Completion of a measure

Remarks

The family $\Sigma_\mu$ is a σ-algebra that contains $\Sigma$. The value $\bar{\mu}(A \cup S)$ does not depend on the choice of $A$ and $S$. Indeed, if $A \cup S = A' \cup S'$ with $S \subseteq N$ and $S' \subseteq N'$, then $A$ and $A'$ differ only inside the null set $N \cup N'$, so $\mu(A) = \mu(A')$. Then $\bar{\mu}$ is a complete measure that extends $\mu$, and it is the smallest complete extension. For example, let $X = \set{1, 2, 3}$, $\Sigma = \set{\varnothing, \set{1, 2}, \set{3}, X}$, and $\mu(A) = 1$ if $3 \in A$ and $0$ otherwise. The null set $\set{1, 2}$ gives $\Sigma_\mu$ equal to the power set of $X$, with $\bar{\mu}(A) = 1$ if $3 \in A$ and $0$ otherwise. With the same $\Sigma$ and $\nu(A) = 1$ if $1 \in A$ and $0$ otherwise, the only null sets are $\varnothing$ and $\set{3}$. So $\Sigma_\nu = \Sigma$, and the power set is not the σ-algebra of the completion of $\nu$. The completion of Lebesgue measure on the Borel sets of $\R$ is Lebesgue measure on the Lebesgue measurable sets.