Completion Theorem (Measure Spaces)

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.


Then there exists a completion $\left({X, \Sigma^*, \bar \mu}\right)$ of $\left({X, \Sigma, \mu}\right)$.


Proof

We give an explicit construction of $\left({X, \Sigma^*, \bar \mu}\right)$.


To this end, define $\mathcal N$ to be the collection of subsets of $\mu$-null sets:

$\mathcal N := \left\{{N \subseteq X: \exists M \in \Sigma: \mu \left({M}\right) = 0, N \subseteq M}\right\}$

Now, we define:

$\Sigma^* := \left\{{E \cup N: E \in \Sigma, N \in \mathcal N}\right\}$

and assert $\Sigma^*$ is a $\sigma$-algebra.


By Empty Set is Null Set, $\varnothing \in \mathcal N$, and thus by Union with Empty Set:

$\forall E \in \Sigma: E \cup \varnothing = E \in \Sigma^*$

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:

\(\displaystyle X \setminus \left({E \cup N}\right)\) \(=\) \(\displaystyle \left({X \setminus E}\right) \cap \left({X \setminus N}\right)\) De Morgan's Laws: Difference with Union
\(\displaystyle \) \(=\) \(\displaystyle \left({X \setminus E}\right) \cap \left({\left({X \setminus M}\right) \cup \left({M \setminus N}\right)}\right)\) Union of Relative Complements of Nested Subsets
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({X \setminus E}\right) \cap \left({X \setminus M}\right)}\right) \cup \left({\left({X \setminus E}\right) \cap \left({M \setminus N}\right)}\right)\) Intersection Distributes over Union
\(\displaystyle \) \(\) \(\displaystyle \)
\(\displaystyle \left({\left({X \setminus E}\right) \cap \left({X \setminus M}\right)}\right)\) \(\in\) \(\displaystyle \Sigma\) $E, M \in \Sigma$, Sigma-Algebra Closed under Intersection
\(\displaystyle \left({X \setminus E}\right) \cap \left({M \setminus N}\right)\) \(\subseteq\) \(\displaystyle M\) Set Difference is Subset, Set Intersection Preserves Subsets
\(\displaystyle \implies \ \ \) \(\displaystyle X \setminus \left({E \cup N}\right)\) \(\in\) \(\displaystyle \Sigma^*\)


Finally, let $\left({E_n}\right)_{n \in \N}$ and $\left({N_n}\right)_{n \in \N}$ be sequences in $\Sigma$ and $\mathcal N$, respectively.

Let $\left({M_n}\right)_{n \in \N}$ be a sequence of $\mu$-null sets such that:

$\forall n \in \N: N_n \subseteq M_n$

Then, compute:

\(\displaystyle \bigcup_{n \mathop \in \N} \left({E_n \cup N_n}\right)\) \(=\) \(\displaystyle \left({\bigcup_{n \mathop \in \N} E_n}\right) \cup \left({\bigcup_{n \mathop \in \N} N_n}\right)\) Union Distributes over Union: Families of Sets
\(\displaystyle \bigcup_{n \mathop \in \N} N_n\) \(\subseteq\) \(\displaystyle \bigcup_{n \mathop \in \N} M_n\) Set Union Preserves Subsets

From Null Sets Closed under Countable Union, also:

$\displaystyle \mu \left({\bigcup_{n \mathop \in \N} M_n}\right) = 0$

hence it follows that:

$\displaystyle \bigcup_{n \mathop \in \N} N_n \in \mathcal N$


Next, as $\Sigma$ is a $\sigma$-algebra, it follows that:

$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \Sigma$

and finally, we conclude:

$\displaystyle \bigcup_{n \mathop \in \N} \left({E_n \cup N_n}\right) \in \Sigma^*$


Therefore, we have shown that $\Sigma^*$ is a $\sigma$-algebra.


Next, define $\bar \mu: \Sigma^* \to \overline{\R}_{\ge 0}$ by:

$\bar \mu \left({E \cup N}\right) := \mu \left({E}\right)$

It needs verification that this well-defines $\bar \mu$.


Lemma

The mapping $\bar \mu$ is well-defined, i.e.:

$\forall E, F \in \Sigma: \forall N, M \in \mathcal N: E \cup N = F \cup M \implies \mu \left({E}\right) = \mu \left({F}\right)$

$\Box$


Next, let us verify that $\bar \mu$ is a measure.

From Union with Empty Set, we have $\varnothing \cup \varnothing = \varnothing$, so by Empty Set is Null Set:

$\bar \mu \left({\varnothing}\right) = \mu \left({\varnothing}\right) = 0$


For a sequence of pairwise disjoint sets $\left({E_n \cup N_n}\right)_{n \in \N}$ in $\Sigma^*$, compute:

\(\displaystyle \bar \mu \left({\bigcup_{n \mathop \in \N} \left({E_n \cup N_n}\right)}\right)\) \(=\) \(\displaystyle \bar \mu \left({\left({\bigcup_{n \mathop \in \N} E_n}\right) \cup \left({\bigcup_{n \mathop \in \N} N_n}\right)}\right)\) Union Distributes over Union: Families of Sets
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E_n}\right)\) Definition of $\bar \mu$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \mu \left({E_n}\right)\) $\mu$ is a measure
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N} \bar \mu \left({E_n \cup N_n}\right)\) Definition of $\bar \mu$

Thus, $\bar \mu$ is a measure.


Since for all $E \in \Sigma$ trivially:

$\bar \mu \left({E}\right) = \mu \left({E}\right)$

if $\left({X, \Sigma^*, \bar \mu}\right)$ is a complete measure space, it also completes $\left({X, \Sigma, \mu}\right)$.


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 $0 = \bar \mu \left({E \cup N}\right) = \mu \left({E}\right)$, $E$ 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 $E' \in \mathcal N$, and this means that (by Union with Empty Set):

$\bar \mu \left({E'}\right) = \bar \mu \left({\varnothing \cup E'}\right) = \mu \left({\varnothing}\right) = 0$


So, any subset of $E \cup N$ is again a $\bar \mu$-null set.

That is, $\left({X, \Sigma^*, \bar \mu}\right)$ is complete.


It follows that $\left({X, \Sigma^*, \bar \mu}\right)$ completes $\left({X, \Sigma, \mu}\right)$.

$\blacksquare$


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