Complete measure

Remarks

Counting measure on the power set of a set is complete, since every subset is measurable. Let $X = \set{1, 2, 3}$, let $\Sigma = \set{\varnothing, \set{1, 2}, \set{3}, X}$, and let $\mu(A) = 1$ if $3 \in A$ and $\mu(A) = 0$ otherwise. Then $\set{1, 2}$ is a null set, but its subset $\set{1}$ is not in $\Sigma$, so $\mu$ is not complete. For a complete measure, a subset of a null set is again a null set, by monotonicity. Lebesgue measure on the σ-algebra of Lebesgue measurable sets is complete. Its restriction to the Borel sets is not complete: the Cantor set is a Borel null set with more subsets than there are Borel sets.