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/Part 4: Measure Theory and Integration/Chapter 14: Positive Measures

14.2 Complete Measures

Definition 14.2.1 (Complete Measure). Let $(X, \cm, \mu)$ be a measure space, then $\mu$ is complete if for any null set $E \in \cm$ and $F \subset E$, $F \in \cm$.

Definition 14.2.2 (Completion). Let $(X, \cm, \mu)$ be a measure space, $\cn = \bracs{N \in \cm| \mu(N) = 0}$, and

\[\ol{\cm}= \bracs{E \cup F| E \in \cm, F \subset N, N \in \cn}\]

then

  1. For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.

  2. $\ol{\cm}\supset \cm$ is a $\sigma$-algebra.

  3. There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\cm$.

  4. $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.

  5. For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_{\cm} = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}}= \ol{\mu}$.

and space $(X, \ol{\cm}, \mu)$ is the completion of $(X, \cm, \mu)$.

Proof. (1): Let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Define $E' = E \setminus N$ and $F' = F \cup (E \cap N)$, then $E' \cap N = \emptyset$ and $A = E \cup F = E' \sqcup F'$.

(2): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \sqcup F$, then

\begin{align*}A^{c}&= (A^{c} \cap N^{c}) \cup (A^{c} \cap N) = (N^{c} \setminus A) \cup (N \setminus A) \\&= \underbrace{(N^c \setminus E)}_{\in \cm}\cup \underbrace{(N \setminus F)}_{\subset N}\in \cm\end{align*}

Let $\seq{A_n}\subset \cm$. For each $n \in \natp$, let $E_{n} \in \cm$, $N_{n} \in \cn$, and $F_{n} \subset N$ such that $A_{n} = E_{n} \cup F_{n}$, then

\[\bigcup_{n \in \natp}A_{n} = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm}\cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n}\in \ol{\cm}\]

(3): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Since $A \setminus N = E \setminus N \in \cm$ and $A \cup N = E \cup N \in \cm$,

\[\mu(A \setminus N) \le \mu(E) \le \mu(A \cup N)\]

Let $E' \in \cm$, $N' \in \cn$, and $F' \subset N'$ such that $A = E' \cup F'$, then

\[\mu(A \setminus (N \cup N')) \le \mu(A \setminus N') \le \mu(E') \le \mu(A \cup N') \le \mu(A \cup (N \cup N'))\]

Now, since $N$ and $N'$ are null,

\[\mu(A \cup (N \cup N')) = \mu(A \setminus (N \cup N')) + \mu(N \cup N') = \mu(A \setminus (N \cup N'))\]

so

\[\mu(A \setminus (N \cup N')) = \mu(E) = \mu(E') = \mu(A \cup (N \cup N'))\]

Thus for any decomposition $A = E \cup F$ where $E \in \cm$ and $F$ is contained in a null set, $\ol{\mu}(A) = \mu(E)$ is well-defined.

To see that $\ol{\mu}$ is a measure, let $\seq{A_n}\subset \ol{\cm}$ be pairwise disjoint. For each $n \in \nat$, let $E_{n} \in \cm$, $N_{n} \in \cn$, and $F_{n} \subset N_{n}$ such that $A_{n} = E_{n} \cup F_{n}$, then

\[\ol{\mu}\paren{\bigsqcup_{n \in \natp}A_n}= \mu\paren{\bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu(E_{n}) = \sum_{n \in \natp}\mu(A_{n})\]

(4): Let $A \in \ol{\cm}$ with $\ol{\mu}(A) = 0$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$, then $\mu(E) = \mu(E \cup N) = 0$. By definition of $\ol{\cm}$, any subset of $E \cup N$ is also in $\ol{\cm}$. Thus $\ol{\mu}$ is complete.

(U): Let $N \in \cn \subset \cm \subset \cf$, then for any $F \subset N$, $\nu(F) \le \nu(N) = \mu(N)$, and $F \in \cf$ by completeness of $(X, \cf, \nu)$. Thus $\cf \supset \ol{\cm}$.

For any $A \in \ol{\cm}$, let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup N$, then

\[\nu(A) = \nu(E \cup F) = \nu(E) = \mu(E) = \ol{\mu}(A)\]
$\square$

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