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 Subset of All, $\varnothing \in \mathcal N$, 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: