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: {{begin-eqn}} {{eqn | l = X \setminus \left({E \cup N}\right)       | r = \left({X \setminus E}\right) \cap \left({X \setminus N}\right)       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \left({X \setminus E}\right) \cap \left({\left({X \setminus M}\right) \cup \left({M \setminus N}\right)}\right)       | c = Union of Relative Complements of Nested Subsets }} {{eqn | r = \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)       | c = Intersection Distributes over Union }} {{eqn|o =}} {{end-eqn}} {{begin-eqn}} {{eqn | l = \left({\left({X \setminus E}\right) \cap \left({X \setminus M}\right)}\right)       | o = \in       | r = \Sigma       | c = $E, M \in \Sigma$, Sigma-Algebra Closed under Intersection }} {{eqn | l = \left({X \setminus E}\right) \cap \left({M \setminus N}\right)       | o = \subseteq       | r = M       | c = Set Difference is Subset, Set Intersection Preserves Subsets }} {{eqn | ll= \implies       | l = X \setminus \left({E \cup N}\right)       | o = \in       | r = \Sigma^* }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \bigcup_{n \mathop \in \N} \left({E_n \cup N_n}\right)       | r = \left({\bigcup_{n \mathop \in \N} E_n}\right) \cup \left({\bigcup_{n \mathop \in \N} N_n}\right)       | c = Union Distributes over Union: Families of Sets }} {{eqn | l = \bigcup_{n \mathop \in \N} N_n       | o = \subseteq       | r = \bigcup_{n \mathop \in \N} M_n       | c = Set Union Preserves Subsets }} {{end-eqn}} 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 === {{:Completion Theorem (Measure Spaces)/Lemma}}{{qed|lemma}} 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: {{begin-eqn}} {{eqn | l = \bar \mu \left({\bigcup_{n \mathop \in \N} \left({E_n \cup N_n}\right)}\right)       | r = \bar \mu \left({\left({\bigcup_{n \mathop \in \N} E_n}\right) \cup \left({\bigcup_{n \mathop \in \N} N_n}\right)}\right)       | c = Union Distributes over Union: Families of Sets }} {{eqn | r = \mu \left({\bigcup_{n \mathop \in \N} E_n}\right)       | c = Definition of $\bar \mu$ }} {{eqn | r = \sum_{n \mathop \in \N} \mu \left({E_n}\right)       | c = $\mu$ is a measure }} {{eqn | r = \sum_{n \mathop \in \N} \bar \mu \left({E_n \cup N_n}\right)       | c = Definition of $\bar \mu$ }} {{end-eqn}} 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 $\left({X, \Sigma, \mu}\right)$. {{qed}}