Completion Theorem (Measure Space)
Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists a completion $\struct {X, \Sigma^*, \bar \mu}$ of $\struct {X, \Sigma, \mu}$.
Proof
We give an explicit construction of $\struct {X, \Sigma^*, \bar \mu}$.
To this end, define $\NN$ to be the collection of subsets of $\mu$-null sets:
- $\NN := \set {N \subseteq X: \exists M \in \Sigma: \map \mu M = 0, N \subseteq M}$
Now, we define:
- $\Sigma^* := \set {E \cup N: E \in \Sigma, N \in \NN}$
and assert $\Sigma^*$ is a $\sigma$-algebra.
By Empty Set is Null Set, $\O \in \NN$, and thus by Union with Empty Set:
- $\forall E \in \Sigma: E \cup \O = E \in \Sigma^*$
that is to say, $\Sigma \subseteq \Sigma^*$.
As a consequence, $X \in \Sigma^*$.
Now, suppose that $E \cup N \in \Sigma^*$, and $N \subseteq M, M \in \Sigma$. Then:
| \(\ds X \setminus \paren {E \cup N}\) | \(=\) | \(\ds \paren {X \setminus E} \cap \paren {X \setminus N}\) | De Morgan's Laws: Difference with Union | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {X \setminus E} \cap \paren {\paren {X \setminus M} \cup \paren {M \setminus N} }\) | Union of Relative Complements of Nested Subsets | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\paren {X \setminus E} \cap \paren {X \setminus M} } \cup \paren {\paren {X \setminus E} \cap \paren {M \setminus N} }\) | Intersection Distributes over Union | |||||||||||
| \(\ds \) | \(\) | \(\ds \) |
| \(\ds \paren {\paren {X \setminus E} \cap \paren {X \setminus M} }\) | \(\in\) | \(\ds \Sigma\) | $E, M \in \Sigma$, Sigma-Algebra Closed under Intersection | |||||||||||
| \(\ds \paren {X \setminus E} \cap \paren {M \setminus N}\) | \(\subseteq\) | \(\ds M\) | Set Difference is Subset, Set Intersection Preserves Subsets | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds X \setminus \paren {E \cup N}\) | \(\in\) | \(\ds \Sigma^*\) |
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ and $\sequence {N_n}_{n \mathop \in \N}$ be sequences in $\Sigma$ and $\NN$, respectively.
Let $\sequence {M_n}_{n \mathop \in \N}$ be a sequence of $\mu$-null sets such that:
- $\forall n \in \N: N_n \subseteq M_n$
Then, compute:
| \(\ds \bigcup_{n \mathop \in \N} \paren {E_n \cup N_n}\) | \(=\) | \(\ds \paren {\bigcup_{n \mathop \in \N} E_n} \cup \paren {\bigcup_{n \mathop \in \N} N_n}\) | Set Union is Self-Distributive: Families of Sets | |||||||||||
| \(\ds \bigcup_{n \mathop \in \N} N_n\) | \(\subseteq\) | \(\ds \bigcup_{n \mathop \in \N} M_n\) | Set Union Preserves Subsets |
From Null Sets Closed under Countable Union, also:
- $\ds \map \mu {\bigcup_{n \mathop \in \N} M_n} = 0$
hence it follows that:
- $\ds \bigcup_{n \mathop \in \N} N_n \in \NN$
Next, as $\Sigma$ is a $\sigma$-algebra, it follows that:
- $\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
and finally, we conclude:
- $\ds \bigcup_{n \mathop \in \N} \paren {E_n \cup N_n} \in \Sigma^*$
Therefore, we have shown that $\Sigma^*$ is a $\sigma$-algebra.
Next, define $\bar \mu: \Sigma^* \to \overline{\R}_{\ge 0}$ by:
- $\map {\bar \mu} {E \cup N} := \map \mu E$
It needs verification that this well-defines $\bar \mu$.
Lemma
The mapping $\bar \mu$ is well-defined, i.e.:
- $\forall E, F \in \Sigma: \forall N, M \in \NN: E \cup N = F \cup M \implies \map \mu E = \map \mu F$
$\Box$
Next, let us verify that $\bar \mu$ is a measure.
From Union with Empty Set, we have $\O \cup \O = \O$, so by Empty Set is Null Set:
- $\map {\bar \mu} \O = \map \mu \O = 0$
For a sequence of pairwise disjoint sets $\sequence {E_n \cup N_n}_{n \mathop \in \N}$ in $\Sigma^*$, compute:
| \(\ds \map {\bar \mu} {\bigcup_{n \mathop \in \N} \paren {E_n \cup N_n} }\) | \(=\) | \(\ds \map {\bar \mu} {\paren {\bigcup_{n \mathop \in \N} E_n} \cup \paren {\bigcup_{n \mathop \in \N} N_n} }\) | Set Union is Self-Distributive: Families of Sets | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \mu {\bigcup_{n \mathop \in \N} E_n}\) | Definition of $\bar \mu$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map \mu {E_n}\) | $\mu$ is a measure | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N} \map {\bar \mu} {E_n \cup N_n})\) | Definition of $\bar \mu$ |
Thus, $\bar \mu$ is a measure.
Since for all $E \in \Sigma$ trivially:
- $\map {\bar \mu} E = \map \mu E$
if $\struct {X, \Sigma^*, \bar \mu}$ is a complete measure space, it also completes $\struct {X, \Sigma, \mu}$.
So suppose that $E \cup N \in \Sigma^*$ is a $\bar \mu$-null set.
Suppose that $N \subseteq M$, with $M$ a $\mu$-null set.
Then by Set Union Preserves Subsets, we have:
- $E \cup N \subseteq E \cup M$
and from $0 = \map {\bar \mu} {E \cup N} = \map \mu E$, $E$ is also a $\mu$-null set.
Hence by Null Sets Closed under Union, $E \cup M$ is a $\mu$-null set.
Therefore, for any $E' \in \Sigma^*$ with $E' \subseteq E \cup N$, we also have by Subset Relation is Transitive:
- $E' \subseteq E \cup M$
whence $E' \in \NN$, and this means that (by Union with Empty Set):
- $\map {\bar \mu} {E'} = \map {\bar \mu} {\O \cup E'} = \map \mu \O = 0$
So, any subset of $E \cup N$ is again a $\bar \mu$-null set.
That is, $\struct {X, \Sigma^*, \bar \mu}$ is complete.
It follows that $\struct {X, \Sigma^*, \bar \mu}$ completes $\struct {X, \Sigma, \mu}$.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 4$: Problem $13$