Complete measure

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 zero). More formally, (X, Σ, μ) is complete if

Failed to parse (Missing texvc executable; please see math/README to configure.): S \subseteq N \in \Sigma \mbox{ and } \mu(N) = 0 \implies S \in \Sigma.

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 (R, B, λ). We now wish to construct two-dimensional Lebesgue measure λ² on the plane R² as a product measure. Naïvely, we would take the σ-algebra on R² to be B ⊗ B, the smallest σ-algebra containing all measurable "rectangles" A₁ × A₂ for A_i ∈ B.

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

Failed to parse (Missing texvc executable; please see math/README to configure.): \lambda^{2} ( \{ 0 \} \times A ) = \lambda ( \{ 0 \} ) \cdot \lambda (A) = 0

for "any" subset A of 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} × A is not defined, but

Failed to parse (Missing texvc executable; please see math/README to configure.): \{ 0 \} \times A \subseteq \{ 0 \} \times \mathbb{R},

and this larger set does have λ²measure zero. So, "two-dimensional Lebesgue measure" as just defined is not complete, and some kind of completion procedure is required.

Construction of a complete measure

Given a (possibly incomplete) measure space (X, Σ, μ), there is an extension (X, Σ₀, μ₀) of this measure space that is complete. The smallest such extension (i.e. the smallest σ-algebra Σ₀) is called the completion of the measure space.

The completion can be constructed as follows:

• let Z be the set of all subsets of μ-measure zero subsets of X (intuitively, those elements of Z that are not already in Σ are the ones preventing completeness from holding true);
• let Σ₀ be the σ-algebra generated by Σ and Z (i.e. the smallest σ-algebra that contains every element of Σ and of Z);
• there is a unique extension μ₀ of μ to Σ₀ given by the infimum

Failed to parse (Missing texvc executable; please see math/README to configure.): \mu_{0} (C) := \inf \{ \mu (D) | C \subseteq D \in \Sigma \}.

Then (X, Σ₀, μ₀) is a complete measure space, and is the completion of (X, Σ, μ).

In the above construction it can be shown that every member of Σ₀ is of the form A ∪ B for some A ∈ Σ and some B ∈ Z, and

Failed to parse (Missing texvc executable; please see math/README to configure.): \mu_{0} (A \cup B) = \mu (A).

Examples

• Borel measure as defined on the Borel σ-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.pl:Miara zupełna