Compact set instead of closed set for $m(E\setminus A)<\epsilon$

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?

• Consider ther sets $A_n=A\cap [-n,n]$. – Tomás Feb 16 '16 at 19:59
• You can iff $m(E) < \infty.$ – zhw. Feb 16 '16 at 20:08

Let $E$ be the real numbers. Then whatever compact set $A$ you remove from $E$, $m(E\backslash A)=\infty$ right?