Outer measure satisfies countable subadditivity (λ*(∪ᵢ Aᵢ) ≤ Σᵢ λ*(Aᵢ)) but not countable additivity. For carefully constructed pathological sets — the kind Banach-Tarski-type constructions use — disjoint sets can fail to have λ*(A ∪ B) = λ*(A) + λ*(B). This failure of additivity is why outer measure alone is not a measure; the Carathéodory condition identifies the subclass of sets on which additivity holds.
Question 2 Multiple Choice
A student argues: 'Lebesgue outer measure and Lebesgue measure are the same thing — both defined on all subsets of ℝⁿ.' What is the key error?
AThe student is correct; they are defined identically on all subsets
BOuter measure is defined on all subsets but lacks additivity; Lebesgue measure is its restriction to the Carathéodory-measurable sets, where countable additivity holds — these are different objects with different domains
CLebesgue measure is defined only on intervals, not all subsets
DOuter measure is strictly larger than Lebesgue measure on every set
This conflation is the central misconception. Outer measure λ* is defined universally (on all subsets) as the infimum of box covers — this universality is necessary for the Carathéodory criterion to function. But λ* is not additive on all subsets. Lebesgue measure is the restriction of λ* to the σ-algebra of Carathéodory-measurable sets, where additivity is guaranteed. The two are defined on different collections and have different properties.
Question 3 True / False
The Lebesgue outer measure of a countable dense set (such as the rational numbers in [0,1]) is positive, because it contains infinitely many points spread throughout the interval.
TTrue
FFalse
Answer: False
A countable set has outer measure zero. Each point can be enclosed in an open interval of length ε/2ⁿ; the total cover has volume at most ε × Σ(1/2ⁿ) = ε, which can be made arbitrarily small. Countability, not density, is what matters: no matter how densely packed the points are, a countable collection can always be covered by open intervals with total length approaching zero. This is one of the most counterintuitive consequences of the definition.
Question 4 True / False
A set E is Carathéodory-measurable if for every test set A, λ*(A) = λ*(A ∩ E) + λ*(A ∩ Eᶜ) — that is, E correctly splits every set into non-interacting pieces.
TTrue
FFalse
Answer: True
This is exactly the Carathéodory criterion. Intuitively, E is measurable if it acts like a clean partition boundary for every possible test set A — with no 'leakage' between E and its complement. When this holds, outer measure on the pieces adds correctly, and E is in the σ-algebra on which λ* is a genuine (countably additive) measure. The condition is stated for all A — including non-measurable sets — which is why outer measure must be defined universally first.
Question 5 Short Answer
Why must Lebesgue outer measure be defined on all subsets of ℝⁿ before the Carathéodory criterion is applied, rather than defining it only on the measurable sets from the start?
Think about your answer, then reveal below.
Model answer: Because the Carathéodory criterion tests whether a set E correctly splits every test set A — and A must range over all subsets, including non-measurable ones. If outer measure were only defined on the measurable sets, you couldn't evaluate the condition λ*(A) = λ*(A ∩ E) + λ*(A ∩ Eᶜ) for arbitrary A. The outer measure must exist everywhere as a preliminary construction; then Carathéodory selects the subcollection of sets on which it behaves as a genuine measure.
This logical ordering is non-negotiable: universality first, selection second. The outer measure is a pre-measure that sacrifices additivity for universality; the Carathéodory theorem restores additivity by restricting the domain. Understanding this order also explains why outer measure is defined via an infimum (approximation from outside) rather than some other construction — it must be computable from arbitrary sets, not just nice ones.