EX3-13

Problem: Define the inner measure of E by |E|_{i}=\text{sup}|F|, where the supremum is taken over all closed subset F ofE. Show that
(i)|E|_{i}\leq|E|_{e}, and
(ii) If |E|_{e}<+\infty, then E is measurable if and only if |E|_{i}=|E|_{e}.
Proof:
(i) By monotonicity of outer measure,|F|=|F|_{e}\leq|E|_{e} for all closed subset F\subseteq E. Thus, we have \text{sup}|F|=|E|_{i}\leq|E|_{e}.
(ii) Suppose E is measurable: then for every \epsilon>0, there exists a closed subset F\subseteq E such that |E-F|_{e}<\epsilon. Since E is measurable and |F|\leq|E|<+\infty,|E-F|_{e}=|E|-|F|<\epsilon. Thus, |E|_{e}=|E|\leq\text{sup}|F|+\epsilon=|E|_{i}+\epsilon for all \epsilon>0, so |E|_{e}\leq|E|_{i}, which yields the equality by part (i).
Conversely, assume that |E|_{e}\doteq|E|_{i}.Then given \varepsilon > 0, there exist an open set G and a closed set F with which F \subseteq E \subseteq G and such that|G| < \varepsilon + |E|_e \qquad \text{and} \qquad |E|_e < |F| + \varepsilon,or equivalently,|G| - |E|_e < \varepsilon \qquad \text{and} \qquad |E|_e - |F| < \varepsilon.Then since (E \setminus F) \subseteq (G \setminus F), the monotonicity of outer measure says that|E \setminus F|_e \le |G \setminus F| = |G| - |F| < 2\varepsilon,where in the second equality we have employed Corollary 3.25. We therefore employ Lemma 3.22 to conclude that E is measurable.