Completion Theorem (Measure Space)

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:

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:

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


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

Proof of Lemma
Let $N_0, M_0 \in \Sigma$ be null sets such that $N \subset N_0, M \subset M_0$.

Then:
 * $E \subset E \cup N = F \cup M \subset F \cup M_0$

so that:
 * $\mu \left({E}\right) \le \mu \left({F \cup M_0}\right) \le \mu \left({F}\right) + \mu \left({M_0}\right) = \mu \left({F}\right) + 0$

Analogously:
 * $F \subset F \cup M = E \cup N \subset E \cup N_0$

so that:
 * $\mu \left({F}\right) \le \mu \left({E \cup N_0}\right) \le \mu \left({E}\right) + \mu \left({N_0}\right) = \mu \left({E}\right) + 0$

In total:
 * $\mu \left({E}\right) = \mu \left({F}\right)$

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:

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 $\left({X, \Sigma, \mu}\right)$.