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:

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:

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
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:

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 $\struct {X, \Sigma, \mu}$.