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1. Sigma-algebra and Lebesgue measure

1. Sigma-Algebra

For the source of definition, it's from wiki [1], as is given in Definition. 1.1, The sigma-algebra ==describe the property that a set is always self-closed==.

Sigma-algebra If X=a,b,c,d, then an example of σ-algebra on X is Σ={,{a,b},{c,d},{a,b,c,d}}, where is the empty set.

In general, a finite algebra is always a σ-algebra if it :

  1. Including whole set (XΣ)
  2. Close for complementary set (if AΣ,then XAΣ)
  3. Close for and if A1,A2,Σ, then i=1AiΣ

2. Lebesgue Measure

Lebesgue measure defines the, see [2] for details.

the Lebesgue measurable is the standard way of assigning a measure to subsets of higher dimensional Euclidean n-spaces.

(1) Length of the set

we describe the measure of Lebesgue measurable set A as :

(2.1.1)λ(A)

for I=[a,b] or I=(a,b), we denote the length of set as l(I)=ba

Lebesgue Outer Measure Lebesgue Outer Measure: For subset ER , we define the Lebesgue outer measure as the minimum discontinuous summation :

(lebesgue-measure)λ(E)=inf{k=1l(Ik):(Ik)is a sequence of open interval with Ek=1Ik}

We take E=Q[0,1] as a 2-element set, Ik as a open set that contains each element of E, then we set the length of I is :

l(Ik)=ϵ2k

Then the sum of length should be :

k=1l(Ik)=ϵ

then the λ(E)ϵ, i.e., we have λ(E)=0

📐 Advanced Mathematics/combinatorics/assets/1. Sigma-algebra and Lebesgue measure 2026-07-13 11.04.34.excalidraw.png

The equivalent statement of is, for any ε>0, there exists a covering S of Ω with closed boxes. We note (lebesgue-measure) can also be stated as :

(2.1.2)m(Ω):=inf{σ(S)|S covering of Ω}

For a rectangular cuboid C=I1×In with vol(C)=l(I1)×l(I2)×l(In)

(2.1.3)λ(E)=inf{k=1vol(Ck):(Ck)kN is a sequence of products of open intervals with Ek=1Ck}

This gives, the set and be convered with limited ranged boxes. The classic example: a single point, then a finite set. Firstly, We Let E=0,1,2 (three points on the real line). Intuitively, three isolated points have "zero length." The outer measure confirms this. For any ε > 0, cover each point with a tiny open interval:

  • 0 ∈ (−ε/6, ε/6), length = ε/3
  • 1 ∈ (1−ε/6, 1+ε/6), length = ε/3
  • 2 ∈ (2−ε/6, 2+ε/6), length = ε/3

For a range set E=[a,b] , Lebesgue measure is :

λ([a,b])=inf(aϵ,b+ϵ)=l([a,b])=ba

(2) Useful properties

  1. Given any ==countable collection of sets== from an algebra (or σ‑algebra) A, we can replace it with a ==disjoint countable collection== of sets that are also in A and have the same union.

For every σ algebra A, We let {En}n1 be an arbitrary sequence of a set A (no-σ-algebra needed), then there exists a sequence of pairwise disjoint collection, {Fn}n, or a countable collection. satisfying for every n :

(2.2.1)k=1nFk=k=1nEk

To prove this, we can take the F as following part :

(2.2.2)F1=G1Fn=k=1n+1Ekk=1nEkk=1nFn=k=1nGn

so any countable algebra can be represented as disjoint collections.

(3) Lebesgue measurable set

Caratheodory criterion : if a set E has the following relation : for every ARn

(2.3.1)λ(E)=λ(AE)+λ(AEc)

where Ec is the complement set of E, We say it satisfy the Caratheodory criterion, or say it is Lebesgue measurable. where λ is the Lebesgue outer measure

Borel σ algebra We note Borel σ algebra is the smallest σ algebra containing all open sets.

The Lebesgue measure of such set E define as Lebesgue outer measure :

(2.3.2)λ(E)=λ(E)

3. Axiom of Choice and Vitali Sets

Axiom of Choice This axiom states that : for every set I, and every I-indexed family (Si)iI of non-empty sets, there exists an I-indexed set (xi)iI of elements of iI such that xiSi for every iI

To be short, This axiom says you can get at 1 element from Si each if its non-empty

Vitali Sets[3] is not Lebesgue measurable. A typical Vitali set is to choose exactly one representative from the each equivalence class under the following relation:

(3.1)xyiffxy is rational

or set the Vitali set as V such that :

(3.2)ifu,vVandvuQthen v=u

from wiki, we know that we can find uncountable pairwise-disjoint set such that :

(3.3)Vk[1,2]

If Vk is measurable, the Lebesgue measure for pairwise-disjoint set yields :

(3.4)λ(k=1Vk)=k=1λ(Vk)=3

This will create a contradiction that 0=3 or =3. Such contradiction is because, the Vitali sets are not measurable.

4. Reference


  1. https://en.wikipedia.org/wiki/Σ-algebra ↩︎

  2. https://en.wikipedia.org/wiki/Lebesgue_measure ↩︎

  3. https://en.wikipedia.org/wiki/Vitali_set ↩︎