqslant1$.

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How about taking $N=1$ and $E=]0,1[$, $E_{n}=]0,\frac{1}{n}[$ for all $n\in\mathbb{N}$. Then \begin{equation*} \lim_{n\to\infty}L^{1}(E_{n})=L^{1}(\bigcap_{n=1}^{\infty}E_{n})=L^{1}(\emptyset)=0, \end{equation*} but there exists no compact set $K_{n}$ (in $\mathbb{R}$ they are closed and bounded sets) so that $E_{n}\subset K_{n}\subset E$ for any $n\in\mathbb{N}$, because if $E_{n}\subset K_{n}$ for some compact $K_{n}$ then $\bar{E_{n}}\subset K_{n}$, but $0\in \bar{E_{n}}$ and $0\notin E$.

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