1. Sigma-algebra and Lebesgue measure

A sigma-algebra is the collection of sets a measure is allowed to touch, and the Lebesgue measure extends length to almost every subset of the real line. Use them as the foundation before measure-theoretic probability, integration and Monte-Carlo arguments.

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Sep 15, 2026

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#math/axiom-of-choice#math/caratheodory-criterion#math/lebesgue-measure#math/sigma-algebra#math/vitali-set

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 $\sigma$-algebra on $X$ is $\Sigma =\{\varnothing ,\{a,b\},\{c,d\},\{a,b,c,d\}\}$, where $\mathrm{\varnothing }$ is the empty set.

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

1. Including whole set $(X\in \mathrm{\Sigma })$
2. Close for complementary set $(\text{if }A\in \mathrm{\Sigma },\text{then }X\mathrm{\setminus }A\in \mathrm{\Sigma })$
3. Close for and $\text{if }{A}_1,A_2,\cdots \in \mathrm{\Sigma }$, then ${\cup }_{i=1}^{\mathrm{\infty }}{A}_i\in \mathrm{\Sigma }$

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 :

$$\begin{array}{}\text{(2.1.1)}& \lambda (A)\end{array}$$

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

Lebesgue Outer Measure Lebesgue Outer Measure: For subset $E\subseteq R$ , we define the Lebesgue outer measure as the minimum discontinuous summation :

$$\begin{array}{}\text{(lebesgue-measure)}& \boxed{{\displaystyle {\lambda }^{\ast }(E)=\text{inf}\{\sum _{k=1}^{\mathrm{\infty }}l(I_k):(I_k)\text{is a sequence of open interval with }E\subset \bigcup _{k=1}^{\mathrm{\infty }}{I}_k\}}}\end{array}$$

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

$$l(I_k)=\frac{ϵ}{2^k}$$

Then the sum of length should be :

$$\sum _{k=1}^{\mathrm{\infty }}l(I_k)=ϵ$$

then the ${\lambda }^{\ast }(E)\le ϵ$, i.e., we have ${\lambda }^{\ast }(E)=0$

The equivalent statement of is, for any $\epsilon >0$, there exists a covering $S$ of $\mathrm{\Omega }$ with closed boxes. We note (lebesgue-measure) can also be stated as :

$$\begin{array}{}\text{(2.1.2)}& m^{\ast }(\mathrm{\Omega }):=inf\{\sigma (S)|S\text{ covering of }\mathrm{\Omega }\}\end{array}$$

For a rectangular cuboid $C=I_1\times \dots I_n$ with $\text{vol}(C)=l(I_1)\times l(I_2)\times \dots l(I_n)$

$$\begin{array}{}\text{(2.1.3)}& {\lambda }^{\ast }(E)=inf\{\sum _{k=1}^{\mathrm{\infty }}\mathrm{vol}(C_k):(C_k{)}_{k\in \mathbb{N}}\text{ is a sequence of products of open intervals with }E\subseteq \bigcup _{k=1}^{\mathrm{\infty }}{C}_k\}\end{array}$$

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 :

$${\lambda }^{\ast }([a,b])=inf(a-ϵ,b+ϵ)=l([a,b])=b-a$$

(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 $\sigma -$ algebra $A$, We let $\{E_n{\}}_{n\ge 1}$ be an arbitrary sequence of a set $A$ (no-$\sigma$-algebra needed), then there exists a sequence of pairwise disjoint collection, $\{F_n{\}}_n$, or a countable collection. satisfying for every $n$ :

$$\begin{array}{}\text{(2.2.1)}& \boxed{{\displaystyle \bigcup _{k=1}^n{F}_k=\bigcup _{k=1}^n{E}_k}}\end{array}$$

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

$$\begin{array}{}\text{(2.2.2)}& F_1=G_1{F}_n={\cup }_{k=1}^{n+1}{E}_k\mathrm{\setminus }{\cup }_{k=1}^n{E}_k\to \bigcup _{k=1}^n{F}_n=\bigcup _{k=1}^n{G}_n\end{array}$$

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 $A\subset R^n$

$$\begin{array}{}\text{(2.3.1)}& \boxed{{\displaystyle {\lambda }^{\ast }(E)={\lambda }^{\ast }(A\cap E)+{\lambda }^{\ast }(A\cap E^c)}}\end{array}$$

where $E^c$ is the complement set of $E$, We say it satisfy the Caratheodory criterion, or say it is Lebesgue measurable. where $\lambda$ is the Lebesgue outer measure

Borel $\sigma$ algebra We note Borel $\sigma$ algebra is the smallest $\sigma$ algebra containing all open sets.

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

$$\begin{array}{}\text{(2.3.2)}& \lambda (E)={\lambda }^{\ast }(E)\end{array}$$

3. Axiom of Choice and Vitali Sets

Axiom of Choice This axiom states that : for every set $I$, and every $I$-indexed family $(S_i{)}_{i\in I}$ of non-empty sets, there exists an I-indexed set $(x_i{)}_{i\in I}$ of elements of ${\cup }_{i\in I}$ such that $x_i\in S_i$ for every $i\in I$

To be short, This axiom says you can get at 1 element from $S_i$ 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:

$$\begin{array}{}\text{(3.1)}& x\sim y\text{iff}x-y\text{ is rational}\end{array}$$

or set the Vitali set as $V$ such that :

$$\begin{array}{}\text{(3.2)}& \text{if}u,v\in V\text{and}v-u\in Q\text{then }v=u\end{array}$$

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

$$\begin{array}{}\text{(3.3)}& V_k\in [-1,2]\end{array}$$

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

$$\begin{array}{}\text{(3.4)}& \lambda (\bigcup _{k=1}^{\mathrm{\infty }}{V}_k)=\sum _{k=1}\lambda (V_k)=3\end{array}$$

This will create a contradiction that $0=3$ or $\mathrm{\infty }=3$. Such contradiction is because, the Vitali sets are not measurable.

4. Reference

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

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

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