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Let $m$ be Lebesgue measure and $A$ a Lebesgue measurable subset of $\mathbb{R}$ with $m(A) < \infty$. Let $\epsilon > 0$. Does there exist $G$ open and $F$ closed such that $F \subset A \subset G$ and $m(G - F) < \epsilon$?

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Yes. Note that if $X\subset Y$ then $m(X-Y)=m(X)-m(Y)$. By regularity of Lebesgue measure you have open $G$ and closed $F$ such that $m(G)-m(A)<\varepsilon/2$ and $m(A)-m(F)<\varepsilon/2$. Then add the two inequalities to get that $m(G-F)=m(G)-m(F)<\varepsilon/2+\varepsilon/2=\varepsilon$.

  • What was wrong with the @mathguy counterexample? I can't seem to find where his is wrong. – layman Jul 01 '16 at 02:48
  • @user46944 There is an open set containing $[0,1]\cap \Bbb{Q}$ that does not have measure $1$. –  Jul 01 '16 at 02:49
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    @user46944, An open set can be both dense and 'highly porous'. – Sangchul Lee Jul 01 '16 at 02:51
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    @user46944 Enumerate the rationals $q_i$ and take the intervals $(q_i-\varepsilon 2^{-i-1},q_i+\varepsilon 2 ^{-i-1})$. See http://math.stackexchange.com/questions/201410/open-measurable-sets-containing-all-rational-numbers –  Jul 01 '16 at 02:51
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    @SangchulLee Your description made me think of swiss cheese lol – layman Jul 01 '16 at 02:53
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    @user46944 Oops, :s the description 'porous' may be misleading... My mental image is that such a set looks like a swiss cheese with 'infinitesimal holes'. – Sangchul Lee Jul 01 '16 at 02:58