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    Approximation of measurable sets by open, closed, and compact sets

    T-IIKSW

    Measurable sets are arbitrarily close in measure to open supersets and closed subsets, and finite-measure ones to compact subsets.Suggest a slogan

    Let \(E\subseteq\RR^n\) be Lebesgue measurable and let \(\varepsilon>0\).

    1. There exists an open set \(O\supseteq E\) with \(m(O\setminus E) < \varepsilon\).

    2. There exists a closed set \(F\subseteq E\) with \(m(E\setminus F) < \varepsilon\).

    3. If \(m(E)<\infty\), there exists a compact set \(K\subseteq E\) with \(m(E\setminus K) < \varepsilon\) [@Fol13].

    The finiteness hypothesis in (3) cannot be dropped. For \(E=\RR^n\), every compact \(K\subseteq\RR^n\) is bounded, hence has finite measure, so \(m(\RR^n\setminus K)=\infty\).

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