- Show that if \(E\subseteq \RR^n\) is measurable with \(m(E) < \infty\), then \[m(E) = \sup \{ m(K) \mid K \subseteq E \text{ is compact}\}\] if and only if for all \(\eps > 0\), there exists a compact set \(K \subseteq E\) such that \(m(K) \geq m(E) - \eps\).
Solution
Let \(E \subseteq \RR^n\) be a Lebesgue measurable set with \(m(E) < \infty\), and consider the set of real numbers: \[ \mathcal{S} = \{ m(K) \mid K \subseteq E \text{ is compact} \}. \] By monotonicity of the Lebesgue measure, for every compact \(K \subseteq E\), we have \(m(K) \leq m(E)\). Thus \(m(E)\) is an upper bound for the set \(\mathcal{S}\).
\((\Longrightarrow)\) Suppose \(m(E) = \sup \mathcal{S}\): By the definition of the supremum of a set of real numbers: For any \(\eps > 0\), \(m(E) - \eps\) is strictly less than the supremum \(\sup \mathcal{S}\). Therefore, \(m(E) - \eps\) cannot be an upper bound for \(\mathcal{S}\). Hence, there exists an element \(m(K) \in \mathcal{S}\) (where \(K \subseteq E\) is compact) such that: \[ m(K) > m(E) - \eps \implies m(K) \geq m(E) - \eps. \]
\((\Longleftarrow)\) Suppose that for all \(\eps > 0\), there exists compact \(K_\eps \subseteq E\) with \(m(K_\eps) \geq m(E) - \eps\): 1. Since \(m(K) \leq m(E)\) for all compact \(K \subseteq E\), \(m(E)\) is an upper bound of \(\mathcal{S}\). Therefore \(\sup \mathcal{S} \leq m(E)\). 2. For each \(\eps > 0\), since \(m(K_\eps) \in \mathcal{S}\), we have: \[ \sup \mathcal{S} \geq m(K_\eps) \geq m(E) - \eps. \] 3. Taking the limit as \(\eps \to 0^+\) (or because the inequality holds for all \(\eps > 0\)): \[ \sup \mathcal{S} \geq m(E). \] Combining both inequalities gives: \[ m(E) = \sup \{ m(K) \mid K \subseteq E \text{ is compact} \}. \]