Measure Theory 22 | Outer measures - Part 3: Proof [dark version]
Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.
Define Φ-measurable sets A by requiring Φ(Q)=Φ(Q∩A)+Φ(Q∩Aᶜ) for all Q⊆X.
Briefing
A key step in turning an outer measure into a genuine measure is proving that the collection of “measurable” sets built from the outer measure forms a σ-algebra. Once that σ-algebra is in place, defining μ(A)=Φ(A) on it automatically yields a countably additive measure—exactly what measure theory needs for later results like Carathéodory’s extension theorem.
The proof starts by naming the σ-algebra candidate: all subsets A⊆X that satisfy the Carathéodory measurability condition relative to the outer measure Φ. Concretely, a set A is declared Φ-measurable if for every subset Q⊆X the outer measure splits additively across A and its complement: Φ(Q)=Φ(Q∩A)+Φ(Q∩Aᶜ). With that definition, the first σ-algebra checks are straightforward. The empty set is measurable because Q decomposes into Q∩∅ and Q∩X, and Φ(∅)=0 for any outer measure. Similarly, X is measurable since its complement is ∅, making the same additivity relation collapse in the same way.
Next comes closure under complements. If A is measurable, then for every Q the defining identity holds. Swapping A with Aᶜ in the identity and using that (Aᶜ)ᶜ=A recovers the same additivity formula for Aᶜ. That establishes that the measurable sets are closed under taking complements.
The hard part is closure under countable unions. The argument first proves closure under finite unions by taking two measurable sets A1 and A2 and showing A1∪A2 is measurable. The proof uses the measurability identities for A1 and A2 with a generic set Q, then combines them using subadditivity of Φ and a careful set-theoretic rewrite. A Venn-diagram style decomposition of intersections with Q, A1, and A2 shows that the inequality produced by subadditivity tightens to the exact additivity form required for measurability of A1∪A2.
For countably many sets, the proof assumes A1, A2, … are measurable and sets A=⋃_{j=1}^∞ A_j. A standard trick is applied: replace the sequence by a pairwise disjoint one without losing the union, which also aligns perfectly with how countable additivity is proved for measures. Using a generic set Q̂, the measurability condition is applied to the disjoint pieces so that intersections like Q̂∩A_j isolate each term. An induction establishes the finite-union identity: Φ(Q̂∩(A1∪…∪A_n)) equals the sum of Φ(Q̂∩A_j) for j=1…n.
To pass from finite to infinite unions, monotonicity of outer measures is used. As n increases, the complements shrink, letting the proof take limits and obtain an inequality involving Φ(Q̂∩A) and the infinite series Σ_{j=1}^∞ Φ(Q̂∩A_j). Finally, σ-subadditivity of Φ is invoked again to convert the series bound into an equality, forcing the Carathéodory additivity identity for A. Since Q̂ was arbitrary, this shows A is measurable, completing closure under countable unions.
As a payoff, the same identity with Q̂=A itself yields countable additivity for μ(A)=Φ(A) on the resulting σ-algebra. That completes the proposition’s Part A (σ-algebra construction) and Part B (μ becomes a measure), setting up the path to Carathéodory’s extension theorem in the next step.
Cornell Notes
The proof constructs a σ-algebra from an outer measure Φ by declaring a set A measurable when Φ splits additively across A and its complement: for every Q⊆X, Φ(Q)=Φ(Q∩A)+Φ(Q∩Aᶜ). It verifies the σ-algebra axioms: ∅ and X are measurable, complements of measurable sets are measurable, and finite unions of measurable sets are measurable. The main work is countable unions: measurable sets A1,A2,… are replaced by a pairwise disjoint sequence with the same union, enabling an induction for finite unions and then a limit argument using monotonicity. The final step uses σ-subadditivity to upgrade inequalities to equalities, proving the union is measurable and that μ(A)=Φ(A) is countably additive on this σ-algebra.
What exactly does it mean for a set A to be “Φ-measurable” in this construction?
Why are ∅ and X automatically measurable under this definition?
How does closure under complements follow without redoing the whole argument?
What is the key idea for proving closure under countable unions?
Where do monotonicity and σ-subadditivity of an outer measure enter the countable-union step?
Review Questions
- In the Carathéodory measurability condition Φ(Q)=Φ(Q∩A)+Φ(Q∩Aᶜ), why must it hold for every Q⊆X rather than just for Q=A or Q=Aᶜ?
- How does replacing a sequence (A_j) by a pairwise disjoint sequence with the same union simplify the proof of countable additivity?
- What role does σ-subadditivity play in turning inequalities into equalities at the end of the countable-union argument?
Key Points
- 1
Define Φ-measurable sets A by requiring Φ(Q)=Φ(Q∩A)+Φ(Q∩Aᶜ) for all Q⊆X.
- 2
Show ∅ is measurable because Φ(∅)=0 makes the identity hold immediately.
- 3
Show X is measurable because its complement is ∅, again collapsing the identity.
- 4
Prove complements preserve measurability by swapping A and Aᶜ in the defining identity and using (Aᶜ)ᶜ=A.
- 5
Prove finite unions of measurable sets are measurable by combining the measurability identities with Φ’s subadditivity and a set-theoretic intersection decomposition.
- 6
For countable unions, convert the sequence into pairwise disjoint sets with the same union to enable induction for finite unions.
- 7
Use monotonicity plus σ-subadditivity to pass from finite to infinite unions and force equality, yielding countable additivity for μ(A)=Φ(A) on the resulting σ-algebra.