Show that \(E\) is measurable iff \(E\) is regular.
Solution
Setup note. For Lebesgue measure on \(\RR^n\), “regular” means: \(\mu^*(E) = \inf\{\mu(G) : G \supseteq E,\ G \text{ open}\}\) (outer) and \(\mu_*(E) = \sup\{\mu(F) : F \subseteq E,\ F \text{ compact}\}\) (inner, with \(\mu_*(E) = \mu^*(E)\)). We prove measurability is equivalent to regularity.
1. Measurable \(\Rightarrow\) outer regular.
for measurable \(E\) with \(\mu(E) < \infty\), by definition of \(\mu^*\) there is a cover by boxes with total measure \(< \mu(E) + \eps\); the open set \(G\) obtained by enlarging the boxes slightly contains \(E\) and has \(\mu(G) \le \mu(E) + \eps\).
2. Measurable \(\Rightarrow\) inner regular.
for measurable \(E\) with \(\mu(E) < \infty\), first approximate by closed sets: \(E^c\) is measurable, so by 1 there is open \(G \supseteq E^c\) with \(\mu(G \setminus E^c) < \eps\), giving closed \(F = G^c \subseteq E\) with \(\mu(E \setminus F) = \mu(G \setminus E^c) < \eps\).
3. Regular \(\Rightarrow\) measurable.
outer regularity gives open \(G_k \supseteq E\) with \(\mu(G_k) \le \mu^*(E) + 1/k\); let \(V = \cap_k G_k\), a \(G_\delta\) containing \(E\), with \(\mu(V) = \mu^*(E)\).
4. Q.E.D.