Completion (6). Combine Problems 16.2 and 11.6 to show that the completion \(\bar{\lambda}^{n}\) of \(n\)-dimensional Lebesgue measure is again inner and outer regular.

Data from problem 16.2

Prove the following corollary to Lemma 16.12: Lebesgue measure \(\lambda^{n}\) on \(\mathbb{R}^{n}\) is outer regular, i.e.

\[
\lambda^{n}(B)=\inf \left\{\lambda^{n}(U): U \supset B, U \text { open }ight\} \quad \forall B \in \mathscr{B}\left(\mathbb{R}^{n}ight)
\]

and inner regular, i.e.

\[
\begin{aligned}
\lambda^{n}(B) & =\sup \left\{\lambda^{n}(F): F \subset B, F \text { closed }ight\} & & \forall B \in \mathscr{B}\left(\mathbb{R}^{n}ight) \\
& =\sup \left\{\lambda^{n}(K): K \subset B, K \text { compact }ight\} & & \forall B \in \mathscr{B}\left(\mathbb{R}^{n}ight) .
\end{aligned}
\]

Data from problem 11.6

 Completion (4). Inner measure and outer measure. Let \((X, \mathscr{A}, \mu)\) be a finite measure space. Define for every \(E \subset X\) the outer resp. inner measure

\[
\begin{aligned}
& \mu^{*}(E):=\inf \{\mu(A): A \in \mathscr{A}, A \supset E\}, \\
& \mu_{*}(E):=\sup \{\mu(A): A \in \mathscr{A}, A \subset E\}
\end{aligned}
\]

(i) Show that for all \(E, F \subset X\)

\[
\begin{array}{cc}
\mu_{*}(E) \leqslant \mu^{*}(E), & \mu_{*}(E)+\mu^{*}\left(E^{c}ight)=\mu(X), \\
\mu^{*}(E \cup F) \leqslant \mu^{*}(E)+\mu^{*}(F), & \mu_{*}(E)+\mu_{*}(F) \leqslant \mu_{*}(E \cup F) .
\end{array}
\]

(ii) For every \(E \subset X\) there exist sets \(E_{*}, E^{*} \in \mathscr{A}\) such that \(\mu\left(E_{*}ight)=\mu_{*}(E)\) and \(\mu\left(E^{*}ight)=\mu^{*}(E)\).

[ use the definition of ' \(\infty\) ' to find sets \(E^{n} \supset E\) with \(\mu\left(E^{n}ight)-\mu^{*}(E) \leqslant \frac{1}{n}\) and consider \(\bigcap_{n} E^{n} \in \mathscr{A}\).]

(iii) Show that \(\mathscr{A}^{*}:=\left\{E \subset X: \mu_{*}(E)=\mu^{*}(E)ight\}\) is a \(\sigma\)-algebra and that it is the completion of \(\mathscr{A}\) w.r.t. \(\mu\). Conclude, in particular, that \(\left.\mu^{*}ight|_{\mathscr{A}^{*}}=\left.\mu_{*}ight|_{\mathscr{A}^{*}}=\bar{\mu}\) if \(\bar{\mu}\) is the completion of \(\mu\).