Complete measure

A measure $ \mu $ on a $ \sigma $- algebra $ \Sigma $ for which $ A \in \Sigma $ and $ | \mu | ( A) = 0 $ imply $ E \in \Sigma $ for every $ E \subset A $. Here $ | \mu | $ is the total variation of $ \mu $( $ | \mu | = \mu $ for a positive measure).

Comments

Complete measures arise as follows (cf. [a1]). Let $ X $ be a set, $ \Sigma $ a $ \sigma $- algebra of subsets of it and $ \mu $ a positive measure on $ \Sigma $. It may happen that some set $ E \in \Sigma $ with $ \mu ( E) = 0 $ has a subset $ N $ not belonging to $ \Sigma $. It is natural, then, to define the measure $ \mu $ on such a set $ N $ as $ \mu ( N) = 0 $.

In general, let $ \Sigma ^ {*} $ be the collection of all sets $ N $ for which there exists sets $ E , F \in \Sigma $ such that $ E \subset N \subset F $, $ \mu ( F - E ) = 0 $. In this situation, define $ \mu ( N) = 0 $. Then $ \Sigma ^ {*} $ is a $ \sigma $- algebra and $ \mu $ becomes a complete measure on it (this process is called completion). $ ( X , \Sigma ^ {*} , \mu ) $ is then called a complete measure space.

References

[a1]	W. Rudin, "Real and complex analysis" , McGraw-Hill (1974) pp. 24
[a2]	E. Hewitt, K.R. Stromberg, "Real and abstract analysis" , Springer (1965)