Mathematics.

measure construction

Lebesgue Measure

Measure Theory80 minDifficulty8 out of 10

You should know: sigma algebras, measure spaces

Overview

Lebesgue measure is the canonical extension of length, area, and volume to a far larger class of subsets of ℝⁿ. On ℝ it assigns to each interval (a, b) the value b − a, and it extends this consistently to all Borel sets and beyond (to the full Lebesgue σ-algebra). The construction via outer measure and Carathéodory's criterion is one of the most elegant in modern mathematics.

Intuition

The idea is to approximate any set E ⊆ ℝ from the outside using a countable cover of open intervals. The infimum of the total length of all such covers is the outer measure λ*(E). Sets that 'split' the outer measure additively — the Carathéodory criterion — are declared measurable. This recovers the familiar length formula on intervals while extending consistently to a vast class of sets.

Formal Definition

Definition

Lebesgue outer measure on ℝ is defined, and Lebesgue measurable sets are those satisfying Carathéodory's criterion.

λ(E)=inf ⁣{n=1(bnan):En=1(an,bn)}\lambda^*(E) = \inf\!\left\{\sum_{n=1}^{\infty}(b_n - a_n) : E \subseteq \bigcup_{n=1}^{\infty}(a_n, b_n)\right\}
Lebesgue outer measure on ℝ
E is Lebesgue measurable    AR,  λ(A)=λ(AE)+λ(AEc)E \text{ is Lebesgue measurable} \iff \forall A \subseteq \mathbb{R},\; \lambda^*(A) = \lambda^*(A \cap E) + \lambda^*(A \cap E^c)
Carathéodory criterion for measurability
L={ER:E is Lebesgue measurable}\mathcal{L} = \{E \subseteq \mathbb{R} : E \text{ is Lebesgue measurable}\}
Lebesgue σ-algebra ℒ
λ=λL:L[0,]\lambda = \lambda^*\big|_{\mathcal{L}} : \mathcal{L} \to [0,\infty]
Lebesgue measure λ (restriction to ℒ)

Notation

NotationMeaning
λ,  m\lambda,\; mLebesgue measure on ℝ
λn\lambda^nLebesgue measure on ℝⁿ
L\mathcal{L}Lebesgue σ-algebra on ℝ
λ(E)\lambda^*(E)Lebesgue outer measure of E
E|E|Sometimes used for λ(E)

Properties

Extension of length

λ((a,b))=λ([a,b])=baa<b\lambda((a,b)) = \lambda([a,b]) = b - a \quad \forall a < b

Translation invariance

λ(E+t)=λ(E)tR,  EL\lambda(E + t) = \lambda(E) \quad \forall t \in \mathbb{R},\; E \in \mathcal{L}

Scaling

λ(cE)=cλ(E)cR\lambda(cE) = |c|\,\lambda(E) \quad \forall c \in \mathbb{R}

Countable sets have measure zero

E countable    λ(E)=0E \text{ countable} \implies \lambda(E) = 0

The Cantor set has measure zero

λ(C)=0 where C is the standard Cantor set\lambda(\mathcal{C}) = 0 \text{ where } \mathcal{C} \text{ is the standard Cantor set}

Worked Examples

  1. Let E = {x₁, x₂, x₃, …} be countable. For each ε > 0 and each n, cover xₙ by the open interval (xₙ − ε/2ⁿ⁺¹, xₙ + ε/2ⁿ⁺¹) of length ε/2ⁿ.

    In=(xnε2n+1,  xn+ε2n+1),In=ε2nI_n = \left(x_n - \frac{\varepsilon}{2^{n+1}},\; x_n + \frac{\varepsilon}{2^{n+1}}\right),\quad |I_n| = \frac{\varepsilon}{2^n}
  2. Then E ⊆ ∪ₙ Iₙ and the total length is Σ ε/2ⁿ = ε.

    n=1ε2n=ε\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n} = \varepsilon
  3. Since ε > 0 is arbitrary, λ*(E) = 0, so E is measurable with λ(E) = 0.

    λ(E)=0\lambda(E) = 0

Answer: Every countable set has Lebesgue measure zero.

Practice Problems

Difficulty 7/10

Prove that the Lebesgue outer measure is countably subadditive.

Difficulty 8/10

Show that every Borel set in ℝ is Lebesgue measurable (i.e., ℬ(ℝ) ⊆ ℒ).

Difficulty 9/10

Describe the construction of a Vitali set (a non-Lebesgue-measurable subset of [0,1]).

Common Mistakes

Common Mistake

Every subset of ℝ is Lebesgue measurable

Non-measurable sets exist (e.g., Vitali sets), assuming the Axiom of Choice. The Lebesgue σ-algebra is strictly larger than ℬ(ℝ) but still does not contain all subsets of ℝ.

Common Mistake

Lebesgue measure and Lebesgue outer measure are the same

Lebesgue outer measure λ* is defined on all subsets of ℝ but is only countably subadditive, not additive. Lebesgue measure is the restriction of λ* to the measurable sets, where it is countably additive.

Common Mistake

A set of measure zero is 'small' in every sense

The Cantor set has measure zero but is uncountable and has cardinality of the continuum. 'Measure zero' is a quantitative smallness condition, not a topological or cardinality statement.

Quiz

What is the Lebesgue measure of the set ℚ ∩ [0,1]?
The standard Cantor set is:
A Vitali set exists because of:

Historical Background

Before Lebesgue, the Riemann integral could only handle functions with at most countably many discontinuities. Henri Lebesgue's 1902 PhD thesis 'Intégrale, longueur, aire' introduced his measure and integral, which could integrate every bounded pointwise limit of Riemann-integrable functions. The Carathéodory extension theorem (1914) later provided a clean general framework for constructing measures from pre-measures.

  1. 1902

    Lebesgue's thesis introduces Lebesgue measure and integral

    Henri Lebesgue

  2. 1905

    Vitali constructs a non-measurable set using the Axiom of Choice

    Giuseppe Vitali

  3. 1914

    Carathéodory formulates the outer-measure extension theorem

    Constantin Carathéodory

  4. 1924

    Banach and Tarski prove the Banach–Tarski paradox, highlighting limits of measure

    Stefan Banach, Alfred Tarski

Summary

  • Lebesgue measure λ on ℝ extends length from intervals to a vast σ-algebra ℒ ⊇ ℬ(ℝ).
  • It is constructed via outer measure λ*(E) = inf{Σ|Iₙ| : E ⊆ ∪Iₙ} and Carathéodory's criterion.
  • Key properties: translation invariance, scaling, countable additivity, and λ((a,b)) = b − a.
  • Countable sets, including ℚ and the standard Cantor set, have measure zero.
  • Non-measurable sets exist (Vitali sets) and require the Axiom of Choice to construct.

References

  1. BookRoyden & Fitzpatrick — Real Analysis, 4th ed. (2010), Chapters 2–3
  2. BookFolland — Real Analysis, 2nd ed. (1999), §1.2–§1.5