Let A be a real set then is it true that $A$ is measurable if and only if
$\forall\epsilon$, $\exists$ open set $G$ and closed set $H$
such that $H\subset A\subset G$ and $\mu(G|H)<\epsilon$.
2026-05-05 17:55:36.1778003736
$A$ is measurable if and only if $\forall\epsilon$, $\exists$ open set $G$ and closed set $H$ such that $H\subset A\subset G$ and $\mu(G|H)<\epsilon$
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It is when $\mu$ is Lebesgue measure.
If $E$ is measurable then the open and closed set approximation is a consequence of the regularity of Lebesgue measure.
Conversely, since Lebesgue measure is complete and all Borel sets are measurable, if $E$ satisfies regularity condition, we can find $F\in F_\sigma,G\in G_\delta$ such that $\mu(G-F)=0,F\subset E\subset G$. Now $E=F\cup (E-F)$ and $E-F$ is contained in a measure zero set and thus measurable, so $E$ is also measurable.