Tricky measure theory question

[AxiomOfChoice]

One possible definition of measurability is this: A set [tex]E \subseteq \mathbb R^d[/tex] is (Lebesgue) measurable if for every [tex]\epsilon > 0[/tex] there exists an open set [tex]\mathcal O \supseteq E[/tex] such that [tex]m_*(\mathcal O \setminus E) < \epsilon[/tex]. Here, [tex]m_*[/tex] indicates Lebesgue outer measure.

Apparently, an equivalent definition is this: "For every [tex]\epsilon > 0[/tex] there exists a closed set [tex]F \subseteq E[/tex] such that [tex]m_*(E\setminus F) < \epsilon[/tex]."

Showing the equivalence of these definitions was a practice problem recently for the final exam in my real analysis class. But I couldn't get it, and even though I'm on break now, it's bugging me. Can someone help? Thanks! (This is also apparently a problem in Stein-Shakarchi's textbook, Real Analysis.)

 

[quasar987]

==>: Suppose E is measurable. Then E^c is measurable. Let O be the open set associated to E^c as in the definition of measurability. Then use F=O^c.

<==: Same thing, just use the open/closed duality in the same way.