Math 2.0

Measure and integration Definition

Measure algebra

Builds on Boolean algebraMeasure

Definition

A measure algebra is a pair (𝔄,μ)(\mathfrak{A}, \mu) such that:

  1. 𝔄\mathfrak{A} is a Boolean algebra in which every countable subset has a supremum.
  2. μ ⁣:𝔄→[0,∞]\mu \colon \mathfrak{A} \to [0, \infty] satisfies μ(0)=0\mu(0) = 0.
  3. μ(⋁kak)=∑kμ(ak)\mu\left(\bigvee_{k} a_k\right) = \sum_k \mu(a_k) for every sequence (ak)k∈ℕ(a_k)_{k \in \N} of pairwise disjoint elements, that is, aj∧ak=0a_j \wedge a_k = 0 for j≠kj \ne k.
  4. μ(a)>0\mu(a) > 0 for every a≠0a \ne 0.

Remarks

Example: let (X,Σ,ν)(X, \Sigma, \nu) be a measure space and 𝒩={E∈Σ:ν(E)=0}\mathcal{N} = \set{E \in \Sigma : \nu(E) = 0}. Call E,F∈ΣE, F \in \Sigma equivalent when ν(E△F)=0\nu(E \mathbin{\triangle} F) = 0. The classes form a Boolean algebra Σ/𝒩\Sigma / \mathcal{N} with countable suprema given by countable unions, and μ([E])=ν(E)\mu([E]) = \nu(E) is well defined. Clauses 2 and 3 come from the measure ν\nu, and clause 4 holds since [E]≠0[E] \ne 0 means ν(E)>0\nu(E) > 0.

Non-example: the sigma-algebra of Borel subsets of ℝ\R with Lebesgue measure satisfies clauses 1 to 3 but not 4. The set {0}\set{0} is a nonzero element of measure 00.

The quotient in the example is called the measure algebra of (X,Σ,ν)(X, \Sigma, \nu).

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Further reading

  1. Web https://en.wikipedia.org/wiki/Measure_algebra
About this page
TagKST0YNAP55
URIm2://KST0YNAP55
Kinddefinition
WritingW0 stub
MSC28A60, 06E10
DefinesMeasure algebra
UsesBoolean algebra, Measure
Changed2 November 2026

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