Completion (3). Let \((X, \overline{\mathscr{A}}, \bar{\mu})\) be the completion of \((X, \mathscr{A}, \mu)\), see Problems 4.15 , and 6.4.

(i) Show that for every \(f^{*} \in \mathcal{E}^{+}(\overline{\mathscr{A}})\) there are \(f, g \in \mathcal{E}^{+}(\mathscr{A})\) with \(f \leqslant f^{*} \leqslant g\) and \(\mu(f eq g)=0\) as well as \(\int f d \mu=\int f^{*} d \bar{\mu}=\int g d \mu\).

(ii) Show that \(u^{*}: X ightarrow \mathbb{R}\) is \(\overline{\mathscr{A}}\)-measurable if, and only if, there exist \(\mathscr{A}\)-measurable functions \(u, w: X ightarrow \overline{\mathbb{R}}\) with \(u \leqslant u^{*} \leqslant w\) and \(u=w \mu\)-a.e.

(iii) Show that \(u^{*} \in \mathcal{L}^{1}(\bar{\mu})\), then \(u, w\) from (ii) can be chosen from \(\mathcal{L}^{1}(\mu)\) such that \(\int u d \mu=\) \(\int u^{*} d \bar{\mu}=\int w d \mu\).

[(i) use Problem 4.15(v). (ii) for ' \(\Rightarrow\) ' consider \(\left\{u^{*}>\alphaight\}\) and use Problem 4.15(v). The other direction is harder. For this consider first step functions using again Problem 4.15(v) and then general functions by monotone convergence. (iii) by Problem 4.15(iii), \(\mu=\bar{\mu}\) on \(\mathscr{A}\), and thus \(\int f d \mu=\int f d \bar{\mu}\) for \(\mathscr{A}\)-measurable \(f\).]

Data from problem 4.15

Data from problem 6.4

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Completion (1). We have seen in Problem 4.12 that measurable subsets of null sets are again null sets: MEA, MCNEA, (N)=0 then (M)= 0; but there might be subsets of N which are not in. This motivates the following definition: a measure space (X, A, ) (or a measure ) is complete if all subsets of u-null sets are again in A. In other words, it holds if all subsets of a null set are null sets. The following exercise shows that a measure space (X, ,) which is not yet complete can be completed.