Complete measure

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In mathematics, a complete measure (or, more precisely, a complete measure space) is a measure space in which every subset of every null set is measurable (having measure 0).

[edit] Motivation

The need to consider questions of completeness can be illustrated by considering the problem of product spaces.

Suppose that we have already constructed Lebesgue measure on the real line: denote this measure space by (\mathbb{R}, \mathcal{B}, \lambda). We now wish to construct two-dimensional Lebesgue measure λ2 on the plane \mathbb{R}^{2} as a product measure. Naïvely, we would take the sigma algebra on \mathbb{R}^{2} to be \mathcal{B} \otimes \mathcal{B}, the smallest sigma algebra containing all measurable "rectangles" A_{1} \times A_{2} for A_{i} \in \mathcal{B}.

While this approach does define a measure space, it has a flaw. Since every singleton set has one-dimensional Lebesgue measure zero,

\lambda^{2} ( \{ 0 \} \times A ) = \lambda ( \{ 0 \} ) \cdot \lambda (A) = 0 for "any" A \subseteq \mathbb{R}.

However, suppose that A is a non-measurable subset of the real line, such as the Vitali set. Then the λ2-measure of \{ 0 \} \times A is not defined, but

\{ 0 \} \times A \subseteq \{ 0 \} \times \mathbb{R},

and this set does have measure zero. So, "Lebesgue measure" as just defined is not complete.

[edit] Construction of a complete measure

Given a (possibly incomplete) measure space (X, \mathcal{F}, \mu), there is an extension (X, \mathcal{F}_{0}, \mu_{0}) of this measure space that is complete. The smallest such extension (i.e. the smallest sigma algebra \mathcal{F}_{0}) is called the completion of the measure space.

The completion can be constructed as follows:

  • let \mathcal{N} be the set of all subsets of μ-null subsets of X;
  • let \mathcal{F}_{0} be the sigma algebra generated by \mathcal{F} and \mathcal{N};
  • there is a unique extension μ0 of μ to \mathcal{F}_{0}:
\mu_{0} (C) := \inf \{ \mu (D) | C \subseteq D \in \mathcal{F} \}.

Then (X, \mathcal{F}_{0}, \mu_{0}) is a complete measure space, and is the completion of (X, \mathcal{F}, \mu).

In the above construction it can be shown that every member of \mathcal{F}_{0} is of the form A \cup B for some A \in \mathcal{F} and some B \in \mathcal{N}, and

\mu_{0} (A \cup B) = \mu (A).

[edit] Examples

  • Borel measure as defined on the Borel sigma algebra generated by the open intervals of the real line is not complete, and so the above completion procedure must be used to define the complete Lebesgue measure.
  • n-dimensional Lebesgue measure is the completion of the n-fold product of the one-dimensional Lebesgue space with itself. It is also the completion of the Borel measure, as in the one-dimensional case.
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