Let \(m_*(E)\) denote the Lebesgue outer measure of a set \(E \subseteq \RR^n\).
Prove using the definition of Lebesgue outer measure that \[ m \qty{ \Union_{j=1}^{\infty } E_j } \leq \sum_{j=1}^{\infty } m_*(E_j) .\]
Prove that for any \(E \subseteq \RR^n\) and any \(\epsilon> 0\) there exists an open set \(G\) with \(E \subseteq G\) and \[ m_*(E) \leq m_*(G) \leq m_*(E) + \epsilon .\]
Solution
Recall \(m_*(E) = \inf\sum_k |Q_k|\), the infimum over countable covers of \(E\) by closed boxes \(Q_k\).
1. \(m_*(\bigcup_{j\ge 1} E_j) \le \sum_{j\ge 1} m_*(E_j)\).
If \(\sum_j m_*(E_j) = \infty\) there is nothing to prove. Otherwise fix \(\eps > 0\) and, for each \(j\), choose a cover \((Q_{j,k})_k\) of \(E_j\) by boxes with \(\sum_k |Q_{j,k}| \le m_*(E_j) + \eps 2^{-j}\). Then \((Q_{j,k})_{j,k}\) is a countable cover of \(\bigcup_j E_j\), so \(m_*(\bigcup_j E_j) \le \sum_{j,k} |Q_{j,k}| \le \sum_j m_*(E_j) + \eps\).
2. For every \(E \subseteq \RR^n\) and \(\eps > 0\) there is an open \(G \supseteq E\) with \(m_*(E) \le m_*(G) \le m_*(E) + \eps\).
If \(m_*(E) = \infty\) take \(G = \RR^n\). Otherwise choose a cover \((Q_k)\) of \(E\) by boxes with \(\sum_k |Q_k| \le m_*(E) + \eps/2\), and open boxes \(U_k \supseteq Q_k\) with \(|U_k| \le |Q_k| + \eps 2^{-k-1}\). Then \(G = \bigcup_k U_k\) is open and contains \(E\), so \(m_*(E) \le m_*(G)\) by monotonicity. The closures \(\overline{U_k}\) cover \(G\), so \(m_*(G) \le \sum_k |U_k| \le m_*(E) + \eps\).