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For Lebesgue measure, we know that it is regular, and for any $\epsilon$ and Borel set $E$, there exists a closed set $A\subset E$ s.t. $m(E\setminus A)<\epsilon$ Can we replace closed set with compact set?

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  • $\begingroup$ Consider ther sets $A_n=A\cap [-n,n]$. $\endgroup$ – Tomás Feb 16 '16 at 19:59
  • $\begingroup$ You can iff $m(E) < \infty.$ $\endgroup$ – zhw. Feb 16 '16 at 20:08
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Let $E$ be the real numbers. Then whatever compact set $A$ you remove from $E$, $m(E\backslash A)=\infty$ right?

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